Instructors' Manual: Theory of Machines and Mechanisms 5th Ed

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Instructors’ Manual to accompany
THEORY OF MACHINES
AND MECHANISMS
Fifth Edition
John J. Uicker, Jr.
Professor Emeritus of Mechanical Engineering
University of Wisconsin – Madison
Gordon R. Pennock
Associate Professor of Mechanical Engineering
Purdue University
Joseph E. Shigley
Late Professor Emeritus of Mechanical Engineering
The University of Michigan
Oxford University Press
NEW YORK OXFORD
2016
2
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Library of Congress Cataloging-in-Publication Data
ISBN: 978-0-19-537124-6
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3
PART 1
KINEMATICS AND MECHANISMS
4
5
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1
Chapter 1
The World of Mechanisms
1.1
Sketch at least six different examples of the use of a planar four-bar linkage in practice.
These can be found in the workshop, in domestic appliances, on vehicles, on agricultural
machines, and so on.
Since the variety is unbounded no standard solutions are provided here.
1.2
The link lengths of a planar four-bar linkage are 1 in, 3 in, 5 in, and 5 in. Assemble the
links in all possible combinations and sketch the four inversions of each. Do these
linkages satisfy Grashof's law? Describe each inversion by name, for example, a crankrocker linkage or a drag-link linkage.
s  1 in, l  5 in, p  3 in, q  5 in;
since 1 in  5 in  3 in  5 in .
these
linkages
all
satisfy
Grashof’s
law
Ans.
Drag-link linkage
Drag-link linkage
Ans.
Crank-rocker linkage
Crank-rocker linkage
Ans.
Double-rocker linkage
Crank-rocker linkage
Ans.
2
1.3
1.4
A crank-rocker linkage has a 100-mm frame, a 25-mm crank, a 90-mm coupler, and a 75mm rocker. Draw the linkage and find the maximum and minimum values of the
transmission angle. Locate both toggle postures and record the corresponding crank
angles and transmission angles.
Extremum transmission angles:  min   1  53.1 max   3  98.1
Ans.
Toggle postures: 2  40.1 2  59.14  228.6 4  90.9
Ans.
Plot the complete path of coupler point C.
3
1.5
Find the mobility of each mechanism.
(a) n  6, j1  7, j2  0;
m  3  6  1  2  7   1 0   1
Ans.
(b) n  8, j1  10, j2  0;
m  3 8  1  2 10   1 0   1
Ans.
(c) n  7, j1  9, j2  0;
Ans.
m  3  7  1  2  9   1 0   0
Note that the Kutzbach criterion fails in the case of part (c); the true mobility is m=1.
The exception is due to a redundant constraint. The assumption that the rolling contact
joint does not allow links 2 and 3 to separate duplicates the constraint of the fixed link
length O2O3 .
m  3  4  1  2  3  1 2   1
(d) n  4, j1  3, j2  2;
Ans.
Note in part (d) that each pair of coaxial sliding ground joints is counted as only a
single prismatic pair.
4
1.6
Use the Kutzbach criterion to determine the mobility of the mechanism.
n  5, j1  5, j2  1;
m  3  5  1  2  5  11  1
Note that the double pin is counted as two single pin j1 joints.
Ans.
5
1.7
Sketch a planar linkage with only revolute joints and a mobility of m=1 that contains a
moving quaternary link. How many distinct variations of this linkage can you find?
To have at least one quaternary link, a planar linkage must have at least eight links. The
Kutzbach criterion then indicates that ten single-freedom joints are required for mobility
of m = 1. According to H. Alt, 1955. “Die Analyse und Synthese der achtgleidrigen
Gelenkgetriebe”, VDI-Berichte, 5, pp. 81-93, there are a total of sixteen distinct eight-link
planar linkages having ten revolute joints, seven of which contain a quaternary link.
These seven are illustrated here:
Ans.
6
1.8
Use the Kutzbach criterion to detemine the mobility of the mechanism. Clearly number
each link and label the lower pairs (j1 joints) and higher pairs (j2 joints).
n  5, j1  5, j2  1;
1.9
m  3  5  1  2  5  11  1
Ans.
Determine the number of links, the number of lower pairs, and the number of higher pairs.
Use the Kutzbach criterion to determine the mobility of the mechanism. Is the answer
correct? Briefly explain.
n  4, j1  3, j2  2;
m  3  4  1  2  3  1 2   1
Ans.
If it is not evident visually that link 3 can be incremented upward without jamming, then
consider incrementing link 3 downward. Since it is clear visually that this determines the
position of all other links, this verifies that mobility of one is correct.
7
1.10
Use the Kutzbach criterion to detemine the mobility of the mechanism. Clearly number
each link and label the lower pairs and higher pairs.
n  5, j1  5, j2  1;
1.11
m  3  5  1  2  5  11  1
Ans.
Determine the number of links, the number of lower pairs, and the number of higher pairs.
Treat rolling contact to mean rolling with no slipping. Using the Kutzbach criterion
determine the mobility. Is the answer correct? Briefly explain.
n  7, j1  8, j2  1;
Ans.
m  3  7  1  2 8  11  1
This result appears to be correct. If all parts remain assembled (connected), then within
the limits of travel of the joints illustrated, it appears that when any one joint is locked the
total system becomes a structure.
8
1.12
Does the Kutzbach criterion provide the correct result for this mechanism?
explain why or why not.
Briefly
n  4, j1  2, j2  3;
Ans.
m  3  4  1  2  2   1 3  2
The joints at A and B are both assumed to allow slipping and are j2 joints. This results in
m = 2 which appears to be correct. If any part except wheel 4 is moved, all other parts are
required to follow. However, after all other parts are in a certain posture, wheel 4 is still
able to rotate while slipping against the frame at A.
1.13
The mobility of the mechanism is m = 1. Use the Kutzbach criterion to determine the
number of lower pairs and the number of higher pairs. Is the wheel rolling without
slipping, or rolling and slipping, at point A on the wall?
Suppose that we identify the number of independent freedoms at A by the symbol k. Then
if we account for all links and all other joints as follows, the Kutzbach criterion gives
n  5; j1  4; j2  1; jk  1;
m  3 5  1  2  4   11   3  k 1  k;
Therefore, to have mobility of m  1 , we must have k  1 independent freedom at A. The
wheel must be rolling without slipping. Then, n  5; j1  5; j2  1; and m  1;
Ans.
9
1.14
Devise a practical working model of the drag-link linkage
Ans.
1.15
Find the advance-to-return ratio of the linkage of Prob. 1.3.
From the values of  2 and  4 we find   188.5 and   171.5 .
Then, from Eq. (1.5),
1.16
Q     1.099 .
Ans.
Plot the complete coupler curve of Roberts' linkage illustrated in Fig. 1.24b. Use AB =
CD = AD = 2.5 in and BC = 1.25 in.
10
1.17
If the handle of the differential screw in Fig. 1.11 is turned 15 revolutions clockwise, how
far and in what direction does the carriage move?
Screw and carriage move by (15 rev)/(16 rev/in) = 0.937 50 in to the left.
Carriage moves (15 rev)/(18 rev/in) = 0.833 33 in to the right with respect to the screw.
Net motion of carriage = 15/16 in – 15/18 in = 15/144 = 0.104 17 in to the left.
Ans.
1.18
Show how the linkage of Fig. 1.15b can be used to generate a sine wave.
With the length and angle of crank 2 designated as R and 2, respectively, the horizontal
motion of link 4 is x4  R cos2  R sin 2  90 .
11
1.19
Devise a crank-rocker four-bar linkage, as in Fig. 1.14c, having a rocker angle of 60. The
rocker length is to be 0.50 m.
Distances shown are in meters.
Ans.
12
1.20
A crank-rocker four-bar linkage is required to have an advance-to-return ratio Q = 1.2.
The rocker is to have a length of 2.5 in and oscillate through a total angle of 60.
Determine a suitable set of link lengths for the remaining three links of the four-bar
linkage.
Following the procedure of Example 1.4, the required advance-to-return ratio gives
Q  180    180     1.2 and, therefore, we must have   16.36 . Then, with the
X-line chosen at 30°, the drawing shown below gives measured distances of
RO4O2  r1  4.34 in , RB2O2  r3  r2  6.42 in , and RB1O2  r3  r2  4.44 in . From these we
get
one
possible
solution,
which
has
link
lengths
RAO2  r2  0.99 in, RBA  r3  5.43 in, and RBO4  r4  2.50 in.
Distances shown are in inches
of
RO4O2  r1  4.34 in,
Ans.
13
1.21
Determine the mobility of the mechanism. Number each link and label the lower pairs and
the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 7, j1 = 8, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(7  1)  2(8)  11  1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 3 would be suitable as inputs since these are pinned to
the ground link. Translation of the slider (link 4) would also be suitable as an input.
Other choices for the input are not particularly practical.
Ans.
14
1.22
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 5 links and the joint types are illustrated in the figure below.
Ans
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 5, j1 = 5, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
m  3(5  1)  2(5)  11  1
Ans.
This is the correct answer for this mechanism, that is, for a single input value there is a
unique posture.
Rotation of link 3 would be a suitable input since it is pinned to the ground link.
Translation of the slider (link 2) would also be suitable as an input. Other choices for the
input are not particularly practical.
Ans.
15
1.23
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 5 links and the joint types are illustrated in the figure below.
Ans
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 8, j1 = 10, and j2 = 0.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(8  1)  2(10)  1 0   1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 5 would be suitable inputs since they are pinned to the
ground link. Translation of the slider (link 4) would also be a suitable input. Other
choices for the input are not particularly practical.
Ans.
16
1.24
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 7 links and the joint types are illustrated in the figure below.
Ans
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 7, j1 = 8, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(7  1)  2(8)  11  1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 4 would be suitable inputs since they are pinned to the
ground link. Other choices for the input are not particularly practical.
Ans.
17
1.25
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 7 links and the joint types are illustrated in the figure below.
Ans.
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 7, j1 = 8, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(7  1)  2(8)  11  1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 4 would be suitable inputs since they are pinned to the
ground link. Translation of the slider (link 7) would also be suitable as an input. Other
choices for the input are not particularly practical.
Ans.
18
1.26
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 8 links and the joint types are illustrated in the figure below.
Ans
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 8, j1 = 10, and j2 = 0.
Note that the double-pin joint between links 5, 6, and 7 is counted as 2 j1 joints.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility
of the mechanism is
m  3(8  1)  2(10)  1 0   1
Ans.
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 6 would be suitable inputs since they are pinned to the
ground link. Translation of either of the sliders (link 4 or link 8) would also be suitable as
inputs. Other choices for the input are not particularly practical.
Ans.
19
1.27
Determine the mechanical advantage of the four-bar linkage in the posture illustrated.
The mechanical advantage, Eq. (3.41), can be written as
RBO4 sin
b RBO4 sin
MA  

a RAO2 sin
RAO2 sin 
(1)
where the angles  ,  , and  are as shown in Fig. P1.27.
O2O4  120 mm, O2 A  60 mm, AB  100 mm, and O4 B  130 mm.
.
To determine the angles  ,  , and  , the law of cosines for the triangle O2AO4 can be
written as
2
2
AO4  RO24O2  RAO
 2 RO4O2 RAO2 cos( O4O2 A)
2
 120 mm    60 mm   2 120 mm  60 mm  cos 30  5529.234 m 2
2
2
Therefore,
AO4  74.359 mm
The law of sines for the triangle O2AO4 can be written as
20
RO2O4
AO4

sin( AO2O4 ) sin( O2 AO4 )
Rearranging this equation gives
RO O
120 mm
sin( O2 AO4 )  2 4 sin( AO2O4 ) 
sin 30  0.8069
AO4
74.359 mm
Therefore, the angle is either
O2 AO4  53.79 or
O2 AO4  126.21
Note that O2 AO4  53.79 can be eliminated since it is not a physically possible result
for the open configuration of the four-bar linkage. Therefore, the correct result for the
angle is
O2 AO4  126.21
(2)
The law of cosines for the triangle ABO4 can be written as
2
2
2
RBO
 RBA
 AO4  2RBA AO4 cos( O4 AB)
4
Rearanging this equation gives
100 mm    74.359 mm   130 mm   0.092 17
cos( O4 AB ) 
2 100 mm  74.359 mm 
2
Therefore, the angle is either
O4 AB  95.29
or
2
2
O4 AB  84.71
Note that the value O4 AB  84.71 can be eliminated since it is not physically possible
for the open configuration of the four-bar linkage. Therefore, the correct result is
O4 AB  95.29
(3)
The angle
  O2 AO4  O4 AB
(4)
Substituting Eqs. (2) and (3) into Eq. (4) gives
(5)
  126.21  95.29  221.50
The law of cosines for the triangle ABO4 can be written as
2
2
2
AO4  RBA
 RBO
 2 RBARBO cos 
4
4
Rearanging this equation gives
2
2
2
100 mm   130 mm    74.359 mm 

cos  
 0.821 95
2 100 mm 130 mm 
Therefore, the transmission angle is
  34.72
Substituting Eqs. (5) and (6) into Eq. (1), the mechanical advantage is
130 mm  sin 34.72o

74.044 mm
MA  

 1.86
o
39.757 mm
 60 mm  sin 221.50
(6)
Ans.
21
1.28
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 7 links and the joint types are indicated in the figure below
Ans.
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 7, j1 = 8, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(7  1)  2(8)  11  1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2, link 4, or link 7 would be suitable inputs since they are each
pinned to the ground link. Other choices for the input are not particularly practical. Ans.
22
1.29
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 7 links and the joint types are illustrated in the figure below.
Ans.
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 5, j1 = 5, and j2 = 1.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(5  1)  2(5)  1 1  1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
Rotation of either link 2 or link 4 would be suitable inputs since they are each pinned
to the ground link. Other choices for the input are not particularly practical.
Ans.
23
1.30
The rocker of a crank-rocker four-bar linkage is required to have a length of 6 in and
swing through a total angle of 30°. Also, the advance-to-return ratio of the linkage is
required to be 1.75. Determine a suitable set of link lengths for the remaining three links.
The advance-to-return-ratio must be
Q     1.75
(1)
where
and
(2)
  180o  
  180o  
Substituting Eqs. (2) into Eq. (1) gives
180    1.75 180   
Therefore,
  49.09
o
  229.09
  130.91
and
Note that the dimensions of the synthesized four-bar linkage to satisfy the given design
constraints are not unique. However, the graphic procedure follows the steps shown in
Example 1.4. (For this problem, refer to the figure below):
(1)
Draw the rocker r4  6 in of the crank-rocker four-bar linkage to a suitable scale,
for example, full scale. Draw the rocker in the two extreme positions, that is, show
the swing angle of the rocker of 30 degrees. Label the ground pivot O4 and label
the pin B in the two positions B1 and B2.
(2)
Through point B1 draw an arbitrary line (labeled the X-line). Through B2 draw a
line parallel to the X-line.
(3)
Lay out the angle   49.09o counterclockwise from the X-line through point B1.
The intersection of this line with the line parallel to the X-line through point B2 is
the input crank pivot O2.
(4)i
The length O2O4 of the ground link can be measured from the drawing. That is
r1  O2O4  8.94 in
Ans.
The other measurements are
O2 B1  r3  r2  4.02 in and O2 B2  r3  r2  3.18 in
Therefore, the length of the input link and the length of the coupler link,
respectively, must be
r2  O2 A  0.42 in and r3  AB  3.60 in
Ans.
This solution of the synthesized four-bar linkage is shown in the figure below.
24
The synthesized four-bar linkage.
25
1.31
Determine a suitable set of link lengths for a slider-crank linkage such that the stroke will
be 500 mm and the advance-to-return ratio will be 1.8.
The advance-to-return ratio must be
Q     1.8
(1)
where
and
(2)
  180o  
  180o  
Substituting Eqs. (2) into Eq. (1) gives
180    1.8 180   
Therefore,
  51.43
o
and
  128.57
  231.43
Note that the dimensions of the synthesized slider-crank linkage to satisfy the given
design constraints are not unique. However, the graphic procedure follows the steps
shown in Example 1.5. (For this problem, refer to the figure below):
(1)
Draw the stroke of the slider-crank linkage to a suitable scale, for example, 1 in ~
100 mm. The length of the stroke of this mechanism is specified as r4  500 mm .
Label the pin B in its two extreme positions as B1 and B2.
(2)
Through point B2 draw an arbitrary line (labeled the X-line). Through point B1
draw a line parallel to the X-line.
(3)
Lay out the angle   51.43o clockwise from the X-line. The intersection of this
line with the line parallel to the X-line through point B1 is the ground pivot O2.
(4)
From the scale drawing below, the ground link; i.e., the offset (the vertical
distance), is measured as
r1  102.5 mm
Ans.
The other measurements are
O2 B2  r3  r2  562.5 mm and O2 B1  r3  r2  118.5 mm
Therefore, the length of the input link and the length of the coupler link,
respectively, are
r2  O2 A  222 mm and r3  AB  340.5 mm
Ans.
The solution of the synthesized slider-crank linkage is shown in the figure below.
26
The synthesized slider-crank linkage.
27
1.32
Determine the transmission angle and the mechanical advantage of the four-bar linkage in
the posture illustrated. What type of four-bar linkage is this?
Grashof’s law for a planar four-bar linkage, Sec. 1.9, Eq. (1.6), states that in order for the
four-bar linkage to be a Grashof chain, the dimensions must satisfy
s l  p q
(1)
The lengths of the four links of the four-bar linkage are
s  20 mm, l  90 mm, p  60 mm, and q  70 mm. .
Substituting these dimensions into Eq. (1) gives
20 mm  90 mm  60 mm  70 mm
or
110 mm  130 mm
Since this inequality is satisfied, the four-bar linkage is a Grashof chain. Also, since the
shortest link s is adjacent to the ground link, the shortest link is a crank; see Sec. 1.9.
Therefore, this is a crank-rocker four-bar linkage.
Ans.
r2  20 mm, r3  70 mm, r4  90 mm, and r1  60 mm.
28
In the posture illustrated, the distance between points A and O4 can be found from the
triangle AO2O4 . The law of cosines can be written as
2
AO4  r22  r12  2r2 r1 cos2
Substituting the given dimensions gives
AO4  ( 20 mm)2  (60 mm)2  2(20 mm)(60 mm)cos 30  1 921.54  mm 
2
2
Therefore, AO4  43.84 mm .
To determine the transmission angle  consider the triangle ABO4 . The law of
cosines can be written as
2
AO4  r32  r42  2r3 r4 cos
Substituting the given dimensions gives
2
1921.54  mm  (70 mm)2  (90 mm)2  2(70 mm)(90 mm)cos
This equation gives cos  0.879 24
Therefore, the transmission angle is   28.45
Ans.
(2)
The angle  can be found from the triangle O2O4 A. From the law of sines,
  20 mm  sin 30 
  13.19
 43.84 mm 
To determine the angle  , consider the triangle BO4 A. From the law of sines, this
angle is
  70 mm  sin 28.45 
  sin 1 
  49.52
43.84 mm


The angle between link 3 (link AB) and the ground link O2O4 can be written as
  sin 1 
3  180o       180o  28.45o  49.52o  13.19o  88.84o
The angle  can be written as
  3  30  88.84  30  58.84
(3)
The mechanical advantage of a four-bar linkage, Sec.1.10, can be written as
r sin 
MA  4
r2 sin 
Substituting Eqs. (2) and (3), and the given link lengths, the mechanical advantage of the
four-bar linkage (in the given posture) is
90 mm  sin 28.45o

MA 
 2.51
Ans.
 20 mm  sin 58.84o
29
1.33
Determine the mobility of the mechanism. Number each link and label the lower pairs
and the higher pairs. Identify a suitable input, or inputs, for the mechanism.
The mechanism has 6 links and the joint types are illustrated in the following figure Ans.
The link numbers and joint types of the mechanism.
Ans.
The number of links, lower pairs, and higher pairs, respectively, are
n = 6, j1 = 6, and j2 = 2.
Substituting these values into the Kutzbach mobility criterion, Eq. (1.1), the mobility of
the mechanism is
Ans.
m  3(6  1)  2(6)  1 2   1
This is the correct answer for this mechanism; that is, for a single input value there is a
unique posture.
The rotation of either link 2 or link 4 would be suitable inputs since they are pinned to
the ground link. Other choices for the input are not particularly practical.
Ans.
30
1.34
A crank-rocker four-bar linkage is illustrated in one of its two toggle postures. Find
2 and 4 corresponding to each toggle posture. What is the total rocking angle of link 4?
What are the transmission angles at the extremes?
94B
(a)
r2  8 in, r3  20 in, and r4  r1  16 in.
From isosceles triangle O4O2 B we can calculate
 r2  r3  2  cos1 8 in  20 in  28.955 ,   57.910,
4
r1
2 16 in 
Ans.
2  cos1 3
r  r2
20 in  8 in
 cos1
 247.976 , 4  135.951 .
2r1
2 16 in 
Ans.
(b)
Then 4  4  4  78.041
Ans.
(c)
Finally, from isosceles triangle O2 BO4 ,   28.955 and    67.976 .
Ans.
 2  cos1
31
1.35
Find  2 and  4 corresponding to a dead-center posture. Is there a toggle posture?
r2  110 mm, r3  100 mm, r4  240 mm, and r1  280 mm.
For the given dimensions, there are two dead-center postures, and they correspond to the
two extreme travel postures of crank O2 A . From O4 AO2 using the law of cosines, we
can find 2  114.05, 4  162.82 and, symmetrically, 2 114.05, 4 162.82. There are
also two toggle postures; these occur at 2  56.50, 4  133.14 and, symmetrically, at
2  56.50, 4  133.14 .
Ans.
32
1.36
Determine the advance-to-return ratio for the slider-crank linkage with the offset e. Also,
determine in which direction the crank should rotate to provide quick return.
An offset slider-crank linkage in the two dead-center postures.
From the figure we can see that e   r3  r2  sin  2   r3  r2  sin  2  180  or

e 
,
 r3  r2 
 2  sin 1 

e 

 r3  r2 
 2  180  sin 1 
 e 
e 
1 
 drive   2   2  180  sin 1 
  sin 

 r3  r2 
 r3  r2 
 e 
e 
1 
 return   2  360   2  180  sin 1 
  sin 

 r3  r2 
 r3  r2 
The advance-to-return ratio is
 e 
e 
1 
180  sin 1 
  sin 

r3  r2 
r3  r2 


Q
Ans.
e 
e 
1 
1 
180  sin 
  sin 

 r3  r2 
 r3  r2 
Assuming driving is when B is sliding to the right, the crank should rotate clockwise. Ans.
33
Chapter 2
Position and Displacement
2.1
Describe and sketch the locus of a point A which moves according to the equations
RAx  atcos  2t  , RAy  atsin  2 t  , and RAz  0 .
The locus is the spiral shown.
2.2
Ans.
Find the position difference from point P to point Q on the curve
y  x2  x  16 , where RPx  2 and RQx  4 .
2
RPy   2   2  16  10 ; R P  2ˆi  10ˆj
RQ  4ˆi  4ˆj
RQy   4   4  16  4 ;
R  R  R  2ˆi  14ˆj  14.14281.87
2
QP
Q
P
Ans.
34
2.3
The path of a moving point is defined by the equation y  2 x2  28 . Find the
position difference from point P to point Q if RPx  4 and RQx  3 .
RPy  2  4   28  4 ;
2
R P  4ˆi  4ˆj
2
RQy  2  3  28  10 ; RQ  3ˆi  10ˆj
R  R  R  7ˆi  14ˆj  15.652243.43
QP
2.4
Q
P
Ans.
The path of a moving point P is defined by the equation y  60  x3 / 3 . What
is the displacement of the point if its motion begins at RPx  0 and ends at
RPx  3 ?
3
RPy  0   60   0  / 3  60 ; R P  0   60ˆj
3
RPy  3  60   3 / 3  51 ; R P  3  3ˆi  51ˆj
R  R (3)  R (0)  3ˆi  9ˆj  9.487  71.57
P
2.5
P
P
Ans.
If point A moves on the locus of Problem 2.1, find its displacement from t = 2 to t = 2.5.
R A  2.0   2.0a cos 4 ˆi  2.0a sin 4 ˆj  2.0aˆi
R  2.5  2.5a cos5 ˆi  2.5a sin 5 ˆj  2.5aˆi
A
ΔR A  R A  2.5  R A  2.0   4.5aˆi
2.6
Ans.
The position of a point is given by the equation R  100e j 2t . What is the path of the
point? Determine the displacement of the point from t = 0.10 to t = 0.40.
The point moves on a circle of radius 100 units with center at the origin.
R  0.10  100e j 0.628  80.902ˆi  58.779ˆj
R  0.40  100e j 2.513  80.902ˆi  58.779ˆj
ΔR  R  0.40  R  0.10  161.804ˆi  161.804180
Ans.
Ans.
35
2.7


The equation R  t 2  4 e jt /10 defines the position of a point. In which direction is the
position vector rotating? Where is the point located when t = 0? What is the next value t
can have if the orientation of the position vector is to be the same as it is when t = 0?
What is the displacement from the first position of the point to the second?
Since the polar angle for the position vector is
   t /10 , then d / dt is negative and therefore
the position vector is rotating clockwise.
Ans.


R  0   02  4 e j 0  40
The position vector will next have the same
orientation when  t /10  2 , that is, when t=20. Ans.
R  20    202  4  e j 2  4040
R  R  20   R  0   4000
2.8
Ans.
The location of a point is defined by the equation R   4t  2  e jt / 30 , where t is time in
2
seconds. Motion of the point is initiated when t = 0. What is the displacement during the
first 3 s? Find the change in angular orientation of the position vector during the same
time interval.
R  0    0  2  e j 0  20  2ˆi
R  3  12  2  e j9 / 30  1454  8.229ˆi  11.326ˆj
ΔR  R  3  R  0   6.229ˆi  11.326ˆj  12.92661.19
Ans.
  54  0  54 ccw
Ans.
36
2.9
Link 2 rotates according to the equation    t / 4 . Block 3 slides outward on link 2
according to the equation r  t 2  2 . What is the absolute displacement R P from t = 1
3
to t = 2? What is the apparent displacement R P ?
3/ 2
R P3  re j   t 2  2  e j t / 4
R P3 1  345  2.121ˆi  2.121ˆj
R  2   690  6ˆj
P3
ΔR P3  R P3  2  R P3 1  2.121ˆi  3.879ˆj  4.421118.67
Ans.
R P3 / 2  re j 0   t 2  2  ˆi 2
R P3 / 2 1  3ˆi2
R  2   6ˆi
P3 / 2
2
ΔR P3 / 2  R P3 / 2  2   R P3 / 2 1  3ˆi2
2.10
Ans.
A wheel with center at O rolls without slipping on the ground at point P. If point O is
displaced 10 in to the right, determine the displacement of point P during this interval.
Since the wheel rolls without slipping,
RO   RPO .
  RO / RPO
 10 in / 6 in  1.667 rad  95.51
       270  95.51  174.49
RPO  6 in174.49  5.972ˆi  0.576ˆj in
ΔR P  ΔR O   RPO  R PO 
 10ˆi  5.972ˆi  0.576ˆj  6ˆj in
ΔR P =4.028ˆi  6.576ˆj  7.712 in58.51 Ans.
2.11
A point Q moves from A to B along link 3 while link 2 rotates from 2  30 to 2  120
Find the absolute displacement of Q.
RQ3  3 in30  2.598ˆi  1.500ˆj in
R  3 in120  1.500ˆi  2.598ˆj in
Q3
ΔRQ3  RQ3  RQ3  4.098ˆi  1.098ˆj in
ΔR
 R  6.000ˆi in
Q5 /3
RAO  RBO  3 in and RBA  RO O  6 in
2
4
4 2
BA
ΔRQ5  ΔRQ3  ΔRQ5 / 3
ΔRQ5  1.902ˆi  1.098ˆj in  2.196 in30
Ans.
37
2.12
The double-slider linkage is driven by moving sliding block 2. Write the loop-closure
equation. Solve analytically for the position of sliding block 4. Check the result
graphically for the posture where   45 ..
The loop-closure equation is
?
I
?
R A  R B  R AB
j  
RAe j /12  RB  RAB e  
Ans.
 RB  RAB e j
Taking the imaginary components of this, we get
RAB  200 mm and   15
RA sin15   RAB sin 
sin 
sin  45
RA   RAB
 200 mm
 546.4 mm
sin15
sin15
2.13
Ans.
The offset slider-crank linkage is driven by crank 2. Write the loop-closure equation.
Solve for the position of slider 4 as a function of  2 .
RAO  1 in , RBA  2.5 in , and RCB  7 in .
?
I
RC  R A  R BA  R CB
RC  RAe j / 2  RBAe j2  RCB e j3
Taking real and imaginary parts,
RC  RBA cos 2  RCB cos3 and 0  RA  RBA sin  2  RCB sin 3
and, solving simultaneously, we get
  R  RBA sin  2 
 3  sin 1  A
 with 90  3  90
RCB


2
RC  RBA cos  2  RCB
  RA  RBA sin  2 
2
 2.5cos  2  48  5sin  2  6.25sin 2  2 in
Ans.
38
2.14
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector. Write the vector
loop equation(s) for the mechanism. Identify suitable input(s), known quantities,
unknown variables, and any constraints. If you identify constraints then write the
constraint equation(s).
One suitable set of two vector loop equations is

?
?
?

R 2  R3  R 4  R5  R1  0
I
?
 C1
C 2
C 3
and R 2  R3  R 44  R 24  R 22  0
The angle  2 is a suitable input. Three constraint equations are required.
Ans.
Ans.
24  4   (C2)
22  2   (C3)
The known quantities are R11, R2, R3, R4, R5, R22, R44,  and .
The unknown quantities are 3 , 4 , 5 , 22 , 24 , 44 , and R24 .
Ans.
44  4 (C1)
Ans.
Ans.
39
2.15
Define a set of vectors that is suitable for a complete kinematic analysis of the rackpinion mechanism. Label and show the sense and orientation of each vector. Assuming
rolling with no slip between rack 4 and pinion 5, write the vector loop equation(s) for the
mechanism. Identify suitable input(s), known quantities, unknown variables, and any
constraints. If you identify constraints then write the constraint equation(s).
One suitable set of vectors is as shown in the figure. The vector loop equation is
?
?
R 2  R 3  R 4  R 6  R 15  R 1  0 with 55  R4 (C1)
The angle  2 is a suitable input.
The known quantities are R1, 1=180°, R2, R3, 4=90°, 5, R6, 6=0, R15, 15=90°.
The unknown variables are,  3 , R4, and 5 .
Ans.
Ans.
Ans.
Ans.
40
2.16
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector. Assuming rolling
with no slipping between gears 2 and 5, write the vector loop equation(s) for the
mechanism. Identify suitable input(s), known quantities, unknown variables, and any
constraints. If you identify constraints then write the constraint equation(s).
One suitable set of vectors is as shown in the figure. The vector loop equation is
?
C
R 2  R 3  R 4  R 5  R 1  0 with 2 2  55  0 (C1)
The angle  2 is a suitable input.
The known quantities are R1, 1=0, R2, 2, R3, R4, 5, and R5.
The unknown variables are 3, 4, and 5.
Ans.
Ans.
Ans.
Ans.
41
2.17
Gear 3, which is pinned to link 4 at point B, is rolling without slipping on semi-circular
ground link 1. The radius of gear 3 is 3 and the radius of the ground link is 1. Define
a set of vectors that are suitable for a complete kinematic analysis of the mechanism.
Label and show the sense and orientation of each vector. Write the vector loop
equation(s) for the mechanism. Identify suitable input(s), known quantities, unknown
variables, and any constraints. If you identify constraints then write the constraint
equation(s).
One suitable set of vectors is shown in the figure. The vector loop equation is
?
R 2  R 4  R5  R 1  0 with R2 2  33 (C1)
The angle  2 is a suitable input.
The known quantities are R1, 1=0, R2, 3, R4, and R5.
The unknown variables are 3, 4, and 5.
Ans.
Ans.
Ans.
Ans.
42
2.18
For the mechanism in Figure P1.6, define a set of vectors that is suitable for a complete
kinematic analysis of the mechanism. Label and show the sense and orientation of each
vector. Write the vector loop equation(s) for the mechanism. Identify suitable input(s),
known quantities, unknown variables, and any constraints. If you identify constraints
then write the constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown below.
The corresponding vector loop equations are

I
?
?
?
?
?
C1C 2
R1  R 2  R 3  R13a  0 and R 3  R 4  R15  R13b  0
with the constraint equation(s) R13a  R13b  constant. (C1) and 13b = 13a (C2).
Angle 2 is a suitable input.
Known quantities are R1, 1=90°, R2, R3, 13a, R4, and 15=0.
Unknown variables are 3, R13a, R13b, 13b, 4, and R15.
Ans.
Ans.
Ans.
Ans.
Ans.
43
2.19
For the mechanism in Figure P1.8, define a set of vectors that is suitable for a complete
kinematic analysis of the mechanism. Label and show the sense and orientation of each
vector. Write the vector loop equation(s) for the mechanism. Identify suitable input(s),
known quantities, unknown variables, and any constraints. If you identify constraints
then write the constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The corresponding set of vector loop equations is

?
?
I

C1
?
?
R1  R 2  R 4  R 5  0 and R1  R 22  R 3  R 35  0
with the constraint equation 22  2   (C1).
Angle 5 is a suitable input.
Known quantities are R, 1, R2, R3, R4, R5, R22, and R35.
Unknown variables are 2, 3, 4, 22, and 35.
Ans.
Ans.
Ans.
Ans.
Ans.
44
2.20
For the mechanism in Figure P1.9, define a set of vectors that is suitable for a complete
kinematic analysis of the mechanism. Label and show the sense and orientation of each
vector. Write the vector loop equation(s) for the mechanism. Identify suitable input(s),
known quantities, unknown variables, and any constraints. If you identify constraints
then write the constraint equation(s).
One set of vectors suitable for a complete kinematic analysis of this mechanism is as
shown in this figure.
The corresponding set of vector loop equations is


?
I
R1  R3  R32  R 2  0 and
with the two constraint equations

?
? C1
C 2
?
I
R11  R 4  R34  R33  R32  R 2  0
33  4  180° (C2).
(C1)
and
The angle 2 is a suitable input.
Known quantities are R1, 1=180°, R2, 3=90°, R4, R11, 11, 32=180°, and R33.
Unknown variables are R3, 4, R32, 33, R34, and 34.
34  4  90°
Ans.
Ans.
Ans.
Ans.
Ans.
45
2.21
For the mechanism in Figure P1.10, define a set of vectors that is suitable for a complete
kinematic analysis of the mechanism. Label and show the sense and orientation of each
vector. Write the vector loop equation(s) for the mechanism. Identify suitable input(s),
known quantities, unknown variables, and any constraints. If you identify constraints
then write the constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The corresponding set of vector loop equations is


I


I
 C1
?
?
R1  R11  R 2  R 3  0 and R11  R 2  R34  R 4  R9  0
with the constraint equation 34  3   . (C1)
Ans.
Ans.
Angle 2 is a suitable input.
Ans.
The known quantities are R1, 1, R2, R4, R9, R11, 11, and R34.
Ans.
The unknown variables are R3, 3, 4, 9, and 34.
However, these equations do not analyze the angular displacement of the small wheel,
body 5. In order to do this, we might consider the apparent angular displacement as seen
by an observer fixed on vector 9 and viewing the point of contact between bodies 5 and 1.
The non-slip condition would provide the constraint
11/9  55/9 (C2)
1  1  9    5  5  9 
55   1  5  9  0
55  R9 9  0
where 5 is the radius of wheel 5 and 5 is the angular displacement of body 5.
Ans.
46
2.22
Write a calculator program to find the sum of any number of two-dimensional vectors
expressed in mixed rectangular or polar forms. The result should be obtainable in either
form with the magnitude and angle of the polar form having only positive values.
Because the variety of makes and models of calculators is vast and no standards are
known for programming them, no solution is shown here.
2.23
Write a computer program to plot the coupler curve of any crank-rocker or double-crank
form of the four-bar linkage. The program should accept four link lengths and either
rectangular or polar coordinates of the coupler point with respect to the coupler.
Again the variety of programming languages makes it impossible to provide a standard
solution. However, one version, written in ANSI/ISO FORTRAN 77, is supplied here as
an example. There are also no universally accepted standards for programming graphics.
Therefore the Tektronix PLOT10 subroutine library, for display on Tektronix 4010 series
displays, is chosen as an old but somewhat recognized alternative. The symbols in the
program correspond to the notation shown in Fig. 2.15 of the text. The required input
data are:
-1
 X5,Y5,
R1, R2, R3, R4, 
 R5,ALPHA, 1
The program can be verified using the data of Example 2.6 and checking the results
against those of Table 2.3.
PROGRAM CCURVE
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A FORTRAN 77 PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK-ROCKER
OR DOUBLE-CRANK FOUR-BAR LINKAGE, GIVEN ITS DIMESNIONS.
ORIGINALLY WRITTEN USING SUBROUTINES FROM TEKTRONIX PLOT10 FOR
DISPLAY ON 4010 SERIES DISPLAYS.
REF:J.J.UICKER,JR, G.R.PENNOCK, & J.E.SHIGLEY, ‘THEORY OF MACHINES
AND MECHANISMS,’FIFTH EDITION, OXFORD UNIVERSITY PRESS, 2015.
EXAMPLE 2.7
WRITTEN BY: JOHN J. UICKER, JR.
ON:
01 JANUARY 1980
READ IN THE DIMENSIONS OF THE LINKAGE.
READ(5,1000)R1,R2,R3,R4,X5,Y5,IFORM
1000 FORMAT(6F10.0,I2)
FIND R5 AND ALPHA.
IF(IFORM.LE.0)THEN
R5=SQRT(X5*X5+Y5*Y5)
ALPHA=ATAN2(Y5,X5)
ELSE
R5=X5
ALPHA=Y5/57.29578
X5=R5*COS(ALPHA)
Y5=R5*SIN(ALPHA)
END IF
INITIALIZE FOR PLOTTING AT 120 CHARACTERS PER SECOND.
CALL INITT(1200)
47
C
C
C
C
C
C
C
C
C
C
C
C
C
C
SET THE WINDOW FOR THE PLOTTING AREA.
CALL DWINDO(-R2,R1+R2+R4,-R4,R4+R4+Y5)
CYCLE THROUGH ONE CRANK ROTATION IN FIVE DEGREE INCREMENTS.
TH2=0.0
DTH2=5.0/57.29578
IPEN=-1
DO 2 I=1,73
CTH2=COS(TH2)
STH2=SIN(TH2)
CALCULATE THE TRANSMISSION ANGLE.
CGAM=(R3*R3+R4*R4-R1*R1-R2*R2+2.0*R1*R2*CTH2)/(2.0*R3*R4)
IF(ABS(CGAM).GT.0.99)THEN
CALL MOVABS(100,100)
CALL ANMODE
WRITE(7,1001)
1001
FORMAT(//’ *** THE TRANSMISSION ANGLE IS TOO SMALL. ***’)
GO TO 1
END IF
SGAM=SQRT(1.0-CGAM*CGAM)
GAM=ATAN2(SGAM,CGAM)
CALCULATE THETA 3.
STH3=-R2*STH2+R4*SIN(GAM)
CTH3=R3+R1-R2*CTH2-R4*COS(GAM)
TH3=2.0*ATAN2(STH3,CTH3)
CALCULATE THE COUPLER POINT POSITION.
TH6=TH3+ALPHA
XP=R2*CTH2+R5*COS(TH6)
YP=R2*STH2+R5*SIN(TH6)
PLOT THIS SEGMENT OF THE COUPLER CURVE.
IF(IPEN.LT.0)THEN
IPEN=1
CALL MOVEA(XP,YP)
ELSE
IPEN=-1
CALL DRAWA(XP,YP)
END IF
TH2=TH2+DTH2
2 CONTINUE
DRAW THE LINKAGE.
CALL MOVEA(0.0,0.0)
CALL DRAWA(R2,0.0)
XC=R2+R3*COS(TH3)
YC=R3*SIN(TH3)
CALL DRAWA(XC,YC)
CALL DRAWA(XP,YP)
CALL DRAWA(R2,0.0)
CALL MOVEA(XC,YC)
CALL DRAWA(R1,0.0)
1 CALL FINITT(0,0)
CALL EXIT
STOP
END
48
2.24
Plot the path of point P for: (a) inverted slider-crank linkage; (b) second inversion of the
slider-crank linkage; (c) Scott-Russell straight-line linkage; and (d) drag-link linkage.
(a)
(b)
(c)
(d)
(a) RCA  2 in , RBA  3.5 in , and RPC  4 in ; (b) RCA  40 mm , RBA  20 mm , and
RPB  65 mm ;
(c)
RBA  RCB  RPB  25 mm ;
RCB  RCD  3 in , and RPB  4 in .
(d)
RDA  1 in ,
RBA  2 in ,
49
2.25
Using the offset slider-crank linkage in Figure P2.13, find the crank angles corresponding
to the extreme values of the transmission angle.
As shown,   90  3 .
Also from the figure
e  r2 sin  2  r3 cos  .
Differentiating with
respect to  2 ;
d
;
r2 cos 2  r3 sin 
d 2
r cos 2
d
.
 2
d 2
r3 sin 
Now, setting d / d 2  0 , we get cos 2  0 .
Therefore, we conclude that  2    2k  1  / 2  90, 270,
2.26
Ans.
Section 1.10 states that the transmission angle reaches an extreme value for the four-bar
linkage when the crank lies on the line between the fixed pivots. Referring to Figure
2.19, this means that  reaches a maximum or minimum when crank 2 is collinear with
the line O2O4 . Show, analytically, that this statement is true.
From O4O2 A :
s 2  r12  r22  2r1r2 cos 2 .
Also, from ABO4 :
s 2  r32  r42  2r3r4 cos  .
Equating these we
differentiate with respect to  2
to obtain
d
or
2r1r2 sin  2  2r3r4 sin 
d 2
r r sin  2
d
.
 12
d 2 r3r4 sin 
Now, for
d
 0 , we have sin  2  0 . Thus,  2  0,  180,  360,
d 2
Q.E.D.
50
2.27
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector. Write the vector
loop equation(s) for the mechanism. Identify suitable input(s), known quantities,
unknown variables, and any constraints. If you identify constraints then write the
constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The three vector loop equations are




?

R11  R 3  R 6  R5  R15  R19  0

C1
 C3
?

 C2
?
?

R 2  R 22  R 55  R15  R19  0

R11  R 33  R 7  R 4  R14  0
with three constraint equations
Ans.
22  2   (C1) 33  3   (C2) and 55  5   (C3).
The angle 2 is a suitable input.
Known quantities are: R2, R3, 4=90°, R5, R6, R7, R11, 11, R14, 14=0, 15=180°,
R19, 19=90°, R22, R33, and R55
Unknown variables are 3, R4, 5, 6, 7, R15, 22, 33, and 55.
Ans.
Ans.
Ans.
Ans.
51
2.28
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector. Write the vector
loop equation(s) for the mechanism. Identify suitable input(s), known quantities,
unknown variables, and any constraints. If you identify constraints then write the
constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The two vector loop equations are



 C1

R 2  R12  R 23  R 32  R 3  0

 C2
?



R 3  R 34  R 4  R5  R15  R1  0
with two constraint equations
Ans.
32  3  90
(C1)
and 34  3  90 (C2).
Ans.
The magnitude R2 is a suitable input.
Ans.
Known quantities are: R1, 1=180°, 2=180°, R3, R4, R5, 5=180°,
R12, 12=90°, 15=90°, 23, R32, and R34.
Ans.
Unknown variables are 3, 4, R15, R23, 32, and 34.
Ans.
Note that the angular displacement of the wheel 5 is related to the distance R15 by rolling
contact. The rolling contact equation between link 5 and the ground link 1 can be written
as
 R15  5  5  15   5 5
The correct sign in this equation is positive because, for a positive (counterclockwise)
rotation of wheel 5, the length of the vector R15 is increasing whereas, for a negative
(clockwise) rotation of wheel 5, the length of the vector R15 is decreasing.
52
2.29
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector. Write the vector
loop equation(s) for the mechanism. Identify suitable input(s), known quantities,
unknown variables, and any constraints. If you identify constraint(s) then write the
constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The three vector loop equations are


?



?

R 2  R 3  R 4  R14  0
R 2  R 9  R 5  R15  0

?
?
C

R 3  R 7  R8  R 6  R 9  0
with one constraint equation

 6 3  9

 3 6  9
where the minus sign is used because gears 3 and 6 have external rolling contact.
The angle 2 is a suitable input.
Known quantities are:
R2, R3, 4=90°, R5, R6, R7,R8, R9=6+3, R14, 14=180°, R15, and 15=0.
Unknown variables are 3, R4, 5, 6, 7, 8, and 9.
Ans.
Ans.
Ans.
Ans.
Ans.
53
2.30
Define a set of vectors that is suitable for a complete kinematic analysis of the
mechanism. Label and show the sense and orientation of each vector.. Write the vector
loop equation(s) for the mechanism. Identify suitable input(s), known quantities,
unknown variables, and any constraints. If you identify constraint(s) then write the
constraint equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The three vector loop equations are


?

 C1

?

?
C 
R 2  R 3  R 4  R1  0
R 26  R 6  R 9  R11  0

?
 C2
 C1
R 2  R 3  R 45  R 5  R 65  R 26  0
There are four constraint equations
26  2   (C1)
45  4    180 (C2)
Ans.
65  6 (C3)

1 7  9

(C4)
 7 1  9
where the minus sign is used because gears 1 and 7 have external rolling contact.
The angle 2 is a suitable input.
Known quantities are: R1, 1=90°, R2, R3, R4, R5, R6, R9, R11, 11=0, R26, and R45.
Unknown variables are 3, 4, 5, 6, 7, 9, 26, 45, R65, and 65.
Ans.
Ans.
Ans.
Ans.
54
2.31
For the input angle 2  300o , measured counterclockwise from the x-axis, determine the
two postures of link 4.
r2  60 mm, r3  140 mm, r4  140 mm, and r1  160 mm.
The vector loop equation can be written
r2  r3  r4  r1  0
The horizontal and vertical components are
r2 cos 2  r3 cos3  r4 cos 4  r1  0
r2 sin 2  r3 sin3  r4 sin 4  0
Squaring, adding, and rearranging these gives us Freudenstein’s equation. That is
A cos4  B sin4  C
A  2r1r4  2r2 r4 cos 2
B  2r2 r4 sin 2
where
C  r32  r42  r12  r22  2r1r2 cos 2
Substituting the known data into these equations gives
A  2 160 mm 140 mm   2  60 mm 140 mm  cos 300  36 400  mm 
B  2  60 mm 140 mm  sin 300  14 549  mm 
2
2
C  140 mm   140 mm   160 mm    60 mm   2 160 mm  60 mm  cos 300
2
 19 600  mm 
2
2
2
2
(1)
55
which reduces Eq. (1) to the form
36 400 cos4  14 549 sin4  19 600  0
To solve this transcendental equation, we define
Z  tan 4 2 
(2)
(3)
which gives
2Z
1 Z2
and
cos


4
1 Z2
1 Z2
Substituting these into Eq. (2), and rearranging, gives
16 800Z 2  29 098Z  56 000  0
which has the solutions
sin4 
Z
14 549 
14 549   16 800  56 000 
2
16 800
 2.886 73 or - 1.154 71
Substituting these two roots back into Eq. (3) gives the two solutions
and 4  98.21
4  141.79
Ans.
56
2.32
For the input angle 2  60o , measured counterclockwise from the x-axis, determine the
two postures of link 4.
r2  80 mm, r3  50 mm, r4  100 mm, and r1  70 mm.
The vector loop equation can be written
r2  r3  r4  r1  0
The horizontal and vertical components are
r2 cos 2  r3 cos3  r4 cos 4  r1  0
r2 sin 2  r3 sin3  r4 sin 4  0
Squaring, adding, and rearranging these gives us Freudenstein’s equation. That is
A cos4  B sin4  C
A  2r1r4  2r2 r4 cos 2
where
B  2r2 r4 sin 2
C  r32  r42  r12  r22  2r1r2 cos 2
Substituting the known data into these equations gives
(1)
57
A  2  70 mm 100 mm   2 80 mm 100 mm  cos 60  6 000  mm 
B  2 80 mm 100 mm  sin 60  13 856.4  mm 
2
2
C   50 mm   100 mm    70 mm   80 mm   2  70 mm 80 mm  cos 60
2
2
 13 200  mm 
2
2
2
which reduces Eq. (1) to the form
6 000 cos4  13 856 sin4  13 200  0
To solve this transcendental equation, we define
Z  tan 4 2 
(2)
(3)
which gives
2Z
1 Z2
and
cos


4
1 Z2
1 Z2
Substituting these into Eq. (2), and rearranging, gives
7 200Z 2  27 712Z  19 200  0
which has the solutions
sin4 
Z
13 856 
 13 856    7 200 19 200 
2
7 200
 2.942 69 or 0.906 20
Substituting these two roots back into Eq. (3) gives the two solutions
and 4  84.37 .
4  142.46
The angular position of link 4, for the open configuration shown is 4  84.37 .
Ans.
58
2.33
Consider a four-bar linkage for which ground link 1 is 14 in, input link 2 is 7 in, coupler
link 3 is 10 in, and output link 4 is 8 in. The fixed x and y axes are specified as horizontal
and vertical, respectively. The origin of this reference frame is coincident with the
ground pivot of link 2, and the ground link is aligned with the x axis. For the input angle
2  60 (counterclockwise from the x axis): (a) Using a suitable scale, draw the linkage
in the open and crossed postures and measure the values of the variables  3 and  4 for
each posture. (b) Use trigonometry (that is, the laws of sines and cosines) to determine
 3 and  4 for the open posture. (c) Use Freudenstein's equation to determine  3 and  4
for both postures. (d) Use the Newton-Raphson iteration procedure to determine  3 and
 4 for the open posture. Using the measurements in (a) as initial estimates for  3 and  4 ,
iterate until the two variables converge to within 0.01°.
(a) Graphic Method. The link dimensions and angles are specified as:
Ground Link:
Input Link:
Coupler Link:
Output Link:
R1 = 14.0 in
R2 = 7.0 in
R3 = 10.0 in
R4 = 8.0 in
θ1 = 0
θ2 = 60˚
θ3 = ?
θ4 = ?
For the specified input angle  2  60  , the two possible configurations of the four-bar
linkage are as shown in the following figure. The loop O2ABO4 is the open posture of the
four-bar linkage and the loop O2AB’O4 is the closed (crossed) posture.
Graphic solution.
59
From measurements of the drawing, the answers for the coupler angle and the output angle are:
The Open Posture:
and
3  11
4  95
The Crossed Posture:
3   71 or  289
4   155 or  205
and
(b) Trigonometry. The notation for the open and crossed configurations of the four-bar
linkage are shown in the following figure.
From the triangle AO2O4, the law of cosines gives
2
AO 4  R12  R22  2 R1R2 cos2
AO 4 
14 in    7 in   2 14 in  7 in  cos 60  12.124 in
2
2
From the triangle O2O4A, the law of cosines gives
14 in   12.124 in    7 in   30
R 2  AO 4  R22
  cos 1
 cos1
2 14 in 12.124 in 
2 R1 AO 4
2
2
2
2
1
From the triangle O4AB, the law of cosines gives
10 in   12.124 in   8 in   41
R 2  AO 4  R42
  cos 3
 cos1
2 10 in 12.124 in 
2 R3 AO 4
2
2
2
2
1
8 in   12.124 in   10 in   55.10
R 2  AO 4  R32
  cos 4
 cos1
2 8 in 12.124 in 
2 R4 AO 4
2
2
2
1
From these, the angles for the open posture are
3      41  30  11
Ans.
4        180  30  55.10  94.90
Ans.
For the crossed posture, the angles are
3      30  41  71  289
Ans.
60
4        180  30  55.10  205.10  154.90
(c) The vector loop equation can be written
R 2  R3  R 4  R1  0
The horizontal and vertical components are
R2 cos 2  R3 cos 3  R4 cos 4  R1  0
R2 sin2  R3 sin3  R4 sin4  0
Squaring, adding, and rearranging these gives us Freudenstein’s equation. That is,
A cos4  B sin4  C
Ans.
(1)
(2)
A  2 R1R4  2 R2 R4 cos 2
B  2 R2 R4 sin2
where
C  R32  R42  R12  R22  2 R1R2 cos 2
Substituting the known data into these equations gives
A  2 14 in 8 in   2  7 in 8 in  cos 60  168 in 2
B  2  7 in 8 in  sin 60  97 in 2
C  10 in   8 in   14 in    7 in   2 14 in  7 in  cos 60  111 in 2
2
2
2
2
which reduces Eq. (2) to the form
168 cos4  97 sin4  111  0
To solve this transcendental equation, we define
Z  tan 4 2 
(3)
(4)
which gives
2Z
1 Z2
and
cos


4
1 Z2
1 Z2
Substituting these into Eq. (3), and rearranging, gives
57Z 2  194  279  0
which has the solutions
sin4 
Z
97 
 97   57  279 
2
57
 1.089 43 or
 4.492 94
Substituting these two roots back into Eq. (4) gives the two solutions
4  94.90
and 4  154.90  205.10 .
The angular position of link 4, for the open posture shown is 4  94.90 .
Ans.
From this, Eqs. (1) gives 3  11.00 .
The crossed posture gives 4  154.90 and 3  71.00  289.00 .
Ans.
Ans.
(d)
For the Newton-Raphson technique, the vector loop equation can be written
f  R 2  R3  R 4  R1  0
From this, the horizontal and vertical components are
61
f x  R2 cos 2  R3 cos 3  R4 cos4  R1  0
f y  R2 sin2  R3 sin3  R4 sin4  0
Expanding these to first order in Taylor series we find
R2 cos 2  R3 cos 3  R3 sin33  R4 cos 4  R4 sin4  4  R1  0
R2 sin2  R3 sin3  R3 cos 33  R4 sin4  R4 cos 4 4  0
and, writing this in matrix format gives
 R3 sin3  R4 sin4   3   R2 cos 2  R3 cos 3  R4 cos 4  R1 
  R cos  R cos        R sin  R sin  R sin

3
4
4 
2
2
3
3
4
4
4
 3


Substituting the given data this becomes
 10.0 sin3 8.0 sin4   3   10.50000  10.0 cos 3  8.0 cos 4 
 10.0 cos  8.0 cos        6.06218  10.0 sin  8.0 sin 
3
4 
3
4
4



(5)
Using the graphic solution of part (a) as an estimate, 3  11 and 4  95, the first
iteration equations are
 1.90809 7.96956  3  0.01352
 9.81627 0.69725      0.00071

 4 

which give corrections of 3  0.000 0471 rad  0.002 70 , 4  0.001 68 rad  0.096 26 .
Therefore, after one iteration, we have 3  11.002 70 and 4  94.903 74.
Substituting again into Eqs. (5) gives
 1.90855 7.97072  3   0.00000099
 9.81618 0.68382      0.000011312 

 4 

This gives corrections of 3  1.142 10 rad  0.000 065 4 , 4  1.49 10 rad  0.000 008 54. .
7
7
Therefore, after two iteration , we have 3  11.002 64 and 4  94.903 73.
Ans.
62
2.34
A crank-rocker four-bar linkage is illustrated in two different postures for which
2  150 and 2  240 . Determine  3 and  4 for the open posture and  3 and  4 for the
crossed posture..
RO4O2  600 mm, RAO2  140 mm, RBA  690 mm, and RBO4  400 mm.
For the first input angle 2  150 , to the following figure
and observe that
R A  0.140 m150,
RO4  0.600 m0.
and
Therefore,
S  RO4  R A  0.600 m0  0.140 m150  0.725 m  5.54.
Referring again to the figure, we note that the vectors in the triangle ABO4 are related by
the equation

?
?
S  R BA  R BO4
(a)
There are two unknown orientations in this equation, and so we identify this as case 4.
Using Eq. (2.49) and substituting S for C, RBA for A, RBO4 for B, S for C, and 4 for B
63
gives
 4   S  cos
1
2
2
S 2  RBO
 RBA
4
2 SRBO4
 0.725 m    0.400 m    0.690 m 
 5.54  cos
2  0.725 m  0.400 m 
2
2
2
1
Ans.
 5.54  111.18  105.64 or  116.72
We note that we could have substituted RBO4=+0.400 m for B and we would have
obtained 4 + 180° for the final result.
Next, using Eq. (2.50) and substituting 3 for A gives
2
2
S 2  RBA
 RBO
4
3   S cos 1
2 SRBA
 0.725 m    0.690 m    0.400 m 
 5.54 cos
2  0.725 m  0.690 m 
2
2
2
1
 5.54 32.72  38.26
or
Ans.
27.18
We follow the same procedure for the second input angle 2  240 . Using the figure
above yields
S  RO4  RA  0.600 m0  0.140 m240  0.681 m10.26.
 0.681 m    0.400 m    0.690 m 
4  10.26  cos
2  0.681 m  0.400 m 
2
2
2
1
 10.26  105.73  115.99
 95.47
 0.681 m    0.690 m    0.400 m 
2  0.681 m  0.690 m 
2
3  10.26 cos1
or
2
Ans.
2
 10.26 33.92  23.66 or 44.18
We recognize the positive angles as the solutions of interest in all cases.
Ans.
64
Page intentionally blank.
65
Chapter 3
Velocity
3.1
The position vector of a point is given by the equation R  100e jt , where R is in inches.
Find the velocity of the point at t  0.40 s.
R  t   100e j t in
R  t   j 100e j t in/s
R  0.40s   j 100e j 0.40 in/s
 j 100  cos0.40  j sin 0.40  in/s
 100 sin 72 in/s  j100 cos72 in/s
R  0.40s  298.783  j97.080 in/s  314.159 in/s162
3.2
Ans.
The path of a point is defined by the equation R   t 2  4  e-j t / 10 , where R is in meters.
Find the velocity of the point at t  20 s.
R  t    t 2  4  e j t /10
R  t   2te j t /10   j /10   t 2  4  e j t /10
R  20 s   40e j 20/10   j /10   202  4  e j 20/10
 40e j 2  j 40.4e j 2
R  20 s   40.000  j126.920 m/s  133.074 m/s  72.51
Ans.
66
3.3
Automobile A is traveling south at 55 mi/h and automobile B is travelling north 60 east
at 40 mi/h. Find the velocity difference between B and A and the apparent velocity of B
to the driver of A?
VA  55 mi/h  90  55ˆj mi/h
V  40 mi/h30  34.641ˆi  20ˆj mi/h
B
VBA  VB  VA  34.641ˆi  75ˆj mi/h
VBA  82.613 mi/h65.2 =82.613 mi/h N 24.8 E
Ans.
Naming B as car 3 and A as car 2, we have VB2  VA since car
2 is translating. Then VB3 / 2  VB3  VB2  VBA
VB3 /2  82.613 mi/h65.2 =82.613 mi/h N 24.8 E
3.4
Ans.
Wheel 2 rotates at 600 rev/min cw and drives wheel 3 without slipping. Find the velocity
difference between points B and A.
2 
 600 rev/min  2 rad/rev 
60 s/min
 20 rad/s cw
VAO2  2 RAO2  80  251 in/s
VBA  VB  VA
Construct the velocity polygon
VB  223 in/s90
VBA  251ˆi  223ˆj in/s
Ans.
 335.75 in/s138.4
67
3.5
The distance between points A and B, located along the radius of a wheel, is
RBA  300 mm. The speeds of points A and B are VA = 80 m/s and VB = 140 m/s,
respectively. Find the diameter of the wheel; the velocities VAB and VBA ; and the angular
velocity of the wheel.
   
V  V  V   140ˆj   80ˆj  60ˆj m/s
VAB  VA  VB  80ˆj  140ˆj  60ˆj m/s
BA
B
Ans.
Ans.
A
VAB
60 m/s

 200 rad/s cw
RAB 0.300 m
VBO2
140 m/s
RBO2 

 0.700 m  700 mm
2 200 rad/s
3.6
2 
Ans.
Dia  2 RBO2  2  0.7 m  1.4 m  1 400 mm
Ans.
An airplane takes off from point B and flies east at 350 mi/h. Simultaneously, another
airplane at point A, 200 miles southeast, takes off and flies northeast at 390 mi/h. (a)
How close will the airplanes come to each other if they fly at the same altitude? (b) If
both airplanes leave at 6:00 p.m., at what time will this occur?
VA  39045 mi/h  276ˆi  276ˆj mi/h ; VB  350ˆi mi/h
V  V  V  74ˆi  276ˆj mi/h
BA
B
A
At initial time R BA  0   200 mi120  100ˆi  173ˆj mi
Later R (t )  R  0  V t   100  74t  ˆi  173  276t  ˆj mi
BA
BA
BA
2
  100  74t   173  276t 
To find the minimum of this: RBA
2
2
2
dRBA
dt  2  100  74t  74   2 173  276t  276   0
163 120t 110 376  0 ; t  0.677 h  41 min or 6:41p.m.
Ans.
R BA  0.677 h   49.8ˆi  13.4ˆj  51.5 mi  165
Ans.
RAB  200 mi
68
3.7
Include a wind of 30 mi/h from the west with the data of Problem 3.6. (a) If airplane A
flies the same heading, what is its new path? (b) What change does the wind make in the
results of Problem 3.6?
With the added wind VA  306ˆi  276ˆj mi/h  412 mi/h42 ; VB  380ˆi mi/h
V  V  V  74ˆi  276ˆj mi/h
BA
B
Ans.
A
Since the velocities are constant, the new path is a straight line at N 48º E.
Since the velocities of both planes change by the same amount, the velocity difference
Ans.
VBA does not change. Therefore the results of Problem 3.6 do not change.
3.8
For the double-slider linkage in the posture illustrated, the velocity of point B is 40 m/s.
Find the velocity of point A and the angular velocity of link 3.
RAB  400 mm
?
?
V A  V B  V AB
VA  48.99 m/s  165
VAB  14.64 m/s  120
V
14.64 m/s
3  AB 
 36.60 rad/s ccw
RAB
0.400 m
Ans.
Ans.
69
3.9
The four-bar linkage in the posture illustrated is driven by crank 2 at 2  45 rad/s ccw.
Find the angular velocities of links 3 and 4.
RAO  4 in, RBA  10 in, RO O  10 in, and RBO  12 in.
2
VAO2  2 RAO2   45 rad/s  4 in   180 in/s
?
4 2
4
?
VB  V A  V BA  VO4  V BO4
VBA  14.34 in/s ; VBO4  184.76 in/s .
VBA 14.34 in/s

 1.43 rad/s ccw
RBA
10 in
V
184.76 in/s
4  BO4 
 15.40 rad/s ccw
RBO4
12 in
3 
Ans.
Ans.
70
3.10
The four-bar linkage in the posture illustrated is driven by crank 2 at 2  60 rad / s cw.
Find the angular velocities of links 3 and 4 and the velocity of pin B and point C on link
3.
RAO  150 mm, RBA  300 mm, RO O  75 mm, RBO  300 mm, RDA  150 mm, and RCD  100 mm.
2
4 2
4
VAO2  2 RAO2   60 rad/s  0.150 m   9.0 m/s
?
?
VB  V A  V BA  VO4  V BO4
VBA  13.020 m/s ; VB  11.360 m/s41
?
Ans.
?
VC  V A  VCA  V B  VCB
VC  3.830 m/s60
V
13.020 m/s
3  BA 
 43.40 rad/s cw
RBA
0.300 m
V
11.360 m/s
4  BO4 
 37.87 rad/s cw
RBO4
0.300 m
Ans.
Ans.
Ans.
71
3.11
The four-bar linkage in the posture illustrated is driven by crank 2 at 2  48 rad / s ccw.
Find the angular velocity of link 3 and the velocity of point C on link 4
RAO  8 in, RBA  32 in, RO O  16 in, RBO  16 in, and RCO  12 in.
2
4 2
4
4
VAO2  2 RAO2   48 rad/s 8.0 in   384.0 in/s
?
?
VB  V A  V BA  VO4  V BO4
VBA  10.7 in/s
?
?
VC  VO4  VCO4  V B  VCB
VC  284.7 in/s  75.8
V
10.7 in/s
3  BA 
 0.335 rad/s ccw
RBA
32.0 in
Ans.
Ans.
72
3.12
For the parallelogram four-bar linkage, demonstrate that 3 is always zero and that
4  2 . How would you describe the motion of link 4 with respect to link 2?
Referring to Fig. 2.19 and using r1  r3 and r2  r4 , we compare Eqs. (2.26) and (2.27) to
see that    . Then Eq. (2.29) gives 3  0 and its derivative is 3  0 .
Ans.
Next we substitute Eq. (2.25) into Eq. (2.33) to see that    2 .
Then Fig. 2.19 shows that, since link 3 is parallel to link 1 3  0  ,
then  4     2 . Finally, the derivative of this gives 4  2 .
Ans.
Since 4/ 2  4  2  0 , link 4 is in curvilinear translation with respect to link 2.
Ans.
73
3.13
The antiparallel, or crossed, four-bar linkage in the posture illustrated is driven by link 2
at 2  1 rad/s ccw. Find the velocities of points C and D.
RAO  RBO  300 mm , RBA  RO O  150 mm , and RCA  RDB  75 mm
2
4
4 2
VAO2  2 RAO2  1 rad/s  0.300 m   0.300 m/s
?
?
VB  V A  V BA  VO4  V BO4
Construct the velocity image of link 3.
VC  0.402 m/s151
VD  0.290 m/s  111
Ans.
Ans.
74
3.14
For the four-bar linkage in the posture illustrated, link 2 has an angular velocity of 60
rad/s ccw. Find the angular velocities of links 3 and 4 and the velocity of point C.
RAO  RBA  6 in, RO O  RBO  10 in, and RCA  8 in.
2
4 2
4
VAO2  2 RAO2   60 rad/s  6.0 in   360.0 in/s
?
?
VB  V A  V BA  VO4  V BO4
Construct the velocity image of link 3:
VC  507.1 in/s156.9
VBA  169.4 in/s ;
VBO4  317.6 in/s
VBA 169.4 in/s

 28.24 rad/s cw
RBA
6.0 in
V
317.6 in/s
4  BO4 
 31.76 rad/s ccw
RBO4
10.0 in
3 
Ans.
Ans.
Ans.
75
3.15
Crank 2 of the inverted slider-crank linkage, in the posture illustrated, is driven at
2  60 rad/s ccw. Find the angular velocities of links 3 and 4 and the velocity of point
B.
RAO  75 mm, RBA  400 mm, and RO O  125 mm.
2
4 2
VAO2  2 RAO2   60 rad/s  0.075 m   4.500 m/s
?
?
VP3  V A  V P3 A  VP4  V P3 /4
Construct the velocity image of link 3:
VB  4.789 m/s96.5
VP A 4.259 m/s
4  3  3 
 22.0 rad/s ccw
RP3 A
0.194 m
Ans.
Ans.
76
3.16
For the four-bar linkage in the posture illustrated, crank 2 has an angular velocity of 30
rad/s cw. Find the velocity of the coupler point C and the angular velocities of links 3
and 4.
RAO  3 in, RBA  RCB  5 in, RO O  10 in, and RBO  6 in.
2
4 2
4
VAO2  2 RAO2   30 rad/s  3.0 in   90.0 in/s
?
?
VB  V A  V BA  VO4  V BO4
Construct the velocity image of link 3:
VC  90.0 in/s126.9
V
90.0 in/s
3  BA 
 18.00 rad/s ccw ;
RBA
5.0 in
V
0
4  BO4 
0
RBO4 8.0 in
Ans.
Ans.
Ans.
77
3.17
For the modified slider-crank linkage in the posture illustrated, crank 2 has an angular
velocity of 10 rad/s ccw. Find the angular velocity of link 6 and the velocities of points
B, C, and D.
RAO  2.5 in, RBA  10 in, RCB  8 in, RCA  RDC  4 in, RO O  8 in, and RDO  6 in.
2
6
2
6
VAO2  2 RAO2  10 rad/s  2.5 in   25.0 in/s
?
?
V B  V A  V BA
VB  11.57 in/s180
Construct velocity image of link 3:
??
?
?
VC  V A  VCA  V B  VCB
VC  24.22 in/s207.6
??
?
Ans.
Ans.
?
V D  V C  V DC  V
 O6  V DO6
VD  24.18 in/s206.2
VDO6 24.18 in/s
6 

 4.03 rad/s ccw
RDO6
6.0 in
Ans.
Ans.
78
3.18
For the four-bar linkage illustrated, the angular velocity of crank 2 is a constant 16 rad/s
cw. Plot a polar velocity diagram for the velocity of point B for all crank positions.
Check the positions of maximum and minimum velocities by using Freudenstein’s
theorem.
RAO  350 mm, RBA  425 mm, RO O  100 mm, and RBO  400 mm.
2
4 2
4
The graphic construction is shown in the posture where  2  135 , where the result is
VB  5.76 m/s  7.2 . It is repeated at increments of  2  15 . The maximum and
minimum velocities are VB,max  9.13 m/s  146.6 at  2  15 and VB,min  4.59 m/s63.7
at  2  225 , respectively. Within graphic accuracy these two positions approximately
verify Freudenstein’s theorem.
A numeric solution for the same problem can be found from Eq. (3.22) using Eqs. (2.25)
through (2.33) for position values. The accuracy of the values reported above have been
verified in this manner.
79
3.19
For the four-bar linkage in the posture illustrated, link 2 is driven at 2  36 rad/s cw.
Find the angular velocity of link 3 and the velocity of point B.
RAO  5 in, RBA  RBO  8 in, and RO O  7 in.
2
4
4 2
VAO2  2 RAO2   36 rad/s  5.0 in   180.0 in/s
?
?
VB  V A  V BA  VO4  V BO4
VBA 25.9 in/s

 3.23 rad/s ccw
RBA
8.0 in
VB  202.8 in/s  56.3
3 
Ans.
Ans.
80
3.20
For the four-bar linkage in the posture illustrated, the angular velocity of the input link 2
is 8 rad/s ccw. Find the velocity of point C and the angular velocity of link 3.
RAO  150 mm, RBA  RBO  250 mm, RO O  75 mm, RCA  300 mm, and RCB  100 mm.
2
4
4 2
VAO2  2 RAO2  8 rad/s  0.150 m   1.200 m/s
?
?
VB  V A  V BA  VO4  V BO4
Construct velocity image of link 3:
??
?
?
VC  V A  VCA  V B  VCB
VC  3.848 m/s  136.8
V
3.784 m/s
3  BA 
 15.14 rad/s ccw
RBA
0.250 m
Ans.
Ans.
81
3.21
For the four-bar linkage in the posture illustrated, link 2 has an angular velocity of 56
rad/s ccw. Find the velocity of point C.
RAO2  150 mm, RBA  RBO  250 mm, RO O  100 mm, and RCA  300 mm.
4
4 2
VAO2  2 RAO2   56 rad/s  0.150 m   8.400 m/s
?
?
VB  V A  V BA  VO4  V BO4
Construct the velocity image of link3:
??
?
?
VC  V A  VCA  V B  VCB
VC  9.028 m/s137.8
Ans.
82
3.22
For the double-slider linkage in the posture illustrated, the angular velocity of the input
crank 2 is 42 rad/s cw. Find the velocities of points B, C, and D.
RAO  2 in, RBA  10 in, RCA  4 in, RCB  7 in, and RDC  8 in.
2
VAO2  2 RAO2   42 rad/s 2.00 in   84.00 in/s
?
?
V B  V A  V BA
VB  65.36 in/s180
Construct velocity image of link 3:
??
?
?
VC  V A  VCA  V B  VCB
VC  67.86 in/s154.2
?
Ans.
Ans.
?
V D  V C  V DC
VD  21.23 in/s90
Ans.
83
3.23
For the linkage used in a two-cylinder 60 V-engine consisting, in part, of an articulated
connecting rod, crank 2 rotates at 2 000 rev/min cw. Find the velocities of points B, C,
and D.
RAO  2 in, RBA  RCB  6 in, RCA  2 in, and RDC  5 in.
2
2000 rev/min  2 rad/rev 
 209.4 rad/s
60 s/min
VAO2   2 RAO2
2 
  209.4 rad/s  2.0 in   418.9 in/s
?
?
V B  V A  V BA
VB  425.9 in/s  120
Construct velocity image of link 3:
??
?
?
VC  V A  VCA  V B  VCB
VC  486.9 in/s  93.3
?
Ans.
Ans.
?
V D  V C  V DC
VD  378.7 in/s  60
Ans.
84
3.24
For the inverted slider-crank linkage in the posture illustrated, the angular velocity of the
crank is 2  24 rad/s cw. Make a complete velocity analysis of the linkage. What is the
absolute velocity of point B? What is its apparent velocity to an observer moving with
link 4?
RAO  8 in and RO O  20 in.
2
4 2
VAO2  2 RAO2   24 rad/s 8.00 in   192.0 in/s
Using the path of P3 on link 4, we write
?
?
VP3  V A  V P3 A  VP4  V P3 /4
3 
VP3 A
RP3 A

161.8 in/s
 6.16 rad/s cw
26.27 in
From this, or from the velocity image of link 3, we find
VB3  157.8 in/s  39.7
Ans.
Then, since link 4 remains perpendicular to link 3, we have 4 = 3 and we find the
velocity image of link 4:
VB3 /4  103.3 in/s  12.4
Ans.
85
3.25
For the linkage in the posture illustrated, the velocity of point A is 1iˆ ft/s. Find the
velocity of coupler point B.
VA  12 in/s
Using the path of P3 on link 4, we write
?
?
VP3  V A  V P3 A  VP4  V P3 /4
Next we construct the velocity image of link 3
?
?
VB  V P3  V BP3  V A  V BA
VB  12.50 in/s  23.0
Ans.
86
3.26
A variation of the Scotch-yoke linkage in the posture illustrated, is driven by crank 2 at
2  36 rad/s ccw. Find the velocity of the crosshead, link 4.
RAO  250 mm.
2
VAO2  2 RAO2   36 rad/s  0.250 m   9.0 m/s
Using the path of A2 on link 4, we write
?
?
V A2  V A4  V A2 /4
(Note that the path is unknown for VA4 / 2 !)
VA4  4.658 m/s180
All other points of link 4 have this same velocity since link 4 is in translation.
Ans.
87
3.27
Perform a complete velocity analysis of the modified four-bar linkage for 2  72 rad/s
ccw.
RAO  RDC  1.5 in, RBA  10.5 in, RO O  6 in, RBO  5 in, RO O  7 in, and REO  8 in.
2
4
2
4
6
2
6
VAO2  2 RAO2   72 rad/s 1.50 in   108.0 in/s
?
?
VB  V A  V BA  VO4  V BO4
Construct the velocity image of link 3
?
?
VC  V A  VCA  V B  VCB
VC3  76.32 in/s203.2
Ans.
Using the path of C3 on link 6, we next write
?
?
V C3  VC6  VC3 /6 and VC6  VC6O6
from which
VC6  43.07 in/s241.4
Then we can complete the velocity image of link 6
VE  77.39 in/s  98.9
Since link 5 remains perpendicular to link 6,
VC O
5  6  6 6  9.67 rad/s cw
RC6O6
Ans.
Ans.
Ans.
From these we can find
VDC  5 RDC   9.67 rad/s 1.5 in   14.505 in/s
??
V D5  V C5  V D5C5  64.49 in/s210.1
Ans.
88
3.28
For the mechanism in the posture illustrated, the velocity of point C is VC = 10 in/s to the
left. There is rolling contact between links 1 and 2, but slip is possible between links 2
and 3. Determine the angular velocity of link 3.
1  6 in, 2  0.75 in, R AD  2.5ˆi  1.75ˆj in, and RED  1 in.
(Note that the path of C3 on link 2 for VC3 /2 is unknown!)
Using the path of C2 on link 3, we write
??
3 
VC3D3
RC3D3
??
?
V C2  VC3  VC2 /3

and
?
VC3  VD3  VC3D3
4.226 in/s
 1.569 rad/s ccw
2.69 in
Ans.
89
3.29
For the circular cam in the posture illustrated, the angular velocity of the cam is 2  15
rad/s ccw. There is rolling contact between the cam and the roller, link 3. Find the
angular velocity of the oscillating follower, link 4.
VB  VBA  2 RBA  15 rad/s1.25 in   18.75 in/s
Construct the velocity image of link 2.
?
?
VD4  V D2  V D4 /2  VE  V D4 E
4 
VD4 E
RD4 E

15.24 in/s
 4.355 rad/s ccw
3.500 in
Ans.
90
3.30
The mechanism in the posture illustrated is driven by link 2 at 10 rad/s ccw. There is
rolling contact at point F. Determine the velocities of points E and G and the angular
velocities of links 3, 4, 5, and 6.
VBA  2 RBA  10 rad/s 1.000 in   10.00 in/s
?
?
?
?
VC  V B  VCB  VD  VCD
Construct velocity image of link 3:
VE3  V B  V E3B  V C  V E3C
VE3  10.06 in/s220.9
Ans.
Using the path of E3 on link 6,
??
??
?
?
V E3  V E6  V E3 /6 and V E6  VH  V E6H
Construct velocity image of link 6:
?
?
VG  V E6  VGE6  VH  VGH
VG  11.93 in/s  57.1
V
13.33 in/s
3  CB 
 3.333 rad/s ccw
RCB
4.000 in
V
6.667 in/s
4  CD 
 3.333 rad/s ccw
RCD
2.000 in
VF E 12.78 in/s
5  5 
 25.56 rad/s cw
RF5 E
0.500 in
6 
VE6 H
RE6 H

4.858 in/s
 3.774 rad/s cw
1.288 in
Ans.
Ans.
Ans.
Ans.
Ans.
91
3.31
The two-piston pump, in the posture illustrated, is driven by a circular eccentric, link 2, at
2  25 rad/s ccw. Find the velocities of the two pistons, links 6 and 7.
VF2  VF2 E  2 RFE   25 rad/s 1.0 in   25.00 in/s
Using the path of F2 on link 3, we write
??
?
?
V F2  V F3  V F2 /3
VF3  VG  V F3G
and
Construct velocity image of link 3:
?
?
?
?
VC  V F3  VCF3  VG  VCG
VD  V F3  V DF3  VG  V DG . Then
?
?
V A  V C  V AC
VA  1.302 in/s180
?
Ans.
?
V B  V D  V BD
VB  5.777 in/s180
Ans.
92
3.32
The epicyclic gear train is driven by the arm, link 2, at 2  10 rad/s cw. Determine the
angular velocity of the output shaft, attached to gear 3.
VB  VBA  2 RBA  10 rad/s  3.000 in   30.00 in/s
Using VD4  0 construct the velocity image of link 4.
VC  60 in/s0 .
V
60.00 in/s
3  CA 
 30.00 rad/s cw
RCA
2.000 in
Ans.
93
3.33
The diagram illustrates a planar schematic approximation of an automotive front
suspension. The roll center is the term used by the industry to describe the point about
which the auto body seems to rotate (roll) with respect to the ground. The assumption is
made that there is pivoting but no slip between the tires and the road. After making a
sketch, use the concepts of instant centers to find a technique to locate the roll center.
By definition, the “roll center” (of the vehicle body, link 2, with respect to the road, link
1,) is the instantaneous center I12. It can be found by the repeated application of
Kennedy’s theorem as shown.
In the automotive industry it has become common practice to use only half of this
graphic construction, assuming, by symmetry, that I12 must lie on the vertical centerline
of the vehicle. Note that this is true only when the right and left suspension arms are
symmetrically positioned. It is not true once the vehicle begins to roll, as in a turn.
Having lost sight of the relationship to instantaneous centers and Kennedy’s theorem,
and remembering only the shortened graphic construction on one side of the vehicle,
many in the industry are now confused and believe that the movement of the roll center,
as roll progresses, is vertical, along the centerline of the vehicle (sometimes called the
“jacking coefficient”!). They should be thinking about the fixed and moving centrodes
(Sec. 3.21), which are more horizontal than vertical!
94
3.34
Locate all instant centers for the linkage of Problem 3.22.
Primary instant centers I12 ,
I 23 , I 34 , I14 (at infinity),
I 35 , I 56 , and I16 (at infinity)
are found by inspection.
All others are found by
repeated applications of
Kennedy’s theorem except
I 46 .
One line can be found by
Kennedy’s theorem for I 46 ;
however, no second line
can be found by Kennedy’s
theorem since no line can
be drawn (in finite space)
between I14 and I16 .
However, we can see that
I 46 must be infinitely
remote because the relative
motion between links 4
and 6 is translation; that is,
the angle between lines on
links 4 and 6 remains
constant.
95
3.35
Locate all instant centers for the mechanism of Problem 3.25.
Primary instant centers I12 (at infinity), I 23 , I 34 (at infinity), and I14 are found by
inspection. All others are found by repeated applications of Kennedy’s theorem.
96
3.36
Locate all instant centers for the mechanism of Problem 3.26.
Primary instant centers I12 , I 23 , I 34 (at infinity), and I14 (at infinity) are found by
inspection. All others are found by repeated applications of Kennedy’s theorem except
I13 .
One line ( I12 I 23 ) can be found for I13 ; however, no second line can be found by
Kennedy’s theorem since no line can be drawn (in finite space) between I14 and I 34 .
However, it may be seen that I13 must be infinitely remote because the relative motion
between links 1 and 3 is translation; that is, the angle between links 1 and 3 is constant.
97
3.37
Locate all instant centers for the mechanism of Problem 3.27.
Primary instant centers I12 , I 23 , I 34 , I14 , I 35 , I 56 (at infinity), and I16 are found by
inspection. All others are found by repeated applications of Kennedy’s theorem.
98
3.38
Locate all instant centers for the mechanism of Problem 3.28.
Primary instant centers I12 and I13 are found by inspection.
One line for I 23 is found by Kennedy’s theorem. The other is found on the perpendicular
to the relative velocity of slipping at the point of contact between links 2 and 3.
99
3.39
Locate all instant centers for the mechanism of Problem 3.29.
Primary instant centers I12 , I 23 , I 34 , and I14 are found by inspection. The other two are
found by Kennedy’s theorem.
100
3.40
The posture of the input link 2 is R AO4  120ˆi mm and the velocity of point A is
V  15ˆi m/s . Determine the first-order kinematic coefficients for the mechanism. Find
A
the angular velocities of links 3 and 4.
RBO4  RBA  120 mm
The loop-closure equation is
I
?
?
r2  r3  r4  0 .
The two scalar position equations are
r2  r3 cos 3  r4 cos  4  0
r3 sin 3  r4 sin 4  0
With the given data, at the position r2  120 mm , the solution is 3  60 and 4  120 .
Taking the derivative of the position equations with respect to input r2 gives
1  r3 sin 33  r4 sin  4 4  0
r3 cos 33  r4 cos  4 4  0
 r3 sin 3 r4 sin  4  3   1
or, in matrix format, 
    
 r3 cos 3 r4 cos  4   4   0 
The determinant of the Jacobian is   r3r4 sin 4  3  and goes to zero when 3   4 or
when 3  4  180 .
The solutions for the first-order kinematic coefficients are
r2  1 m/m, 3  r4 cos 4   4.811 rad/m , and 4  r3 cos 3   4.811 rad/m .
The input velocity is given as r2  15.0 m/s .
3  3r2  72.17 rad/s (ccw) and 4  4r2  72.17 rad/s (cw)
Ans.
Ans.
101
3.41
For the rack-and-pinion mechanism in the posture illustrated, link 2 is the input and
pinion 3 is rolling without slipping on rack 4 at point D. Determine the first-order
kinematic coefficients of the links 3 and 4. If the constant input velocity is VG  3ˆi in/s,
determine the angular velocities of rack 4 and pinion 3.
RGO4  10 in, and RDG  3  5 in .
I
C
?
The loop-closure equation is r2  ρ3  r4  0 with the constraint 3  4  90. The two
scalar position equations are
r2  3 sin  4  cos  4 r4  0
 3 cos  4  sin  4 r4  0
At the position r2  10 in , the solution is 4  150 and r4   3 tan 4  8.660 in .
Taking the derivative of the position equations with respect to input r2 gives
1  3 cos  4 4  sin  4 r4 4  cos  4 r4  0
3 sin  4 4  cos  4 r4 4  sin  4 r4  0
or, simplifying by use of the position equations and putting into matrix format,
 0 cos  4   4  1 
 r sin    r    0
 
2
4 4 
The determinant of the Jacobian is   r2 cos 4 and goes to zero when 4  90 or r2  0 .
The solutions for the first-order kinematic coefficients are
4  sin 4   0.057 7 rad/in and r4  r2   1.154 7 in/in
Ans.
The input velocity is given as r2  3.0 in/s . From this we can get
4  4r2  0.173 2 rad/s (cw)
Ans.
However, we must notice that vector ρ3 is not attached to link 3. To find 3 we start
with the constraint for rolling with no slip. If we designate rotation of link 3 by the angle
 3 then 3  3  4   r4 . Dividing this by t and taking the limit, we get the
angular velocity of the pinion, link 3
3  3  4  r4 3  0.519 6 rad/s (ccw)
Ans.
102
3.42
For the rack-and-pinion mechanism of Example 2.8, (Figures 2.33 and 2.34), the
dimensions are R1  800 mm , R9  550 mm , 34  60 , and 3  500 mm . In the
posture where R  750 mm , the input link 2 has a velocity of V  0.150ˆj m/s .
A
2
Determine the first-order kinematic coefficients to obtain the velocity of the rack 4 and
the angular velocity of pinion 3.
Using the vectors defined in Example 2.8, the complex algebra loop-closure equation is
jR2  jR9e j34  R34e j34  jR1  R4e j  0
and the two scalar position equations are
 R9 sin 34  R34 cos 34  R4  0
R2  R9 cos 34  R34 sin 34  R1  0
At R2  750 mm with the given dimensions these give R34  259.8 mm and R4  606.2 mm
with the no-slip condition that R34  33 .
Taking the derivative of the position equations with respect to input R2 gives
  R4  0
 cos 34 R34
 0
1  sin 34 R34
  33 .
with the condition that R34
From these, the first-order kinematic coefficients are
  1.154 7 m/m , 3  2.309 4 rad/m , and R4  0.577 35 m/m
R34
The velocity of the rack is V   R R ˆi  0.086 6ˆi m/s
Ans.
The angular velocity of the pinion is 3  3R2  0.346 4 rad/s ccw.
Ans.
4
4
2
Ans.
103
3.43
For the mechanism in the posture illustrated, RAO4  10 in and the input velocity is
V  5ˆi in/s . Determine the first-order kinematic coefficients to obtain the angular
A
velocity of link 3 and the slipping velocity between links 3 and 4.
RPA  5 in and APO4  90.
Let the following vectors be defined as R AO4  r2e j 0 , R PA  r3e j3 , R PO4  jr4e j3 . Then
the loop-closure equation is
r2  r3e j3  jr4e j3  0
The two scalar equations are
r2  r3 cos 3  r4 sin 3  0
r3 sin 3  r4 cos 3  0
which, at the input position r2  10 in , has the solution 3  cos1   r3 r2   240 and
r4  r3 tan 3  8.66 0 in .
Taking the derivative of the position equations with respect to input r2 gives
1  r3 sin 33  r4 cos 33  sin 3r4  0
r3 cos 33  r4 sin 33  cos 3r4  0
or, simplifying by use of the position equations and putting into matrix format,
sin 3  3   1
 0
 r  cos    r     0 
 
3  4 
 2
The determinant of the Jacobian is   r2 sin 3 and goes to zero when 3  0 at r2  5 in .
From the solution of these equations, the first-order kinematic coefficients are
3  cos3   0.057 7 rad/in and r4  r2   1.154 7 in/in
Ans.
For the given input velocity of r2  5.0 in/s ,
the angular velocity of link 3 is 3  3  3r2  0.289 rad/s (cw)
and the slipping velocity is V3/4  r4  r4r2  5.774 in/s .
Ans.
Ans.
104
3.44
For the mechanism in the posture illustrated in Figure P3.30, the input crank 2 has an
angular velocity 2  10 rad/s ccw and there is rolling contact between links 5 and 6 at
point F. Determine the first-order kinematic coefficients of links 3, 4, 5, and 6. Find the
angular velocities of links 3, 4, 5, and 6 and the velocities of points E and G.
Let the following vectors be defined:
R BA  r2e j2 ,
RCB  r3e j3 ,
R DA  r1e j 0 ,
RCD  r4e j4 , R EB  ½r3e j3  j1.5e j3 in, R HA  1.0  j1.5 in, and R EH  j 0.5e j6  r6e j6 .
Then there are two loop-closure equations
R BA  R CB  R CD  R DA  0
R BA  R EB  R EH  R HA  0
and four corresponding scalar equations
r2 cos  2  r3 cos 3  r4 cos  4  r1  0
r2 sin  2  r3 sin 3  r4 sin  4  0
r2 cos  2  ½ r3 cos 3  1.5sin 3  0.5sin 6  r6 cos 6  1.0  0
r2 sin  2  ½ r3 sin 3  1.5cos 3  0.5cos  6  r6 sin  6  1.5  0
Numeric solution of these with the dimensions specified at the posture with 2  180
gives the current position as 3  28.955, 4  75.522, 6  14.478, r6  1.186 in.
Taking derivatives of these four equations with respect to input  2 gives
r2 sin  2  r3 sin 33  r4 sin  4 4  0
r2 cos  2  r3 cos 33  r4 cos  4 4  0
r2 sin  2  ½ r3 sin 33  1.5cos 33  0.5cos  6 6  r6 sin  6 6  cos  6 r6  0
r2 cos  2  ½ r3 cos 33  1.5sin 33  0.5sin  6 6  r6 cos  6 6  sin  6 r6  0
105
Numeric solution gives the solution as 3  0.333 34 rad/rad, 4  0.333 34 rad/rad,
6  0.377 51 rad/rad, and r6  1.089 56 in/rad.
The no-slip condition gives the displacement constraint r6  5  5  6  from which
we find r6  5 5  6  , which gives 5  6  r6 5  2.556 63 rad/rad.
Therefore the first-order kinematic coefficients are
3  0.333 3 rad/rad, 4  0.333 3 rad/rad, 5  2.556 6 rad/rad, 6  0.377 5 rad/rad. Ans.
The angular velocities are
3  32  3.333 rad/s ccw, 4  42  3.333 rad/s ccw, 5  52  25.566 rad/s (cw),
and 6  62  3.775 rad/s (cw).
Ans.
The positions of points E and G are
xE  r2 cos  2  ½ r3 cos 3  1.5sin 3
yE  r2 sin  2  ½ r3 sin 3  1.5cos 3
xG  1.0  1.0sin  6  3.0 cos  6
yG  1.5  1.0 cos  6  3.0sin  6
The derivatives of these give the first-order kinematic coefficients
xE   r2 sin  2  ½r3 sin 33  1.5cos 33  0.760 26 in/rad
yE   r2 cos  2  ½ r3 cos 33  1.5sin 33  0.658 72 in/rad
xG  1.0cos 66  3.0sin 66  0.648 65 in/rad
yG  1.0sin 66  3.0cos 66  1.002 19 in/rad
And the velocities are
VE  xE  2ˆi  yE  2 ˆj  7.603ˆi  6.587ˆj  10.059 in/s  139.09
V  x  ˆi  y  ˆj  6.487ˆi  10.022ˆj  11.938 in/s  57.09
G
G 2
G 2
Ans.
106
3.45
For the mechanism in the posture illustrated, where RAO4  40 mm, the 4 is rolling
without slipping on rack 3 at point B. Determine the first-order kinematic coefficients of
rack 3 and pinion 4. If VA  150ˆi mm/s, determine the angular velocities of rack 3 and
pinion 4 and the velocity of point E. Also, determine the velocity along rack 3 of the
point of contact between links 3 and 4 (that is, point B).
Let the following vectors be defined: R AO4  jr2 ,
R BA  r3e j3 , and R BO4  r4e j4  jr4e j3 .
Then
the
loop-closure
equation
R AO4  R BA  R BO4  0 and the scalar equations are
is
r3 cos 3  r4 sin 3  0
r2  r3 sin 3  r4 cos 3  0
The position solution for the given data at
r2  40.0 mm is 3  120, r3  34.641 mm .
Taking derivatives of these equations with respect to
input r2 gives
cos 3r3  r3 sin 33  r4 cos 33  0
4  20 mm and REB  RBA .
1  sin 3r3  r3 cos 33  r4 sin 33  0
or, simplifying by use of the position equations and putting into matrix format,
cos 3 r2   r3   0 
 sin  0       1
3

  3  
The determinant of the Jacobian is   r2 sin 3 and goes to zero when 3  0 or 180.
From these, the first-order kinematic coefficients are r3  1 sin 3 and 3  1  r2 tan 3 
The no-slip condition gives the displacement constraint r3  4  4  3  from
which we find r3  4 4  3  , which gives 4  3  r3 4 .
Therefore the first-order kinematic coefficients are
r3  1.1547 m/m, 3  14.43 rad/m, 4  43.30 rad/m.
Ans.
For r2  0.150 m/s , the angular velocities of links 3 and 4 are
3  3r2  2.165 rad/s ccw and 4  4r2  6.495 rad/s (cw) .
Ans.
Given that REA = 2r3 = 69.3 mm, the position of point E is R E  xE  jyE  jr2  69.3e j3
xE  69.3cos 3  34.65 mm and yE  r2  69.3sin 3  20.02 mm
The derivative with respect to input r2 gives
xE  69.3sin 33  0.8663 m/m
yE  1  6.93cos 33  0.4999 m/m
and
xE  xE r2  0.129 95 m/s and
yE  yE r2  0.074 99 m/s
The velocity of point E is VE  0.150 04 m/s150.01
Ans.
The velocity along rack 3 of the point of contact between links 3 and 4 is
VB4 /3  r3  r3r2   1.1547 m/m  0.15 m/s   0.173 20 m/s
VB4 /3  0.173 20 m/s  60
Ans.
107
3.46
For the mechanism in the posture illustrated, where 2  150, RPA  RAO4 , and
RPB  RBA , determine the first-order kinematic coefficients of links 3, 4, and 5. If the
angular velocity of the input link 2 is ω2 = 5 rad/s cw, determine: (a) the angular
velocities of links 3 and 4, (b) the velocity of link 5, and (c) the velocity of point P fixed
in link 4.
RAO2  10 in and RPO4  20 in
Using instant centers, the first-order kinematic coefficients for links 3, 4, and 5 are
RI I
10.000 in
3  23 12 
 0.500 rad/rad
RI23 I13
20.000 in
 4 
RI24 I12
RI24 I14

5.774 in
 0.500 rad/rad
11.548 in
xB  0, and rB  yB  RI25 I12  8.660 in/rad
From these, with 2  5 rad/s (cw),
3  32  2.50 rad/s ccw, 4  42  2.50 rad/s ccw
(a)
V  r   43.30ˆj in/s
(b)
B
(c)
B
2
Ans.
Ans.
Ans.
Ans.
Ans.
rP  4 RPI14   0.500 rad/rad  20.000 in   10.000 in/rad
VP  rP2  50.00 in/s120
Ans.
108
3.47
For the inverted slider-crank linkage in the posture illustrated, where θ4 = 60º, the input
link 2 is moving parallel to the x-axis. Determine the first-order kinematic coefficients of
links 3 and 4. Also, determine the conditions for the determinant of the coefficient matrix
to become zero. If VB  15 in/s constant to the right, determine the angular velocities
of links 3 and 4.
(a)
(b)
RBA  RAO4  4 in.
The two scalar loop-closure equations are
xB  RBA cos 3  RAO4 cos 4  0
yB  RBA sin 3  RAO4 sin 4  0
The solution at the current position, with yB  7.464 in, is xB  2.000 in and 3  90 .
Taking the derivative with respect to input r2 gives
1  RBA sin 33  RAO4 sin  4 4  0
 RBA cos 33  RAO4 cos  4 4  0
which in matrix format becomes
 RBA sin 3

  RBA cos 3
RAO4 sin  4  3   1


 RAO4 cos  4   4   0 
The determinant of the Jacobian matrix is   RBA RAO4 sin 4  3   8.00 in 2 .
The first-order kinematic coefficients for links 3 and 4 are
3  RAO4 cos 4   0.25 rad/in and 4   RBA cos3   0
The angular velocities are
3  3r2  3.75 rad/s (cw)
Ans.
and 4  4r2  0
Ans.
The conditions for which   0 are that 3   4 or 3  4  180 ; e.g., this will happen
when 3  4  68.907 and xB  2.879 in as shown in part (b) of the figure above.
109
3.48
For the rack-and-pinion mechanism in the posture illustrated in Figure P2.15, the input
link 2 is vertical and BAO2 is 150°.
The dimensions are 5  2.5 in,
R
 8ˆi  4ˆj in, R  2 in, and R  6 in. (a) Show the locations of all instant
O5O2
AO2
BA
centers. (b) Using instant centers, determine the first-order kinematic coefficients of link
3, rack 4, and pinion 5. (c) If 2  10 rad/s cw, determine the angular velocity of link 3,
the velocity of rack 4, and the angular velocity of pinion 5.
(a) The instant centers are shown in the following figure:
(b) The first-order kinematic coefficients are
RI I
2.00 in
3  23 12 
 0.385 rad/rad
RI23 I13 5.20 in
r4  RI24 I12  1.15 in/rad
5 
RI25 I12
RI25 I15

5.85 in
 0.463 rad/rad
12.63 in
Ans.
Ans.
Ans.
(c) The requested velocities are
3  32   0.385 rad/rad  10 rad/s   3.85 rad/s (ccw)
Ans.
V4  r42  1.15 in/rad  10rad/s   11.5 in/s    90 
Ans.
5  52   0.463 rad/rad  10 rad/s   4.63 rad/s (cw)
Ans.
110
3.49
For the mechanism in the posture illustrated in Fig. P2.16, 2  1 in, 5  2 in,
RBA  7.071 in, and RBC  6 in. Determine the first-order kinematic coefficients of links
3, 4, and 5. If link 2 is driven at 2  5 rad/s ccw, determine the angular velocities of
links 3, 4, and 5.
The two scalar loop closure equations are
RO5O2  5 cos 5  RBC cos  4  RBA cos 3  2 cos  2  0
5 sin 5  RBC sin  4  RBA sin 3  2 sin  2  0
with the rolling contact constraint equation
55   2 2 .
At the position 2  90 with the given
dimensions the solution is 3  45, 4  90, 5  0.
The derivatives of these equations with respect to
input  2 are
 5 sin 55  RBC sin  4 4  RBA sin 33  2 sin  2  0
5 cos 55  RBC cos44  RBA cos33  2 cos 2  0
with the constraint 55   2 .
In matrix form, these appear as
 RBA sin 3  RBC sin  4  3   5 sin 55  2 sin  2    2  sin 5  sin  2  



  R cos 
RBC cos  4   4    5 cos 55  2 cos  2   2  cos 5  cos  2  
3
 BA
The determinant is   RBA RBC sin 3  4   30.000 in 2 .
At 2  90 with the given dimensions the first-order kinematic coefficients are
3  2 RBC sin 4  5   sin 4  2    0.200 rad/rad
Ans.
4  2 RBC sin 3  5   sin 3  2    0
Ans.
5   2 5  0.500 rad/rad
Ans.
The angular velocities are
3  32  0.200 rad/rad  5 rad/s   1.00 rad/s (cw)
4  42  0
5  52  0.500 rad/rad  5 rad/s   2.50 rad/s (cw)
Ans.
Ans.
Ans.
111
3.50
For the mechanism in the posture illustrated in Figure P2.17, link 4 is parallel to the xaxis and link 5 is coincident with the y-axis. The radius of wheel 3 is 3  0.75 in,
RO2O5  7.0 in,
RBA  5.5 in, and RAO5  2.6 in.
Determine the first-order kinematic
coefficients of links 3, 4, and 5. If the input link 2 has an angular velocity of
2  15 rad/s cw, determine the angular velocities of links 3, 4, and 5.
The two scalar equations for loop closure are
RO2O5  RBO2 cos  2  RBA cos  4  RAO5 cos 5  0
RBO2 sin  2  RBA sin  4  RAO5 sin 5  0
with the rolling contact constraint equation 3  3  2    1  1  2  , and, since
1  0 , this reduces to 33   1  3  2 .
At the posture where 4  0, 5  90 and with the given dimensions, satisfaction of the loop
closure equation requires that RBO2  3.001 67 in, 1  2.251 67 in, and 2  119.982.
The derivatives of the loop-closure equations with respect to input  2 are
 RBO2 sin  2  RBA sin  4 4  RAO5 sin 55  0
RBO2 cos  2  RBA cos  4 4  RAO5 cos 55  0
In matrix form, these appear as
 RBA sin  4

  RBA cos  4
RAO5 sin 5   4   RBO2 sin  2 



 RAO5 cos 5  5    RBO2 cos  2 
and the constraint equation derivative gives 33   1  3  .
The determinant is   RBA RAO5 sin 5  4   14.3 in 2 .
At 2  120 with the given dimensions the first-order kinematic coefficients are
3   1  3  3  4.002 rad/rad
Ans.
4  RBO RAO sin 5  2    0.272 73 rad/rad
Ans.
5  RBA RBO sin 2  4    1.000 rad/rad
Ans.
2
5
2
The angular velocities are
3  32  4.002 rad/rad  15 rad/s  60.033 rad/s (cw)
Ans.
4  42  0.272 73 rad/rad  15 rad/s  4.091 rad/s (ccw)
Ans.
5  52  1.000 rad/rad  15 rad/s  15.000 rad/s (cw)
Ans.
112
3.51
For the mechanism in the posture illustrated, wheel 3 is rolling without slipping on the
ground link at point C while sliding in the slot in link 2. Write the vector loop equation
and determine the first-order kinematic coefficients of the mechanism. If the angular
velocity of the input is 2  30 rad/s ccw , determine: the angular velocity of the wheel;
and the apparent velocity of the center of the wheel, point A, with respect to the slot in
link 2.
1  60 mm, 3  15 mm, RCO  60ˆi  60ˆj mm, and RAO  75 mm.
2
2
Using the vectors shown in the figure above, the vector loop equation can be written as
jr1  r13e j13  r2e j2  0
Ans.
From this, the two scalar equations are
r13 cos 13  r2 cos 2  0
r1  r13 sin13  r2 sin2  0
At the current posture, r13  45 mm and 13  180. Therefore, these equations give
r2  75 mm and 2  tan1  45 mm 60 mm   126.87.
The derivatives of the above loop-closure equations with respect to input  2 give
113
r1313 sin13  r2 cos 2  r2 sin2  0
r1313 cos 13  r2 sin2  r2 cos 2  0
which in matrix format are
  r13 sin13  cos 2  13    r2 sin2 

 r cos 
 sin2   r2   r2 cos 2 
13
 13
At the current posture the determinant of the Jacobian is
  r13 cos 13  2   27.0 mm
We note that this determinant becomes zero when r13 is perpendicular to r2 and that this
occurs at 2  138.59 and at 2  41.41.
Using Cramer’s rule, the solution of the above equations for the first-order kinematic
coefficients at the current position are
13  r2   75 mm 27 mm  2.778 rad/rad
Ans.
r2  r2 tan 13   2   0.100 m/rad
The first-order kinematic coefficient of the wheel can be obtained from the rolling contact
equation. The rolling contact condition can be written as
1
3  13

3
 1  13
Note that the positive sign must be chosen because the contact between gears 1 and 3 is
internal rolling. Differentiating this equation with respect to the input position  2 (and
using the positive sign) gives
1 3  13

3
13
which gives
3  1  1 3 13  8.333 rad/rad
Ans.
Therefore the angular velocity of the wheel is
3  32   8.333 rad/rad  30 rad/s  250 rad/s (cw)
Ans.
and the apparent velocity of the center of the wheel, point A, with respect to the slot in
link 2 is
VA3 /2  r22   0.100 m/rad  30 rad/s   3.000 m/s
Ans.
114
3.52
For the linkage in the posture illustrated, link 2 is the input, link 3 is horizontal, and link
4 is vertical. Write the vector loop equation and determine the kinematic coefficients of
the mechanism. If the angular velocity of link 2 is 2  30 rad/s ccw, determine the
angular velocity of links 3 and 4.
Defining the vectors shown in the
figure, the vector loop equation can
be written as
Ans.
r2  r3  r4  r1  0
The corresponding two scalar loopclosure equations are
r2 cos 2  r3 cos 4  90  r4 cos 4  r1  0
r2 sin 2  r3 sin 4  90  r4 sin 4  0
At the posture shown, these equations
have a solution of 4  90 and
r4  3 3 in  5.196 in
RO4O2  12 in, RAO2  6 in, and RBA  9 in.
Taking the derivatives of the scalar loop-closure equations with respect to the input angle
 2 gives
 r2 sin 2  r34 sin 4  90  r44 sin 4  r4 cos4  0
r2 cos 2  r34 cos 4  90  r44 cos 4  r4 sin 4  0
Writing these equations in matrix form, they become
 r3 sin 4  90  r4 sin 4  cos 4  4   r2 sin 2 

   

r
cos


90


r
cos


sin



3
4
4
4
4

  r4    r2 cos 2 
For the current posture, using the data found from the loop-closure equations, these
become
5.196 in 0  4  5.196 in 

 9 in
1  r4   3 in 

The determinant of the Jacobian is   5.196 in and Cramer’s rule gives the solution as
4  1.000 rad/rad and r4  12.000 in/rad
Since we know that the angle of link 3 is 3  4  90 , the derivative of this constraint
equation gives 3  4  1.000 rad/rad. Therefore the angular velocity of link 3 is
3  3  32  1.000 rad/rad  30 rad/s  30 rad/s
Ans.
115
3.53
For the mechanism in the posture illustrated, determine: the first-order kinematic
coefficients of the mechanism. If the velocity of the link 2 is VA2  0.30 m/s in the
direction shown, determine (a) the angular velocity of link 3; (b) the apparent velocity of
pin A2 with respect to the slot in link 3; and (c) the velocity of point B.
RA2O3  750 mm and RBO3  1 250 mm
Defining the vectors shown in the figure, the vector loop equation can be written as
I
??
r1  r2  r3  0
and the corresponding two scalar equations are
r2 cos  30  r3 cos 3  0
r1  r2 sin  30  r3 sin 3  0
In a real situation, the distance r1 would be known and r2 would be given as input; the
length and angle of vector r3 would both be unknown.
As given here, however, the two scalar equation can be solved for
r1  836.52 mm
and, at the posture shown,
r2  612.37 mm .
To find the kinematic coefficients of the mechanism, we take the derivative of the scalar
loop-closure equations above with respect to the input variable r2. This gives the
116
following two equations
cos  30   r3 cos 3  r33 sin 3  0
sin  30  r3 sin 3  r33 cos 3  0
which, in matrix format, appear as follows
  cos 3 r3 sin 3   r3    cos  30  
  sin   r cos         sin  30 
3
3
3  3



The determinant of the Jacobian is   r3 and never becomes zero.
The solutions for these two first-order kinematic coefficients at the given posture are
r3  r3 cos  30  3    0.258 82 m/m
3  sin  30  3    1.287 90 rad/m
Ans.
(a) The input velocity is r2  VA2  0.3 m/s . Using this, the angular velocity of link 3 is
3  3  3r2   1.287 90 rad/m 0.3 m/s  0.386 rad/s (cw)
Ans.
(b) The apparent velocity of pin A2 with respect to the slot in link 3 is
VA2 /3  r3  r3r2   0.258 82 m/m 0.3 m/s   0.0776 m/s45
Ans.
(c) For finding the velocity of point B we first write the position vector
RB  RBO3 e j3
Taking the derivative of this with respect to input r2 gives the first-order kinematic
coefficient
RB  jRBO33e j3  j 1.25 m 1.2879 rad/m e j 4  1.6099e j 4 m/m
Therefore, the velocity of point B is
VB  RB  RB r2  1.6099e j 4 rad/rad   0.3 m/s  0.483 m/s - 45
Ans.
117
3.54
For the mechanism in the posture illustrated, the internal track of input gear 2 is in rolling
contact with gear 3 at point C and the external track is in rolling contact with rack 5 at
point F. Gear 3 is also in rolling contact with the fixed gear (link 1) at point E.
Determine the first-order kinematic coefficients of gear 3 and the rack. Also, determine
the angular velocities of gear 3 and link 4; and the velocity of the rack if gear 2 has an
angular velocity 2  77 rad/s ccw. .
1  REO  4 in, 2  RCO  18 in, 3  RCD  7 in, and RFA  RBF  20 in.
1
2
The rolling contact equation for the contact between gear 1 and gear 3 can be written as
1
 
 3 4
3
1  4
where the negative sign is used since the gears are in external contact. Rearranging this,
it can be put into the form
33   1  3  4
(1)
The rolling contact equation between gear 2 and gear 3 can be written as
2
 
 3 4
3
2  4
Here the gears are in internal contact; therefore a positive sign must be used. Noting that,
gear 2 is the input, we have 2  1, and we can rearrange this equation as follows
2   3  2  4  33
Solving this equation simultaneously with Eq. (1) gives
2
18 in 9
4 


rad/rad
 1  2  22 in 11
118
11
9
Ans.
4  rad/rad
7
7
If we take the position of link 5 to be measured from point A (fixed in link 1) to point F
(fixed in link 5), the position of link 5 can be defined as R5  RFA . Then the rolling
contact constraint between link 2 and link 5 can be written as
R5   2 2  5   2 1  0  2  18 in/rad ,
Ans.
and, from Eq. (1)
3 
where the positive sign is used since a positive rotation of link 2 causes link 5 to move to
the right which causes R5  RFA to become larger.
The angular velocities of gears 3 and 4 can now be written as
Ans.
3  32   9 7 rad/rad  77 rad/s   99 rad/s ccw
and
4  42   9 11 rad/rad  77 rad/s  63 rad/s ccw
The velocity of rack 5 is
VF5 /1  R52  18 in/rad  77 rad/s   1386 in/s
Ans.
Ans.
The velocity of rack 5 is to the right since a positive result shows that the distance
R5  RFA is increasing.
119
3.55
For the mechanism in the posture illustrated, the input arm, link 2, is pinned to the ground
at O1 and is pinned to the center of gear 3 at A. The center of gear 4 is also pinned to the
ground at O1 and gear 5 is pinned to the ground at O5 . Gears 3, 4, and 5 are all in rolling
contact at point B. Determine the first-order kinematic coefficients of gears 3, 4, and 5.
If the angular velocity 2  15 rad/s cw, use the kinematic coefficients to determine the
angular velocities of gears 3, 4, and 5.
1  100 mm, 3  100 mm, 4  300 mm, and 5  500 mm.
The rolling contact equation between fixed gear 1 and gear 3 can be written as
1
 
 3 2
3
1  2
where the negative sign is used because the gears are in external contact. Rearranging
this, and recognizing that 2  1 because arm 2 is the input, this equation can be solved
for  3 .
120
3 
 1  3   200 mm  2 rad/rad
3
100 mm
(1)
The rolling contact equation between gear 3 and gear 4 can be written as
3
 
 4 2
4
3   2
where the positive sign is used since the gears are in internal contact. Rearranging this,
recognizing that 2  1 , and using Eq. (1), this equation can be solved for  4 .
4 
 1  4   400 mm  4 rad/rad
4
300 mm
3
(2)
The rolling contact equation between gear 4 and gear 5 can be written as
5  1
4

5
 4  1
where the positive sign is used since the gears are in internal contact. Rearranging this,
and using Eq. (2), this equation can be solved for  5 .
4
300 mm  4

4 
(3)
 rad/rad   0.8 rad/rad
5
500 mm  3

Substituting the known input angular velocity, 2  15 rad/s, into Eqs. (1), (2), and (3),
3  32   2 rad/rad  15 rad/s   30 rad/s (cw)
Ans.
5 
4

rad/rad   15 rad/s   20 rad/s (cw)
3

5  52   0.8 rad/rad  15 rad/s   12 rad/s (cw)
4  42  
Ans.
Ans.
121
3.56
For the mechanism in the posture illustrated, the radius of the wheel (link 5) is rolling on
the circular ground link.
Determine the first-order kinematic coefficients of the
mechanism. If the input link has a constant velocity of VA2  5i m/s, determine the
angular velocities of links 3, 4, and 5.
RBO4  120 mm, RCO4  180 mm, RBA  120 mm, and 5 = 20 mm.
One suitable set of vectors for the mechanism is shown in the figure. Using these, the
vector loop closure equation can be written as
R4e j4  R3e j3  R2e j  0
with the two scalar component equations
R4 cos 4  R3 cos 3  R2 cos180  0
R4 sin4  R3 sin3  0
For the posture shown the values are 3  240 , 4  120 , and R2  R3  R4  120 mm.
Note that the magnitude of R2 has a positive value, but is at an angle of 180°. Note also
that this magnitude is the independent input for the mechanism.
The derivative of the scalar loop-closure equations with respect to the input are
122
 R44 sin4  R33 sin3  cos180  0
R44 cos 4  R33 cos 3  0
And, in matrix format, these become
  R3 sin3  R4 sin4  3   1

 R cos 
R4 cos 4  4   0 
3
 3
(1)
The determinant of the coefficient matrix is
  R3R4 sin 4  3 
and we see that this becomes zero when 4   , 3   or when 4   2 , 3  3 2 .
Therefore, this mechanism can only be operated with R2 as input between these limits, or
difficulty will be encountered. At the posture shown, however, there is no difficulty.
Eqs. (1) can be solved by Cramer’s rule
 cos 120 
 R4 cos 4
3 

 4.811 rad/m
Ans.
R3 R4 sin 4  3   0.120 m  sin  120 
4 
cos  240 
R3 cos 3

 4.811 rad/m
R3 R4 sin 4  3   0.120 m  sin  120 
Ans.
The negative sign for  3 indicates that link 3 is rotating clockwise for a positive change
in the input. The positive sign for  4 indicates that link 4 is rotating counterclockwise for
a positive change in the input.
The rolling constraint equation between link 5 and link 1 can be written as
   4
1
 5
5
1  4
where the positive sign is used because of the internal rolling contact. Substituting
known data this can be solved for
   1     180 mm 4.811 rad/m  43.299 rad/m
Ans.
5  5


4
5
20 mm
For the given input velocity VA2  5 i m/s. we can see that the positive value of R2 must
be decreasing in size and, therefore, that we must have R2  5 m/s.
Using this and the first-order kinematic coefficients we can find the angular velocities of
the other links.
3  3R2   4.811 rad/m  5 m/s   24.055 rad/s (ccw)
Ans.
4  4R2   4.811 rad/m  5 m/s   24.055 rad/s (cw)
5  5R2   43.299 rad/m  5 m/s   216.495 rad/s (ccw)
Ans.
Ans.
123
3.57
For the mechanism in the posture illustrated, the pinion (link 3) rolls without slip on rack
4 at point B. Determine the first-order kinematic coefficients of the mechanism. If the
velocity of input link 2 is VA2  4.8 in/s constant upward, determine the angular velocity
of the pinion, and the velocity of the rack.
3  2 in
A suitable set of vectors for this mechanism is shown in the figure. From these, the loopclosure equation can be written as
R4  jR1  R43e j 2 3  R24e j 6  jR2  0
where R1 has constant length, but R2 varies as the input. R4 is also variable, showing the
horizontal motion of link 4. The corresponding two scalar equations are
R4  R43 cos120  R24 cos 30  0
(1)
R1  R43 sin120  R24 sin 30  R2  0
In addition, we can write the rolling contact constraint between the pinion 3 and the rack
4 as
124
R43   33
(2)
where the negative sign is used since a positive (counterclockwise) movement of  3
corresponds to a decrease in the length of R43.
Differentiating Eqs. (1) and constraint Eq. (2) with respect to input motion R2 we get
 cos120  0
R4  R43
 sin120  1  0
R43
    33
R43
The solutions to these three equations give the first-order kinematic coefficients of the
mechanism.
  1 sin120  1.154 70 in/in
R43
Ans.
R4  1 tan120  0.577 35 in/in
1
1
3 

 0.577 35 rad/in
Ans.
3 sin120  2 in  0.866 03
With the input velocity of VA given, the angular velocity of the pinion is
3  3R2   0.577 35 rad/in  4.8 in/s   2.771 rad/s (cw)
Ans.
and the velocity of the rack is
V4  R4 R2   0.577 35 in/in  4.8 in/s   2.771 in/s (  ) .
Ans.
125
3.58
For the mechanism in the posture illustrated, determine the first-order kinematic
coefficients of links 3, 4, 5, and 6. If the constant input velocity VA2  0.090 ĵ m/s,
determine the angular velocities of links 3 and 5 and the velocity of point P.
RCO4  RDO6  RBC  RBD  RAB  40 mm, RPD  20 mm, and R A  104 mm ˆi  yAˆj
There are 8 primary instant centers, namely, I12, I14, I16, I23, I34, I35, I46, and I56, and there
are 7 secondary instant centers, namely, I13, I15, I24, I25, I26, I36, and I46. The locations of
the 15 instant centers are shown on the figure The procedure to locate the secondary
instant centers is marked on the Kennedy circle shown in the following figure.
The velocity of point A fixed in link 3 can be written as
VA3  I13 I 233  VA2  R2
Rearranging this gives the first-order kinematic coefficient of link 3; that is,

1
1
3  3 

 9.62 rad/m
R2 I13 I 23 104 mm
(1) Ans.
where the measured distance is I13 I 23  104 mm. Since the input R2 = yA is defined by the
vertical distance from the x-axis to point A (that is, the vector is negative and its change is
126
positive with the given upward input velocity VA2  0.090 ĵ m/s ), therefore, the correct
sign for  3 is positive. In other words, link 3 is rotating counterclockwise about the
instant center I13.
The velocity of point B fixed in link 4 can be written as
VB4  I14 I 244  VB2  R2
Rearranging this equation gives the first-order kinematic coefficient of link 4; that is,

1
1
Ans.
4  4 

 14.4 rad / m
R2 I14 I 24 69 mm
where the measured distance is I14 I 24  69 mm. Note that the correct sign is positive
because link 4 is rotating counterclockwise about the instant center I14.
The velocity of point E fixed in link 5 can be written as
VE5  I15 I 255  VE2  R2
Rearranging this equation gives the first-order kinematic coefficient of link 5; that is,

1
1
5  5 

 14.4 rad / m
(2) Ans.
R2 I15 I 25 69 mm
where the measured distance is I15 I 25  69 mm. Note that the correct sign is positive
because link 5 is rotating counterclockwise about the instant center I15.
The velocity of point F fixed in link 6 can be written as
VF6  I16 I 266  VF2  R2
Therefore, the first-order kinematic coefficient of link 6 can be written as

1
1
6  6 

 9.62 rad / m
R2 I16 I 26 104 mm
Ans.
where the measured distance is I16 I 26  104 mm. Note that the correct sign is positive
because link 6 is rotating counterclockwise about the instant center I16.
Check: Note that the mechanism is a pantograph. Due to the symmetry of the linkage, it
is expected that the kinematic coefficient of link 4 is the same as that of link 5 and that
the kinematic coefficient of link 3 is the same as that of link 6.
Since we are given the input velocity as VA2  R2  90ˆj mm/s, we can use Eqs. (1) and
(2) above to find the angular velocities of links 3 and 5.
3  3R2   9.62 rad/m  0.090 m/s   0.866 rad/s (ccw)
Ans.
5  5R2  14.4 rad/m  0.090 m/s   1.30 rad/s (ccw)
Ans.
The velocity of point P fixed in link 5 can be written as
VP  I15 P5   34.5 mm 1.30 rad/s   0.045 m/s
Ans.
where the measured distance is I15 P  34.5 mm.
Check: Note that the given mechanism is a pantograph. Due to the link lengths and the
symmetry of the mechanism, point P traces a path of the same shape as point A but at a
scale of one half. Therefore, it is expected that point P should have a velocity in the same
direction as that of point A but with half the magnitude.
127
3.59
For the mechanism in the posture illustrated, the line AB is vertical and the line CD is
horizontal. Determine the first-order kinematic coefficients of the mechanism. If the
angular velocity of the input link 2 is 2  10 rad/s cw, determine the angular velocities
of links 3 and 4; and the velocity of point D fixed in link 4 with respect to point C fixed
in link 3.
RAO2  4 in, RBA  8 in, RDO4  4 in, and RO2O4  2 in.
A suitable set of vectors for this mechanism is shown in the figure. From these, the
complex vector form of the loop-closure equation can be written as
j   4
Ans.
R1  R2e j2  jR22e j2  R3e j3  R34e  3   R4e j4  0
The corresponding two scalar equations are
R1  R2 cos 2  R22 sin 2  R3 cos 3  R34 cos 45 cos 3  R34 sin 45 sin 3  R4 cos 4  0
(1)
R2 sin 2  R22 cos2  R3 sin 3  R34 sin 45 cos3  R34 cos 45 sin 3  R4 sin 4  0
with the additional constraint equation
3  135  4  0 .
(2)
In these equations  2 is the input variable and  3 ,  4 , and R34 are response variables.
Differentiating Eqs (1) and (2) with respect to input variable  2 , we obtain
128
  R34 cos 45 sin 33
 R2 sin  2  R22 cos  2  R3 sin 33  cos 45 cos 3 R34
  R34 sin 45 cos 33  R4 sin  4 4  0
 sin 45 sin 3 R34
  R34 sin 45 sin 33
R2 cos  2  R22 sin  2  R3 cos 33  sin 45 cos 3 R34
  R34 cos 45 cos 33  R4 cos  4 4  0
 cos 45 sin 3 R34
3   4  0
The third of these equations can be used to eliminate  4 from the other two. This leaves
two equations which can be put into matrix form as follows
  R3 sin 3  R34 sin 3  45  R4 sin 4 cos 3  45    3   R2 sin 2  R22 cos2 

    

   R2 cos 2  R22 sin 2 
 R3 cos 3  R34 cos 3  45  R4 cos 4 sin 3  45    R34
In the posture illustrated, the given data shows that R2  4 in, R22  8 in, R3  5.657 in,
R34  2 in, R4  4 in, 2  0, 3  135, and 4  90.
above matrix equation becomes
 8 in 1  3   8 in 
 6 in 0   R    4 in 

  34  

Substituting these data, the
The determinant of the coefficient matrix is   6 in and Cramer’s rule gives the
solution for the first-order kinematic coefficients as
 3  0.667 rad/rad 
Ans.
 R    2.667 in/rad 

 34  
4  0.667 rad/rad .
and
The angular velocities of links 3 and 4 are
3  32   0.667 rad/rad  10 rad/s  6.667 rad/s (cw)
4  42   0.667 rad/rad  10 rad/s  6.667 rad/s (cw)
Ans.
Ans.
Ans.
The position vector of point D fixed in link 4 with respect to point C fixed in link 3 is
R D4C3  R34e j3  4
  R34 cos 45 cos 3  R34 sin 45 sin 3   j  R34 sin 45 cos 3  R34 cos 45 sin 3 
The derivative with respect to input 2 gives the kinematic coefficient
  R34 cos 45 sin 33  sin 45 sin 3 R34
  R34 sin 45 cos 33 
RD4C3   cos 45 cos 3 R34
  R34 sin 45 sin 33  cos 45 sin 3 R34
  R34 cos 45 cos 33 
 j  sin 45 cos 3 R34
  R34 sin 3  45  3   j sin 3  45  R34
  R34 cos 3  45 3 
 cos 3  45  R34
and with the data for the posture shown this becomes
RD4C3    R334 ˆi    R343  ˆj     2.667 in/rad   ˆi     2 in  0.667 rad/rad   ˆj
  2.667 in/rad  ˆi  1.333 in/rad  ˆj  2.981 in/rad  153.43
Finally, The velocity of point D fixed in link 4 with respect to point C fixed in link 3 is
VD4C3  RD4C32   2.981 in/rad  153.43 10 rad/s   29.81 in/s26.57
Ans.
129
3.60
For the mechanism in the posture illustrated, gear 3 rolls without slipping on link 4 at
point C. Determine the first-order kinematic coefficients of the mechanism. If the
velocity of the input link 2 is a constant VB   0.5ˆj m / s, determine the angular velocity
of gear 3 and the velocity of link 4.
3  25 mm
A suitable set of vectors for this mechanism is shown in the figure. Using these vectors,
the loop-closure equation for the mechanism is
 jR2  R4  R34e j 2 3  R3e j 5 6  0
The corresponding two scalar equations are
 R4  R34 cos120  R3 cos 150  0
(1)
 R2  R34 sin120  R3 sin  150  0
where R2 is the input variable and R4 and R34 are response variables. In addition, there is
130
a displacement constraint for the rolling contact at point C.
R34  3 3  4


(2)
In the posture illustrated, where R2  25 mm, these scalar equations give the positions of
the response variables as R34  14.434 mm and R4  14.434 mm.
Differentiating Eqs. (1) with respect to input variable R2 gives
 cos120  0
 R4  R34
 sin120  0
1  R34
which can be written in matrix form as
   0
cos120 1  R34
 sin120 0   R   1

 4   
The determinant of the Jacobian is   sin120  0.866 and, using Cramer’s rule, the
solutions for the first-order kinematic coefficients are
   1.154 7 m/m 
 R34
Ans.
 R    0.577 m/m 

 4 
and, from Eq. (2), we find
which yields
  3 3  0
R34

R34
1.154 7 m/m
 46.188 rad/m
3
0.025 m
The angular velocity of gear 3 is
3  3R2   46.188 rad/m 0.500 m/s  23.094 rad/s
3 

Ans.
Ans.
and the velocity of link 4 is
R4  R4 R2   0.577 m/m 0.500 m/s   0.2885 m/s
The positive sign indicates that the magnitude of vector R4 is increasing. Therefore, link
4 is moving to the left; that is, the velocity of link 4 is
VC4  0.288 5ˆi m/s
Ans.
131
3.61
For the mechanism in the posture illustrated, link 3 is vertical and link 4 is horizontal.
Determine the first-order kinematic coefficients of the mechanism, and the coupler point
C. If the input link 2 has a constant angular velocity 2 = 15 rad/s ccw, then determine
the velocity of point C..
RO4O2  12 in, RAO2  6 in, and RCA  13 in
Using the vectors shown in the figure, the loop-closure equation for the mechanism is
R2e j2  R3e j3  R4e j4  jR1  0
with the additional constraint equation
3  4  90
(1)
The two scalar equations for loop-closure are
R2 cos 2  R3 cos 3  R4 cos 4  0
(2)
R2 sin 2  R3 sin 3  R4 sin 4  R1  0
At the posture indicated the input variable is 2  30 and the response variables are
3  90, 4  0, and R3  9 in, and R4  5.196 in.
Differentiating Eqs. (2) with respect to input variable 2 we obtain
132
 R2 sin 2  R3 sin 33  R4 cos 4  R4 sin 44  0
(3)
R2 cos 2  R3 cos 33  R4 sin 4  R4 cos44  0
and, from Eq. (1), we get 3  4 , which allows us to put Eqs. (3) into the matrix form
  R3 sin 3  R4 sin 4  cos 4  4   R2 sin 2 

 R cos   R cos 
 sin 4   R4    R2 cos 2 
3
4
4
 3
Using the data for the posture indicated, this becomes
1 4   3 in 
 9 in
 5.196 in 0   R    5.196 in 

 4 

for which, the determinant of the Jacobian is   5.196 in, and the solutions may be
found by Cramer’s rule:
4   1 rad/rad 
Ans.
 R    12 in/rad 

 4 
and 3  4  1 rad/rad.
The scalar equations for the position of point C are
xC  R2 cos 2  CA cos 3
Ans.
yC  R2 sin 2  CA sin 3
which gives first-order kindematic coefficients of
xC   R2 sin  2  CA sin 33  16 in/rad
yC  R2 cos 2  CA cos 33  5.196 in/rad
R  16ˆi  5.196ˆj in/rad  16.823 in/rad162
C
Therefore, the velocity of point C is
VC  RC2  16.823 in/rad16215 rad/s  252.34 in/s162
Ans.
Ans.
133
Chapter 4
Acceleration
4.1
The position vector of a point is defined by the equation

t3  ˆ
R   4t   i  10ˆj
3

where R is in inches and t is in seconds. Find the acceleration of the point at t  2 s.

t3 
R  t    4t   ˆi  10ˆj

3 

R  t   4  t 2 ˆi


R  t   2tˆi
4.2
R  2 s   2  2  ˆi  4ˆi in/s2
Ans.
A point moves according to the equation
 2 t3  ˆ t3 ˆ
R = t  i  j .
6
3

where R is in meters and t is in seconds. Find the acceleration at t = 3 s.

t3   t3 
R  t  =  t 2   ˆi    ˆj

6   3 


t2 
R  t  =  2t   ˆi  t 2ˆj

2 

R  t  =  2  t  ˆi  2tˆj
R  3 s  =  2  3 ˆi  2  3 ˆj
R  3 s  = ˆi  6ˆj m/s2
Ans.
134
4.3
The path of a point is described by the equation
R = (t 2  4)e j t /10
where R is in millimeters and t is in seconds. Find the unit tangent vector to the path, the
normal and tangential components of the absolute acceleration, and the radius of
curvature of the path, at t = 20 s.
R  t  = (t 2  4)e j t /10
j 2
(t  4)e j t /10
10
j t   j t /10 j t  j t /10  2 2

R t  =  2 

e

(t  4)e j t /10
e
5
5
100


Noting, at t = 20 s, that e j t /10  e j 2  1.0 , we find that
R  20 s  = (202  4)  404 mm
R  t  = 2te j t /10 
j
(202  4)  40.00  j126.92  133.07 mm/s  72.5
10
j 20  j 20  2

R  20 s  =  2 

(202  4)  37.873  j 25.133 mm/s 2

5 
5
100

From the direction of the velocity we find the unit tangent and unit normal vectors
uˆ t  1.0  72.5  cos  72.5 ˆi  sin  72.5  ˆj  0.300 71ˆi  0.953 72ˆj
uˆ n  kˆ  uˆ t   sin  72.5 ˆi  cos  72.5  ˆj  0.953 72ˆi  0.300 71ˆj
R  20 s  = 2  20  
From these, the components of the point’s absolute acceleration are
An  uˆ n R  0.953 72ˆi  0.300 71ˆj 37.873ˆi  25.133ˆj mm/s 2  43.678 mm/s2



A  uˆ R   0.300 71ˆi  0.953 72ˆj  37.873ˆi  25.133ˆj mm/s   12.581 mm/s
t
t
2
2
Ans.
Ans.
Ans.
Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is

R
2
133.07 mm/s   405.4 mm

2
Ans.
An
43.678 mm/s2
where the negative sign indicates that the center of curvature is in the negative uˆ n
direction from the point.
135
4.4
The motion of a point is described by the equations
t 3 sin 2 t
6
where x and y are in feet and t is in seconds. Find the acceleration of the point at
t  1.40 s .
x  4t cos  t 3
and
y
 t 3 sin 2 t  ˆ
R  t    4t cos  t 3  ˆi  
j
6


 t 2 sin 2 t  t 3 cos 2 t  ˆ
R  t    4 cos  t 3  12 t 3 sin  t 3  ˆi  
+
j
2
3


  2 2t 2 
ˆ
2
R  t    48 t 2 sin  t 3  36 2t 5 cos  t 3  ˆi  t 1 
 sin 2 t + 2 t cos 2 t  j
3 
 

2
2
R 1.40 s  1 112.620ˆi  19.753ˆj ft/s  1 112.796 ft/s   1.0
4.5
Ans.
Link 2, in the posture illustrated, has an angular velocity 2  120 rad/s ccw and an
angular acceleration 2 = 4 800 rad/ s 2 ccw. Determine the absolute acceleration of point
A..
RAO2  500 mm
A A  AO2  A nAO2  AtAO2
 22 R AO2   2kˆ  R AO2

 
 
2
  120 rad/s  0.500ˆi m  4 800 rad/s 2kˆ  0.500ˆi m
A A  7 200ˆi  2 400ˆj m/s2  7 589 m/s 2161.6

Ans.
136
4.6
The accelerations of points A and B of link 2, which is rotating clockwise, are as given.
Determine the angular velocity, the angular acceleration, and the acceleration of the
midpoint C of link 2.
RBA  20 in, AA  600 ft/s2 , and AB  150 ft/s2 .
A B  A A  AnBA  AtBA
Construct the acceleration polygon.
t
Then, from measurement of ABA
,
2  
n
ABA
594.6 ft/s 2

 18.9 rad/s cw
RBA
20 12 ft
Ans.
Note the ambiguous sign of this square root. The sense of  cannot be determined from
the accelerations, but here is found from the problem statement.
At
170.1 ft/s 2
 2  BA 
 102.1 rad/s2 cw
Ans.
RBA
20 12 ft
AC  309.2 ft/s244.0
Ans.
137
4.7
For the given kinematic data for link 2, find the velocity and acceleration of points B and
C..
RBA  16 in, RCA  10 in, and RCB  8 in
VB  VA  VBA
VBA  2 RBA   24 rad/s 16 12 ft   32 ft/s
Construct the velocity polygon.
VB  12.0 ft/s270
VC  8.367 ft/s12.1
Ans.
Ans.
A B  A A  AnBA  AtBA
n
ABA
 22 RBA   24 rad/s  16 12 ft   768.0 ft/s 2
2
t
ABA
  2 RBA  160 rad/s2  16 12 ft   213.3 ft/s 2
Construct the acceleration polygon.
A B  395.1 ft/s2165
Ans.
AC  210.2 ft/s2240.3
Ans.
138
4.8
The angular velocity and angular acceleration of link 2 of the Scott-Russell linkage in the
posture illustrated, are 2  20 rad/s cw and  2  140 rad/s2 cw , respectively.
Determine the velocity and acceleration of point B and the angular acceleration of link 3.
RAO2  RCA  RBA  100 mm
VA  VO2  VAO2
VAO2  2 RAO2   20 rad/s  0.100 m   2.000 m/s
VC  VA  VCA
Construct the velocity image of link 3.
VB  3.76 m/s  90
A A  AO2  A
n
AO2
A
Ans.
t
AO2
n
AAO
 22 RAO2   20 rad/s   0.100 m   40.0 m/s2
2
2
t
AAO
  2 RAO2  140 rad/s2   0.100 m   14.0 m/s 2
2
n
t
AC  A A  ACA
 ACA
n
2
ACA
 VCA
RCA   2.00 m/s 
2
 0.100 m   40.0 m/s2
Construct the acceleration image of link 3.
A B  53.7 m/s2  90
3 
t
ACA
14.0 m/s2

 140 rad/s 2 ccw
RCA 0.100 m
Ans.
Ans.
139
4.9
In the posture illustrated in Fig. P4.8, the slider 4 is moving to the left with a constant
velocity VC = 2 m/s. Find the angular velocity and angular acceleration of link 2.
VA  VC  VAC  VO2  VAO2 .
Construct the velocity image of link 3.
V
3.864 m/s
2  AO2 
 38.64 rad/s cw
RAO2
0.100 m
Ans.
A A  AC  AnAC  AtAC  AO2  AnAO  AtAO
2
V
n
AC
2
AC
A
2
/ RAC   3.864 m/s  /  0.100 m   149.28 m/s2
2
n
2
AAO
 VAO
/ RAO2   3.864 m/s  /  0.100 m   149.28 m/s2
2
2
2
2 
4.10
t
AAO
2
RAO2

557.13 m/s2
 5 571.3 rad/s 2 cw
0.100 m
Ans.
If the velocity of point B in the posture illustrated in Problem 3.8 is constant, then
determine the acceleration of point A and the angular acceleration of link 3.
A A  A B  AnAB  AtAB
14.64 m/s   535.9 m/s2
V2
 AB 
RAB
0.400 m
2
n
AB
A
A A  757.9 m/s215
3 
t
AB
Ans.
2
A
535.9 m/s

 1 339.7 rad/s 2 cw
RAB
0.400 m
Ans.
140
4.11
If the angular velocity of crank 2 in the posture illustrated in Problem 3.9 is constant, then
determine the angular accelerations of links 3 and 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   45 rad/s   4 in   8 100 in/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA  14.34 in/s  / 10.0 in   20.57 in/s 2 (Ignore compared to other components.)
n
BA
2
BA
2
n
2
ABO
 VBO
/ RBO4  184.8 in/s  / 12.0 in   2 844.8 in/s 2
4
4
2
3 
t
ABA
5 632.7 in/s2

 563.3 rad/s2 ccw
RBA
10.0 in
t
ABO
4
1 484.1 in/s 2
4 

 123.7 rad/s2 ccw
RBO4
12.0 in
Ans.
Ans.
141
4.12
If the angular velocity of crank 2 in the posture illustrated in Problem 3.10 is constant,
then determine the acceleration of point C and the angular accelerations of links 3 and 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   60 rad/s   0.150 m   540.0 m/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA  13.02 m/s  /  0.300 m   565.1 m/s2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4  11.36 m/s  /  0.300 m   430.1 m/s2
4
4
2
Construct the acceleration image of link 3.
AC  209.6 m/s210.6
3 
4 
t
ABA
136.5 m/s 2

 455.0 rad/s 2 ccw
RBA
0.300 m
t
ABO
4
RBO4

46.04 m/s2
 153.5 rad/s2 cw
0.300 m
Ans.
Ans.
Ans.
142
4.13
If the angular velocity of crank 2 in the posture illustrated in Problem 3.11 is constant,
then determine the acceleration of point C and the angular accelerations of links 3 and 4.
t
A A  AO2  AnAO  A AO
2
2
n
AAO
 22 RAO2   48 rad/s  8.000 in   18 432.0 in/s 2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
n
2
ABA
 VBA
/ RBA  10.7 in/s  /  32.0 in   3.6 in/s 2 (Ignore compared to other components.)
2
n
2
ABO
 VBO
/ RBO4   374.6 in/s  / 16.0 in   8 770.8 in/s 2
4
4
2
Construct the acceleration image of link 4.
AC  37 254 in/s2114.4
3 
4 
t
BA
A
55 730 in/s

 1 741.6 rad/s 2 ccw
RBA
32.0 in
t
ABO
4
RBO4
Ans.
2

48 892 in/s 2
 3 055.8 rad/s 2 ccw
16.0 in
Ans.
Ans.
143
4.14
If the angular velocity of crank 2 in the posture illustrated in Problem 3.13 is constant,
then determine the accelerations of points C and D and the angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2  1.0 rad/s   0.300 m   0.300 m/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   0.263 5 m/s  /  0.150 m   0.462 9 m/s2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4   0.227 0 m/s  /  0.300 m   0.171 8 m/s 2
4
4
2
Construct the acceleration image of link 3.
AC  0.515 4 m/s2  119.8
Ans.
A D  0.492 5 m/s221.8
Ans.
4 
t
ABO
4
RBO4

0.221 2 m/s 2
 0.737 4 rad/s 2 cw
0.300 m
Ans.
144
4.15
If the angular velocity of crank 2 in the posture illustrated in Problem 3.14 is constant,
then determine the acceleration of point C and the angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   60.0 rad/s   6.0 in   21 600 in/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA  169.4 in/s  /  6.0 in   4 783 in/s 2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4   317.6 in/s  / 10.0 in   10 090 in/s 2
4
4
2
Construct the acceleration image of link 3.
AC  31 250 in/s2  68.9
4 
t
ABO
4
RBO4

14 950 in/s 2
 1 495 rad/s 2 ccw
10.0 in
Ans.
Ans.
145
4.16
If the angular velocity of crank 2 in the posture illustrated in Problem 3.16 is constant,
then determine the acceleration of point C and the angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   30.0 rad/s   3.0 in   2 700 in/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   90.0 in/s  /  5.0 in   1 620 in/s 2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4   0.0 in/s  /  5.0 in   0.0 in/s 2
4
4
2
Construct the acceleration image of link 3.
AC  5 940 in/s2216.9
4 
t
ABO
4
RBO4

4 320 in/s2
 720 rad/s2 ccw
6.0 in
Ans.
Ans.
146
4.17
If the angular velocity of crank 2 in the posture illustrated in Problem 3.17 is constant,
then determine the acceleration of point B and the angular accelerations of links 3 and 6.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2  10.0 rad/s   2.50 in   250.0 in/s 2
2
2
2
A B  A A  AnBA  AtBA
n
2
ABA
 VBA
/ RBA  18.70 in/s  / 10.0 in   34.99 in/s 2
2
A B  200.9 in/s20
3 
Ans.
t
ABA
174.9 in/s2

 17.49 rad/s2 ccw
RBA
10.0 in
Ans.
n
t
 ACA
 A B  AnCB  AtCB
Construct the acceleration image of link 3, or AC  A A  ACA
A D  AC  AnDC  AtDC  AO6  A nDO  AtDO
6
n
DC
A
V
2
DC
6
/ RDC   0.600 in/s  /  4.0 in   0.090 in/s 2 (Ignore compared to other parts.)
2
n
2
ADO
 VDO
/ RDO6   24.18 in/s  /  6.0 in   97.475 in/s 2
6
6
2
6 
t
ADO
6
RDO6

64.878 in/s 2
 10.81 rad/s 2 cw
6.0 in
Ans.
147
4.18
For the four-bar linkage of Problem 3.18 in the posture illustrated, determine the angular
acceleration of crank 2 to ensure that the angular acceleration of link 4 is zero.
t
A B  AO4  AnBO  A BO
4
n
BO4
A
V
2
BO4
4
/ RBO4   5.764 m/s  /  0.400 m   83.056 m/s2
2
A A  A B  AnAB  AtAB  AO2  A nAO  AtAO
2
2
A  V / RAB   5.004 m/s  /  0.425 m   58.93 m/s2
n
AB
2
AB
2
n
2
AAO
 VAO
/ RAO2   5.600 m/s  /  0.35 m   89.60 m/s2
2
2
2
t
AAO
2
18.54 m/s2
2 

 52.98 rad/s 2 ccw
RAO2
0.350 m
Ans.
148
4.19
For the four-bar linkage of Problem 3.19 in the posture illustrated, determine the angular
acceleration of crank 2 to ensure that the angular acceleration of link 4 is 100 rad/ s 2 cw.
A B  AO4  AnBO  AtBO
4
4
n
2
ABO
 VBO
/ RBO4   202.8 in/s  / 8.0 in   5 140.7 in/s 2
4
4
2
t
ABO
  4 RBO4  100 rad/s2  8.0 in   800.0 in/s 2
4
A A  A B  AnAB  AtAB  AO2  A nAO  AtAO
2
2
A  V / RAB   25.87 in/s  / 8.0 in   83.63 in/s 2 (Ignore compared to other components.)
n
AB
2
2
AB
n
2
AAO
 VAO
/ RAO2  180.0 in/s  /  5.0 in   6 480.0 in/s 2
2
2
2
2 
t
AAO
2
RAO2

20 903 in/s 2
 4 180.7 rad/s 2 ccw
5.0 in
Ans.
149
4.20
If the angular velocity of crank 2 in the posture illustrated in Problem 3.20 is constant,
then determine the acceleration of point C and the angular acceleration of link 3.
t
A A  AO2  AnAO  A AO
2
2
n
AAO
 22 RAO2  8.0 rad/s   0.150 m   9.600 m/s 2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
n
2
ABA
 VBA
/ RBA   3.783 9 m/s  /  0.250 m   57.270 m/s2
2
n
2
ABO
 VBO
/ RBO4   3.483 7 m/s  /  0.250 m   48.544 m/s2
4
4
2
Construct the acceleration image of link 3.
AC  63.41 m/s2  22.9
3 
t
BA
Ans.
2
A
15.647 m/s

 62.59 rad/s 2 cw
RBA
0.250 m
Ans.
150
4.21
If the angular velocity of crank 2 in the posture illustrated in Problem 3.21 is constant,
then determine the acceleration of point C and the angular acceleration of link 3.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   56.0 rad/s   0.150 m   470.40 m/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   6.966 m/s  /  0.250 m   194.09 m/s2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4  11.380 m/s  /  0.250 m   518.06 m/s2
4
4
2
Construct the acceleration image of link 3.
AC  450.6 m/s2255.6
3 
t
BA
Ans.
2
A
18.52 m/s

 74.08 rad/s2 cw
RBA
0.250 m
Ans.
151
4.22
If the angular velocity of crank 2 in the posture illustrated in Problem 3.22 is constant,
then determine the accelerations of points B and D.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   42.0 rad/s   2.00 in   3 528.0 in/s 2
2
2
2
n
2
ABA
 VBA
/ RBA   42.64 in/s  / 10.0 in   181.9 in/s 2
A B  A A  AnBA  AtBA
2
A B  2117 in/s 20
Ans.
n
t
 ACA
 A B  AnCB  AtCB
Construct the acceleration image of link 3, or AC  A A  ACA
A D  AC  AnDC  AtDC
n
2
ADC
 VDC
/ RDC   61.66 in/s  / 8.0 in   475.2 in/s 2
2
A D  1 976 in/s 290
Ans.
152
4.23
If the angular velocity of crank 2 in the posture illustrated in Problem 3.23 is constant,
then determine the accelerations of points B and D.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   209.4 rad/s   2.00 in   87 730 in/s 2
2
2
2
n
2
ABA
 VBA
/ RBA   218.75 in/s  /  6.0 in   7 975.4 in/s 2
A B  A A  AnBA  AtBA
2
A B  29 287 in/s2240
Ans.
Construct the acceleration image of link 3, or AC  A A  A
A D  AC  A
n
DC
A
n
CA
A
t
CA
 A B  A CB  A CB
n
t
t
DC
n
2
ADC
 VDC
/ RDC   269.0 in/s  /  5.0 in   14 473 in/s 2
2
A D  48 372 in/s2120
Ans.
153
4.24 to 4.30 The nomenclature for a the four-bar linkage is illustrated in Figure. P4.24; the
dimensions and data are given in Table P4.24 to P4.30. The angular velocity  2 is
constant for each problem; (a negative sign indicates that the direction is clockwise). The
dimensions of even-numbered problems are inches; and odd-numbered problems are
millimeters. For each problem, determine 3 , 4 , 3 , 4 , 3 , and  4 .
Table P4.24 to P4.30
Prob.
r1
r2
r3
r4
2 , deg
2 , rad/s
P4.24
4
6
9
10
240
1
P4.25
100
150
250
250
-45
56
P4.26
14
4
14
10
0
10
P4.27
250
100
500
400
70
-6
P4.28
8
2
10
6
40
12
P4.29
400
125
300
300
210
-18
P4.30
16
5
12
12
315
-18
This group of problems was solved on a programmable calculator. The position solution
values were found from Eqs. (2.25) through (2.32). The velocity values were found from
Eqs. (3.22). The acceleration values were found from Eqs. (4.31) and (4.32).
Prob. 3 , deg  4 , deg
3 , rad/s 4 , rad/s
 4 , rad/s2
3 , rad/s2
4.24
4.25
4.26
4.27
4.28
4.29
4.30
105.29
171.01
45.57
28.32
24.17
38.42
73.16
159.60
195.54
91.15
55.88
63.73
155.60
138.51
0.809 3
70.452 7
-4.000 0
-0.632 6
-1.516 7
-6.855 2
-0.505 1
0.525 0
47.566 8
-4.000 0
-2.155 7
1.712 9
-1.234 5
7.275 3
0.230 28
3 196.657 49
-1.120 22
7.822 40
41.414 93
62.500 44
-206.384 28
0.008 09
3 330.841 37
54.890 98
6.704 18
74.975 93
-96.514 21
-94.122 01
154
4.31
For the inverted slider-crank linkage in the posture illustrated, crank 2 has a constant
angular velocity of 60 rev/min ccw. Find the velocity and acceleration of point B and the
angular velocity and acceleration of link 4.
RO4O2  12 in, RAO2  7 in, and RBO4  28 in
 rev  rad  1 min 
 2

  6.283 rad/s
 min  rev  60 s 
VA2  VO2  VA2O2  VA4  VA2 / 4
2   60
VA2O2  2 RA2O2   6.283 rad/s  7.0 in   43.98 in/s
Construct the velocity polygon.
V
36.43 in/s
4  A4O4 
 6.572 rad/s cw
RA4O4
5.543 in
Ans.
VB  VO4  VBO4
VBO4  4 RBO4   6.572 rad/s  28.0 in   184.0 in/s
VB  184.0 in/s  19.1
Since we know the path of A2 on link 4, we write
A A2  AO2  ΑnA2O2  Α At 2O2  A A4  AcA2 A4  AnA2 / 4  AtA2 / 4 ; A A4  AO4  ΑnA4O4  ΑtA4O4
Ans.
AAn2O2  22 RA2O2   6.283 rad/s   7.0 in   276.35 in/s 2
2
AcA2 A4  24 × VA2 /4  2  6.572 rad/s  24.64 in/s   323.93 in/s219.1
n
A2 / 4
A

VA22 / 4
A /4
2
 24.64 in/s   0

2

Construct the acceleration polygon.
At
478.78 in/s 2
 4  A4O4 
 86.38 rad/s 2 ccw
RA4O4
5.543 in
Construct the acceleration image of link 4.
A B  2702.7 in/s2187.5
Ans.
Ans.
155
4.32
For the modified Scotch-yoke linkage in the posture illustrated in Problem 3.26,
determine the acceleration of link 4.
Since we know the path of A2 on link 4, we write
A A2  AO2  ΑnA2O2  Α At 2O2  A A4  AcA2 A4  AnA2 / 4  AtA2 / 4
AAn2O2  22 RA2O2   36.0 rad/s   0.250 m   324.0 m/s2
2
A
c
A4 A2
 24 × VA2 / 4  2  0.0 rad/s  6.588 m/s   0 ; A
n
A4 / 2

VA22 / 4
A /4
2
Next we construct the acceleration polygon.
A A4  290.5 m/s2180.0
 6.588 m/s   0

2

Ans.
156
4.33
For the linkage in the posture illustrated in Problem 3.27, determine the acceleration of
point E.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   72.0 rad/s  1.50 in   7 776 in/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA  151.2 in/s  / 10.5 in   2 177.3 in/s2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4   72.0 in/s  /  5.0 in   1 036.8 in/s 2
4
4
2
Construct the acceleration image of link 3.
Since we know the path of C3 on link 6, we write
AC3  AC6  ACc 3C6  ACn 3 / 6  AtC3 / 6 and AC6  AO6  ΑCn 6O6  ΑCt 6O6
ACc 3C6  26 × VC3 / 6  2  9.673 rad/s  50.15 in/s   970.16 in/s2
n
C3 / 6
A

VC23 / 6
C / 6
3
n
C6O6
A

VC26O6
RC6O6
 50.15 in/s   0

2

 43.07 in/s   416.6 in/s2

2
4.453 in
Construct the acceleration image of link 6.
A E  7 232.9 in/s2252.8
Ans.
157
4.34
For the inverted slider-crank linkage in the posture illustrated in Problem 3.24, determine
the acceleration of point B and the angular acceleration of link 4.
n
AAO
 22 RAO2   24.0 rad/s  8.0 in   4 608 in/s 2
2
2
Since we know the path of P3 on link 4, we write
A P3  A A3  ΑnP3 A3  ΑtP3 A3  A P4  AcP3P4  AnP3 / 4  AtP3 / 4
APn3 A3  VP23 A3 / RP3 A3  161.8 in/s  /  26.27 in   996.6 in/s 2
2
AcP3P4  24 × VP3 /4  2  6.159 rad/s 103.7 in/s   1 277.5 in/s2  102.4
n
P3 / 4
A

VP23 / 4
P / 4
3
103.7 in/s   0

2

Construct the acceleration image of link 3.
A B  4 950.6 in/s2  25.7
Since links 3 and 4 remain perpendicular,
APt 3 A3 1 202.8 in/s 2
 4  3 

 45.78 rad/s 2 ccw
RP3 A3
26.27 in
Ans.
Ans.
158
4.35
For the linkage in the posture illustrated in Problem 3.25, determine the acceleration of
point B and the angular acceleration of link 3.
Since we know the path of P3 on link 4, we write
A P3  A A3  ΑnP3 A3  ΑtP3 A3  A P4  AcP3P4  AnP3 /4  AtP3 /4
APn3 A3  VP23 A3 / RP3 A3   2.735 in/s  /  9.0 in   0.831 in/s 2
2
AcP3P4  24 × VP3 /4  2  0.3039 rad/s 12.31 in/s   7.481 in/s2  102.8
n
P3 / 4
A

VP23 / 4
P / 4
3
12.31 in/s   0

2

Construct the acceleration image of link 3.
A B  20.23 in/s2  102.1
APt 3 A3
11.22 in/s2
3 

 1.247 rad/s2 cw
RP3 A3
9.0 in
Ans.
Ans.
159
4.36
For the linkage in the posture illustrated in Problem 3.31, the input angular velocity is
constant. Determine the accelerations of points A and B.
Since we know the path of F2 on link 3, we write
A F2  A E2  ΑnF2 E2  ΑFt 2 E2  A F3  AcF2 F3  AnF2 / 3  AtF2 / 3 and A F3  AG3  AnF3G3  AtF3G3
AFn2 E2  22 RF2 E2   25.0 rad/s  1.00 in   625 in/s 2
2

n
F3G3
A
VF23G3
RF3G3
 4.110 in/s   2.777 in/s2

2
6.083 in
AcF2 F3  23 × VF2 /3  2  0.676 rad/s  24.66 in/s   33.34 in/s2  80.5
n
F2 / 3
A

VF22 / 3
F / 3
 24.66 in/s   0

2
2

Construct the acceleration image of link 3.
A A  AC  AnAC  AtAC
A B  A D  AnBD  AtBD
and
n
2
AAC
 VAC
/ RAC   7.807 in/s  /  6.0 in   10.16 in/s 2
2
A A  169.2 in/s20
n
BD
A
V
2
BD
Ans.
/ RBD   7.622 in/s  /  6.0 in   9.68 in/s
A B  806.1 in/s20
2
2
Ans.
160
4.37
For the mechanism in the posture illustrated in Problem 3.32, crank 2 has an angular
acceleration of 2 rad/ s 2 ccw. Determine the acceleration of point C4 and the angular
acceleration of link 3.
A B  A A  AnBA  AtBA
n
ABA
 22 RBA  10.0 rad/s   3.0 in   300.0 in/s 2
2
t
ABA
  2 RBA   2.0 rad/s2   3.0 in   6.0 in/s 2
A D4  ArD4 /1  A B  AnDB  AtDB
n
2
ADB
 VDB
/ RDB   30.0 in/s  / 1.0 in   900.0 in/s 2
2
Draw the acceleration image of link 4.
AC4  600.1 in/s291.1
AC3  AC4  A
r
C3 / 4
 AA  A  A
n
CA
Ans.
t
CA
n
2
ACA
 VCA
/ RCA   60.0 in/s  /  2.0 in   1 800.0 in/s 2
2
Draw the acceleration image of link 3.
ACt A 12.0 in/s2
3  3 3 
 6.0 rad/s2 ccw
RC3 A3
2.0 in
Ans.
161
4.38
For the mechanism in the posture illustrated in Problem 3.29, the input angular velocity is
constant. Determine the angular accelerations of links 3 and 4.
Since we know the path of D4 on link 2, we write
A D2  A A2  AnD2 A2  A Dt 2 A2
ADn2 A2  22 RD2 A2  15.0 rad/s  1.250 in   281.25 in/s2
2
A D4  A D2  AcD4 D2  AnD4 / 2  AtD4 / 2  A E4  AnD4 E4  AtD4 E4
AcD4 D2  22 × VD4 /2  2 15.0 rad/s  49.41 in/s   1 482.3 in/s280.8
ADn4 / 2  VD24 / 2 D4 / 2   49.41 in/s 
ADn4 E4  VD24 E4
 2.50 in   976.58 in/s2
2
RD E  15.24 in/s   3.50 in   66.39 in/s2
2
4 4
Draw the acceleration images of links 2 and 4.
ADt 4 E4 352.4 in/s 2
4 

 100.7 rad/s 2 ccw
RD4 E4
3.50 in
Ans.
AC3  AC2  ACr 3 / 2  A D3  ACn 3D3  ACt 3D3
ACn3D3  VC23D3 RC3D3   41.91 in/s 
3 
ACt 3 D3
RC3 D3

2
 0.50 in   3 513.0 in/s2
231.7 in/s2
 463.4 rad/s 2 ccw
0.50 in
Ans.
162
4.39
For the mechanism in the posture illustrated in Problem 3.30, the input angular velocity is
constant. Determine the acceleration of point G and the angular accelerations of links 5
and 6.
A B  A A  AnBA  A BAt
n
ABA
 22 RBA  10 rad/s  1.0 in   100.0 in/s 2
2
n
t
n
t
AC  A B  ACB
 ACB
 A D  ACD
 ACD
n
2
ACB
 VCB
/ RCB  13.33 in/s  /  4.0 in   44.44 in/s 2
2
n
2
ACD
 VCD
/ RCD   6.66 in/s  /  2.0 in   22.22 in/s 2
2
Construct the acceleration image of link 3.
Since we know the path of E3 on link 6, we write
A E3  A E6  AcE3E6  AnE3 / 6  AtE3 / 6 and A E6  A H6  ΑnE6 H6  ΑtE6 H6
AcE3E6  26 × VE3 /6  2  3.774 rad/s 10.89 in/s   82.23 in/s2104.5
AEn3 / 6  VE23 / 6 /  E3 / 6  10.89 in/s  /   0
2
AEn6 H6  VE26 H6 RE6 H6   4.86 in/s  1.29 in  18.33 in/s 2
2
Construct the acceleration image of link 6.
AG  351.4 in/s2  64.5
AEt 6 H6
141.9 in/s 2
6 

 110.2 rad/s 2 cw
RE6 H6
1.29 in
Ans.
Ans.
A F5  A F6  ArF5 / 6  A E5  AnF5E5  AtF5E5
AFn5 E5  VF25 E5 RF5 E5  12.78 in/s 
5 
AFt 5 E5
RF5 E5

0 in/s 2
0
0.50 in
2
 0.50 in   326.7 in/s2
Ans
163
4.40
Continue Problem 3.40 and find the second-order kinematic coefficients of links 3 and 4.
Assuming an input acceleration of AA2  5 m/s2 find the angular accelerations of links 3
and 4.
RBO4  RBA  120 mm
From the solution of Prob. 3.40 we have the derivative of the loop-closure equations with
respect to the input r2. In matrix form this is
 r3 sin 3 r4 sin  4  3  1 
 r cos  r cos       0
 
3
4
4 4
3
From these we found the determinant of the Jacobian and the first-order kinematic
derivatives. At the posture indicated these are   r3r4 sin 4  3   0.012 471 m2 ,
3  r4 cos 4   4.811 rad/m , and 4  r3 cos3   4.811 rad/m .
The next derivative of the above equations with respect to input r2, in matrix form, gives
 r3 sin 3 r4 sin 4  3  r3 cos 332  r4 cos  4 42 
 r cos  r cos      
2
2 
3
4
4 4
3
 r3 sin 33  r4 sin  4 4 
The solution to this set of equations is
3 1  r4 cos  4
     r cos 
3
 4
 3

r4 sin  4   r3 cos 3  r4 cos  4  32 
 
r3 sin 3   r3 sin 3  r4 sin  4   42 
 32 
r42
1  r3r4 cos 3   4 

 
r32
r3r4 cos 3   4    42 

For the specified posture these give values of 3  13.363 rad/m2
Ans.
and 4  13.363 rad/m .
From these we find angular accelerations of
3  3r2  3r22  3 030.7 rad/s2 (cw)
Ans.
4  4r2   r  3 030.7 rad/s ccw
Ans.
2
and
2
4 2
2
Ans.
164
4.41
Continue Problem 3.49 and find the second-order kinematic coefficients of links 3, 4, and
5. Assuming constant angular velocity for link 2 find the angular accelerations of links 3,
4, and 5.
From the solution of Prob. 3.49 we have the derivative of
the loop-closure equations with respect to the input  2 .
In matrix form this is
 RBA sin 3  RBC sin  4  3    2  sin 5  sin  2  


  R cos 
RBC cos  4   4   2  cos 5  cos  2  
3
 BA
with the constraint 55   2 and the determinant
  RBA RBC sin 3  4  .
At the posture shown these give values of
  30.000 in 2
3  2 RBC sin  (4  5   sin 4  2    0.200 rad/rad
4  2 RBC sin  (3  5   sin 3  2    0
5   2 5  0.500 rad/rad
The next derivative of these equations gives
 RBA sin 3
  R cos 
3
 BA
 RBC sin  4  3   RBA cos 3

RBC cos  4   4   RBA sin 3
RBC cos  4  32 
 
RBC sin  4   42 
   cos 5  2 cos  2  5 
 2
 
   2 sin 5   2 sin  2   1 
with the derivative of the constraint giving 5  0 . The solution of the above equations
gives
2
 32 
RBC
3 1   RBA RBC cos 3   4 
 
    
2
 RBA
RBA RBC cos 3   4    42 
 4


2  RBC cos  4  5  RBC cos  2   4   5 


  RBA cos 3  5  RBA cos  2  3    1 
At the posture shown the values of the second-order kinematic derivatives are
3  0.240 rad/rad 2 , 4  0.150 rad/rad 2 , and 5  0 .
Ans.
With 2  5 rad/s  const , the requested angular accelerations are
3  322  6.0 rad/s2 ccw ,
 4  422  3.75 rad/s2 ccw ,
5  522  0 .
Ans.
165
4.42
Continue Problem 3.50 and find the second-order kinematic coefficients of links 3, 4, and
5. Assuming constant angular velocity for link 2 find the angular accelerations of links 3,
4, and 5.
From the solution of Prob. 3.50 we have the derivative of the loop-closure equations with
respect to the input  2 . In matrix form this is
RAO5 sin 5   4   RBO2 sin  2 
 RBA sin  4

   

  RBA cos  4  RAO5 cos 5  5    RBO2 cos  2 
with the constraint 33   1  3  and the determinant   RBA RAO5 sin 5  4  .
At the posture shown these give values of   14.3 in 2
3   1  3  3  4.000 rad/rad
4  RBO RAO sin 5  2    0.273 rad/rad
2
5
5  RBA RBO sin  (2  4    1.000 rad/rad
2
The next derivative of these equations gives
RAO5 sin 5   4   RBA cos  4  RAO5 cos 5   42   RBO2 cos  2 
 RBA sin  4

   
 2 

  RBA cos  4  RAO5 cos 5  5   RBA sin  4  RAO5 sin 5  5   RBO2 sin  2 
with the derivative of the constraint giving 3  0 . The solution of the above equations
gives
2
  42  1  RAO5 RBO2 cos  2  5 
RAO
 4 1  RBA RAO5 cos  4  5 
5


  

  
2
RBA
 RBA RAO5 cos  4  5   52    RBA RBO2 cos  2   4  
 5

At the posture shown the values of the second-order kinematic derivatives are
3  0 , 4  0.000 35 rad/rad 2 , and 5  0.420 rad/rad 2 .
Ans.
With 2  15 rad/s  const , the requested angular accelerations are
3  322  0 ,
 4  422  0.078 8 rad/s2 ccw ,
5  522  94.41 rad/s2 (cw) . Ans.
166
4.43
Draw the inflection circle for the absolute motion of the coupler link of the double-slider
linkage. Select several points on the centrode normal and find their conjugate points.
Plot portions of the paths of these points to demonstrate for yourself that the conjugates
are indeed the centers of curvature.
RBA  125 mm
167
4.44
Draw the inflection circle for the absolute motion of the coupler of the four-bar linkage.
Find the center of curvature of the coupler curve of point C and generate a portion of the
path of C to verify your findings.
RcA  2.5 in, RAO2  0.9 in, RBO4  3.5 in, RPO4  1.17 in
Since point C is on the inflection circle, its center of curvature is at infinity and its point
path is a straight line in the vicinity of the position shown.
Ans.
168
4.45
For the motion of the coupler relative to the frame, find the inflection circle, the centrode
normal, the centrode tangent, and the centers of curvature of points C and D of the
linkage of Problem 3.13. Choose points on the coupler coincident with the instantaneous
center of velocity and inflection pole and plot nearby portions of their paths.
169
4.46
For the four-bar linkage in the posture illustrated, link 2 is 30 counterclockwise from the
ground link and the angular velocity and angular acceleration of the coupler link are
3  5 rad/s ccw and 3  20 rad/s2 cw , respectively. For the instantaneous motion of
the coupler link show: (a) the velocity pole I, the pole tangent T, and the pole normal N;
(b) the inflection circle and the Bresse circle; and (c) the instantaneous center of
acceleration. Then determine; (d) the radius of curvature of the path of coupler point C;
(e) the velocity of C; (f) the angular velocity of link 2; (g) the velocity of the pole I; (h)
the acceleration of C; and (i) the acceleration of the velocity pole.
RO4O2  2.5 in, RAO2  1 in , RBA  3.15 in, RBO4  1.5 in, and RCB  1 in.
(a) The pole I is coincident with the instant center I13 shown in the figure below. The
instant center I 24 and the collineation axis are as shown in the figure. From Bobillier's
theorem, the angle from the collineation axis to the first ray (say link 2) is measured as
84° cw. This is equal to the angle from the second ray (link 4) to the pole tangent T; that
is, 84° cw. Therefore, the pole tangent T is as shown in the figure and the pole normal N,
which is perpendicular to the pole tangent T, is also shown.
(b) The inflection point J A for point A on link 3 can be obtained from the Euler-Savary
equation; that is,
 0.69 in   0.47 in
R2
RAJ A  AI 
RAOA
1.00 in
2
The location of the inflection point J A is shown in the figure.
Similarly, the inflection point J B for point B on link 3 can be obtained from the EulerSavary equation; that is,
 2.81 in   5.27 in
RBI2
RBJ B 

RBOB
1.50 in
2
The location of the inflection point J B is shown on the figure.
Knowing the pole normal and the two inflection points, the inflection circle can be drawn.
The inflection circle for the motion of link 3 with respect to 1, the inflection pole J, and
the center of the inflection circle (denoted as point O) are shown on the figure. Note that
the pole normal N points from the pole I toward the inflection pole J and the pole tangent
T is 90 clockwise from the pole normal. The diameter of the inflection circle for the
motion 3/1 is measured as
170
RJI  2.46 in
Ans.
The diameter of the Bresse circle is
2
5 rad/s 

32
b
R 
2.46 in  3.07 in
Ans.
 3 JI 20 rad/s 2
Since the angular acceleration of the coupler link is clockwise (that is, negative), the
Bresse circle must lie on the positive side of the pole tangent, as shown on the figure.
(c) The point of intersection of the inflection circle and the Bresse circle (other than pole I) is
the acceleration center  of the coupler link; see the figure.
(d) From the Euler-Savary equation, the radius of curvature of the coupler point C is
1.87 in   0.82 in
R2
C  RCOC  CI 
RCJC
4.26 in
2
Ans.
The location of the center of curvature of the path of point C (that is, OC ) is as shown on
the figure.
(e) The velocity of coupler point C is
171
VC  3 RCI   5 rad/s 1.87 in   9.35 in/s
Ans.
The direction of the velocity vector of point C is as shown on the figure.
(f) The angular velocity of link 2 can be written as
RI I
0.686 in
2  23 13 3 
5 rad/s  3.43 rad/s (cw)
RI23 I12
1.00 in
(g) The velocity of the pole I is
v  3 RJI   5 rad/s  2.46 in  12.30 in/s  1.03 ft/s
Ans.
Ans.
Since the angular velocity of the coupler link is positive (counterclockwise) the velocity
of the pole must be negative; that is, in the direction opposite to the pole tangent T (as
shown on the figure).
(h) The acceleration of coupler point C can be written as
AC  RC 34   32
  3.68 in 
 5 rad/s    20 rad/s2 
4
2
 117.68 in/s2  9.81 ft/s2
Ans.
The angle from the line  I to the pole normal N is measured as 38.25 ccw, as shown in
the figure. Therefore, the direction of the acceleration vector of point C is 38.25 ccw
from the line connecting  to C, as shown in the figure.
(i) The acceleration of the pole I can be written as
2
AI  v3  32 RJI   5 rad/s   2.46 in   61.44 in/s2  5.12 ft/s2
The acceleration of the pole is directed along the pole normal N as shown on the figure.
The angle from the horizontal axis to the pole normal N is measured as 33.0.
As a check, the acceleration of the pole can be written as
AI  RI  34   32
 1.90 in 
 5 rad/s    20 rad/s2 
4
=60.89 in/s2 = 5.07 ft/s2
2
Ans.
172
4.47
Consider the double-slider linkage in the posture given in Problem 3.8. Point B moves
with a constant velocity VB  40 m/s to the left as illustrated in the figure. The angular
velocity and angular acceleration of coupler link AB are 3  36.6 rad/s ccw and
3  1 340 rad/s2 cw, respectively. For the absolute motion of coupler link AB in the
specified posture, draw the inflection circle and the Bresse circle. Then determine: (a) the
radius of curvature of the path of point C which is a point of link 3 midway between
points A and B; and (b) the velocity of the velocity pole I. Using the instantaneous center
of acceleration determine: (c) the acceleration of the pole I; and (d) the accelerations of
points A and C.
The pole I is the point coincident with the instant center I13 at the intersection of the
vertical line through point B and the line perpendicular to the direction of motion of slider
2 through point A. Since point A moves on a straight line, the center of curvature for
point A is at infinity. Hence, the inflection point J A is coincident with point A.
Similarly, the inflection point J B is coincident with point B since point B also moves on
a straight line and the center of curvature of point B is at infinity. Knowing the two
inflection points J A and J B and the pole I, the center O of the inflection circle is obtained
as the intersection of the perpendicular bisectors of IJ A and IJ B as shown in the figure
below. The centrode normal passes through I and O and intersects the inflection circle at
inflection point J. The diameter of the inflection circle is measured as RJI  1 545 mm .
We keep in mind that the centrode normal N points from I toward J and the centrode
tangent is 90 clockwise from the centrode normal. The diameter of the Bresse circle is
173
 36.6 rad/s  (1.545 m)  1.545 m
32
RJI 
3
1340 rad/s 2
Since the angular acceleration of link 3 (that is,  3 ) is clockwise (negative), the Bresse
circle must lie on the positive side of the centrode tangent. The Bresse circle is
positioned as shown in the figure.
2
b
As a check, note that point B, fixed in link 3, moves on a straight line with a constant
velocity. Hence point B must be the acceleration center for the absolute motion of link 3.
With the above construction of the inflection circle and the Bresse circle, the acceleration
center  (the point of intersection of the inflection circle and the Bresse circle) coincides
with point B.
The angle from the line I  to the centrode normal N is measured as 45 ccw . To
check, in Eq. (4.48),

1340 rad/s 2
  tan 1 32  tan 1
 45 ccw
2
3
 36.6 rad/s 
(a) The radius of curvature of point C, from the Euler-Savary equation, is
RCI2
(1 205.33 mm)2
C  RCC 

 43 779.23 mm  43.78 m
RCJC
33.19 mm
Ans.
Since the center of curvature C  for point C does not lie on the paper, the direction is
indicated by an arrow on the figure.
(b) The magnitude of the velocity of the pole I is
  RJI 3  (1.545 m)  36.6 rad/s   56.55 m/s
Ans.
Since the angular velocity of link 3 is positive (counterclockwise) the pole velocity  is
in the negative pole tangent direction as shown in the figure.
(c) The magnitude of the acceleration of the pole I is
AI  3  (56.55 m/s)  36.6 rad/s   2 070 m/s 2
Ans.
The acceleration of the pole points along the positive pole normal as shown in the figure.
(d) The magnitudes and directions of the velocities of points A and C are found as:
Since point A is fixed in link 3 the velocity of point A is
Ans.
VA  3 RAI   36.6 rad/s  (1.338 m)  48.971 m/s
The direction of the velocity of A is perpendicular to the line RAI as shown in the figure.
Since point C is fixed in link 3 the velocity of point C is
VC  3 RCI   36.6 rad/s  (1.205 m)  44.103 m/s
Ans.
The direction of the velocity of C is perpendicular to the line RCI as shown in the figure.
(e) The magnitudes and directions of the accelerations of points A and C are found as:
From Eq. (4.45), the acceleration of point A is given by
174
AA  RA 34   32
  0.400 m  (36.6 rad/s) 4  (1 340 rad/s 2 ) 2
Ans.
 757.89 m/s 2
The angle between the line  I and the centrode normal is 45 ccw. Therefore, the
acceleration of point A is directed at an angle of 45 ccw from the line  A as shown in
the figure.
The acceleration of point C can be written as
AC  RC 34   32
  0.200 m  (36.6 rad/s) 4  (1 340 rad/s 2 ) 2
Ans.
 378.95 m/s 2
The acceleration of point C is at an angle of 45 ccw from line  C as shown in the
figure.
175
4.48
For the linkage of Problem 3.17, link 2 is rotating with an angular velocity
For the
2  15 rad/s ccw and an angular acceleration  2  320.93 rad/s2 cw .
instantaneous motion of the connecting rod 3, find: (a) the inflection circle and the Bresse
circle; (b) the location of the instantaneous center of acceleration; (c) the center of
curvature of the path traced by the coupler point C; (d) the accelerations of points A, B,
and C; and (e) the acceleration of the inflection pole J.
(a) The velocity pole I for the connecting rod 3 is coincident with the instant center I13 .
Since point B on link 3 travels on a straight line, it is an inflection point; that is, JB
coincides with point B (the inflection circle for the motion 3/1 has to pass through point
B). The inflection point for point A on link 3 can be obtained from the Euler-Savary
equation; that is,
2
13.37 in   71.50 in
RAI
RAJ A 

RAOA
2.50 in
2
The inflection circle is drawn through points I, JA, and JB as shown in the figure below.
The diameter of the inflection circle is measured as
RJI  88.55 in
Ans.
The centrode tangent T and the centrode normal N are also shown in the figure. Note that
176
the centrode normal passes through the pole I and the center of the inflection circle O and
intersects the inflection circle at the inflection pole J. The positive centrode normal
points from I to J and the positive centrode tangent is 90° clockwise from the centrode
normal.
In order to draw the Bresse circle, the angular velocity and angular acceleration of link 3
must be known. The method of kinematic coefficients (see Sections 3.11 and 4.11) is
used here to determine 3 and  3 .
The vectors for the slider-crank portion of the mechanism are shown in the figure. The
vector loop equation can be written as
R 2  R3  R 4  R1  0
The X and Y components can be written as
R2 cos 2  R3 cos 3  R4  0
R2 sin 2  R3 sin 3  R1  0
Differentiating these with respect to the input  2 gives
 R2 sin 2  R3 sin 33  R4  0
R2 cos 2  R3 sin 33  0
where 3  d3 d2 and R4  dR4 d2 are the first-order kinematic coefficients of links
3 and 4, respectively. Writing these equations in matrix form gives
  R3 sin 3

 R3 cos 3
1  3   R2 sin  2 
   

0   R4    R2 cos  2 
The determinant of the coefficient matrix is   R3 cos 3 . The length of the input link is
R2  2.5 in and the length of the coupler link is R3  10 in.
The slider offset is
R1  1.50 in . For the given input position  2  135o , the coupler angle (found from
trigonometry) is 3  19.07 . Substituting the known data gives
3.267 1  3   1.768 
 9.451 0   R    1.768

  4  

Therefore, the first-order kinematic coefficients of link 3 and link 4 are
3  0.187 1 rad/rad and R4  1.15 7 in/rad
The angular velocity of link 3 is
3  32  (0.187 1 rad/rad)(15 rad/s)  2.81 rad/s (ccw)
Differentiating the above matrix equation with respect to the input variable  2 gives
  R3 sin 3

 R3 cos 3
1  3   R cos   R cos   2 
2
2
3
3 3

   
0   R4  R sin   R sin   2 
2
2
3
3
3


177
Then substituting the numerical data gives
3.267 1  3   1.437 
 9.451 0   R    1.653 

  4  

Using Cramer's rule, the second-order kinematic coefficients of the mechanism are
3  0.175 rad/rad 2 and R4  2.01 in/rad 2
The angular acceleration of link 3 can be written (see Table 4.2) as
       2
3
3
2
3
2
 (0.187 1 rad/rad)(320.93 rad/s 2 )  (0.175 rad/rad 2 )(15 rad/s)2
 20.67 rad/s 2 (cw)
The diameter of the Bresse circle for the motion 3/1 can now be written as
 2.81 rad/s   33.83 in
32
Ans.
  88.55 in 
3
20.67 rad/s 2
Since the angular acceleration of link 3 is clockwise (negative) the Bresse circle must lie
on the positive side of the centrode tangent as shown in the figure.
2
b  RJI
(b) The acceleration center  for the absolute motion of link 3 is the point of
intersection of the inflection circle and the Bresse circle. The angle from the line  I to
the centrode normal N is measured as 69.09 ccw . As a check, from Eq. (4.44), the angle
 is given by
2
3
1 20.67 rad/s
  tan
 tan
 69.09  ccw 
2
32
 2.81 rad/s 
1
(c) From the Euler-Savary equation, the radius of curvature of point C can be written as
R2
(12.95 in)2
C  RCOC  CI 
 3.46 in
RCJC
48.41 in
The center of curvature OC of point C is as shown in the figure.
(d) The velocity of point A is
VA  3 RAI  (2.81 rad/s)(13.37 in)  37.51 in/s
As a check, the velocity of point A can also be found as
VA  2 RAO2  (15 rad/s)(2.5 in)  37.50 in/s.
The direction of velocity of point A is 135° as shown in the figure.
The velocity of point B is
VB  3 RBI  (2.81 rad/s)(6.18 in)  17.35 in/s
The direction of velocity of point B is 180° as shown in the figure.
The velocity of point C is
VC  3 RCI  (2.81 rad/s)(12.95 in)  36.34 in/s
The direction of velocity of point C is 152.39° as shown in the figure.
From Eq. (4.45), the acceleration of point A can be written as
AA  R A 34  32  44.39 in  2.81 rad/s   (20.67 rad/s 2 ) 2  982.01 in/s2
4
Ans.
178
With   69.09o ccw, the direction of the acceleration of point A is 9.94° as shown in the
figure. The acceleration of point B can be written as
AB  R B 34  32  37.30 in  2.81 rad/s   (20.67 rad/s 2 ) 2  825.14 in/s2
4
Ans.
The direction of the acceleration of point B is as shown in the figure. The acceleration of
point C can be written as
AC  R C 34  32  44.54 in  2.81 rad/s   (20.67 rad/s 2 ) 2  985.28 in/s2
4
Ans.
The direction of the acceleration of point C is 4.79° as shown in the figure.
(e) The velocity of the inflection pole J is the same as the velocity of the pole I.
Therefore, the velocity of the inflection pole J is
v  3 RJI   2.81 rad/s 88.55 in   248.51 in/s
The direction of the velocity of inflection pole J is perpendicular to line IJ as shown in
the figure.
The acceleration of the inflection pole J is
AJ  R J 34  32  82.72 in  2.81 rad/s   (20.67 rad/s 2 ) 2  1 829.90 in/s2
4
Ans.
The acceleration of the inflection pole J makes an angle of   69.09 with the line  J .
179
4.49
Figure P3.32 illustrates an epicyclic gear train driven by the arm, link 2, with an angular
velocity 2  3.33 rad/s cw and an angular acceleration  2  15 rad/s2 ccw . Define
point E as a point on the circumference of the planet gear 4 horizontal to the right of point
B such that the angle DBE  90. For the absolute motion of gear 4, draw the inflection
circle and the Bresse circle on a scaled drawing of the epicyclic gear train. Then
determine: (a) the location of the instantaneous center of acceleration of the planet gear;
(b) the radii of curvature of the paths of points B and E; (c) the locations of the centers of
curvature of the paths of points B and E; and (d) the accelerations of points B and E and
pole I.
Since the problem is for the motion 4/1 where 4 is the planet gear which is in internal
rolling contact with the fixed ring gear 1 then the pole I is coincident with the instant
center I14 which is the point of contact between the two gears. Note that the fixed
centrode is gear 1 and the moving centrode is gear 4. Recall that the centrode normal N
points from the fixed centrode toward the moving centrode (in the neighborhood of the
pole I). Therefore, the centrode normal N points vertically downward. Also, recall that
the Euler-Savary equation can be written as
1
1
1


RJI RIOF RIOM
The center of curvature of the fixed centrode OF is coincident with the center of the fixed
gear, that is, point A, and the center of curvature of the moving centrode OM is coincident
with the center of the planet gear; that is, point B. Recall that I with respect to OF and I
with respect to OM are both vertically upward; therefore, the radii of curvature of the
fixed and the moving centrodes, respectively, are
1  RIOF  4 in and 4  RIOM  1 in
Substituting into the Euler-Savary equation gives
1
1
1
3



RJI 4 in 1 in 4 in
Therefore, the diameter of the inflection circle is
RJI  4 3  1.33 in
The inflection circle is shown in the figure below. Recall that the centrode tangent T is
90 clockwise from the centrode normal N and is, therefore, horizontal and is positive to
the left.
As a check: The inflection point for point C is coincident with the inflection pole J;
therefore, the radius of curvature of the path of point C, from the Euler-Savary equation,
is
R2
(2.00 in) 2
C  RCOC  CI 
 6 in
RCJC
0.66 in
To determine the angular velocity and the angular acceleration of the planet gear 4, the
rolling contact equation between the planet gear 4 and the fixed ring gear 1 (see Chapter
3, Example 3.9) can be written as
180
 4   2
1   2
 
1
4 in

 4
4
1 in
We use the positive sign here since there is internal rolling contact between the planet
gear 4 and the ring gear 1. Differentiating this equation with respect to the input position,
the rolling contact equation, in terms of first-order kinematic coefficients, is written as

 4   2
  1 4
 

1
2
4
Since the input is the arm (link 2) then  2  1 , and since the ring gear 1 is fixed then
1  0 . Therefore, the first- and second-order kinematic coefficients of the planet gear 4
from the above equation are
4   3 rad/rad
and
 4  0
The angular velocity of the planet gear 4 is
4  4 2  (3 rad/rad)(3.33 rad/s)  10 rad/s (ccw)
The angular acceleration of the planet gear 4 is
 4  4 2  422  (3 rad/rad) 15 rad/s   0  45 rad/s2 (cw)
Therefore, the diameter of the Bresse circle for the motion 4/1 is
10 rad/s   2.96 in
2
b  RJI 4  1.33 in 
4
45 rad/s 2
Since the angular acceleration of planet gear 4 is clockwise (negative), the Bresse circle
lies on the positive side of the centrode tangent T. The Bresse circle is as shown in the
figure.
2
181
(a) The acceleration center  for the absolute motion of planet gear is the intersection
of the inflection circle and the Bresse circle. The acceleration center  is shown in the
figure. The angle from the line I  to the centrode normal N is measured as
  24.23 ccw
Check: The angle  is given by the relation
  tan 1
2
4
1 45 rad/s

tan
 24.23o ccw
2
2
4
10 rad/s 
(b) and (c) The radius of curvature of the path of point B, from the Euler-Savary
equation, is
1.00 in   3 in
R2
 B  RBOB  BI 
RBJ B
0.33 in
2
Ans.
Note that the center of curvature OB of point B is coincident with point A and the
inflection point J B is coincident with the inflection pole J. The radius of curvature of the
path of point E, from the Euler-Savary equation, is
1.41 in   4.24 in
REI2
 E  REOE 

REJ E
0.47 in
2
Ans.
The center of curvature OE of point E is shown on the figure.
(d) The acceleration of point B can be written as
AB  R B 44   42   0.51 in 
10 rad/s   (45 rad/s2 )2  56.00 in/s2
4
Ans.
With   24.23o ccw the direction of the acceleration of point B is as shown in the figure.
The acceleration of point E can be written as
AE  R E 44   42  1.51 in 
10 rad/s   (45 rad/s2 )2  165.58 in/s 2
4
Ans.
With   24.23 ccw the direction of the acceleration of point E is as shown in the figure.
The acceleration of the pole I is
Ans.
AI  4  (13.33 in/s) 10 rad/s   133.33 in/s 2
The acceleration of the pole is directed along the positive pole normal as shown in the
figure.
Check: The acceleration of the pole I can also be obtained from the equation
AI  R I 44   42  1.216 in 
10 rad/s   (45 rad/s2 )2  133.3 in/s2
4
182
4.50
On 18×24-in paper, draw the four-bar linkage full size, placing A 6 in up from the lower
edge and 7 in left of the right edge. (Better utilization of the paper is obtained by tilting
the frame through about 15 as indicated.) For the coupler link draw the inflection circle,
and .the cubic of stationary curvature. Choose a coupler point C coincident with the
cubic and plot a portion of its coupler curve in the vicinity of the cubic. Find the
conjugate point C  . Draw a circle through C with center at C  and compare this circle
with the actual path of C. Find Ball’s point. Locate a point D on the coupler at Ball’s
point and plot a portion of its path. Compare the result with a straight line.
RAA  1 in, RBA  5 in, RBA  1.75 in, and RBB  3.25 in
Drawn with a precise CAD system above, the circle around center C  matches the
coupler curve near C to better than visual comparison for the 30 of crank rotation
shown. Similarly, Ball’s point D follows an almost perfect straight line over the same
range as shown.
183
4.51
For the mechanism in the posture illustrated in Figure P3.51, the first- and second-order
kinematic coefficients are 3   8.333 rad/rad, r2   100 mm/rad, 3   8.642 rad/rad2 ,
and r2   237.033 mm/rad2 (where  2 is the input and r2 is the vector from the ground
pin O2 to pin A). The wheel 3 is rolling without slipping on the ground link at point C
and sliding in the slot that is cut in link 2. The radius of the ground link is 1  60 mm
and the radius of the wheel is 3  15 mm. Determine: (a) the unit normal vector to the
path of point D; (b) the radius of curvature of the path of this point; and (c) the x and y
coordinates of the center of curvature of this path. If the angular velocity of link 2 is a
constant 2  30 rad/s ccw then determine the acceleration of point D.
Suitable vectors for the path of point D are shown in the figure. The vector loop equation
can be written as follows
rDe jD  xD  jyD  r2e j2  r23e j23
where r23  3  15 mm and 23  0. Therefore the position coordinates of point D are
xD  r2 cos 2   3 cos  23  30 mm
y D  r2 sin 2  3 sin 23  60 mm
184
The first-order kinematic coefficients of point D are
 0
xD  r2 cos 2  r2 sin  2   3 sin  23 23
yD  r2 sin  2  r2 cos  2   3 cos  23 23  250 mm/rad
rD  x2  y2  250 mm/rad
The second-order kinematic coefficients of point D are
2   3 sin 23 23
  694.62 mm/rad 2
xD  r2cos 2  2r2 sin 2  r2 cos 2   3 cos 2323
2   3 cos  23 23
  259.33 mm/rad 2
yD  r2sin 2  2r2 cos 2  r2 sin 2   3 sin 2323
The velocity of point D is
VD  xD ˆi  yD ˆj 2  7.500ˆj m/s  7.500 m/s  90


(a) The unit tangent vector to the path of point D is
x ˆi  yD ˆj 250ˆj mm/rad
uˆ tD  D

 ˆj
rD
250 mm/rad
The unit normal vector to the path of point D is
uˆ nD  kˆ × utD  kˆ × ˆj  ˆi
 
Ans.
(b) The radius of curvature of the path of point D can be written as
rD3
D 
xD yD  yD xD
 250 mm/rad 

 90 mm
2
 0  259.33 mm/rad    250 mm/rad   694.62 mm/rad 2 
3
Ans.
Since the radius of curvature of the path of point D is negative, the unit normal vector uˆ nD
must point away from the center of curvature
(c) The x and y coordinates of the center of curvature of the path of point D can be
written as
 y
250 mm/rad
xCC  xD   D D  30 mm+  90 mm 
 120 mm
Ans.
rD
250 mm/rad
x
0
Ans.
yCC  yD   D D  60 mm+  90 mm 
 60 mm
rD
250 mm/rad
The velocity of point D can be written as
VD  xD ˆi  yD ˆj 2  0ˆi  7.500ˆj m/s  7.500 m/s  90


The acceleration of point D can be written as
A D  xD ˆi  yD ˆj 22  xD ˆi  yD ˆj 2




 625 158ˆi  233 397ˆj m/s2  667 306 m/s2  159.53
Ans.
185
4.52
For the mechanism in the posture illustrated, the first and second-order kinematic
coefficients are 3  0.50 rad/ft, R4   1.00 ft/ft, 3  0.25 rad/ft 2 , and R4   1.00 ft/ft 2 .
The roller 4 is pinned to link 3 at B and is rolling without slipping on the vertical ground
link at C. Determine: (a) the first and second-order kinematic coefficients of point P; (b)
the unit tangent vector and the unit normal vector to the path traced by P; (c) the radius of
curvature of this path; and (d) the x and y coordinates of the center of curvature of this
path. If the constant velocity of the input is V2   10 ˆi ft/s, then determine the
acceleration of P.
R 2  R AO  2 ˆi ft, R 4  R BO  2 ˆj ft, RPA  4 ft, and 4  0.375 ft.
The vectors that are chosen for the mechanism are shown in the figure. From the
geometry of the right-angle triangle O A B , the length of link 3 is R3 = 2.828 ft.
(a) The x and y coordinates of point P are
xP  R2 cos  2  RPA cos 3   0.828 43 ft
yP  R2 sin  2  RPA sin 3   2.828 43 ft
Differentiating with respect to the input position R2 gives
xP  cos  2  RPA sin 3 3   0.414 21 ft/ft
yP  sin  2  RPA cos 33   1.414 21 ft/ft
Ans.
Differentiating again with respect to the input position R2 gives
xP   RPA cos 3 32 RPA sin 33  0
yP   RPA sin  32  RPA cos 33   1.414 21 ft/ft 2
(b) The unit tangent vector to the path at point P can be written as
Ans.
186


utP  xP ˆi  yP ˆj rP
where
rP  
xP2  yP2   1.473 63 ft/ft
The negative sign is chosen here because the input is negative (that is, the input vector
R 2 is decreasing in length). Therefore, the unit tangent vector to the path at P is
Ans.
ut  x ˆi  y ˆj r  0.281 08ˆi  0.959 68ˆj

P
P
P

P
The unit normal vector to the path at P is
unP   yP ˆi  xP ˆj rP  0.959 68ˆi  0.281 08ˆj


Ans.
(c) The radius of curvature of the path of P is
rP3
P 
xP y P  xP y P

  1.473 63 ft/ft 
3
Ans.
  5.462 89 ft
(  0.414 21 ft/ft)(  1.414 21 ft/ft)  ( 1.414 21 ft/ft)(  0)
The negative sign implies that the unit normal vector to the path at P is pointing away
from the center of curvature of the path of P.
(d) The x and y coordinates of the center of curvature of the path of P can be written as
xCC  xP   P   yP rP   4.414 21 ft
yCC  yP   P   xP rP   1.292 91 ft
Ans.
The center of curvature of the path of P is shown in the figure. The radius of curvature
and the approximate path of P are also shown in the figure.
The velocity of point P can be written as
VP  ( xP ˆi  yP ˆj) R2 = 4.1421 ˆi  14.1421 ˆj ft/s  14.7362 ft/s73.67
The acceleration of point P is
A P  ( xP ˆi  yP ˆj) R2  ( xP ˆi  yP ˆj) R22  0 ˆi  141.421 ˆj ft/s2  141.21 ft/s2  90
Ans.
187
4.53
For the rack-pinion mechanism in the posture illustrated, the first and second-order
kinematic
coefficients
are
R2   400 mm/rad,
R4   346.41 mm/rad,
R2  1 385.6 mm/rad 2 , and R4  1 400 mm/rad 2 (where R 2 is the vector from the ground
pin O2 to the point of contact C of link 3). Determine: (a) the first and second-order
kinematic coefficients of point B; (b) the unit tangent vector and the unit normal vector to
the path traced by point B; (c) the radius of curvature of this path; and (d) the x and y
coordinates of the center of curvature of this path. If the constant angular velocity of the
input link is ω2   12 kˆ rad/s then determine the acceleration of point B.
3  100 mm
(a) Using the vectors shown in the figure, the loop equation for the mechanism is
R2e j2  R32e j32  R4  0
which has the two scalar component equations
R2 cos 2  R32 cos 32  R4  0
R2 sin 2  R32 sin 32  0
with the constraint
2  32  90
The first and second-order kinematic coefficients of link 3 (that is,  3 and  3 ) can be
obtained from the rolling constraint equation between links 2 and 3. This rolling
constraint equation can be written in terms of first-order kinematic coefficients as
 R2  3 3  2 
Note that the positive sign must be used in this equation because the vector R 2 is
increasing in length for counterclockwise rotation of link 3.
Substituting 2  1, since the angle 2 is the input, and rearranging this equation, the firstorder kinematic coefficient of link 3 is
188
R2
400 mm/rad
 1  3 rad/rad
3
100 mm
Differentiating again with respect to the input angle θ2, the second-order coefficient of
link 3 can be written as
R 1 385.6 mm/rad 2
3  2 
  13.856 rad/rad 2
3
100 mm
From the figure we can write an equation for R4 as
3
100 mm
R4 

 200 mm
sin 2
sin 30
The vector equation for point B fixed in link 3 can be written as
RBe jB  R4  3e j3
with two scalar components
xB  R4   3 cos 3  200 mm
3 
1 
y B  3 sin 3  100 mm
Differentiating these equations with respect to the input angle θ2, the first-order kinematic
coefficients of point B are
xB  R4  3 sin 33  646.41 mm/rad
Ans.
yB  3 cos 33  0
rB   x2  y2  646.41 mm/rad
where the correct sign is negative because the input is negative (that is, the input angular
velocity is clockwise).
Differentiating again with respect to the input angle θ2, the second-order kinematic
coefficients of point B are
xB  R4   3 cos 332  3 sin 33  2 785.6 mm/rad 2
Ans.
yB    3 sin 332  3 cos 33  900.0 mm/rad 2
(b) The unit tangent vector to the path of point B can be written as
uˆ tB  xB ˆi  yB ˆj rB  ˆi
Ans.
The unit normal vector to the path of point B can be written as
uˆ nB  uˆ bB  uˆ tB  kˆ  ˆi  ˆj
Ans.


(c) The radius of curvature of the path of point B can be written as
rB3
B 
 464.27 mm
Ans.
xB yB  yB xB
Since the radius of curvature of the path of point B is a positive value, the unit normal
vector uˆ nB must point toward the center of curvature.
(d) The x-coordinate of the center of curvature of the path of point B can be written as
  y 
xCC  xB   B  B   200 mm
 rB 
189
This equation indicates that the x-coordinate of the center of curvature of the path of point
B is to the right of the y-axis directly above point B.
The y-coordinate of the center of curvature of the path of point B is
 x 
yCC  y B   B  B   364.27 mm
 rB 
This result indicates that the y-coordinate of the center of curvature of the path of point B
is above the x-axis. The location of the center of curvature of the path of point B is
shown in the following figure.
The velocity of point B can be written as
VB  ( xB ˆi  yB ˆj) 2  7 756.9ˆi mm/s  7 756.9 mm/s0
The acceleration of point B can be written as
A B  ( xB ˆi  yB ˆj)2 2  ( xB ˆi  yB ˆj) 2
 401 126ˆi  129 600ˆj mm/s2  421 543 mm/s217.91
Ans.
Ans.
190
4.54
For the gear train in Prob. P3.54, the angular velocity and acceleration of input gear 2 are
2  77 rad/s ccw and 2  5 rad/s2 cw, respectively. Determine: (a) the second-order
kinematic coefficients of gear 3 and rack 5; (b) the angular accelerationss of gear 3 and
link 4; and (c) the acceleration of the rack.
1  REO  4 in, 2  RCO  18 in, 3  RCD  7 in, and RFA  RBF  20 in.
1
2
(a) The rolling contact equation for the contact between gear 1 and gear 3 can be written
as
     
  
and  1  3 4
 1 3 4
 3 1 4
 3 1  4
Since the gears are in external contact the negative sign must be used. Therefore,
rearranging these equations, the first- and second-order kinematic coefficients of link 3
can be written as
      11 in 
  3  1   11 in 
3  4  3 1   

 4 and 3  4 
 4
  3   7 in 
  3   7 in 
Rearranging these equations, the kinematic coefficients of link 4 can be written as
 3 
 3 
 7 in 
 7 in 
4  3 
(1)
  3
  3 

 and 4  3
 11 in 
 11 in 
 1   3 
 1   3 
The rolling contact between gear 2 and gear 3 can be written as
  
     
 2  3 4
and  2  3 4
 3  2   4
 3 2  4
Since the gears are in internal contact the positive sign must be used. Rearranging these
equations, the kinematic coefficient of link 3 can be written as
   2  2
  11 in  18 in
3  4  3
 4 
(2a)


 7 in  7 in
 3  3
191
 3  2  2
  11 in  18 in
 4 
(2b)


 7 in  7 in
 3  3
Solving Eqs. (1) and (2) simultaneously gives the first- and second-order kinematic
coefficients of links 3 and 4.
9
9
and 4  rad/rad
3  rad/rad
7
11
9
9
and 4  rad/rad 2
Ans.
3  rad/rad 2
7
11
The rolling contact constraint between link 2 and link 5 can be written as
 R5  2 2  5 
3  4 
There are two possible cases:
(i) If the vector R5 is defined as R5  AF then the positive sign must be used, because a
counterclockwise rotation of the input causes link 5 to move to the right which moves
point F further from point A; in this case the first-order kinematic coefficient of link 5 is
R5  18 in 1 rad/rad  0  18 in/rad
(ii) If the vector R5 is defined as R5  FB then the negative sign must be used because a
counterclockwise rotation of the input causes link 5 to move to the right which moves
point F closer to point B; in this case the first-order kinematic coefficient of link 5 is
R5  18 in 1 rad/rad  0  18 in/rad
Ans.
Note that both scenarios will give the same answers for the kinematic analysis of the
mechanism. Also note that the second-order kinematic coefficient of link 5 can be
written as
 R5  2 2  5  0
(b) The angular velocity of gear 3 can be written as
3  32  ( 9 / 7)( 77 rad/s)  99 rad/s
The positive sign indicates that link 3 is rotating counterclockwise.
The angular acceleration of gear 3 can be written as
3  32  322  (9 / 7)( 5 rad/s2 )  (9 / 7)(77 rad/s)2  7 616.6 rad/s2
The angular velocity of link 4 can be written as
4  42  ( 9 / 11)(77 rad/s)  63 rad/s
The angular acceleration of link 4 can be written as
4  42  422  (9 / 11)( 5 rad/s2 )  (9 / 11)(77 rad/s)2  4 846.9 rad/s2
(c) The acceleration of link 5 can be written as
2
R5  R52  R522  18 in/rad   5 rad/s2    0 77 rad/s   90 in/s2
The negative sign indicates that link 5 is accelerating to the left.
Ans.
Ans.
Ans.
192
4.55
For the gear train in Problem 3.55, the angular velocity and acceleration of the input arm
2, are 2  50 rad/s cw and 2  15 rad/s2 cw, respectively. Using the method of
kinematic coefficients determine the angular accelerations of gears 3, 4, and 5.
The rolling contact equation for gear 3 rolling on the ground link 1 can be written as
       
   
 1   3 2  and  1   3 2 
 3  1   2 
3  1  2 
The correct sign on the left hand side is negative since there is external contact.
Therefore, since link 2 is the input,
1  3  0 
     1 rad/rad 

 1  3

 and 
3  0  0 
3  0  1 rad/rad 
Substituting the known radii for gears 1 and 3 and rearranging, gives
100 mm  0  1 rad/rad   100 mm 3  1 rad/rad 
and
which give solutions of
100 mm  0  0  100 mm 3  0
3  2 rad/rad
and 3  0
The rolling contact equation between gear 3 and gear 4 can be written as
      
       
 3   4 2  and  3   4 2 
4  3   2 
4  3   2 
The correct sign on the left hand side is positive since there is internal contact.
Therefore, following a similar strategy,
100 mm 3  1 rad/rad   300 mm 4  1 rad/rad 
and
100 mm 3  0  300 mm 4  0
(1)
193
which give solutions of
4  1.333 rad/rad
and 4  0
The rolling contact equation between gear 4 and gear 5 can be written as
      
    
 4   5 1  and  4   5 1 
5   4  1
5   4  1 
(2)
The correct sign on the left hand side is positive since there is internal contact.
Therefore, following a similar strategy,
300 mm 4  0  500 mm 5  0
and
300 mm 4  0  500 mm 5  0
which give solutions of
5  0.800 rad/rad
and 5  0
The angular velocities of gears 3, 4, and 5 can be written as
3  32   2 rad/rad  50 rad/s  100 rad/s (cw)
(3)
4  42  1.333 rad/rad  50 rad/s  66.67 rad/s (cw)
5  52   0.8 rad/rad  50 rad/s  40 rad/s (cw)
The angular accelerations of gears 3, 4, and 5 can be written as
3  32  322
  2 rad/rad   15 rad/s2    0 50 rad/s   30 rad/s2 (cw)
2
Ans.
 4   4 2   422
 1.333 rad/rad   15 rad/s 2    0  50 rad/s   20 rad/s 2 (cw)
2
Ans.
3  32  322
 1.333 rad/rad   15 rad/s2    0 50 rad/s   20 rad/s2 (cw)
2
Ans.
194
4.56
For the mechanism in the posture illustrated, link 4 is rolling without slipping on the
ground at point C and the first- and second-order kinematic coefficients are
3   1.341 rad/rad, R4   3.097 in/rad, 3   2.475 rad/rad 2 , and R4   7.372 in/rad2
(where R4 is the vector from the origin to pin B which connects links 3 and 4).
Determine the radius of curvature of the path of point D, and the x and y coordinates of
the center of curvature of this path. If the constant input angular velocity of link 2 is
ω2   9 kˆ rad/s , determine the acceleration of point D.
RO2  1.00 in, RAO2  1.50 in, RBA  2.00 in, RB  2.267 in, RCB  0.50 in, and RDB  1.50 in.
The vectors for point D are shown in the figure. The x and y components of the equation
are
xD  R4  RDB cos 4  2.267 in
yD 
RDB sin 4  1.50 in
Differentiating these equations with respect to the input position 2 gives
xD  R4  RDB sin 44
yD 
RDB cos  4 4
The kinematic coefficient of link 4 is obtained from the rolling contact equation; that is,
 R4  4 4  1  44
A counterclockwise rotation of link 4 causes R4 to become shorter; therefore, a negative
sign must be used on the left-hand side of this equation. Solving for the first-order
kinematic coefficient of link 4 gives
4   R4 4  6.194 rad/rad
Therefore,
xD  R4  RDB sin 44  12.388 in/rad
yD 
RDB cos 44  0
and
RD   xD2  yD2  12.388 in/rad
195
Note that a negative sign must be used here because the input is given as negative (that is,
clockwise)
Taking another derivative of these equations with respect to the input 2 we find the
second-order kinematic coefficients
 4   R4  4  14.744 rad/rad 2
xD  R4  RDB cos 4 42  RDB sin  4 4  29.488 in/rad 2
yD   RDB sin  4 42  RDB cos  4 4  57.548 in/rad 2
The radius of curvature of the path of point D can be written as
RD 3
Ans.
D 
 2.667 in
xD yD  yD xD
The negative sign indicates that the unit normal vector is pointing away from the center of
curvature of the path of point D. The coordinates of the center of curvature of the path of
point D can be written as
xcc  xD   D   yD RD   2.267 in
Ans.
ycc  y D   D  xD RD   1.167 in
The velocity of point D can be written as
VD  ( xD ˆi  yD ˆj) 2  111.5ˆi in/s  111.5 in/s0
The acceleration of point D can be written as
A D  ( xD ˆi  yD ˆj) 22  ( xD ˆi  yD ˆj) 2
 2388.5ˆi  4661.4ˆj in/s2  5237.7 in/s2  117.1
Ans.
196
4.57
For the linkage in the posture illustrated, the first- and second-order kinematic
coefficients are 3   3.0 rad/rad, R4  86.6 mm/rad, 3   13.856 rad/rad2 , and
R4  150 mm/rad 2 (where R4 is the vector from the ground pivot O2 to pin B).
Determine the radius of curvature of the path of point C, and the x and y coordinates of
the center of curvature of this path. If the input angular velocity of link 2 is a constant
2  22 rad/s ccw then determine the acceleration of point C.
RAO2  43.3 mm, RBA  25 mm, and RCA  75 mm.
A suitable set of vectors for analysis of the linkage are shown in the figure.
The component equations for the position of point C can be written as
xC  R2 cos 2  RCA cos 3  86.6 mm
yC  R2 sin 2  RCA sin 3  0
The first-order kinematic coefficients of point C are
xC   R2 sin 2  RCA sin 33  150 mm/rad
yC  R2 cos 2  RCA cos 33  173.2 mm/rad
rC   xC 2  yC 2    150 mm/rad    173.2 mm/rad   229.1 mm/rad
2
2
The positive sign is used here because the motion of the input link is positive, that is,
counterclockwise.
The second-order kinematic coefficients of point C are
xC   R2 cos 2  RCA cos 332  RCA sin 33  1 125.8 mm/rad 2
yC   R2 sin 2  RCA sin 332  RCA cos 33  600.0 mm/rad 2
The radius of curvature of the path of point C can be written as
197
rC 3
C 
xC yC  yC xC
 229.1 mm/rad 

 150 mm/rad   600 mm/rad 2    173.2 mm/rad   1 125.8 mm/rad 2 
3
C  114.53 mm
Ans.
The negative sign indicates that the unit normal vector is pointing away from the center of
curvature of the path of point C (see figure).
The coordinates of the center of curvature of the path of point C can be written as
xCC  xC  C   yC rC   0
yCC  yC  C  xC rC   75 mm
The velocity of point C can be written as
VC  ( xC ˆi  yC ˆj) 2  3 300ˆi  3 810ˆj mm/s  504.1 mm/s  130.9
The acceleration of point C can be written as
AC  ( xC ˆi  yC ˆj) 22  ( xC ˆi  yC ˆj) 2
 544.89ˆi  290.40ˆj m/s2  617.44 m/s2  151.94
Ans.
Ans.
198
4.58
For the mechanism in the posture illustrated, the first- and second-order kinematic
coefficients are 3   3.464 rad/rad, 4  1 rad/rad, 3  5.464 rad/rad 2 , and
4  7.732 rad/rad 2 , respectively. The line AC in link 3 is parallel to the x-axis. The
circular wheel, link 4, is rolling on the ground link at point E and rolling on link 3 at
point B. Determine: (a) the first- and second-order kinematic coefficients of point C; (b)
the unit tangent vector and the unit normal vector to the path traced by point C; (c) the
radius of curvature of this path; and (d) the x and y coordinates of the center of curvature
of this path. If the input angular velocity of link 2 is a constant ω2  15 kˆ rad/s then
determine the acceleration of point C.
RAO2  4 in, RBA  1 in, RCA  2 in, and  4  1 in.
The two scalar component equations for point C can be written as
xC  R2 cos 2  R3 cos 3  5.464 in
yC  R2 sin 2  R3 sin 3  2.000 in
Differentiating with respect to input  2 gives the first-order kinematic coefficients
xC   R2 sin 2  R3 sin 33  2.000 in/rad
Ans.
yC  R2 cos 2  R3 cos 33  3.464 in/rad
rC  
xC2  yC2 
 2.000 in/rad    3.464 in/rad   4.000 in/rad
2
2
The positive sign must be used here because the input is positive (that is, the input link 2
is rotating counterclockwise).
Differentiating again with respect to  2 gives the second-order kinematic coefficients
xC   R2 cos 2  R3 cos 332  R3 sin 333  27.464 in/rad 2
Ans.
yC   R2 sin 2  R3 sin 332  R3 cos 33  8.928 in/rad 2
199
The unit tangent vector to the path at point C can be written as
uˆ tC  xC ˆi  yC ˆj rC  0.500ˆi  0.866ˆj  1.000  120
Ans.
The unit normal vector to the path at point C is
uˆ Cn   yC ˆi  xC ˆj rC  0.866ˆi  0.500ˆj  1.000  30
Ans.




The radius of curvature of the path of point C can be written as
rC3
C 
xC yC  xC yC

 4.000 in/rad 
3
 2.000 in/rad  8.928 in/rad 2    27.464 in/rad 2   3.464 in/rad 
C  0.566 in
Ans.
The negative sign implies that the unit normal vector is pointing away from the center of
curvature of the path of point C. The radius of curvature and the approximate path of
point C are shown on the figure.
The x and y coordinates of the center of curvature of the path of point C can be written as
  xC  C  yC rC   5.464 in   0.566 in  0.866 in/rad 1.000 in/rad   4.974 in
xCC
  yC  C  xC rC   2.000 in   0.566 in  0.500 in/rad 1.000 in/rad   2.283 in
yCC
The center of curvature of the path of point C is shown in the figure.
The acceleration of point C can be written as
AC  ( xC ˆi  yC ˆj) 22  ( xC ˆi  yC ˆj)  2
 6179.4ˆi  2 008.8ˆj in/s 2  6 497.7 in/s 2161.99
Ans.
200
4.59
For the linkage in the posture iillustrated, the first- and second-order kinematic
  0, and R34
   50 mm/rad 2
coefficients are 3  4  0.5 rad/rad, 3  4  0, R34
(where R34 is the vector from point B fixed in link 3 to point C fixed in link 4).
Determine: (a) the radius of curvature of the path of point B; and (b) the center of
curvature of the path of this point. If the input angular velocity of link 2 is a constant
2  10 rad/s clockwise then determine the acceleration of point B.
RAO2  100 mm, RBA  141.4 mm, RCB  RO2O4  100 mm, and RCO4  100 mm.
The x and y components for point B are
xB  R2 cos 2  R3 cos 3  0
yB  R2 sin 2  R3 sin 3   100 mm
Differentiating these equations with respect to the input position θ2, the first-order
kinematic coefficients of point B are
xB   R2 sin 2  R3 sin 33  50 mm/rad
yB  R2 cos 2  R3 cos 33  50 mm/rad
rB   xB 2  yB 2  
50 mm/rad   50 mm/rad    70.71 mm/rad
2
2
The negative sign must be used here because the input angular velocity is given as
negative (clockwise).
Differentiating these equations again with respect to the input position θ2, the secondorder kinematic coefficients of point B are
xB   R2 cos 2  R3 cos 332  R3 sin 33   75 mm/rad 2
yB   R2 sin 2  R3 sin 332  R3 cos 33  25 mm/rad 2
The unit tangent vector for the path of point B is
(50 mm/rad) ˆi  (50 mm/rad) ˆj
uˆ tB 
 70.71 mm/rad
  0.7071 ˆi  0.7071 ˆj  1.000  135
201
The unit normal vector is 90 counterclockwise from the unit tangent vector; that is,
x ˆi  yB ˆj  yB ˆi  xB ˆj
uˆ nB  kˆ ×uˆ tB  kˆ × B

rB
rB
 0.7071 ˆi  0.7071 ˆj  1.000  45
The directions of the unit tangent and unit normal vectors to the path of point B are
shown in the figure. Note that the unit tangent vector must point in the same direction as
the velocity of point B and the unit normal vector must be 90o counterclockwise from the
unit tangent vector.
The radius of curvature of the path of point B can be written as
rB 3
B 
xB yB  yB xB
(  70.71 mm/rad)3
Ans.
  70.71 mm
(50 mm/rad)(25 mm/rad)  (50 mm/rad)(  75 mm/rad)
The negative sign indicates that the unit normal vector to the path of point B is pointing
away from the center of curvature of the path of point B; see the figure.
The coordinates of the center of curvature of the path of point B can be written as
  yB 
  50 mm/rad 
xCC  xB   B 
 0  (  70.71 mm/rad) 

   50 mm
  70.71 mm/rad 
 rB 
Ans.
 xB 
 50 mm/rad 
yCC  yB   B     100 mm  (  70.71 mm/rad) 
   50 mm
  70.71 mm/rad 
 rB 
The coordinates of the center of curvature of the path of point B are shown in the figure.
B 
The acceleration of point B can be written as
AC  ( xC ˆi  yC ˆj) 22  ( xC ˆi  yC ˆj)  2
 7.500ˆi  2.500ˆj m/s 2  7.906 m/s 2161.57
Ans.
202
4.60
For the mechanism in the posture illustrated, the first- and second-order kinematic
coefficients are 3   4.333 rad/rad, 4  0, R4  26 in/rad, 3  0, 4  0.813 rad/rad 2 ,
and R4  4.875 in/rad 2 (where the rotation of link 2 is the input and R 4 is the vector from
point O4 to point C on link 3). The angle between the line AB in link 3 and the line O2 A
is a right angle. Determine: (a) the first- and second-order kinematic coefficients of point
B; (b) the unit tangent vector and the unit normal vector to the path traced by this point;
(c) the radius of curvature of the path traced by this point; and (d) the x and y coordinates
of the center of curvature of the path traced by this point. If the input angular velocity of
link 2 is a constant ω2  15 kˆ rad/s, then determine the acceleration of point B.
RAO2  26 in and RBA  RCA  3  6 in.
The scalar equations for the position coordinates of point B are
xB  R2 cos 2  R3 cos 3  22.627 in
yB  R2 sin 2  R3 sin 3  14.142 in
Differentiating these equations with respect to the input angle θ2, the first-order kinematic
coefficients of point B are
xB   R2 sin 2  R3 sin 3 3   36.770 in/rad
Ans.
yB  R2 cos 2  R3 cos 3 3  0
rB   xB 2  yB 2  36.770 in/rad
The correct sign is positive since the input is specified as positive (that is, the input
angular velocity is counterclockwise).
Differentiating these equations again with respect to the input angle θ2, the second-order
kinematic coefficients of point B are
xB   R2 cos 2  R3 cos 332  R3 sin 33   98.052 in/rad 2
Ans.
yB   R2 sin 2  R3 sin 3 32  R3 cos 33  61.282 in/rad 2
The unit tangent vector to the path of point B can be written as
uˆ tB  xB ˆi  yB ˆj rB  1.000ˆi
Ans.


The unit normal vector to the path of point B (which is 90o counterclockwise from the
203
unit tangent vector) can be written as
uˆ nB   yB ˆi  xB ˆj rB  1.000ˆj


The unit tangent vector uˆ tB and the unit normal vector uˆ nB are shown in the figure.
The radius of curvature of the path of point B can be written as
rB3
B 
xB yB  yB xB
B 
 36.770 in/rad 
3
 36.770 in/rad   61.282 in/rad 2   0  98.052 in/rad 2 
 22.062 in
Ans.
Since the radius of curvature of the path of point B is a negative value the unit normal
vector uˆ nB must point away from the center of curvature. This result is in complete
agreement with the figure.
The x-coordinate of the center of curvature of the path of point B can be written as
xCC  xB   B   yB rB   22.627 in
Ans.
The y-coordinate of the center of curvature of the path of point B is
yCC  yB   B  xB rB   36.204 in
Ans.
The location of the center of curvature of the path of point B is shown in the figure.
The velocity of point B can be written as
VB  ( xB ˆi  yB ˆj) 2
 (  36.770 in/rad ˆi  0ˆj) (15 rad/s)   551.55 ˆi in/ s  551.55 in/s180
The acceleration of point B can be written as
A B  ( xB ˆi  yB ˆj)2 2  ( xB ˆi  yB ˆj)2
 (  98.052 in/rad 2 ˆi  61.282 in/rad 2 ˆj) (15 rad/s) 2  (  36.770 in/rad ˆi  0 ˆj) (0 rad/s2 )
A B  22 061.7ˆi  13 788.5ˆj in/s2  26 016.1 in/s2148
Ans.
204
4.61
For the linkage of Prob. 3.61 in the posture illustrated, the first and second-order
kinematic coefficients are 3  4  1 rad/rad, R4   12 in/rad, 3  4   2.309 rad/rad2 ,
   20.785 in/rad 2 (where R34 is the vector from O4 to the point B fixed in link
and R34
3). Determine the first and second-order kinematic coefficients of the coupler point C.
Then determine: (a) the unit tangent vector and the unit normal vector to the path traced
by point C; (b) the radius of curvature of this path; and (c) the x and y coordinates of the
center of curvature of this path. If the angular velocity of input link 2 is a constant
ω2  15 kˆ rad/s then determine the acceleration of point C.
Fig. P3.61 RO4O2  12 in, RAO2  6 in, RCA  13 in.
The vector equation for point C can be written as
RC  R2e j2  RCAe j3
which has horizontal and vertical components of
xC  R2 cos 2  RCA cos 3  5.196 in
yC  R2 sin 2  RCA sin 3  16.000 in
Differentiating with respect to the input 2 gives
xC   R2 sin 2  RCA sin 33  16.000 in/rad
yC  R2 cos 2  RCA cos 33  5.196 in/rad
Ans.
rC  x2  y2  16.823 in/rad
and differentiating again with respect to 2 gives
xC   R2 cos 2  RCA cos 332  RCA sin 33  35.218 in/rad 2
yC   R2 sin 2  RCA sin 332  RCA cos 33  16.00 in/rad 2
Ans.
205
The unit tangent and unit normal vectors to the path of point C can be written as


uˆ Ct  xC ˆi  yC ˆj rC  0.951 10 ˆi  0.308 88ˆj
Ans.
uˆ Cn  kˆ × uˆ Ct  0.308 88ˆi  0.951 10ˆj
The radius of curvature of the path of point C can be written as
C  rC3  xC yC  xC yC   10.844 in
Ans.
Ans.
The positive sign implies that the unit normal vector is pointing towards the center of
curvature of the path of point C.
The x and y coordinates of the center of curvature of the path of point C can be written as
xCC  xC  C yC rC  1.847 in
yCC  yC  C xC rC  5.687 in
The center of curvature of the path of point C is as shown in the figure.
Ans.
Ans.
Check: Note that the instant center I13 must lie on the unit normal vector uˆ Cn . From
Kennedy’s theorem, the instant center I13 can be shown to be coincident with the instant
center I12 which is coincident with the ground pivot O2. Also note that the instant center
I24 lies at infinity. Therefore, 4 = 2.
The velocity of point C can be written as
VC = (xC iˆ  yC ˆj) 2  240ˆi  77.942ˆj in/s  252.339 in/s162.01
Since 2 is constant, the acceleration of point C can be written as
AC = (xC ˆi  yC ˆj) 22  7 925.85ˆi  3 600ˆj in/s2  8 705.12 in/s 2  155.57
Ans.
206
4.62
For the mechanism in the posture illustrated, the first- and second-order kinematic
coefficients are 3  2.165 rad/rad, 4   7.143 rad/rad, 3   9.369 rad/rad 2 , and
4  26.784 rad/rad 2 . Determine: (a) the first- and second-order kinematic coefficients
for point C; (b) the radius of curvature of the path of this point; and (c) the x and y
coordinates of the center of curvature of this path. If the input angular velocity of link 2
is a constant 2  50 rad/s counterclockwise, then determine the acceleration of point C.
RAO2  500 mm, RBA  400 mm, RCA  346.4 mm, and 4  140 mm.
The x and y components of point C are
xC  R2 cos 2  R33 cos 33   596.41 mm
yC  R2 sin 2  R33 sin 33  433.01 mm
Differentiating these equations with respect to the input position θ2, the first-order
kinematic coefficients of point C are
xC   R2 sin 2  R33 sin 333   433.01 mm/rad
Ans.
yC  R2 cos 2  R33 cos 333   999.98 mm/rad
rC   xC2  yC2  1 089.71 mm/rad
The positive sign must be used here because the input angular velocity is
counterclockwise.
Differentiating again with respect to the input position θ2, the second-order kinematic
coefficients of point C are
xC   R2 cos 2  R33 cos 3332  R33 sin 333  1 873.7 mm/rad 2
Ans.
yC   R2 sin 2  R33 sin 3332  R33 cos 333  2 812.5 mm/rad 2
The unit tangent vector for point C can be written as
uˆ Ct  xC ˆi  yC ˆj rC   0.3974 ˆi  0.9177 ˆj  1.000  113.4


The unit normal vector can be written as
uˆ Cn   yC ˆi  xC ˆj rC  0.9177 ˆi  0.3974 ˆj  1.0000  23.4


The radius of curvature of the path of point C can be written as
207
rC 3
Ans.
 1 973.1 mm  1.973 m
xC yC  yC xC
The positive sign indicates that the unit normal vector to the path of point C is pointing
toward the center of curvature of the path of point C,
The coordinates of the center of curvature of the path of point C can be written as
 y 
xcc  xC  C  C 
 rC 
Ans.
  999.98 mm/rad 
  596.41 mm  (1 973.1 mm) 
  1 214.2 mm
 1 089.7 mm/rad 
C 
 x 
y yy  yC  C  C 
 rC 
Ans.
  433.01 mm/rad 
 433.01 mm  (1 973.1 mm) 
   351.0 mm
 1 089.7 mm/rad 
The coordinates of the center of curvature of the path of point C are as shown in the
figure.
The velocity of point C can be written as
VC  ( xC ˆi  yC ˆj) 2
 (  433.01 mm/rad ˆi  999.98 mm/rad ˆj )(50 rad / s)
VC   21 651 ˆi  49 999 ˆj mm / s  54485 mm/s-113.4
The acceleration of point C can be written as
A C  ( xC ˆi  yC ˆj) 22  ( xC ˆi  yC ˆj)  2
 (1 873.7 mm/rad 2 ˆi  2 812.5 mm/rad 2 ˆj) (50 rad/s) 2
 (  3 678.9 mm/rad ˆi  999.98 mm/rad ˆj) (0)
AC  4 684 250 ˆi  7 031 250 ˆj mm/s2  8 448 708 mm/ s2 56.3
Ans.
208
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209
Chapter 5
Multi-Degree-of-Freedom Mechanisms
5.1
The slotted links 2 and 3 are driven independently at constant speeds of 2  30 rad/s cw
and 3  20 rad/s cw, respectively. Find the absolute velocity and acceleration of the
center of the pin P4 carried in the two slots.
xB = 100 mm, and yB = 25 mm.
Identifying the pin as separate body 4 and, noting the two paths it travels on bodies 2 and
3, we write
VP2  2 RP2 A   30 rad/s  0.054 9 m   1.647 m/s
VP3  3 RP3B   20 rad/s  0.102 6 m   2.052 m/s
VP4  VP2  VP4 /2  VP3  VP4 /3
Construct the velocity polygon
VP4  2.355 m/s15.6
A P2  AO2  AnP O  A Pt O ;
2 2
n
P2O2
A
Ans.
A P3  AO3  AnP O  A Pt O
2 2
3 3
3 3
  RP2O2   30.0 rad/s   0.054 9 m   49.410 m/s .
2
2
2
2
APn3O3  32 RP3O3   20.0 rad/s   0.102 6 m   41.039 m/s2
2
A P4  A P2  AcP4 P2  AnP4 / 2  AtP4 / 2  A P3  AcP4 P3  AnP4 / 3  AtP4 / 3
210
AcP4 P2  22 × VP4 /2  2  30.0 rad/s 1.683 m/s   100.98 m/s2  30.0
AcP4 P3  23 × VP4 /3  2  20.0 rad/s 1.155 m/s   46.22 m/s2225.0
n
P4 / 2
A

VP24 / 2
P / 2
4
1.683 m/s   0 ;

2

n
P4 /3
A

VP24 /3
 P /3
4
1.155 m/s   0

2

Construct the acceleration polygon.
A P4  125.73 m/s2  66.6
Ans.
For comparison, let us now solve the same problem by use of kinematic coefficients. The
loop-closure constraint equations can be written as
r2 cos  2  r3 cos 3  xB  0
r2 sin  2  r3 sin 3  yB  0
Recognizing that  2 and  3 are the two independent degrees of freedom, and that r2 and
r3 are dependent position unknowns. Since these appear linearly (which is not true in
other problems), the loop-closure equations can be written in matrix form as follows:
cos  2  cos 3   r2   xB 

 sin 
 sin 3   r3    yB 
2

For the given position 2  60 and 3  135 , and the dimensions are xB  100 mm and
yB  25 mm. The determinant of this set is    sin 3  2   0.966 , and the solutions
for the two unknown position values are r2  54.90 mm and r3  102.60 mm .
position coordinates of the center of the pin are
xP  r2 cos  2  xB  r3 cos 3  27.452 mm
The
yP  r2 sin  2   yB  r3 sin 3  47.548 mm
Taking the derivatives of the loop-closure equations with respect to both  2 and  3 , in
turn, we find the following two sets of equations for the first-order kinematic coefficients.
r3 sin 3 
cos  2  cos 3   r22 r23   r2 sin  2

 sin 
 sin 3   r32 r33   r2 cos 2 r3 cos 3 
2

and the solutions for these are
 r22 r23  1  r2 cos 3   2 
r r    
r2
 32 33 



r3 cos 3  2  
At the current position, the numeric values of the first-order kinematic coefficients are
 r22 r23   0.014 711 m/rad 0.106 218 m/rad 
 r  r     0.056 841 m/rad 0.027 491 m/rad 

 32 33  
r3
Using these, the first-order kinematic coefficients for the center of the pin are
xP 2  r22 cos  2  r2 sin  2  r32 cos 3  0.040 192 m/rad
yP 2  r22 sin  2  r2 cos  2  r32 sin 3  0.040 192 m/rad
xP 3  r23 cos 2  r33 cos 3  r3 sin 3  0.053 109 m/rad
yP 3  r23 sin  2  r33 sin 3  r3 cos 3  0.091 987 m/rad
211
With the given independent input velocities of 2  30 rad/s cw and 3  20 rad/s cw,
the velocity of the pin P4 is
xP  xP 22  xP 33  2.268 m/s
yP  yP 22  yP 33  0.634 m/s
V  x ˆi  y ˆj  2.268ˆi  0.634ˆj m/s  2.355 m/s15.62
P
P
P
Ans.
Taking the second derivatives of the loop-closure equations with respect to both  2 and
 3 , in turn, we find the following three sets of equations for the second-order kinematic
coefficients:
 r223
 r233
 
cos  2  cos 3   r222
 sin 


 r323
 r333
 
 sin 3   r322
2

r23 sin  2  r32 sin 3 2r33 sin 3  r3 cos 3 
 2r  sin  2  r2 cos  2
  22

 2r22 cos  2  r2 sin  2 r23 cos  2  r32 cos 3 2r33 cos 3  r3 sin 3 
and the solutions for these are
 r223
 r233
  1 2r22 cos 3  2   r2 sin 3  2  r23 cos 3  2   r32
2r33

r222

r r r    
2r22
r23  r32 cos 3  2  2r33 cos 3  2   r3 sin 3  2 
 322 323 333 

At the current position, the numeric values of the second-order kinematic coefficients are
 r223
 r233
  0.062 788 m/rad 2 0.087 307 m/rad 2 0.056 922 m/rad 2 
 r222
 r  r  r    
2
2
2
 322 323 333   0.030 461 m/rad 0.125 195 m/rad 0.117 331 m/rad 
The second-order kinematic coefficients for the center of the pin are
 cos  2  2r22 sin  2  r2 cos  2  r322
 cos 3  0.021538 m/rad 2
xP 22  r222
 sin  2  2r22 cos  2  r2 sin  2  r322
 sin 3  0.021538 m/rad 2
yP 22  r222
 cos  2  r23 sin  2  r323
 cos 3  r32 sin 3  0.048 334 m/rad 2
xP 23  r223
 sin  2  r23 cos  2  r323
 sin 3  r32 cos 3  0.128 719 m/rad 2
yP 23  r223
 cos  2  r333
 cos 3  2r33 sin 3  r3 cos 3  0.028 461 m/rad 2
xP33  r233
 sin  2  r333
 sin 3  2r33 cos 3  r3 sin 3  0.049 296 m/rad 2
yP33  r233
With the given independent input velocities and accelerations of 2  30 rad/s cw,
3  20 rad/s cw, and  2  3  0 , the acceleration of the pin P4 is
xP  xP 2 2  xP 33  xP2222  2 xP2323  xP3332  50.00 m/s 2
yP  yP 2 2  yP 3 3  yP2222  2 yP 2323  yP3332  115.36 m/s
A  x ˆi  y ˆj  50.00ˆi  115.36ˆj m/s2  125.73 m/s2  66.6
Ans.
It should be noted that the exact match between the graphic and the analytic solutions
achieved for this problem is not at all typical, nor can such matches be expected. Graphic
results are usually far less accurate than the analytic. The reason for the agreement
achieved here is that a very precise CAD system was used for the graphic constructions.
P
P
P
212
5.2
For the five-bar linkage in the posture illustrated, the angular velocity of link 2 is 15 rad/s
cw and the angular velocity of link 5 is 15 rad/s cw. Determine the angular velocity of
link 3 and the apparent velocity VB4 /5 .
RO2O5  200 mm23.1 , RAO2  300 mm , and RBA  200 mm
Let us define r1  RO2O5  200 mm, 1  23.1, r2  RAO2  300 mm, r3  RBA  200 mm,
and r4  RBO5 . Then the loop-closure equation can be written as
r11  r22  r33  r45  0
with horizontal and vertical components of
r1 cos 1  r2 cos  2  r3 cos 3  r4 cos 5  0
r1 sin 1  r2 sin  2  r3 sin 3  r4 sin 5  0
Solution of these position equations give two unknowns, r4  300 mm and 3  203.1 .
Derivatives of the loop-closure equations with respect to each of the independent degrees
of freedom,  2 and  5 , give the first-order kinematic coefficients
 r3 sin 3  cos 5  32 35   r2 sin  2 r4 sin 5 
 r cos 
  r  r     r cos 

sin

r4 cos 5 
3
5   42
45 
2
 3
 2
The determinant of the Jacobian is   r3 cos 3  5  and this goes to zero whenever
3  5   2k  1  2 ; that is, whenever link 3 is perpendicular to link 5. At the current
position,   159.94 mm. The solutions for the first-order kinematic coefficients are
r4

32 35  1  r2 cos 5   2 

 r  r      r r sin   
 42 45 
 2 3  3 2  r3r4 sin 3  5  
At the current posture, with the given data, the values for the first-order kinematic
coefficients are
32 35  1.875 74 rad/rad 1.875 74 rad/rad 
 r  r     0.225 25 m/rad 0.225 25 m/rad 

 42 45  
Therefore, with 2  5  15 rad/s, we have
3  32 2  35 5  0 and VB4 /5  r42 2  r45 5  0
Ans.
213
5.3
For the five-bar linkage in the posture illustrated in Figure P5.2, the angular velocity of
link 2 is 2  25 rad/s ccw and the apparent velocity VB4 /5 is 5 m/s upward along link 5.
Determine the angular velocities of links 3 and 5.
Here we can continue the solution of Prob. 5.2. However, the problem is now expressed
in terms of two different input variables,  2 and r4 , as independent degrees of freedom.
Therefore, we can use the same vectors and the same loop-closure equations. However,
we must now take derivatives with respect to  2 and r4 to find new first-order kinematic
coefficients. The result, in matrix form, is
 r3 sin 3 r4 sin 5  32 34   r2 sin  2 cos 5 

 r cos 
r4 cos 5  52 54   r2 cos  2 sin 5 
3
 3
The determinant of the Jacobian is now   r3r4 sin 3  5  , which goes to zero whenever
link 3 is aligned with link 5. The solutions for the first-order kinematic coefficients are
r4
 
0
8.327 50 rad/m 
32 34  1  r2 r4 sin 5   2 

        r r sin   

 52 54 
 2 3  3 2  r3 cos 3  5   1.000 00 rad/rad 4.439 58 rad/m 
With the given input velocities, 2  25 rad/s and r4  5 m/s , the requested velocities are
3  32 2  34 r4  41.64 rad/s (cw) and 5  52 2  54 r4  47.20 rad/s ccw
Ans.
214
5.4
For Problem 5.2, assuming that the two given input velocities are constant, determine the
angular acceleration of link 3 at the instant indicated.
Starting with the equations of Prob. 5.2 for the first-order kinematic coefficients,
 r3 sin 3  cos 5  32 35   r2 sin  2 r4 sin 5 

 r cos 
 sin 5   r42 r45   r2 cos  2 r4 cos 5 
3
 3
we can take derivatives with respect to each independent variable to find equations for the
second-order kinematic derivatives
 325
 355
 
 cos 5  322




 
 sin 5   r422
r425
r455
 r3 cos 3322  r2 cos  2 r3 cos 332 35  sin 5 r42 r3 cos 3352  2sin 5 r45  r4 cos 5 


2
2
 r3 sin 332  r2 sin  2 r3 sin 332 35  cos 5 r42 r3 sin 335  2 cos 5 r45  r4 sin 5 
With satisfaction, we notice that the Jacobian is identical with that of Prob. 5.2. The
solution to these equations gives
 r3 sin 3
 r cos 
3
 3
 325
 355
 
322
 r r r  
 422 425 455 
2

r3 sin 3  5 32 35  r42
r3 sin 3  5 352  2r45
1  r3 sin 3  5  32  r2 sin 5   2 


2 2
2
2 2

r3 32  r2 r3 cos 3   2 
r3 cos 3  5 32 35  r3 sin 3  5  r42 r3 35  2r3 sin 3  5  r45  r3 r4 cos 3  5 
Substituting the numeric data, including results of Prob. 5.2, we get
 325
 355
   2.641 69 rad/rad 2 4.050 06 rad/rad 2 5.45843 rad/rad 2 
322

 r 

2


r
r
0.534 56 m/rad 2
1.518 19 m/rad 2 
 422 425 455   0.579 95 m/rad
Therefore, given that 2  5  15 rad/s (cw) and  2  5  0, the angular acceleration
of link 3 is
 22  2325
 25  355
 52  0
Ans.
3  32  2  35 5  322
215
5.5
Link 2 rotates at a constant angular velocity of 10 rad/s ccw while the sliding block 3
slides toward point A on link 2 at the constant rate of 5 in/s. Find the absolute velocity
and absolute acceleration of point P of block 3.
RAO2  3.0 in, RBA  6.0 in, and RPA  4.0 in
With the dimensions given, O2AP is a 3-4-5 right triangle and R PO2  5.0ˆj in .
For velocity, we write
VP3  VP2  VP3 /2
VP2  VO2  ω 2  R P2O2

 

 10kˆ rad/s  5ˆj in  50ˆi in/s
Also, we are given VP3 /2  5 in/s . From these we construct the velocity polygon shown in
the figure, and from this we measure
VP3  47.17 in/s  175.13
Ans.
For acceleration, we write
A P3  A P2  A cP3P2  A nP3 /2  A tP3 /2
A P2  AO2  A nP2O2  A Pt2O2


2
APn2O2   10 rad/s  5ˆj in  500ˆj in/s 2
APc3 P2  22VP3 /2  2 10 rad/s  5 in/s   100 in/s 2
APn3 /2  VP23 /2   VP23 /2   0
APt 3 /2  0
From these we construct the acceleration polygon shown in the figure, and we measure
A P3  447.21 in/s2  79.70
Ans.
216
5.6
For Problem 5.5, determine the value of the sliding velocity VP3 /2 that minimizes the
absolute velocity of point P of block 3. In addition, find the value of VP3 /2 that minimizes
the absolute acceleration of point P of block 3.
By careful inspection of the velocity polygon of Prob. 5.5 we can see that the absolute
velocity VP3 is minimized when it becomes perpendicular to VP3 /2 . Reconstructing the
velocity polygon in this condition, as illustrated in the figure, we find
VP3  40.00 in/s  143.13
and
VP3 /2  30.00 in/s  53.13
Ans.
Similarly, the absolute acceleration of P3 is minimized when it becomes perpendicular to
A cP3 P2 . Reconstructing the acceleration polygon in this condition, as illustrated in the
figure, we find A P3  400.00 in/s2  53.13 and AcP3P2  300.00 in/s2  143.13 . Then,
from this, we can calculate
APc3 P2
300.00 in/s2
 15.00 in/s
Ans.
22 2 10.00 rad/s 
Note how visualization of the inherent geometry has dramatically simplified this problem,
compared to a totally mathematical approach. Note also that VP3 /2 must increase in both
VP3 /2 
cases.

217
5.7
The left two two-link planar robot is attempting to transfer a small object labeled P to the
similar right robot. At the posture indicated, 2  45 and 3/2  15 . (Note that
3/2  3  2 is given because that is the angle controlled by the motor in joint A.)
Determine  4 and 5/4 to allow the second robot to take over possession of the object P.
RO4O2  1 m, RAO2  RBO4  0.3 m, and RPA  RPB  0.4 m .
The loop-closure constraint equations at this instant allow us to write
1.0  0.3cos  4  0.4cos 5  0.4cos 30  0.3cos 45  0
0.3sin  4  0.4sin 5  0.4sin 30  0.3sin 45  0
These can be rearranged to read
cos 5  0.750cos  4  1.10364
sin 5  0.750sin  4  1.03033
Now, by squaring and adding, we eliminate the variable 5.
1.0  0.5625  1.65547cos 4  1.54550sin 4  2.27960
or
1.65547cos 4 1.54550sin 4  1.84210  0
Next, by defining Z  tan 4 2  , and by use of the standard identities, this becomes
1.65547 1  Z 2   1.54550  2Z   1.84210 1  Z 2   0
or
0.18663Z 2  3.09100Z  3.49757  0
The roots of this equation give
Z  15.34054 and Z  1.22164
and from the definition of Z these give two values of 4
4  172.54 and 4  101.39
Now, returning these to the above equations, we can solve for values of 5
5  111.10 and 5  162.84
Of these, the second value of each pair fits our figure. Therefore,
4  101.39 and 5/4  61.45
Ans.
218
5.8
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  15 rad/s ccw, what angular velocities must be used for 4
and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  15 rad/s (ccw)  5 rad/s (ccw)
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  0.3 m   3.0 m/s
VPA  3 RPA   5 rad/s  0.4 m   2.0 m/s
From these data and equations, we can construct the velocity polygon shown in the figure.
This allows us to find data for the following calculations:
4 
VBO4
5 
VPB 0.687 m/s

 1.72 rad/s ccw
RPB
0.4 m
RBO4

1.351 m/s
 4.50 rad/s cw
0.3 m
5/4  5  4  1.72 rad/s    4.50 rad/s   6.22 rad/s (ccw)
Ans.
Ans.
219
If an analytical solution is preferred, we start with the robot on the left, where we find
3  2  3/2  10 rad/s (cw)  15 rad/s (ccw)  5 rad/s (ccw)

 

V  ω × R   5.0kˆ rad/s  ×  0.4cos 30ˆi  0.4sin 30ˆj m   1.000 00ˆi  1.732 05ˆj m/s
VA  ω2 × R AO2  10.0kˆ rad/s × 0.3cos 45ˆi  0.3sin 45ˆj m  2.121 32ˆi  2.121 32ˆj m/s
PA
3
PA
VP  VA  VPA  1.121 32ˆi  0.389 27ˆj m/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  ×  0.4 cos162.84ˆi  0.4 sin162.84ˆj m 
VB  ω 4 × R BO4  4kˆ rad/s × 0.3cos101.39ˆi  0.3sin101.39ˆj m
VPB  ω5 × R PB
5
VP  VB  VPB   0.294 09i  0.059 25 j m  4   0.118 02i  0.382 18 j m  5
Next, by setting the two equations for VP equal, and then separating the î and ĵ
components, we obtain a set of two equations for 4 and 5
 0.294 09 m 0.118 02 m  4   1.121 32 m/s 
 0.059 25 m 0.382 18 m      0.389 27 m/s 

 5 

The solutions to these equations give
4   4.502 rad/s (cw) 
    1.716 rad/s (ccw) 

 5 
5/4  5  4  1.716 rad/s    4.502 rad/s   6.218 rad/s (ccw)
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphical
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.
Since we are given velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23I13 10.0 rad/s


 2.0 or RI23 I13  2.0 RI23 I12
3 RI23I12
5.0 rad/s
Therefore I13 takes the position shown in the figure below. When the other instant
centers are found through the Aronhold-Kennedy theorem, this results in the instant
centers shown for I 24 , I 34 and for I 25 , I 35 .
Once the remaining instant centers are found, we may find the information requested in
the problem. We find
R
0.818 84 m
4  I24 I12 2 
Ans.
 10 rad/s   4.502 rad/s (cw)
RI24 I14
1.818 84 m
5 
RI25 I12
RI25 I15
2 
0.193 87 m
 10 rad/s   1.716 rad/s (ccw)
1.129 51 m
220
5/4  5  4  1.716 rad/s    4.502 rad/s   6.218 rad/s (ccw)
Ans.
Again, the very high degree of agreement is a consequence of the precision of the CAD
system used in finding the distances between instant centers.
221
5.9
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  10 rad/s ccw, what angular velocities must be used for 4
and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  10 rad/s (ccw)  0
Note that this implies that link 3 is in translation relative to the ground.
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  0.3 m   3.0 m/s
VPA  3 RPA   0 rad/s  0.4 m   0
From these data and equations, we can construct the velocity polygon shown in the figure.
This allows us to find data for the following calculations:
4 
VBO4
5 
VPB 2.845 m/s

 7.11 rad/s ccw
RPB
0.4 m
RBO4

3.020 m/s
 10.07 rad/s cw
0.3 m
5/4  5  4   7.11 rad/s    10.07 rad/s   17.18 rad/s (ccw)
Ans.
Ans.
222
If an analytical solution is preferred, we start with the robot on the left, where we find
3  2  3/2  10 rad/s (cw)  10 rad/s (ccw)  0

 

V  ω × R   0kˆ  ×  0.4cos 30ˆi  0.4sin 30ˆj m   0ˆi  0ˆj
VA  ω2 × R AO2  10.0kˆ rad/s × 0.3cos 45ˆi  0.3sin 45ˆj m  2.121 32ˆi  2.121 32ˆj m/s
PA
3
PA
VP  VA  VPA  2.121 32ˆi  2.121 32ˆj m/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  ×  0.4 cos162.84ˆi  0.4 sin162.84ˆj m 
VB  ω 4 × R BO4  4kˆ rad/s × 0.3cos101.39ˆi  0.3sin101.39ˆj m
VPB  ω5 × R PB
5
VP  VB  VPB   0.294 09i  0.059 25 j m  4   0.118 02i  0.382 18 j m  5
Next, by setting the two equations for VP equal to each other, and then separating the î
and ĵ components, we obtain a set of two equations for 4 and 5
 0.294 09 m 0.118 02 m  4   2.121 32 m/s 
 0.059 25 m 0.382 18 m      2.121 32 m/s 

 5 

The solutions to these equations give
4   10.067 rad/s (cw) 
    7.111 rad/s (ccw) 

 5 
5/4  5  4   7.111 rad/s    10.067 rad/s   17.178 rad/s (ccw)
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphic
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.
Since we are given velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23I13 10.0 rad/s


  or RI23 I13  
3 RI23I12
0.0 rad/s
Therefore, I13 goes to infinity in the direction shown in the figure below. This indicates
that link 3 is in translation with respect to the ground, which we could see when we found
that 3  0 . When the other instant centers are found through the Aronhold-Kennedy
theorem, this results in the instant centers shown for I 24 , I 34 and for I 25 , I 35 . However, we
find that the lines toward instant center I 24 are essentially parallel, and that instant center
I 24 appears to also be at infinity. This implies that link 4 is in translation with respect to
link 2, which means that 4  2  10 rad/s (cw) .
223
Once the remaining instant centers are found, we may find the information requested in
the problem. We find
R
Ans.
4  I24 I12 2  10.000 rad/s (cw)
RI24 I14
5 
RI25 I12
RI25 I15
2 
0.462 59 m
 10 rad/s   7.112 rad/s (ccw)
0.650 43 m
5/4  5  4   7.112 rad/s    10.000 rad/s   17.112 rad/s (ccw)
Ans.
Note that the precision is not perfect this time, in spite of the use of a high-precision CAD
system. However, this probably stems from the possibility that the apparent parallelism
and intersections at infinity were likely not perfect. Still, the precision is amazing for a
graphic solution.
224
5.10
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  0, what angular velocities must be used for 4 and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  0 rad/s  10 rad/s (cw)
Notice that this implies that link 3 and link 2 rotate together as a single unit.
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  0.3 m   3.0 m/s
VPA  3 RPA  10 rad/s  0.4 m   4.0 m/s
From these data and equations, we can construct the velocity polygon shown in the figure.
This allows us to find data for the following calculations:
4 
VBO4
RBO4

6.360 m/s
 21.20 rad/s cw
0.3 m
Ans.
225
5 
VPB 7.161 m/s

 17.90 rad/s ccw
RPB
0.4 m
5/4  5  4  17.90 rad/s    21.20 rad/s   39.10 rad/s (ccw)
Ans.
If an analytical solution is preferred, we start with the robot on the left, where we find
V  ω × R  10.0kˆ rad/s × 0.3cos 45ˆi  0.3sin 45ˆj m  2.121 32ˆi  2.121 32ˆj m/s

 

V  ω × R   10.0kˆ rad/s  ×  0.4cos 30ˆi  0.4sin 30ˆj m   2.000 00ˆi  3.464 10ˆj m/s
A
PA
2
AO2
3
PA
VP  VA  VPA  4.121 32ˆi  5.585 42ˆj m/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  ×  0.4 cos162.84ˆi  0.4 sin162.84ˆj m 
VB  ω 4 × R BO4  4kˆ rad/s × 0.3cos101.39ˆi  0.3sin101.39ˆj m
VPB  ω5 × R PB
5
VP  VB  VPB   0.294 09i  0.059 25 j m  4   0.118 02i  0.382 18 j m  5
Next, by setting the two equations for VP equal to each other, and then separating the î
and ĵ components, we obtain a set of two equations for 4 and 5
 0.294 09 m 0.118 02 m  4   4.121 32 m/s 
 0.059 25 m 0.382 18 m      5.585 42 m/s 

 5 

The solutions to these equations give
4   21.198 rad/s (cw) 
    17.901 rad/s (ccw) 

 5 
5/4  5  4  17.901 rad/s    21.198 rad/s   39.099 rad/s (ccw)
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphic
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5. However,
since we have equal velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23I13 10.0 rad/s


 1.0
3 RI23I12 10.0 rad/s
and therefore I13 becomes coincident with I12 as shown in the figure below. When the
other instant centers are found through the Kennedy-Aronhold theorem, this also results
in coincident instant centers for I 24 , I 34 and for I 25 , I 35 as shown in the figure. Under
these input velocity conditions, links 2 and 3 act as a single solid unit.
226
Once the remaining instant centers are found, however, we may still find the information
requested in the problem. We find
R
1.892 86 m
Ans.
4  I24 I12 2 
 10 rad/s   21.20 rad/s (cw)
RI24 I14
0.892 86 m
5 
RI25 I12
RI25 I15
2 
0.694 13 m
 10 rad/s   17.90 rad/s (ccw)
0.387 73 m
5/4  5  4  17.90 rad/s    21.20 rad/s   39.10 rad/s (ccw)
Ans.
Again, the perfect agreement results from the precision of the CAD system used in
finding the distances between instant centers.
227
5.11
To successfully transfer an object between two robots, as described in Problems 5.7 and
5.8, it is helpful if the accelerations are also matched at point P. Assuming that the two
input accelerations are  2  3  0 at this instant for the robot on the left, what angular
accelerations must be given to the two input joints of the robot on the right to achieve
this?
Starting after the solutions of Prob. 5.8 (the velocity analysis) is completed, the condition
for the acceleration of point P is written as
t
t
t
t
A P  A O2  A nAO2  A AO
 A nPA  A PA
 A O4  A nBO4  A BO
 A nPB  A PB
2
4
n
AO2
A

2
VAO
2
RAO2
 3.0 m/s   30.0 m/s 2

2
0.3 m
 2.0 m/s   10.0 m/s 2
V2
A  PA 
RPA
0.4 m
2
n
PA
n
BO4
A

2
VBO
4
RBO4
1.351 m/s   6.084 m/s 2

2
0.3 m
 0.687 m/s   1.180 m/s 2
V2
A  PB 
RPB
0.4 m
2
n
PB
Note that a difficulty arises in the graphic solution of the above acceleration equation in
that the two unknowns do not arise consecutively in the equation. Nevertheless, recalling
that vector addition is commutative (independent of order), we can proceed with the
vectors appearing out of order, as is shown in the dotted lines in the figure above. Once
the solution with the dotted lines is completed, we have obtained the correct magnitudes
and directions of the two unknown tangential components. However, we do not have a
228
valid acceleration polygon unless we now arrange the components in their correct order,
according to the original acceleration equation, as is shown in the solid lines in the
acceleration polygon. If this is not done, the acceleration image point B cannot be
labeled, and the absolute acceleration of B and acceleration images of links 4 and 5
cannot be correctly shown.
Whether or not the vectors are arranged in their correct order, however, we can proceed
with the solution for the two unknown angular accelerations as follows:
t
ABO
4
27.583 m/s2
4 

 91.944 rad/s 2 ccw
RBO4
0.3 m
Ans.
t
APB
15.127 m/s2

 37.818 rad/s 2 ccw
RPB
0.4 m
Ans.
5 
229
5.12
The circular cam is driven by link 2 at a constant angular velocity 2  15 rad/s ccw.
Link 3 is rotating at a constant angular velocity 3  5 rad/s cw, causing slipping at point
C. Determine: (a) the first and second-order kinematic coefficients of the mechanism;
(b) the angular velocity and acceleration of link 4; and (c) the velocity of slipping at
point C.
RKO2  RO4 K  3 in, RBO2  1.25 in, RCB  2  2 in, RCD  3  0.5 in, and RDO4  3.5 in.
The loop-closure equation can be written as
RBO2 e j2  RDBe j23  RDO4 e j4  jRO4K  RKO2  0
which gives the two equivalent scalar equations
RBO2 cos 2  RDB cos 23  RDO4 cos4  RKO2  0
RBO2 sin 2  RDB sin 23  RDO4 sin 4  RO4 K  0
and these can be rearranged to read
RDB cos 23  RDO4 cos 4   RBO2 cos 2  RKO2
RDB sin 23  RDO4 sin 4   RBO2 sin 2  RO4 K
At the current posture, with the given dimensions, these become
(1)
230
 2.5 in  cos23   3.5 in  cos4  3.884 in
 2.5 in  sin 23   3.5 in  sin 4  2.116 in
and these have the solution 23  80.757 and 4  174.239 .
(a)
Taking the derivative of Eqs. (1) with respect to input  2 gives
  RDO4 sin 442  RBO2 sin 2
 RDB sin 23232
  RDO4 cos 442   RBO2 cos 2
RDB cos 23232
and, in matrix form, this becomes
   RBO2 sin 2 
  RDB sin 23 RDO4 sin 4  232

 R cos 


 RDO4 cos 4   42
    RBO2 cos 2 
23
 DB
The determinant of the Jacobian is
  RDB RDO4 sin 23  4   8.734 in 2
(2)
and, by Cramer’s rule, the solution for the first-order kinematic coefficients are
  0.317 rad/rad and 42
  0.290 rad/rad .
Ans.
232
Recognizing that RDB  2  3  2.5 in is a constant, we see that nothing in Eqs. (1) is a
function of the rotation of link 3. Therefore, the derivative of Eqs. (1) with respect to  3
gives
  0
  RDB sin 23 RDO4 sin 4  233

(3)
 R cos 

 RDO4 cos 4   43
  0
23
 DB
with the solution
  0 and 43
  0.
233
Taking the derivatives of Eqs. (2) with respect to  2 gives
2  RDO4 cos442
2 
   RBO2 cos 2  RDB cos 23232
  RDB sin 23 RDO4 sin 4  2322


 R cos 
 RDO4 cos 4   422
2  RDO4 sin 442
2 
   RBO2 sin 2  RDB sin 23232
23
 DB
from which, the second-order kinematic coefficients are
  3.527 rad/rad 2 and 422
  0.090 rad/rad 2 .
2322
Ans.
Ans.
Taking the derivatives of Eqs. (2) with respect to  3 gives
  0
  RDB sin 23 RDO4 sin 4  2323
 R cos 


   0

R
cos



DB
23
DO
4
4

  423   
from which, the second-order kinematic coefficients are
  0 and 423
  0 .
2323
Ans.
Finally, taking the derivatives of Eqs. (3) with respect to  3 gives
  0
  RDB sin 23 RDO4 sin 4  2333

 R cos 


 RDO4 cos 4   433
  0
23
 DB
from which, the second-order kinematic coefficients are
  0 and 433
  0 .
2333
Ans.
231
(b)
The angular velocity of link 4 is
4  42 2  43 3  0.290 rad/rad 15 rad/s   0  5 rad/s   4.35 rad/s ccw
Ans.
The angular acceleration of link 4 is
 22  2 423
 23   433
 32   42  2   43 3
 4   422
 0.090 rad/rad 2 15 rad/s   0  0  0  0
2
Ans.
 20.25 rad/s2 (cw)
(c)
The positions of point C of link 2 and point C of link 3 are given by
RC2  RBO2 e j2   2e j23

 

 RBO2 cos  2   2 cos  23 ˆi  RBO2 sin  2   2 sin  23 ˆj
 0.563 in ˆi  2.858 in ˆj  2.913 in101.14
RC3  RKO2  jRO4 K  RDO4 e j4  3e j23
However, remembering how each segment rotates, the velocities of these two point are
VC2  j 2 RC2  j2 RC2

 15 rad/s  0.563 in ˆj  2.858 in ˆi

 42.869 in/s ˆi  8.440 in/s ˆj  43.692 in/s   168.86
VC3  j4 RDO4 e j4  j3 3e j23


  5.00 rad/s  0.50 in   cos80.76ˆj  sin 80.76i 
  4.35 rad/s  3.50 in  cos174.24ˆj  sin174.24ˆi
 3.996 in/s ˆi  14.747 in/s ˆj  15.278 in/s   105.16
The slipping velocity at point C is
VC3 /2  VC3  VC2  38.873 in/s ˆi  6.307 in/s ˆj  39.381 in/s   9.216
Ans.
232
5.13
The tracing point C of the pantograph linkage is required to follow a prescribed curve,
that is, the two independent input variables are xC and yC. Then point P, which carries a
pen, traces a similar curve, that is, the outputs of the linkage are the xP and yP components
of the motion of the pen. Determine the first-order kinematic coefficients of point P.
RAO2  RCA  RDB  300 mm, RBA  RDC  RPD  200 mm.
The vector loop
RBAe j2  RDBe j3  RDC e j5  RCAe j4  0
under the conditions, with the given data, becomes
RBA  e j2  e j5   RDB  e j3  e j4   0
It is clear that this equation is satisfied when 5  2 and 3  4 .
Recognizing this, two more vector loop equations can be written, namely
RAO2 e j2  RCAe j3  jyC  xC  0
RBO2 e j2  RPB e j3  jy P  xP  0
The first of these can be separated into the following two scalar equations
RAO2 cos  2  RCA cos 3  xC
RAO2 sin 2  RCA sin 3  yC
and the second can be separated into the following two scalar equations
RBO2 cos  2  RPB cos 3  xP
RBO2 sin 2  RPB sin 3  yP
(1)
(2)
233
Taking the derivative of Eqs. (1) with respect to xC, the first independent variable,
substituting the known dimensions, and expressing the result in matrix form yields
 0.300 m sin 2 0.300 m sin 3  2 x  1

 0.300 m cos 
0.300 m cos 3  3x  0
2

The determinant of the Jacobian is   0.090 m2 sin 3  2  , and Cramer’s rule gives
the solution
2 x   3.333 rad/m  cos 3 sin 3  2 
3x   3.333 rad/m  cos 2 sin 3  2 
where the subscript x refers to derivatives with respect to xC.
Taking the derivative of Eqs. (1) with respect to yC, the second independent variable,
substituting the known dimensions, and expressing the result in matrix form yields
 0.300 m sin 2 0.300 m sin 3  2 y  0

 0.300 m cos 
0.300 m cos 3  3y  1
2

The determinant of the Jacobian is   0.090 m2 sin 3  2  as it was before, and
Cramer’s rule gives the solution
2 y   3.333 rad/m  sin 3 sin 3  2 
3y   3.333 rad/m  sin 2 sin 3  2 
where the subscript y refers to derivatives with respect to yC.
Taking the derivative of Eqs. (2) with respect to xC, the first independent variable, and
substituting all known values yields
Ans.
xPx  1.667 m/m and
yPx  0
Taking the derivative of Eqs. (2) with respect to yC, the second independent variable, and
substituting all known values yields
yPy  1.667 m/m
xPy  0 and
Ans.
These results show that any small displacement of the tracing point C will produce a
similar small displacement of pen P, but magnified to 1.667 times the size.
234
5.14
The mechanism has rolling contact at point A, but there can be slipping at point B. Link 2
has a constant angular velocity 2  20 rad/s ccw and link 3 has an angular velocity
3  5 rad/s cw and an angular acceleration of 3  2 rad/s2 ccw . Determine: (a) the
first- and second-order kinematic coefficients of link 5; and (b) the angular velocity and
angular acceleration of link 5.
RO5O2  2.50ˆi  3.75ˆj in, RCO2  2.75 in, RKO5  1.00 in, 3  2 in, and 4  0.75 in.
For the choice of vectors shown in the figure,the loop-closure equation can be written as


j  5  
2
RO5O2  RKO5 e 


j  5  
2
 RBK e j5  RCBe 
 RCO2 e j2
and this can be rearranged into the following form
RBK e j5  j RKO5  RCB e j5  RCO2 e j2  RO5O2


235
Substituting the dimensional data and separating the real and imaginary parts, gives two
equivalent scalar equations as follows
RBK cos 5  1.75 in  sin 5   2.75 in  cos 2  2.50 in
(1)
RBK sin 5  1.75 in  cos 5   2.75 in  sin 2  3.75 in
Multiplying the first of these equations by sin 5 and the second equation by  cos5 , gives
RBK sin 5 cos 5  1.75 in  sin 2 5   2.75 in  cos 2 sin 5   2.50 in  sin 5
 RBK sin 5 cos 5  1.75 in  cos2 5    2.75 in  sin 2 cos 5   3.75 in  cos 5
Adding these equations now eliminates the unknown variable RBK.
 1.75 in    3.75 in    2.75 in  sin 2  cos 5     2.50 in    2.75 in  cos2  sin 5
If we define z  tan 5 2  , then cos 5  1  z 2  1  z 2  , and sin 5  2 z 1  z 2  ,
Then, making these substitutions in the above equation and clearing fractions we get
 1.75 in  1  z 2    3.75 in    2.75 in  sin 2  1  z 2     2.50 in    2.75 in  cos 2  2 z
which can be rearranged to read
 2.00 in    2.75 in  sin 2  z 2   5.00 in   5.50 in  cos 2  z  5.50 in    2.75 in  sin 2   0
For the posture shown in the figure, 2  90, and the equation becomes
 0.75 in  z 2   5.00 in  z   2.75 in   0
Of the two roots, the pertinent one for this posture is
z  6.06178
and, for this root, we find
5  2 tan 1  6.06178  161.265
Taking this value back to either of Eqs. (1) we find the corresponding value of RBK
RBK  2.046 34 in.
(a) To find the first-order kinematic coefficients of link 5, we take the derivative of Eqs.
(1) with respect to each of the independent input variables. First, for input 2,
 2 cos 5  RBK sin 552  1.75 in  cos 552    2.75 in  sin 2
RBK
(2)
 2 sin 5  RBK cos 552  1.75 in  sin 552   2.75 in  cos 2
RBK
In matrix form, this becomes
cos 5   RBK sin 5  1.75 in  cos 5    RBK
 2     2.75 in  sin 2 




 sin 5  RBK cos 5  1.75 in  sin 5    52    2.75 in  cos 2 
The determinant of the Jacobian is
  RBK  2.046 34 in
Then, according to Cramer’s rule,
 2   2.75 in   RBK sin 5  2   1.75cos 5  2    3.359 65 in/rad
RBK
52   2.75 in  cos 5  2    0.431 64 rad/rad
Ans.
To find the kinematic coefficient for the rotation of link 4, we must write the rolling
contact constraint for point B. This gives
236
4  4  2   3  3  2 
(3)
The partial derivative of this with respect to the independent input rotation 2 gives
4 42  1  3
42   3  4  4  3.666 67 rad/rad
(4)
We can see that the rotation of independent input 3 does not affect the rotation of link 5.
Therefore, taking the partial derivative of Eqs. (1) with respect to 3 gives
cos 5   RBK sin 5  1.75 in  cos 5    RBK
 3   0
(5)


 
 sin 5  RBK cos 5  1.75 in  sin 5    53  0
And, by Cramer’s rule, the solutions for these first-order kinematic coefficients are
 3  0 and 53  0.
Ans. (6)
RBK
For the rotation of link 4, we take the partial derivative of Eq. (3).
443   3
43   3 4  2.666 67 rad/rad
(7)
To find the second-order kinematic coefficients, we first take the derivative of Eqs. (2)
with respect to independent input variable 2, In matrix form, this becomes
cos 5   RBK sin 5  1.75 in  cos 5    RBK
 22 



 
 sin 5  RBK cos 5  1.75 in  sin 5    522
   2.75 in  cos  2  2 RBK
 2 sin 552   RBK cos 5  1.75 in  sin 5  522 


 2 cos 552   RBK sin 5  1.75 in  cos 5  522 
   2.75 in  sin  2  2 RBK
We note, with satisfaction, that the Jacobian matrix is the same as that for the first-order
equations.
Substituting all known data, these equations become
 22   2.434 54 in/rad 2 
 0.947 01 1.000 00 in    RBK

     
2
  2.936 31 in/rad 
 0.321 19   2.500 00 in    522
Recalling from above that   2.04634 in , Cramer’s rule gives the solutions for these
second-order kinematic coefficients as
 22  1.539 35 in/rad 2 and 522
  0.976 75 rad/rad2 .
RBK
Ans.
Next we take the derivative of Eqs. (2) with respect to independent input variable 3.
Using Eq. (6) this gives
cos 5   RBK sin 5  1.75 in  cos 5    RBK
 23  0



    0
 sin 5  RBK cos 5  1.75 in  sin 5    523   
The solutions for these second-order kinematic coefficients are
 23  0 and 523
  0 .
RBK
Ans.
From the derivative of Eqs. (5) with respect to independent input variable 3, after using
Eq. (6) this gives us
237
cos 5   RBK sin 5  1.75 in  cos 5    RBK
 33  0



  0
 sin 5  RBK cos 5  1.75 in  sin 5    533
The solutions for these second-order kinematic coefficients are
 33  0 and 533
  0 .
RBK
Also, from derivatives of Eqs. (4) and (7)
  0 , 423
  0 , and 433
  0 .
422
From Eq. (5.4) the angualr velocity of link 5 is
5  52 2  53 3   0.431 64 rad/rad  20 rad/s   0  5 rad/s   8.633 rad/s ccw
Ans.
(b)
Ans.
From Eq. (5.8) the angualr acceleration of link 5 is
 22  2523
 23  533
 32  52  2  53 3
5  522
 22  2  0  23   0  32  52  0    0  3
 522
  0.976 75 rad/rad 2   20 rad/s 
5  390.70 rad/s2 ccw.
2
Ans.
238
5.15
The mechanism has rolling contact at point A. For the current posture 2  60 . Link 2
has constant angular velocity 2  50 rad/s ccw , and link 3 has angular velocity
5  25 rad/s cw and angular acceleration 5  20 rad/s2 ccw . Determine: (a) the firstand second-order kinematic coefficients for links 3 and 4 and (b) the angular velocities
and angular accelerations of links 3 and 4.
RO5O2  300ˆi  190ˆj mm, RAO2  300 mm, RCB  300 mm, 4  60 mm, and 5  130 mm.
With the vectors shown in the figure, the loop-closure equation can be written as
RO5O2   5  4  e j6  RBAe j3  RAO2 e j2
Separating the real and imaginary parts, substituting the dimensional data, and
rearranging, gives two equivalent scalar equations as follows
239
190 mm  cos6   300 mm  cos3   300 mm  cos2  300 mm
190 mm  sin 6   300 mm  sin 3   300 mm  sin 2  190 mm
(1)
There is also a constraint equation describing the rolling contact at point A.
4  4  6   5  5  6 
or
44  55    4  5  6 .
(2)
To obtain an accurate solution for the position variables for this posture, we first solve
both equations for the 3 terms.
 300 mm  cos3  190 mm  cos6   300 mm  cos2  300 mm
 300 mm  sin 3  190 mm  sin 6   300 mm  sin 2  190 mm
Then, by squaring and adding, we eliminate the 3 unknown.
114 000 1  cos 2  cos 6   72 200  114 000sin 2  sin 6
 162 200  180 000cos 2  114 000sin 2   0
For the specific posture shown in the figure, with 2  60 , this becomes
57 000cos6  170 927sin 6  170 927  0
If we define Z  tan 6 2  , then cos6  1  Z 2  1  Z 2  , and sin 6  2Z 1  Z 2  ,
then, making these substitutions in the above equation and clearing fractions we get
113 927Z 2  341 854Z  227 927  0
Of the two roots, the pertinent one for this posture is
Z  1.000 00
and, for this root, we find
6  2 tan 1 1.000 00  90.00
Taking this value back to Eqs. (1) we find the corresponding value of 3  60.
(a) To find the first-order kinematic coefficients we take the partial derivative of Eqs. (1)
with respect to input variable 2. In matrix format, this gives
  190 mm  sin 6  300 mm  sin 3  62     300 mm  sin 2 
(3)

   

 190 mm  cos 6   300 mm  cos 3  32    300 mm  cos  2 
The determinant of the Jacobian is
2
  57 000sin 6  3   28 500  mm 
Then, by Cramer’s rule, we find the first-order kinematic coefficients,
62  90000sin 2  3    2.734 82 rad/rad
and
32  57 000sin 2  3    1.000 00 rad/rad
Ans.
Also, by taking the partial derivative of Eq. (2) with respect to input variable 2, we get
442   4  5  62
(4)
or
42  1  5 4  62  3.166 6762  8.660 26 rad/rad
Ans.
240
In similar fashion, we take the partial derivative of Eqs. (1) with respect to the second
independent input variable 5. In matrix format, since the right-hand side of Eqs. (1) are
not dependent on 5, this gives us
  190 mm  sin 6  300 mm  sin 3  65  0
(5)

     

190
mm
cos


300
mm
cos

0






35
6
3




From these we find
35  0
and 65  0 .
Ans.
Also, by taking the partial derivative of Eq. (2) with respect to input variable 2, we get
445  5   4  5  65
(6)
45  5 4  2.166 67 rad/rad
or
Ans.
To find the second-order kinematic coefficients, first we take the partial derivative of Eqs.
(3) with respect to input variable 2. In matrix format, this gives
 
  190 mm  sin 6  300 mm  sin 3  622

   
 190 mm  cos 6   300 mm  cos 3  322 
190 mm  cos 6622   300 mm  cos 3322   300 mm  cos  2   300.000 mm/rad 2 


2
2
2
 190 mm  sin 662   300 mm  sin 332   300 mm  sin  2  1 421.056 mm/rad 
Then, by Cramer’s rule, we find the second-order kinematic coefficients,
  9.474 rad/rad2 and 622
  14.533 rad/rad 2
Ans.
322
Also, by taking the partial derivative of Eq. (4) with respect to input variable 2, we get
   4  5  622

4422
or
  1  5 4  622
  3.166 67622
  46.021 rad/rad 2
422
Ans.
Similarly, we take the partial derivative of Eqs. (3) with respect to input variable 5. In
matrix format, this gives
  0
  190 mm  sin 6  300 mm  sin 3  625

      
 190 mm  cos 6   300 mm  cos 3  325  0
from which, by Cramer’s rule, we find
  0 and 625
  0
325
Ans.
Also,
by
taking
the
partial
derivative
2
  1  5 4  622
  3.166 67622
  46.021 rad/rad of Eq. (4) with respect to input
422
variable 5, we get
   4  5  625

4425
  0
425
or
Ans.
Finally, we take the partial derivative of Eqs. (5) with respect to input variable 5. This
gives
  0
  190 mm  sin 6  300 mm  sin 3  655

      
 190 mm  cos 6   300 mm  cos 3  355  0
241
from which, by Cramer’s rule, we find
  0 and 655
  0
Ans.
355
and, by taking the partial derivative of Eq. (6) with respect to input variable 5, we get
   4  5  655

4455
  0
455
or
Ans.
(b) The angular velocities of links 3 and 4 are
3  32 2  35 5
  1.000 rad/rad  50 rad/s    0  25 rad/s 
 50 rad/s
3  50 rad/s cw
Ans.
4  42 2   45 5
 8.660 rad/rad  50 rad/s    2.167 rad/rad  25 rad/s 
 487.18 rad/s
4  487.18 rad/s ccw
Ans.
The angular accelerations of links 3 and 4 are
 22  2325
 25  355
 52   32  2   35 5
3  322
  9.474 rad/rad 2   50 rad/s   2  0  50 rad/s  25 rad/s    0  25 rad/s 
2
2
+  1.000 rad/rad  0    0   20 rad/s 2 
 23 685 rad/s2
3  23 685 rad/s2 cw
 22  2 425
 25   455
 52   42
  2   45
 5
 4   422
  46.021 rad/rad 2   50 rad/s   2  0  50 rad/s  25 rad/s    0 25 rad/s 
2
+ 8.660 rad/rad  0    2.167 rad/rad   20 rad/s2 
 115 009 rad/s
2
 4  115 009 rad/s2 ccw
Ans.
2
Ans.
242
5.16
The mechanism has rolling contact between rack 3 and gear 4 at point F, between gears 4
and 5 at point C, and between gears 5 and 6 at point E. Link 2 has constant angular
velocity 2  50 rad/s ccw, and link 6 has angular velocity 6  25 rad/s cw and angular
acceleration 6  20 rad/s2 ccw. Determine: (a) the first- and second-order kinematic
coefficients for gear 5 and the rack; and (b) the velocity and acceleration of the rack and
(c) the angular velocities and accelerations of links 4 and 5.
RFA  RBF  500 mm, 4  RCO4  450 mm, 5  RCD  175 mm, 6  REO2  100 mm,
and RDO2  275 mm.
The rolling contact constraint at point E can be written in terms of angular displacements
seen by an observer on link 2 as follows
6  6  2   5  5  2 
or
55   5  6  2  66
(1)
Similarly, for the rolling contact constraint at point C, we can write
4  4  2   5  5  2 
44  55   4  5  2
(2)
For the rolling contact constraint at point F we can write
r3x  44
(3)
or
243
(a) We can find the first-order kinematic coefficients for the rotations of link 5 by taking
partial derivatives of Eq. (1) with respect to each of the independent input variables.
552   5  6 
(4)
52   5  6  5  275 mm 175 mm  1.571 43 rad/rad
556   6
56   6 5  100 mm 175 mm  0.571 43 rad/rad
Ans.
(5)
Ans.
By taking partial derivatives of Eq. (2) with respect to each of the independent input
variables we find
 442  552    4  5 
(6)
42  1  5  4    5  4  52  1.222 22 rad/rad
 446  556  0
46   5  4  56  0.222 22 rad/rad
(7)
In a similar manner, we can find the first-order kinematic coefficients for the horizontal
motion of link 3 by taking partial derivatives of Eq. (3) with respect to each of the
independent input variables.
   450 mm1.222 22 rad/rad   550 mm/rad
r32x  442
Ans. (8)
   450 mm 0.222 22 rad/rad   100 mm/rad
r36x  446
Ans. (9)
We can find the second-order kinematic coefficients by taking further partial derivatives
of Eqs. (4)-(9) with respect to the independent input variables. These give
  0, 526
  0,
  0,
Ans.
522
566
  0, 426
  0, 466
  0,
422
x  0,
r322
x  0,
r326
x  0.
r366
Ans.
(b) We can find the requested velocities and accelerations as follows
v3x  r32x2  r36x6
  50 mm/rad  50 rad/s    100 mm/rad  25 rad/s 
 5.0 m/s
4  42 2   46 6
 1.222 rad/rad  50 rad/s    0.222 rad/rad  25 rad/s 
 66.66 rad/s ccw
5  52 2  56 6
 1.571 rad/rad  50 rad/s    0.571 rad/rad  25 rad/s 
 92.86 rad/s ccw
x
x 22  2r326
x 26  r366
x 62  r32x2  r36x6
a3  r322
 0  0  0  0   100 mm/rad   20 rad/s2 
 2.0 m/s2
Ans.
Ans.
Ans.
Ans.
244
x 22  2 426
x 26   466
x 62   42x 2   46x 6
 4x  422
 0  0  0  0   0.222 rad/rad   20 rad/s2 
 4.44 rad/s (cw)
x 22  2526
x 26  566
x 62  52x 2  56x6
  522
Ans.
2
x
5
 0  0  0  0   0.571 rad/rad   20 rad/s2 
 11.43 rad/s (cw)
2
Ans.
245
PART 2
DESIGN OF MECHANISMS
246
Page intentionally blank.
247
Chapter 6
Cam Design
6.1
The reciprocating radial roller follower of a plate cam is to rise 2 in with simple harmonic
motion in 180 of cam rotation and return with simple harmonic motion in the remaining
180 . If the roller radius is 0.375 in and the prime-circle radius is 2 in, construct the
displacement diagram, the pitch curve, and the cam profile for clockwise cam rotation.
248
6.2
A plate cam with a reciprocating flat-face follower has the same motion as in Problem
6.1. The prime-circle radius is 2 in, and the cam rotates counterclockwise. Construct the
displacement diagram and the cam profile, offsetting the follower stem by 0.75 in in the
direction that reduces the bending of the follower during rise.
249
6.3
Construct the displacement diagram and the cam profile for a plate cam with an
oscillating radial flat-face follower that rises through 30 with cycloidal motion in 150
of counterclockwise cam rotation, then dwells for 30 , returns with cycloidal motion in
120 , and dwells for 60 . Determine the necessary length for the follower face, allowing
5 mm clearance at the free end. The prime-circle radius is 30 mm, and the follower pivot
is 120 mm to the right.
Note that, with the prime circle radius given, the cam is undercut and the follower will
not reach positions 7 and 8. The follower face length shown is 200 mm but can be made
as short as 195 mm (position 9) from the follower pivot.
Ans.
250
6.4
A plate cam with an oscillating roller follower is to produce the same motion as in
Problem 6.3. The prime-circle radius is 60 mm, the roller radius is 10 mm, the length of
the follower is 100 mm, and it is pivoted at 125 mm to the right of the cam rotation axis.
The cam rotation is clockwise. Determine the maximum pressure angle.
From a graphic analysis, max  39 at   240 ; this is unacceptable.
Ans.
251
6.5
For full-rise simple harmonic motion, write the equations for the velocity and the jerk at
the midpoint of the motion. Also, determine the acceleration at the beginning and the end
of the motion.
Using Eqs. (6.12) and (6.11) we find
 1 L
 L
y    
sin 
2 2
  2  2
 1
 3L 
 3L
y      3 sin   3
2
2
2
  2

  2L
 2L
y   0  
cos
0

2
2 2

 2

  2L
 2L
y   1 
cos



2
2 2

 2
6.6
 1 L
y   

  2  2
 1
 3L
y      3 3
2
  2

  2L 2
y   0 

2

 2


 2L
y   1   2  2
2


Ans.
Ans.
Ans.
Ans.
For full-rise cycloidal motion, determine the values of  for which the acceleration is
maximum and minimum. What are the formulae for the accelerations at these positions?
Find the equations for the velocity and the jerk at the midpoint of the motion.
Using Eqs. (6.13), we know that acceleration is an extremum when jerk is zero. This
occurs when cos 2   0 ; that is, when    1 4 or when    3 4 ,
  1  2 L
 2 L

y     2 sin  2  ymax
2

  4 
  3  2 L
3
2 L

y     2 sin
  2  ymin
2

  4 
 1 L
2L
y     1  cos   

  2 
  1  4 2 L
4 2 L
y    
cos



3
3
  2
Ans.
Ans.
Ans.
Ans.
252
6.7
A plate cam with a reciprocating follower is to rotate clockwise at 400 rev/min. The
follower is to dwell for 60 of cam rotation, after which it is to rise to a lift of 2.5 in.
During 1 in of the return motion, it must have a constant velocity of 40 in/s.
Recommend standard cam motions from Sec. 6.7 to be used for high-speed operation and
determine the corresponding lifts and cam rotation angles for each segment of the cam.
The curves shown are initially only sketches and not drawn to scale. They suggest the
standard curve types that might be chosen. The actual choices are shown in the table
below.
253

 400 rev/min  2 rad/rev   41.888 rad/s cw
 60 s/min 
To match the required velocity condition in segment DE we must have
y4  y4
40.000 in/s  y4  41.888 rad/s 
y4  0.954 930 in/rad   L4  4   1.000 in   4
 4  1.047 198 rad  60.000
Matching the first derivatives at D and E we find
 L3  23   y4  0.954 930 in/rad
L3   0.607 927 in/rad  3
(1)
2L5 5  y4  0.954 930 in/rad
L5   0.477 465 in/rad  5
(2)
Matching the second derivatives at C we find
5.268 30  2.5000 in  22   2 L3 432
22  5.337 90432 L3  8.780 5003 (3)
For geometric continuity, we have
(4)
L1  L2  L3  L4  L5 or
L3  L5  1.500 0 in
(5)
1   2  3   4  5  2 or
 2  3  5  4.188 790 rad
Equations (1) through (5) are now solved simultaneously for  2 , L3 ,  3 , L5 , and  5 .
The results are summarized in the following table:
Seg. Type
Eq.
L, in
rad
, deg
AB dwell
--0
1.047 198
60.000
th
BC 8 order poly.
(6.14)
2.500 0
1.089 824
62.442
CD half harmonic
(6.20)
0.082 4
0.135 268
7.750
DE uniform
--1.000 0
1.047 198
60.000
EA half cycloidal
(6.25)
1.417 6
2.963 698
169.808
6.8
Repeat Problem 6.7 except with a dwell for 20 of cam rotation.
The procedure is the same as for Prob. 6.7. The results are:
Seg. Type
Eq.
L, in
AB
BC
CD
DE
EA
dwell
8th order poly.
half harmonic
uniform
half cycloidal
--(6.14)
(6.20)
--(6.25)
0
2.500 0
0.243 6
1.000 0
1.256 4
rad
, deg
0.349 066
1.870 958
0.398 667
1.047 198
2.617 296
20.000
107.198
22.842
60.000
149.960
254
6.9
If the cam of Problem 6.7 is driven at constant speed, determine the time of the dwell and
the maximum and minimum velocity and acceleration of the follower for the cam cycle.
The duration of the dwell is t  1   1.047 198 rad 41.888 rad/s  0.025 s
Ans.
Working from the equations listed, the maximum and minimum values of the kinematic
coefficients in each segment of the cam are as follows:
Seg.
Eq.




ymin
ymin
ymax
ymax
in/rad
in/rad
in/rad2
in/rad2
AB
--0
0
0
0
BC
(6.14)
12.085 208
0
11.089 138
11.089 138
CD
(6.20)
0
0
0.956 868
11.089 138
DE
--0
0
0.954 929
0.954 929
EA
(6.25)
0
0.507 032
0
0.956 643
   12.085 208 in/rad  41.888 rad/s   506.2 in/s
ymax  ymax
Ans.
    0.956 868 in/rad  41.888 rad/s   40.0 in/s
ymin  ymin
Ans.
  2  11.089 138 in/rad2   41.888 rad/s  19 457 in/s2
ymax  ymax
Ans.
  2   11.089 138 in/rad2   41.888 rad/s  19 457 in/s2
ymin  ymin
Ans.
2
2
255
6.10
A plate cam with an oscillating follower is to rise through 20 in 60 of cam rotation,
dwell for 45 , then rise through an additional 20 , return, and dwell for 60 of cam
rotation. Assuming high-speed operation, recommend standard cam motions from Sec.
6.7 to be used, and determine the lifts and cam-rotation angles for each segment of the
cam.
From the sketches shown (not drawn to scale), the curve types identified in the table
below are chosen.
Next, equating the second derivatives at D, the remaining entries in the table are found.
L
L
5.26830 32  5.26830 42
3
4

L
 4  2.000

L3
2
4
2
3
 4  23


3   4  1  2 3  195
3  80.772
 4  114.228
Seg.
AB
BC
CD
DE
EA
Type
cycloidal
dwell
8th order poly.
8th order poly.
dwell
Eq.
(6.13)
--(6.14)
(6.17)
---
L, deg
20.000
0
20.000
40.000
0
rad
, deg
1.047 198
0.785 399
1.409 731
1.993 661
1.047 198
60.000
45.000
80.772
114.228
60.000
Ans.
256
6.11
Determine the maximum velocity and acceleration of the follower for Problem 6.10,
assuming that the cam is driven at a constant speed of 600 rev/min.
Working from the equations listed, the maximum and minimum values of the kinematic
coefficients in each segment of the cam are as follows:
Seg.
Eq.




ymin
ymin
ymax
ymax
2
rad/rad
rad/rad
rad/rad
rad/rad2
AB
(6.13)
0.666 667
0
2.000 000
2.000 000
BC
--0
0
0
0
CD
(6.14)
0.440 004
0
0.925 344
0.924 344
DE
(6.17)
0
0.925 402
0.622 264
0.925 402
EA
--0
0
0
0
   600 rev/min  2 rad/rev   60 s/min   62.832 rad/s
    0.666 667 rad/rad  62.832 rad/s   41.888 rad/s
ymax  ymax
Ans.
  2   2.000 rad/rad2   62.832 rad/s  7 896 rad/s2
ymax  ymax
Ans.
2
257
6.12
The boundary conditions for a polynomial cam motion are as follows: for   0 , y  0 ,
and y  0 , whereas for    , y  L, and y  0 . Determine the appropriate displacement
equation and the first three derivatives of this equation with respect to cam rotation.
Sketch the corresponding diagrams.
Since there are four boundary conditions, we choose a cubic polynomial
     
3C
C
2C


y 

   
 
y  C0  C1 
2
 C2 
3
 C3 
2
1
3
2
Then from the boundary conditions:
   0  C  0
C
y    0  

 0
y    1.0   C  C  L

3C
2C
y    1.0  


 0
y 
0
1
2
3
3
2
C0  0
C1  0
C2  3L
C3  2 L
Therefore the equation and its three derivatives are:
2
3
2
3


y 
 3L 
 2L 
 L 3 
2 




 

2
2


y 
 6L 
 6L 
 6L  
 

 
 
 
 
6L
6L 
 
12 L 2 
y 
  2 
   2 1  2  

         
     
  
 
 
 
y     12L


3
Ans.
Ans.
Ans.
Ans.
258
6.13
Determine the minimum face width using 0.1-in allowances at each end and determine
the minimum radius of curvature for the cam of Problem 6.2.
Referring to Prob. 6.2 for the data and figure,
R0  2.000 0 in
L  2.000 0 in
  180   rad
From Eqs. (6.12) and (6.15) for simple harmonic motion,
   L  2   1.000 0 in/rad
   L  2   1.000 0 in/rad
ymax
ymin
From Eq. (6.29)
  allowances
Face width  ymax  ymin
Face width  1.000 0 in    1.000 0 in   2  0.1 in   2.200 0 in
Ans.
From Eq. (6.27):
  R0  Y  y
L
L
L
1  cos     cos    R0  (constant)
2
2
2

   2.000 0 in    2.000 0 in  2  3.000 0 in
 R0 
Ans.
259
6.14
Determine the maximum pressure angle and the minimum radius of curvature for the cam
of Problem 6.1.
Referring to Prob. 6.1 for the figure and data,
R0  2.000 0 in
Rr  0.375 in
L  2.000 0 in
  180   rad
For simple harmonic motion, Eq. (6.12) can be substituted into Eq. (6.33) to give
sin 
. This can be differentiated and d d set to zero to find the angle
tan  
3  cos
  70.53 at which max  19.47 . However, it is much simpler to use the nomogram of
Fig. 6.28 to find max  20 directly. For the accuracy needed, the nomogram is
considered sufficient.
Ans.
From Fig. 6.30a, using R0 L  1.0 , we get   min  Rr  R0  1.43 .
This gives
min  1.43R0  Rr  1.43  2.000 0 in    0.375 in   2.49 in
Ans.
260
6.15
A radial reciprocating flat-face follower is to have the motion described in Problem 6.7.
Determine the minimum prime-circle radius if the radius of curvature of the cam is not to
be less than 0.5 in. Using this prime-circle radius, what is the minimum length of the
follower face using allowances of 0.15 in on each side?
  12.085 in/rad , ymin
  0.957 in/rad , ymin
  11.089 in/rad 2
From Prob. 6.9, ymax
Therefore, from Eq. (6.28),
   0.500 in    2.500 in    11.089 in   9.089 in
R0   min  Y  ymin
Ans.
Also, from Eq. (6.29),
  ymin
  allowances  12.085 in    0.957 in   2  0.15 in   13.342 in Ans.
Face width  ymax
261
6.16
Graphically construct the cam profile of Problem 6.15 for clockwise cam rotation.
262
6.17
A radial reciprocating roller follower is to have the motion described in Problem 6.7.
Using a prime-circle radius of 20 in, determine the maximum pressure angle and the
maximum roller radius that can be used without undercutting.
We will use the nomogram of Fig. 6.28 to find the maximum pressure angle in each
segment of the cam. Calculations are shown in the following table. Asterisks are used to
signify values used with the nomogram to adjust half-return curves to equivalent fullreturn curves, and to adjust the prime-circle baseline.
Seg.
max , deg
R0* , in
R0* L*
 * , deg
L* , in
BC
20.000 0
2.500 0
CD
22.335 2
0.164 8
EA
20.000 0
2.835 2
For the total cam, max  12.
8.0
135.5
7.0
62.4
15.5
339.6
12
1
3
Ans.
Also we use Figs. 6.32 and 6.33 to check for undercutting. Again, asterisks are used to
denote values that are adjusted for use with the charts. Note that doubling as was done
for use of the nomogram is not necessary since we have figures for half-harmonic and
half-cycloidal cam segments. Note also that segment EA need not be checked since
undercutting occurs only in segments with negative acceleration.
Seg.
L, in
 , deg
R0* L
R0* , in
 min  Rr  R0* Rrmax , in
BC
CD
20.000 0
22.417 6
2.500 0
0.082 4
8.0
272.1
62.1
7.7
To avoid undercutting for the entire cam, Rr  14.5 in .
0.725
0.680
14.5
15.2
Ans.
263
6.18
Graphically construct the cam profile of Problem 6.17 using a roller radius of 0.75 in.
Cam rotation is to be clockwise.
264
6.19
A plate cam rotates at 300 rev/min and drives a reciprocating radial roller follower
through a full rise of 75 mm in 180° of cam rotation. Find the minimum radius of the
prime-circle if simple harmonic motion is used and the pressure angle is not to exceed
25 . Find the maximum acceleration of the follower.
Using max  25 and   180 , Fig. 6.28 gives R0 L  0.75 . Therefore
R0  0.75L  0.75  75 mm   56 mm

Ans.
 300 rev/min  2 rad/rev   31.416 rad/s
 
ymax
60 s/min
 2  0.075 m 
 L

 0.037 5 m/rad 2
2
2 2
2  rad 
2
  2   0.037 5 m/rad 2   31.416 rad/s   37.0 m/s2
ymax  ymax
2
6.20
Repeat Problem 6.19 except that the motion is cycloidal.
Figure 6.28 gives R0 L  0.95 . Therefore R0  0.95L  0.95  75 mm   71 mm
 
ymax
2 L

2

2  0.075 m 
 rad 
2
2
Ans.
Repeat Problem 6.19 except that the motion is eighth-order polynomial.
Figure 6.28 gives R0 L  0.95 . Therefore R0  0.95L  0.95  75 mm   71 mm
 
ymax
5.2683L

2

5.2683  0.075 m 
 rad 
2
Ans.
 0.040 0 m/rad 2
  2   0.040 0 m/rad 2   31.416 rad/s   39.5 m/s2
ymax  ymax
2
6.22
Ans.
 0.047 7 m/rad 2
  2   0.047 7 m/rad 2   31.416 rad/s   47.1 m/s2
ymax  ymax
6.21
Ans.
Ans.
Using a roller diameter of 20 mm, determine whether the cam of Problem 6.19 is
undercut.
Using R0 L  0.75 and   180 , Fig. 6.30a gives  min  Rr  R0  1.55 .
min  1.55  56 mm   10 mm   76.8 mm  0 ; thus, this cam is not undercut.
Ans.
265
6.23
Equations (6.30) and (6.31) describe the profile of a plate cam with a reciprocating flatface follower. If such a cam is to be cut on a milling machine with cutter radius Rc ,
determine similar equations for the center of the cutter.
In complex polar notation, using Eq. (6.26) and using u and v to denote the local
rectangular part coordinates of the cutter center, the loop-closure equation is
ue j  jve j  jR0  jY  y  jRc
Dividing this by e j
u  jv  j  R0  Rc  Y  e j  ye j
Now separating this into real and imaginary parts we find
u   R0  Rc  Y  sin   y cos
v   R0  Rc  Y  cos  y sin 
Ans.
266
6.24 & 6.25 Since programming languages vary greatly, particularly with the use of graphics,
no attempt is made to show a “standard” solution for these problems.
6.26
A plate cam with an offset reciprocating roller follower has a dwell of 60° and then rises
in 90° to another dwell of 120°, after which it returns in 90° of cam rotation. The radius
of the base circle is 40 mm, the radius of the roller follower is 15 mm, and the follower
offset is 20 mm. For the rise motion 60    150, the equation of the displacement (the
lift) is


y  40   sin  


where y is in millimeters and  is the cam rotation angle in radians. (a) Find equations
for the first- and second-order kinematic coefficients of the lift y for this rise motion. (b)
Sketch the displacement diagram and the first- and second-order kinematic coefficients
for the follower motion described. Comment on the suitability of this rise motion in the
context of the other displacements specified. At the cam rotation angle  = 120°,
determine the following: (c) the location of the point of contact between the cam and
follower, expressed in the moving Cartesian coordinate system attached to the cam; (d)
the radius of curvature of the pitch curve and the radius of curvature of the cam surface;
and (e) the pressure angle of the cam. Is this pressure angle acceptable?
(a)
From the equation given for the lift, the first- and second-order kinematic
coefficients are
1

Ans.
y  40sin  mm/rad 2
y  40   cos   mm/rad and


(b)
Sketches of the first- and the second-order kinematic coefficients of the
displacement diagram are shown here.
The rise motion specified is not suitable between dwells on either side since the first- and
second-order kinematic coefficients are not zero at the beginning and end of the rise. A
cycloidal rise curve would be preferable, but would have higher values of acceleration
and would lead to higher forces at mid-range.
At the cam angle = 120º, we have     60  60   3 rad .


 3

y  40   sin    40 
 sin 60   48.0 mm






267
1

1

y  40   cos    40   cos60   32.7 mm/rad




y  40sin   40sin 60  34.6 mm/rad 2
The global coordinates of the trace point are
X 0    20 mm,
Y0 
 Rb  Rr    2   40 mm  15 mm   20 mm  51.2 mm
2
2
2
X  X 0  20 mm, Y  Y0  y  51.2 mm  48.0 mm  99.2 mm
The cam coordinates of the trace point (the pitch curve) and derivatives are
u   X cos  Y sin     20 mm cos120   99.2 mm  sin120  75.9 mm
v   X sin   Y cos    20 mm sin120   99.2 mm  cos120  66.9 mm
u   X sin  Y cos  y sin  38.6 mm/rad
v   X cos  Y sin  y cos  92.3 mm/rad
w  u2  v2  100.0 mm/rad
u   X cos  Y sin   2 y cos  y sin   13.87 mm/rad2
v   X sin  Y cos  2 y sin  y cos   2.76 mm/rad 2
(c)
From Eq. (6.36), the cam coordinates of the point of contact (the cam surface) are
v
u
Ans.
ucam  u  Rr
 62.1 mm
vcam  v  Rr
 61.1 mm
w
w
(d)
From Eq. (6.37), the radius of curvature of the pitch curve is
w3

 72.2 mm
Ans.
uv  vu
(e)
From Eq. (6.39) the pressure angle is
 v 
 u 
Ans.
cos      sin     cos  0.9919
  7.3
 w 
 w 
This pressure angle is markedly less than 30° and is acceptable.
268
6.27
A plate cam with an offset reciprocating roller follower is to be designed using the input,
the rise and fall, and the output motion shown in Table P6.27. The radius of the base
circle is 30 mm, the radius of the roller follower is 12.5 mm, and the follower offset
(eccentricity) is 15 mm.
Table P6.27 Displacement information for plate cam with reciprocating roller follower
Cam Angle (deg)
0° - 20°
20° - 110°
110° - 120°
120° - 200°
200° - 270°
270° - 360°
Rise or Fall (mm)
0
25
0
5
0
30
Follower Motion
Dwell
Full-rise simple harmonic motion
Dwell
Full-rise cycloidal motion
Dwell
Full-return cycloidal motion
Comment on the suitability of the motions specified. At the cam rotation angle   50,
determine the following: (a) the first-, second-, and third-order kinematic coefficients of
the lift curve, (b) the coordinates of the point of contact between the roller follower and
the cam surface, expressed in the Cartesian coordinate system rotating with the cam, (c)
the radius of curvature of the pitch curve, (d) the unit tangent and the unit normal vectors
to the pitch curve, and (e) the pressure angle of the cam.
The 20°-110° segment of the motion which specifies simple harmonic motion is not
suitable for high-speed operation since there will be discontinuities in the second
derivatives at both ends of that segment where it interfaces with dwells. Cycloidal
motion would correct this problem but would give higher peak acceleration. Still, simple
harmonic motion is specified.
(a)
Eqs. (6.12), with   50  20  30, L  25 mm,   90   2 rad, gives
L
  25 mm 

1  cos   6.25 mm
1  cos

2
 
2 
3
 L 

y 
sin
  25 mm  sin  21.65 mm/rad
2

3
y
 2L


cos
 2  25 mm  cos  25.0 mm/rad 2
2
2

3
3
 L 

y   3 sin
 4  25 mm  sin  86.60 mm/rad3
2

3
y 
Therefore the global coordinates of the tracepoint are
R0  Rb  Rr  30 mm  12.5 mm  42.5 mm,
Y0  R02   2 
 42.5 mm  15 mm  39.76 mm
2
2
X    15 mm, Y  Y0  y  39.76 mm  6.25 mm  46.01 mm
The cam coordinates of the trace point (the pitch curve) and derivatives are
u   X cos  Y sin    15 mm cos50   46.01 mm  sin50  44.89 mm
v   X sin   Y cos   15 mm sin50   46.01 mm  cos50  18.08 mm
Ans.
Ans.
Ans.
269
u   X sin  Y cos  y sin  34.67 mm/rad
v   X cos  Y sin  y cos  30.97 mm/rad
w  u2  v2  46.49 mm/rad
u   X cos  Y sin   2 y cos  y sin   2.10 mm/rad 2
v   X sin  Y cos  2 y sin  y cos  35.18 mm/rad2
(b)
From Eq. (6.36), the cam coordinates of the point of contact (the cam surface) are
v
u
Ans.
ucam  u  Rr
 36.56 mm
vcam  v  Rr
 8.76 mm
w
w
(c)
From Eq. (6.37), the radius of curvature of the pitch curve is
3

  w  uv  vu   87.02 mm
Ans.
(d)
The unit tangent and the unit normal vectors to the pitch curve are
t
uˆ   u w ˆi   v w ˆj  0.746ˆi  0.666ˆj
Ans.
uˆ n   v w ˆi   u w ˆj  0.666ˆi  0.746ˆj
Ans.
(e)
From Eq. (6.39) the pressure angle is
cos     v w sin    u w cos  0.9897
Ans.
  8.2
This pressure angle at the specified cam rotation angle of   50 is much less than 30°
and is very acceptable.
270
6.28
A plate cam with a radial reciprocating roller follower is to be designed using the input,
the rise and fall, and the output motion shown in Table P6.28. The base circle diameter is
3 in and the diameter of the roller is 1 in. Displacements are specified as follows:
Table P6.28 Displacement information for plate cam with reciprocating roller follower.
Input  (deg)
Lift L (in)
Output y
0  90
Cycloidal rise
3.0
90  105
Dwell
0
105  195
Cycloidal fall
3.0
195  210
Dwell
0
210  270
Simple harmonic rise
2.0
270  285
Dwell
0
285  345
Simple harmonic fall
2.0
345  360
Dwell
0
Plot the lift curve (displacement diagram), and the profile of the cam. (a) Comment on
the lift curves at appropriate positions of the cam, (for example, when the cam rotation
angle is   0,   45,   180,   210,   225, and   300 ). (b) Identify
on your cam profile the location(s) and the value(s) of the largest pressure angle. Would
this pressure angle cause difficulties for a practical cam-follower system? (c) Identify on
your cam profile the location(s) of any discontinuities in position, velocity, acceleration,
and/or jerk. Are these discontinuities acceptable (why or why not)? (d) Identify on your
cam profile any regions of positive radius of curvature of the cam profile. Are these
regions acceptable (why or why not)? (e) For the values given in Table P6.28, what
design changes would you suggest to improve this cam design?
The lift curve (displacement diagram) is shown in Fig. 1.
Fig. 1. The lift curve (displacement diagram).
271
The cam profile is shown in Fig. 2.
Fig. 2. The cam profile.
(a)
Because of the choice of simple harmonic motion rise and return curves, there are
discontinuities in acceleration at   210,   270,   285, and   345.
Because of adjacent dwells, cycloidal motion would be preferable, although it would lead
to higher peak accelerations in these segments.
(b)
The pressure angle, see Sec. 6.10, should be less than 30. In this design, the
pressure angle is more than the accepted value at the cam angles
and
  16  64,
131  180,
216  256,
299  341
Therefore, this cam profile is not a good design. The high values of the pressure angle
may be due to the selection of the displacement curves, the diameters of the base and
prime circles, and the diameter of the roller.
(c)
Position discontinuities never occur. Discontinuities in the derivatives occur only
at transitions between dwell segments and lift/return segments of motion. Discontinuities
272
in the derivatives are undesirable. There is an acceleration discontinuity at the beginning
and end of the simple harmonic motions, both rise and return. There is a jerk
discontinuity at the beginning and end of the cycloidal motions, both rise and return.
Whether these discontinuities are acceptable depends on the intended speed of operation,
and on the masses and stiffnesses involved.
(d)
The radius of curvature of a cam profile should always be negative for a good
cam design. Positive curvature means that the cam has a concave section of surface and
there is the possibility that the follower may lose contact with the cam. If the radius of
curvature of the cam profile is positive then the radius of curvature of the cam must be
greater than the radius of the roller. In the proposed design, the positive values of the
radius of curvature of the cam are always greater than 0.5 in (i.e., the radius of the roller).
The radius of curvature of the cam is positive for the cam angles
  3  25 , 170 192 , 215  220 , and 261  270
The radius of curvature is positive, and smaller than the radius of the roller follower, for
the cam angles
215  216
  9  13 , 181 187 , and
Note that the radius of curvature of the cam is zero between the cam rotation angles  =
214 and  = 215 degrees, meaning that pointing occurs. Also, it could imply that
undercutting occurs. Also, with the exception of where the radius of curvature of the cam
goes to zero, there is an inflection point at the boundary of each range of angles for which
the radius of curvature is positive.
(e)
Possible design changes to the cam-follower system: (i) Increase the radius of the
prime circle (with the same lift curve); in general this will reduce the pressure angle. (ii)
Change the profiles to match acceleration at the transistion (blend) points to eliminate
acceleration discontinuities. (iii) Change both simple harmonic motion profiles to
cycloidal; this will make accelerations continuous but will also increase accelerations
(and pressure angle) in the middle parts of the rise and the return profiles. (iv) We may
want to increase the diameter of the roller if the contact stresses are high. (v) We could
numerically explore the effects on forces of changing the offset (eccentricity)  . These
are not obvious from observation.
273
Continue using the same displacement information and the same design parameters as in
Problem 6.28. Use a spreadsheet to determine and plot the following for a complete
rotation of the cam: (a) the first-order kinematic coefficients of the follower center; (b)
the second-order kinematic coefficients of the follower center; (c) the third-order
kinematic coefficients of the follower center; (d) the lift curve (displacement diagram);
(e) the radius of curvature of the cam surface; and (f) the pressure angle of the camfollower system. Is the pressure angle suitable for a practical cam-follower system?
(a)
The first-order kinematic coefficients for the cam design are shown in Fig. 3.
First order Kinematics Coeeficients
6
xf c'
yf c'
4
2
First order KC - in
6.29
0
-2
-4
-6
0
50
100
150
200
250
300
350
 (Cam angle) - degrees
Fig. 3. The first-order kinematic coefficients.
(b)
The second-order kinematic coefficients for the cam design are shown in Fig. 4.
Fig. 4. The second-order kinematic coefficients.
400
274
(c)
The third-order kinematic coefficients for the cam design are shown in Fig. 5.
Third order Kinematics Coeeficients
40
x f c p"'
y f c "'
30
20
Third order KC - in
10
0
-10
-20
-30
-40
0
50
100
150
200
250
300
350
400
 (Cam angle) - degrees
Fig. 5. The third-order kinematic coefficients.
(d)
The lift curve (displacement diagram) for the cam design is shown in Fig. 6.
Pos it ion
Profile
3
2. 5
Lift - in
2
1. 5
1
0. 5
0
0
50
100
150

(Cam
200
angle) -
250
degrees
300
Fig. 6. The lift curve (displacement diagram).
(e)
The radius of curvature of the cam surface is shown in Fig. 7.
Fig. 7. The radius of curvature of the cam surface.
350
400
275
The radius of curvature of a cam profile should always be negative for a proper cam
design. Positive curvature means that the cam has a concave section of surface and there
is the possibility that the follower may lose contact with the cam. If the radius of
curvature of the cam profile is positive then the radius of curvature of the cam must be
greater than the radius of the follower. In the proposed design, the positive values of the
radius of curvature of the cam are always greater than 0.5 in (i.e., the radius of the
follower).
Note that the radius of curvature of the cam surface goes to infinity when the cam rotation
angle  is 4°, 26°, 168°, 191°, 225°, and 330°; i.e., there are six inflection points on the
cam surface. Also, note the discontinuity at the start and the end of the simple harmonic
profile at  = 210°, 270°, 285°, and 345°. These discontinuities are due to the
discontinuity in second derivative between dwells and simple harmonic profiles and for
this reason simple harmonic profiles adjacent to dwells are not generally recommended
for high-speed cam-follower systems.
(f)
The pressure angle for the cam-follower system is shown in Fig. 8.
Pressure Angle
50
45
40
 Pressure angle - degrees
35
30
25
20
15
10
5
0
0
50
100
150
200
250
300
350
400
 (Cam angle) - degrees
Fig. 8. The pressure angle of the cam-follower system.
The recommended value of the pressure angle is that it remain less than 30 degrees. In
this design, the pressure angle is more than the accepted value at the cam angles
  18  62, 134  178, 220  256, and 302  336 . Therefore, this cam profile
is not a good design. The high values of the pressure angle may be due to the selection of
the displacement curves, the dimensions of the cam, and the diameter of the roller.
276
6.30
The cam rotation angle, the rise and fall, and the output motion of a disk cam with a
reciprocating roller follower are given in Table P6.30. The diameter of the base circle of
the cam is 90 mm, the diameter of the roller follower is 30 mm, and the follower
eccentricity is 20 mm.
Table P6.30 Displacement information for plate cam with reciprocating roller follower.
Cam angle(degrees)
Lift L (mm)
Output y
0° - 45°
45° - 120°
120° - 130°
130° - 180°
0
35
0
15
Dwell
Full-rise simple harmonic motion
Dwell
Full-rise cycloidal motion
180° - 210°
210° - 290°
290° - 310°
0
20
0
Dwell
Full-return simple harmonic motion
Dwell
310° - 360°
30
Full-return cycloidal motion
Sketch the displacement diagram and its first two derivatives. At the cam rotation angle
  230, determine: (a) the first- and second-order kinematic coefficients of the
displacement diagram; (b) the coordinates of the point of contact between the cam and
the roller follower, expressed in the moving Cartesian coordinate system attached to the
cam; (c) the radius of the curvature of the cam profile; and (d) the pressure angle of the
cam. Is this pressure angle acceptable for this cam-follower system?
Displacement diagram for the cam-follower system.
The first- and second-order kinematic coefficients.
277
Note that this is a poor cam design. Early failure should be expected because there are
discontinuities in acceleration at   45,   120,   210, and   290. Cycloidal
motion would be superior choices in segments two and six because of the neighboring
dwells.
At the cam rotation angle   230, we have 6  210, Y6  30 mm, L6  20 mm, 6  80,
    6  230  210  20,  6  20 80  1 4
L6 
  20 mm 

 1  cos  
1  cos   17.071 mm
2
6 
2 
4
(a) The first- and second-order kinematic coefficients of the follower displacement are
  20 mm 
L


y   6 sin

sin  15.910 mm/rad
Ans.
26
6
2 80 rad 180
4
y
y  
 2  20 mm 
 2 L6


cos


cos  35.797 mm/rad 2
2
2
2
26
6
4
2 80  rad 180
Ans.
The global coordinates of the pitch curve, measured from the center of rotation, are
X    20 mm
Y  R02  e2  Y6  y 
 45 mm  15 mm    20 mm   30 mm  17.071 mm
2
2
 103.639 mm
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  20 mm  cos 230  103.639 mm  sin 230  92.248 mm
Ans.
v   X sin   Y cos
   20 mm  sin 230  103.639 mm  cos 230  51.297 mm
The derivatives of these give
u '   X sin   Y cos   y 'sin 
   20 mm  sin 230  103.639 mm  cos 230   15.910 mm  sin 230
 39.109 mm/rad
v '   X cos  Y sin   y ' cos
   20 mm  cos 230  103.639 mm  sin 230   15.910 mm  cos 230
 102.475 mm/rad
w   u   v   39.109 mm/rad   102.475 mm/rad   109.684 mm/rad
u   X cos  Y sin   2 y  cos  y  sin 
2
2
2
2
   20 mm  cos 230  103.639 mm  sin 230  2  15.910 mm  cos 230
  35.797 mm  sin 230
 140.123 mm/rad 2
Ans.
278
v  X sin   Y cos  2 y sin   y cos
  20 mm  sin 230  103.639 mm  cos 230  2  15.910 mm  sin 230
  35.797 mm  cos 230
 49.931 mm/rad 2
(b) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
 92.248 mm  15 mm 102.475 mm/rad 109.684 mm/rad   78.234 mm
Ans.
vcam  v  Rr  u w
 51.297 mm  15 mm  39.109 mm/rad 109.684 mm/rad   45.949 mm
Ans.
(c) The radius of curvature of the pitch curve is given by Eq. (6.37) as
3
109.684 mm 

w3


 80.896 mm
uv  vu  39.109 mm  49.931 mm   102.475 mm 140.123 mm 
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  80.896 mm  15 mm  65.896 mm
Ans.
(d) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
  102.475 mm 109.684 mm  sin 230   39.109 mm 109.684 mm  cos 230
 0.944 89
  19.11
The pressure angle is acceptable, that is, the pressure angle is less than 30°.
Ans.
Ans.
279
6.31
The cam angle, the rise and fall, and the output motion of a disk cam with a reciprocating
roller follower are as given in Table P6.31. The diameter of the base circle of the cam is
9.60 in, the diameter of the roller follower is 2.40 in, and the eccentricity (offset) of the
roller follower is 2.80 in.
Table P6.31 Displacement information for plate cam with reciprocating roller follower.
Cam angle (degrees)
Lift L (in)
0° - 5°
5° - 115°
115° - 120°
0
2.00
0
Dwell
Full-rise cycloidal motion
Dwell
120° - 180°
180° - 210°
210° - 310°
310° - 325°
3.40
0
4.80
0
Full-rise simple harmonic motion
Dwell
Full-return eighth-order polynomial motion
Dwell
325° - 360°
0.60
Full-return simple harmonic motion
Output y
Sketch the displacement diagram and its first two derivatives. At the cam rotation angle
  235o , determine: (a) the first and second-order kinematic coefficients of the
displacement diagram; (b) the coordinates of the point of contact between the cam and
the roller follower, expressed in the rotating Cartesian coordinate system attached to the
cam; (c) the radius of the curvature of the cam profile; and (d) the pressure angle of the
cam.
The displacement diagram for the cam-follower system is shown here.
Note that this is a poor cam design. Early failure should be expected because there are
280
discontinuities in acceleration at   120,   180,   210,   325. and   360.
Cycloidal motion would be superior choices in segments four, six, and eight because of
the neighboring dwells.
For the cam rotation angle   235, we have 6  210, Y6  0.6 in, L6  4.8 in,
6  100  1.745 33 rad,     6  235  210  25,  6  25 100  1 4,
Using Eq. (6.17a), we find the lift at this position to be
2
5

 
 
Y  Y6  L6 1.0  2.634 15    2.780 55  

 6 
 6 
6
7
8
 
 
  
3.170 60    6.877 95    2.560 95   
 6 
 6 
  6  
2
5

1
1
 0.6 in  4.8 in 1.0  2.634 15    2.780 55  
4
4

6
7
8
1
1
1 
3.170 60    6.877 95    2.560 95   
4
4
 4  
 4.624 68 in
(a) From Eqs. (6.17b) and (6.17c), the first- and second-order kinematic coefficients are
4
 
 
L6 
y    5.268 30    13.902 75  
6 
 6 
 6 

5
6
7
 
 
  
19.023 60    48.145 65    20.487 60   
 6 
 6 
  6  

4
4.80 in 
1
1
5.268
30

13.902
75

 
 
1.745 33 rad 
4
4
5
6
7
1
1
1 
19.023 60    48.145 65    20.487 60   
4
4
 4  
 3.450 65 in/rad
Ans.
281
 
L 
y    62 5.268 30  55.611 00  
6 
 6 

3
4
5
6
 
 
  
95.118 00    288.873 90    143.413 20   
 6 
 6 
  6  
3

1

5.268 30  55.611 00  
2 
4
1.745 33 rad  
4.80 in
4
5
6
1
1
1 
95.118 00    288.873 90    143.413 20   
4
4
 4  
 6.736 18 in/rad 2
The global coordinates of the pitch curve are
X    2.80 in
Y  R02  e2  Y 
Ans.
 4.80 in  1.20 in    2.80 in   4.624 68 in
2
2
 9.931 28 in
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  2.80 in  cos 235   9.931 28 in  sin 235  9.741 24 in
v   X sin   Y cos
   2.80 in  sin 235   9.931 28 in  cos 235  3.402 72 in
The derivatives of these give
u '   X sin   Y cos   y 'sin 
   2.80 in  sin 235   9.931 28 in  cos 235   3.450 65 in  sin 235
 0.576 12 in/rad
v '   X cos  Y sin   y ' cos
   2.80 in  cos 235   9.931 28 in  sin 235   3.450 65 in  cos 235
 11.720 45 in/rad
w 
 u   v   0.576 12 in/rad   11.720 45 in/rad   11.734 60 in/rad
2
2
2
2
u   X cos  Y sin   2 y  cos  y  sin 
   2.80 in  cos 235   9.931 28 in  sin 235  2  3.450 65 in  cos 235
  6.736 18 in  sin 235
 19.217 62 in/rad 2
v  X sin   Y cos  2 y sin   y cos
  2.80 in  sin 235   9.931 28 in  cos 235  2  3.450 65 in  sin 235
  6.736 18 in  cos 235
 1.613 22 in/rad 2
282
(b) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
 9.741 24 in  1.20 in 11.720 45 in/rad 11.734 60 in/rad   8.543 in
Ans.
vcam  v  Rr  u w
 3.402 72 in  1.20 in 11.720 45 in/rad 11.734 60 in/rad   4.601 in
Ans.
(c) The radius of curvature of the pitch curve is given by Eq. (6.37) as
3
11.734 60 in 

w3


 7.145 in
uv  vu  0.576 12 in 1.613 22 in   11.720 45 in 19.217 62 in 
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  7.145 in  1.20 in  5.945 in
(d) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
Ans.
  11.720 45 in 11.734 60 in  sin 235   0.576 12 in 11.734 60 in  cos 235
 0.846 32
Ans.
  32.19
The pressure angle is not good, that is, the pressure angle is a little greater than 30°.
However, it may be acceptable depending on the application.
Ans.
283
6.32
The cam angle, the rise and fall, and the output motion of a disk cam with a reciprocating
roller follower are as given in Table P6.32. The diameter of the base circle of the cam is
180 mm, the diameter of the roller follower is 80 mm, and the eccentricity of the roller
follower is 40 mm.
Table.P6.32 Displacement information for disk cam and reciprocating roller follower.
Cam angle  (degrees)
Lift L (mm)
Output y
0° - 40°
40° - 100°
100° - 180°
0
60
20
Dwell
Half-rise simple harmonic motion
Half-rise cycloidal motion
180° - 260°
260° - 360°
0
60  20
Dwell
Full return cycloidal motion
Part I. Sketch the displacement diagram and its first two derivatives. At the cam rotation
angle   85o , determine: (a) the first, second, and third-order kinematic coefficients of
the displacement diagram; (b) the radius of curvature of the cam surface; (c) the unit
tangent and normal vectors to the cam at the point of contact with the follower; (d) the
coordinates of the point of contact between the cam and the follower. Express your
answers in the moving Cartesian coordinate system attached to the cam; and (e) the
pressure angle of the cam. Part II. Repeat the problem for the cam angle   120o ,
The displacement diagram for this cam-follower system is shown here.
The first- and second-order kinematic coefficients.
284
Note that this is a poor cam design. Early failure should be expected because there is a
discontinuity in acceleration at   40. Cycloidal motion would be a superior choice in
segment two because of the neighboring zero acceleration values.
Part I. For the cam rotation angle   85, we have 2  40, Y2  0, L2  60 mm, 2  60,
    2  85  40  45,  2  45 60  3 4,

 
3 

Y  Y2  L2  1  cos
  0  60 mm 1  cos
  37.039 mm
2 2 
24 


(a) The first-, second-, and third-order kinematic coefficients of the follower
displacement are:
  60 mm 
L

3
Ans.
y  2 sin

sin
 83.149 mm/rad
2 2
2 2 2  60 rad 180
24
 2  60 mm 
 2 L2

3
y 
cos

cos
 51.662 mm/rad 2
2
2
2
42
2  2 4  60  rad 180
24
 3  60 mm 
 3 L2

3
y  
cos


cos
 77.493 mm/rad 3
3
3
3
8 2
22
24
8  60  rad 180
Ans.
Ans.
The global coordinates of the pitch curve are
X    40 mm
Y  R02  e2  Y 
 90 mm  40 mm    40 mm   37.039 mm
2
2
 160.732 mm
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  40 mm  cos85  160.732 mm  sin 85  163.607 mm
Ans.
v   X sin   Y cos
   40 mm  sin 85  160.732 mm  cos85  25.839 mm
The derivatives of these give
u '   X sin   Y cos  y 'sin 
   40 mm  sin 85  160.732 mm  cos85  83.149 mm  sin 85
 56.994 mm/rad
v '   X cos   Y sin   y ' cos 
   40 mm  cos85  160.732 mm  sin 85  83.149 mm  cos85
 156.360 mm/rad
w 
 u   v  56.994 mm/rad    156.360 mm/rad   166.423 mm/rad
2
2
2
2
u   X cos   Y sin   2 y  cos  y  sin 
   40 mm  cos85  160.732 mm  sin 85  2 83.149 mm  cos85
  51.662 mm  sin 85
 97.647 mm/rad 2
Ans.
285
v  X sin   Y cos  2 y  sin   y  cos
  40 mm  sin 85  160.732 mm  cos85  2 83.149 mm  sin 85
  51.662 mm  cos85
 135.323 mm/rad 2
(b) The radius of curvature of the pitch curve is given by Eq. (6.37) as
3
166.423 mm 

w3


uv  vu  56.994 mm  135.323 mm    156.360 mm  97.647 mm 
 200.575 mm
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  200.575 mm  40 mm  160.575 mm
Ans.
(c) the unit tangent and normal vectors to the cam at the point of contact with the follower
are given by
uˆ t   u w ˆi   v w ˆj
  56.994 mm 166.423 mm  ˆi   156.360 mm 166.423 mm  ˆj
 0.342 46ˆi  0.939 53ˆj
uˆ n   v w ˆi   u w ˆj  0.939 53ˆi  0.342 46ˆj
Ans.
Ans.
(d) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
 92.248 mm  15 mm 102.475 mm/rad 109.684 mm/rad   78.234 mm
Ans.
vcam  v  Rr  u w
 51.297 mm  15 mm  39.109 mm/rad 109.684 mm/rad   45.949 mm
Ans.
(e) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
   0.939 53 sin85   0.342 46 cos85  0.965 80
  15.03
The pressure angle is acceptable, that is, the pressure angle is less than 30°.
Ans.
Ans.
Part II. For the cam rotation angle   120, we have 3  100, Y3  60 mm, L3  20 mm,
3  80,     3  120  100  20,  3  20 80  1 4,
 1
 

1 1
Y  Y3  L3   sin
  60 mm  20 mm   sin   89.850 mm
3 
4
4 
 3 
(a) The first-, second-, and third-order kinematic coefficients of the follower
displacement are:
L 
 
20 mm


y  3  1  cos  
Ans.
1  cos   76.820 mm/rad
3 
 3  80  rad 180 
4
286
 L3 
20 2 mm

sin

sin  71.594 mm/rad 2
2
2
3
3
4
80  rad 180
Ans.
 2 L3

20 3 mm

cos


cos  161.088 mm/rad3
3
3
3
3
3
4
80  rad 180
Ans.
y  
y  
The global coordinates of the pitch curve are
X    40 mm
Y  R02  e2  Y 
 90 mm  40 mm    40 mm   89.850 mm
2
2
 213.543 mm
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  40 mm  cos120   213.543 mm  sin120  164.934 mm
Ans.
v   X sin   Y cos
   40 mm  sin120   213.543 mm  cos120  141.413 mm
The derivatives of these give
u '   X sin   Y cos  y 'sin 
Ans.
   40 mm  sin120   213.543 mm  cos120   76.820 mm  sin120
 74.884 mm/rad
v '   X cos  Y sin   y ' cos 
   40 mm  cos120   213.543 mm  sin120   76.820 mm  cos120
 203.344 mm/rad
w 
 u   v   74.884 mm/rad    203.344 mm/rad   216.694 mm/rad
2
2
2
2
u   X cos  Y sin   2 y  cos  y  sin 
   40 mm  cos120   213.543 mm  sin120  2  76.820 mm  cos120
  71.594 mm  sin120
 303.756 mm/rad 2
v  X sin   Y cos  2 y  sin   y  cos
  40 mm  sin120   213.543 mm  cos120  2  76.820 mm  sin120
  71.594 mm  cos120
 44.153 mm/rad 2
(b) The radius of curvature of the pitch curve is given by Eq. (6.37) as
3
216.694 mm 

w3


uv  vu  74.884 mm  44.153 mm    203.344 mm  303.756 mm 
 156.364 mm
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  156.364 mm  40 mm  116.364 mm
Ans.
287
(c) the unit tangent and normal vectors to the cam at the point of contact with the follower
are given by
uˆ t   u w ˆi   v w ˆj
  74.884 mm 216.694 mm  ˆi   203.344 mm 216.694 mm  ˆj
 0.345 57ˆi  0.938 39ˆj
uˆ n   v w ˆi   u w ˆj  0.938 39ˆi  0.345 57ˆj
Ans.
Ans.
(d) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
 164.934 mm  40 mm  203.344 mm/rad 216.694 mm/rad   127.398 mm
Ans.
vcam  v  Rr  u w
 141.413 mm  40 mm  74.884 mm/rad 216.694 mm/rad   127.590 mm Ans.
(e) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
   0.938 39  sin120   0.345 57  cos120  0.985 45
  9.78
This pressure angle is very acceptable, that is, the pressure angle is less than 30°.
Ans.
Ans.
288
6.33
The cam angle, the rise and fall, and the output motion of a disk cam with a reciprocating
roller follower are as given in Table P6.33. The diameter of the base circle of the cam is
2.80 in, the diameter of the roller follower is 1.20 in, and the follower eccentricity is 0.40
in.
Table.P6.33 Displacement information for disk cam and reciprocating roller follower.
Cam Angle  (degrees)
Lift L (in)
Output y
0° - 30°
30° - 90°
90° - 120°
120° - 180°
0
1.20
0
0.80
Dwell
Full-rise simple harmonic motion
Dwell
Full-rise cycloidal motion
180° - 210°
210° - 270°
270° - 300°
0
0.80
0
Dwell
Full-return cycloidal motion
Dwell
300° - 360°
1.20
Full-return simple harmonic motion
Sketch the displacement diagram and its first two derivatives. At the cam angle
  150, determine: (a) the first-, second-, and third-order kinematic coefficients of the
displacement diagram; (b) the radius of the curvature of the cam surface; (c) the unit
tangent and normal vectors to the cam at the point of contact with the follower; (d) the
coordinates of the point of contact between the cam and the follower. Express your
answers in the moving Cartesian coordinate system attached to the cam; and (e) the
pressure angle of the cam.
Displacement diagram for the cam-follower system.
289
Note that this is a poor cam design. Early failure should be expected because there are
discontinuities in acceleration at   30,   90,   300, and   360. Cycloidal motion
would be a superior choice in segments two and eight because of the neighboring dwells.
For the cam rotation angle   150, we have 4  120, Y4  1.20 in, L4  0.8 in,
4  60  1.047 20 rad,     4  150  120  30,  4  30 60  1 2,
Using Eq. (6.13a), we find the lift at this position to be

1
2 
Y  Y4  L4  
sin
 4 
  4 2
1 1
 360  
 1.20 in  0.8 in  
sin 
   1.600 00 in
 2 
 2 2
(a) From Eqs. (6.13b) - (6.13d), the first-, second-, and third-order kinematic coefficients
are
L 
2 
0.80 in 
 360  
y  4  1  cos
1  cos 
Ans.

   1.527 88 in/rad

4 
 4  1.047 20 rad 
 2 
2 L4
2
2 0.80 in
 360 
Ans.
y 
sin

sin 
0
2
2
4
 4 1.047 20 rad 
 2 
4 2 L4
2
4 2 0.80 in
 360 
3
cos 
  27.501 78 in/rad

 4 1.047 20 rad 
2


The global coordinates of the pitch curve are
X    0.40 in
y 
3
4
cos

Y  R02  e2  Y 
3
1.40 in  0.60 in    0.40 in   1.600 00 in
2
2
 3.559 59 in
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  0.40 in  cos150   3.559 59 in  sin150  1.433 38 in
v   X sin   Y cos
   0.40 in  sin150   3.559 59 in  cos150  3.282 70 in
Ans.
290
The derivatives of these give
u '   X sin   Y cos  y 'sin 
   0.40 in  sin150   3.559 59 in  cos150  1.527 88 in  sin150
 2.518 76 in/rad
v '   X cos  Y sin   y ' cos 
   0.40 in  cos150   3.559 59 in  sin150  1.527 88 in  cos150
 2.756 57 in/rad
w   u   v   2.518 76 in/rad    2.756 57 in/rad   3.734 01 in/rad
u   X cos  Y sin   2 y cos  y sin 
2
2
2
2
   0.40 in  cos150   3.559 59 in  sin150  2 1.527 88 in  cos150   0  sin150
 4.079 75 in/rad 2
v  X sin   Y cos  2 y  sin   y  cos
  0.40 in  sin150   3.559 59 in  cos150  2 1.527 88 in  sin150   0  cos150
 1.754 82 in/rad 2
(b) The radius of curvature of the pitch curve is given by Eq. (6.37) as
w3

uv  vu
 3.734 01 in 
 3.323 in
 2.518 76 in 1.754 82 in    2.756 57 in  4.079 75 in 
3

Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  3.323 in  0.600 in  2.723 in
(c) The unit tangent and unit normal vectors are given by
uˆ t   u w ˆi   v w ˆj
Ans.
  2.518 76 in 3.734 01 in  ˆi   2.756 57 in 3.734 01 in  ˆi
 0.674 55ˆi  0.738 23ˆj
uˆ n   v w ˆi   u w ˆj
Ans.
  2.756 57 in 3.734 01 in  ˆi   2.518 76 in 3.734 01 in  ˆi
 0.738 23ˆi  0.674 55ˆj
(d) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
 1.433 38 in  0.60 in  2.756 57 in/rad 3.734 01 in/rad   0.990 in
Ans.
Ans.
vcam  v  Rr  u w
 3.282 70 in  0.60 in  2.518 76 in/rad 3.734 01 in/rad   2.878 in
(e) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
Ans.
291
cos     v w sin    u w cos
   2.756 57 in 3.734 01 in  sin150   2.518 76 in 3.734 01 in  cos150
 0.953 29
Ans.
  17.58
This pressure angle is acceptable (for the given input cam angle) because pressure angles
up to 30 are commonly used without causing major difficulties.
292
6.34
The cam angle, the rise and fall, and the output motion of a disk cam with a reciprocating
roller follower are given in the table. The diameter of the base circle of the cam is 75
mm, the diameter of the roller follower is 25 mm, and the follower eccentricity is 20 mm.
Table 6.34. Displacement information for disk cam and reciprocating roller follower.
Cam Angle  (degrees)
Lift L (mm)
Output Motion y
0° - 60°
60° - 180°
180° - 240°
240° - 360°
0
90
0
90
Dwell
Full-rise cycloidal motion
Dwell
Full-return cycloidal motion
Sketch the displacement diagram and its first two derivatives. At the cam angle
  300, determine: (a) the first and second-order kinematic coefficients of the
displacement diagram; (b) the coordinates of the point of contact between the cam and
the roller follower. Express your answers in the moving Cartesian coordinate system
attached to the cam; (c) the radius of the curvature of the cam surface; and (d) the
pressure angle of the cam.
Displacement diagram for the cam-follower system.
293
For the cam rotation angle   300, we have 4  240, Y4  0, L4  90 mm, 4  120,
    4  300  240  60,  4  60 120  1 2,


1
2 
2 
 1 1
Y  Y4  L4  1 

sin
sin
  0  90 mm 1  
  45.000 mm
4 
2 
 2 2
  4 2
(a) The first- and second- -order kinematic coefficients of the follower displacement are:
L 
2 
90 mm
2 

y   4  1  cos
Ans.

 1  cos
  85.944 mm/rad
4 
4 
120  rad 180 
2 
y  
2 L4

2
4
sin
2
4

2  90 mm 
120  rad 180
2
2
sin
2
0
2
Ans.
The global coordinates of the pitch curve are
X    20 mm
Y  R02  e2  Y 
 37.5 mm  12.5 mm   20 mm   45.000  90.826 mm
2
2
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  20 mm  cos 300   90.826 mm  sin 300  68.658 mm
Ans.
v   X sin   Y cos
   20 mm  sin 300   90.826 mm  cos 300  62.734 mm
The derivatives of these give
u '   X sin   Y cos  y 'sin 
   20 mm  sin 300   90.826 mm  cos 300   85.944 mm  sin 300
 137.163 mm/rad
v '   X cos  Y sin   y ' cos 
   20 mm  cos 300   90.826 mm  sin 300   85.944 mm  cos 300
 25.686 mm/rad
w 
 u   v  137.163 mm/rad    25,686 mm/rad   139.547 mm/rad
2
2
2
2
u   X cos  Y sin   2 y  cos  y  sin 
   20 mm  cos 300   90.826 mm  sin 300  2  85.944 mm  cos 300
  0 mm  sin 300
 17.286 mm/rad 2
v  X sin   Y cos  2 y  sin   y  cos
  20 mm  sin 300   90.826 mm  cos 300  2  85.944 mm  sin 300
  0 mm  cos 300
 211.593 mm/rad 2
(b) The point of contact between the cam and the roller is given by Eq. (6.36) as
Ans.
294
ucam  u  Rr  v w
 68.658 mm  12.5 mm  25.686 mm/rad 139.547 mm/rad   66.357 mm
Ans.
vcam  v  Rr  u w
 62.734 mm  12.5 mm 137.163 mm/rad 139.547 mm/rad   50.448 mm
Ans.
(c) The radius of curvature of the pitch curve is given by Eq. (6.37) as
3
139.547 mm 

w3


uv  vu 137.163 mm  211.593 mm    25.686 mm  17.286 mm 
 95.086 mm
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  95.086 mm  12.5 mm  82.586 mm
(d) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
Ans.
   25.686 mm 139.547 mm  sin 300  137.163 mm 139.547 mm  cos300
 0.650 86
Ans.
  49.39
This is an unacceptable cam design since the pressure angle is too large, and would
experience a very early failure. The pressure angle should be less than 30 degrees
throughout the cycle of operation.
295
6.35
The cam rotation angle, the rise and fall, and the output motion of a disk cam with a
reciprocating roller follower are as given in Table 6.35. The diameter of the base circle of
the cam is 2.80 in, the diameter of the roller follower is 1.20 in, and the follower
eccentricity is 0.40 in.
Displacement information for disk cam and reciprocating roller follower.
Cam Angle  (degrees)
Lift L (in)
Output motion y
0° - 30°
30° - 90°
90° - 120°
120° - 180°
0
1.20
0
0.80
Dwell
Full-rise simple harmonic motion
Dwell
Full-rise cycloidal motion
180° - 210°
210° - 270°
270° - 300°
0
0.80
0
Dwell
Full-return cycloidal motion
Dwell
300° - 360°
1.20
Full-return simple harmonic motion
Sketch the displacement diagram and its first two derivatives. At the cam rotation angle
  230o , determine: (a) the first-, second-, and third-order kinematic coefficients of the
displacement diagram; (b) the radius of the curvature of the cam surface; (c) the unit
tangent and normal vectors to the cam at the point of contact with the follower; (d) the
coordinates of the point of contact between the cam and the follower. Express your
answers in the moving Cartesian coordinate system attached to the cam; and (e) the
pressure angle of the cam.
Displacement diagram for the cam-follower system.
Note that this is a poor cam design. Early failure should be expected because there are
296
discontinuities in acceleration at   30,   90,   300, and   360. Cycloidal motion
would be a superior choice in segments two and eight because of the neighboring dwells.
For the cam rotation angle   230, we have 6  210, Y6  1.20 in, L6  0.8 in,
6  60  1.047 20 rad,     6  230  210  20,  4  20 60  1 3,
Using Eq. (6.16a), we find the lift at this position to be


1
2 
Y  Y6  L6  1 

sin
 6 
  6 2
 1 1
 360  
 1.20 in  0.8 in 1  
sin 
   1.843 60 in
 3 
 3 2
(a) From Eqs. (6.16b) - (6.16d), the first-, second- and third-order kinematic coefficients
are
L 
2 
0.80 in 
 360 
y   6  1  cos
1  cos 
Ans.

  1.145 91 in/rad

6 
6 
1.047 20 rad 
 3 
2 L
2
2 0.80 in
 360 
2
Ans.
y   2 6 sin

sin 
  3.969 55 in/rad
2
6
6
1.047 20 rad   3 
4 2 L6
2
4 2 0.80 in
 360 
3
cos 
  13.750 89 in/rad

6
3


1.047 20 rad 
The global coordinates of the pitch curve are
X    0.40 in
y  
3
6
cos

Y  R02  e2  Y 
3
1.40 in  0.60 in    0.40 in   1.843 60 in
2
2
 3.803 19 in
The cam coordinates of the pitch curve are
u  X cos  Y sin 
  0.40 in  cos 230   3.803 19 in  sin 230  3.170 53 in
v   X sin   Y cos
   0.40 in  sin 230   3.803 19 in  cos 230  2.138 23 in
The derivatives of these give
u '   X sin   Y cos  y 'sin 
   0.40 in  sin 230   3.803 19 in  cos 230   1.145 91 in  sin 230
 1.260 41 in/rad
v '   X cos  Y sin   y ' cos 
   0.40 in  cos 230   3.803 19 in  sin 230   1.145 91 in  cos 230
 3.907 11 in/rad
w 
 u   v   1.260 41 in/rad   3.907 11 in/rad   4.105 38 in/rad
2
2
2
2
Ans.
297
u   X cos  Y sin   2 y  cos  y  sin 
   0.40 in  cos 230   3.803 19 in  sin 230  2  1.145 91 in  cos 230
  3.969 55 in  sin 230
 7.684 53 in/rad 2
v  X sin   Y cos  2 y  sin   y  cos
  0.40 in  sin 230   3.803 19 in  cos 230  2  1.145 91 in  sin 230
  3.969 55 in  cos 230
 2.934 17 in/rad 2
(b) The radius of curvature of the pitch curve is given by Eq. (6.37) as
w3

uv  vu
 4.105 38 in 

 2.052 in
 1.260 41 in  2.934 17 in    3.907 11 in 7.684 53 in 
3
Therefore, Eq. (6.38) gives the radius of curvature of the cam as
cam    Rr  2.052 in  0.600 in  1.452 in
(c) The unit tangent and unit normal vectors are given by
uˆ t   u w ˆi   v w ˆj
Ans.
  1.260 41 in 4.105 38 in  ˆi   3.907 11 in 4.105 38 in  ˆi
 0.307 01ˆi  0.951 70ˆj
uˆ n   v w ˆi   u w ˆj
Ans.
  3.907 11 in 4.105 38 in  ˆi   1.260 41 in 4.105 38 in  ˆi
 0.951 70ˆi  0.307 01ˆj
(d) The point of contact between the cam and the roller is given by Eq. (6.36) as
ucam  u  Rr  v w
Ans.
 3.170 53 in  0.60 in  3.907 11 in/rad 4.105 38 in/rad   2.599 51 in
Ans.
vcam  v  Rr  u w
 2.138 23 in  0.60 in  1.260 41 in/rad 4.105 38 in/rad   1.954 in
Ans.
(e) The pressure angle at this cam rotation angle is given by Eq. (6.39) as
cos     v w sin    u w cos
   3.907 11 in 4.105 38 in  sin 230   1.260 41 in 4.105 38 in  cos 230
 0.926 39
Ans.
  22.12
This pressure angle is acceptable (for the given input cam angle) because pressure angles
up to 30 are commonly used without causing major difficulties.
298
6.36
The mass m is constrained to move only in the vertical direction. The circular cam has an
eccentricity of 2 in, a speed of 20 rad/s, and a weight of 8 lb. Neglecting friction, find the
angle   t at the instant the cam follower jumps.
 F  W  F  my
my  mg  F
y  e 1  cos t 
y   2e cos t
When in contact, the contact force between the cam and follower is
F  m 2e cos t  mg
Contact is lost and jump begins when F = 0; that is, when
m 2e cos t  mg  0
mg
386 in/s2
cos t  


 0.483
2
m 2e
 20 rad/s  2 in 
  t  cos1  0.483  118.85
Ans.
299
6.37
In Figure P6.36a, the mass m is driven up and down by the eccentric cam and it has a
weight of 10 lb. The cam eccentricity is 1 in. Assume no friction.
(a) Derive the equation for the contact force.
(b) Find the cam velocity  corresponding to the beginning of the cam follower jump.
(a)
(b)
 F  W  F  my
my  mg  F
y  e 1  cos t 
y   2e cos t
F  m 2e cos t  mg
Follower jump begins when cos t  1 and F = 0: that is, when
m 2e  mg  0

g
386 in/s2

 19.65 rad/s
e
1 in 
Ans.
Ans.
300
6.38
In Figure P6.36a, the slider has a mass of 2.5 kg. The cam is a simple eccentric and
causes the slider to rise 25 mm with no friction. At what cam speed in revolutions per
minute will the slider first lose contact with the cam? Sketch a graph of the contact force
at this speed for 360 of cam rotation.
From Prob. 6.37, we have for the contact force
F  m 2e cos t  mg
Follower jump begins when cos t  1 and F = 0; that is, when  2e  g  0
Note that e = L/2 = 12.5 mm.

g

e
9.81 m/s 2
 28.01 rad/s  267.5 rev/min
 0.0125 m 
Ans.
301
6.39
The cam-and-follower system in Figure P6.36b has k  1 kN / m, m  0.90 kg,
Y  15 15cos t mm, and   60 rad / s. The retaining spring is assembled with a preload
of 2.5 N. (a) Compute the maximum and minimum values of the contact force. (b) If the
follower is found to jump off the cam, compute the angle   t corresponding to the
beginning of jump.
(a)
Let Fc = contact force, and P = preload.
mY  kY  P  Fc
 F  Fc  kY  P  mY
Y  0.015  0.015cos 60t m ,
Y  54cos60t m/s2
 17.5  33.6cos 60t N
Fc,max  17.5  33.6 N  51.1 N ,
Fc,min  0
Fc   0.90 kg   54cos 60t m/s 2   1 000 N/m  0.015  0.015cos 60t m   2.5 N
(b)
Jump begins when Fc = 0; that is, when
  60t  cos1  17.5 N 33.6 N   121.39
Ans.
Ans.
302
6.40
Figure P6.36b illustrates the model of a cam-and-follower system. The motion machined
into the cam is to move the mass to the right through a distance of 2 in with parabolic
motion in 150 of cam rotation, dwell for 30, return to the starting position with simple
harmonic motion, and dwell for the remaining 30 of cam rotation. There is no friction
or damping. The spring rate is 40 lb/in, and the spring preload is 6 lb, corresponding to
the Y  0 position. The weight of the mass is 36 lb. (a) Sketch a displacement diagram
showing the follower motion for the entire 360 of cam rotation. Without computing
numeric values, superimpose graphs of the acceleration and cam contact force onto the
same axes. Show where jump is most likely to begin. (b) At what speed in revolutions
per minute would jump begin?
(a) Just as in Prob. 6.39, if we let Fc = contact force and P = preload:
mY  kY  P  Fc
 F  Fc  kY  P  mY
Using second-order kinematic coefficients and assuming that the input shaft speed is
constant, then Y  Y  2 and
Fc   36 lb 386 in/s2  Y  2   40 lb/in  Y  6 lb
Going through the different phases of the motion defined above, we can sketch the
approximate curve shown for the cam contact force
This sketch shows that jump is very possible at point A   t  75 or point B
  t  180 or point C   t  330 , the three points where the contact force
drops discontinuously, depending on whether  is large enough for the contact force to
indicate a negative value.
303
(b) For point A   t  75 , L  2 in ,   150  2.618 rad .
From Eq. (6.6a),
Y  1 in and, from Eq. (6.6c), Y   1.167 in/rad . Therefore,
Fc , A   36 lb 386 in/s2  Y  2   40 lb/in  Y  6 lb
2
  0.109 lb  s2   2  46 lb
Thus, Fc , A  0 for   20.559 rad/s
For point B
  t  180 , L  2 in ,   150  2.618 rad .
From Eq. (6.12a),
Y  2 in and, from Eq. (6.21c), Y   1.440 in/rad . Therefore,
Fc ,B   36 lb 386 in/s2  Y  2   40 lb/in  Y  6 lb
2
  0.109 lb  s2   2  86 lb
Thus, Fc , B  0 for   25.308 rad/s
For point C   t  330 , y  y  0 , and Fc,C  6 lb for all values of  .
Of these cases, jump begins at A when   20.559 rad/s  196.3 rev/min .
Ans.
304
6.41
A cam-and-follower mechanism is illustrated in abstract form in Figure P6.36b. The cam
is cut so that it causes the mass to move to the right a distance of 25 mm with harmonic
motion in 150 of cam rotation, dwell for 30 , then return to the starting position in the
remaining 180 of cam rotation, also with harmonic motion. The spring is assembled
with a 22-N preload and it has a rate of 4.4 kN/m. The follower mass is 17.5 kg.
Compute the cam speed in revolutions per minute at which jump would begin.
Just as in Prob. 6.39, if we let Fc = contact force and P = preload:
mY  kY  P  Fc
 F  Fc  kY  P  mY
Using first-order kinematic coefficients and assuming that the input shaft speed is
constant, then Y  Y  2 and
Fc  17.5 kg  Y  2   4 400 N/m Y  22 N
Going through the different phases of the motion defined above shows that jump is most
likely at the transition from the dwell to the full-return simple-harmonic motion since, at
that position, Y  and Fc suddenly drop. For that position   t  180 , L  0.025 m ,
  180  3.1416 rad .
From Eq. (6.6c), Y  0.025 m and Y   0.036 m/rad 2 .
Therefore,
Fc  17.5 kg  Y  2   4400 N/m  Y  22 N
  0.630 N  s2   2  132 N
Thus, Fc  0 for   14.475 rad/s  138.2 rev/min.
Ans.
305
6.42
Lever OAB is driven by a cam cut to give the roller a rise of 1 in with parabolic motion
and a parabolic return with no dwells. The lever and roller are to be assumed weightless,
and there is no friction. Calculate the jump speed if l  5 in and the mass B weighs 5 lb.
Taking moments about the fixed pivot
2
 M O  lFA  2lmg  m  2l  Y
FA  4mlY  2mg
FA   0.259 lb  s2  Y  10 lb
Going through the different phases of the motion defined above shows that jump is most
likely at the transition from the concave to the convex parabolic rise motion since, at that
position, y and Fc suddenly drop. For that position   t  90 , L  1.000 in ,
  180  3.1416 rad .
From
Eqs.
(6.6a)
and
(6.6c),
Y  0.500 in
and
Y   0.405 in/rad2 . Therefore,
Y  Y  2 l   0.405 in/rad 2 5 in   2  0.0811 2
FA   0.259 lb  s 2   0.0811  2  10 lb
  0.021 lb  s 2   2  10 lb
Thus, FA  0 for   21.8 rad/s  208.4 rev/min.
Ans.
306
6.43
A cam-and-follower system similar to the one in Figure 6.41 uses a plate-cam driven at a
speed of 600 rev/min and employs simple harmonic rise and parabolic return motions.
The events are rise in 150 , dwell for 30 , and return in 180 . The retaining spring has
a rate k = 14 kN/m with a precompression of 12.5 mm. The follower has a mass of 1.6
kg. The external load is related to the follower motion Y by the equation
F  0.325  10.75Y , where Y is in meters and F is in kilonewtons. Dimensions
corresponding to Fig. 6.41 are R = 20 mm, r = 5 mm, lB  60 mm, and lC  90 mm.
Using a rise of L = 20 mm and assuming no friction, plot the displacement, cam-shaft
torque, and radial component of the cam force for one complete revolution of the cam.
  600 rev/min  62.832 rad/s
For simple harmonic rise motion, we use Eqs. (6.12) with L  0.020 m and   150 .
For the first part of the parabolic return motion, following Example 6.1,
2
y  0.020 1  2     m , y    0.080    m , y   0.080  2  0.008 106 m


For the second part of the parabolic return motion,
2
y  0.040 1     m , y    0.080  1     m , y  0.080  2  0.008 106 m
Then we can use Eq. (6.50)
F23y  325  10 750 y  14 000  y  0.0125  1.6  y 2  N
 500  3 250 y  6 317 y N
and Eqs. (6.48) and (6.52)
a tan   Y   Y 
T12  a tan  F23Y  Y F23Y
307
  t , deg
y, m
y , m/s
y , m/s2
F23Y , N
T12 , N·m
0
0
0
0.000 489
0.001 910
0.004 122
0.006 910
0.010 000
0.013 090
0.015 878
0.018 090
0.019 511
0.020 000
0.003 708
0.007 053
0.009 708
0.011 413
0.012 000
0.011 413
0.009 708
0.007 053
0.003 708
0
165
180
0.020 000
0.020 000
0
0
195
210
225
240
255
270
0.019 722
0.018 889
0.017 500
0.015 556
0.013 056
0.010 000
-0.002 122
-0.004 244
-0.006 366
-0.008 488
-0.010 610
-0.012 732
285
300
315
330
345
360
0.006 944
0.004 444
0.002 500
0.001 111
0.000 278
0
-0.010 610
-0.008 488
-0.006 366
-0.004 244
-0.002 122
0
551.2
591.0
588.1
579.8
566.9
550.6
532.5
514.4
498.1
485.2
476.9
474.0
565.0
565.0
565.0
513.8
512.9
510.2
505.7
499.4
491.2
481.3
583.7
573.8
565.6
559.3
554.8
552.1
551.2
591.0
0
15
30
45
60
75
90
105
120
135
150
0.008 106
0.014 400
0.013 695
0.011 650
0.008 464
0.004 450
0
-0.004 450
-0.008 464
-0.011 650
-0.013 695
-0.014 400
0
0
0
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.014 400
-2.181
-4.089
-5.503
-6.284
-6.390
-5.871
-4.836
-3.422
-1.768
0
0
0
1.088
2.165
3.219
4.239
5.212
6.128
7.430
6.088
4.801
3.561
2.355
1.172
0
308
6.44
Repeat Problem 6.43 with a speed of 900 rev/min, and F14  0.110  10.75Y kN, where Y
is in meters, and the coefficient of sliding friction is   0.025.
  900 rev/min  94.248 rad/s
For simple harmonic rise motion, we use Eqs. (6.12) with L  0.020 m and   150 .
For the first part of the parabolic return motion, following Example 6.1,
2
y  0.020 1  2     m , y    0.080    m , y   0.080  2  0.008 106 m


For the second part of the parabolic return motion,
2
y  0.040 1     m , y    0.080  1     m , y  0.080  2  0.008 106 m
Then we can use Eq. (6.50)
110  10 750 y  14 000 Y  0.0125  1.6 Y  2  N
y
F23 
1  1.666 667Y  0.033 333 tan  sgn Y 

285  24 750Y  14 212Y  N
1  1.666 667Y  0.033 333 tan  sgn Y 
and Eqs. (6.48) and (6.52)
a tan   Y   Y 
T12  a tan  F23Y   yF23Y
309
  t , deg
y, m
y , m/s
y , m/s2
F23Y , N
T12 , N·m
0
0
0
0.000 489
0.001 910
0.004 122
0.006 910
0.010 000
0.013 090
0.015 878
0.018 090
0.019 511
0.020 000
0.003 708
0.007 053
0.009 708
0.011 413
0.012 000
0.011 413
0.009 708
0.007 053
0.003 708
0
165
180
0.020 000
0.020 000
0
0
195
210
225
240
255
270
0.019 722
0.018 889
0.017 500
0.015 556
0.013 056
0.010 000
-0.002 122
-0.004 244
-0.006 366
-0.008 488
-0.010 610
-0.012 732
285
300
315
330
345
360
0.006 944
0.004 444
0.002 500
0.001 111
0.000 278
0
-0.010 610
-0.008 488
-0.006 366
-0.004 244
-0.002 122
0
400
490
498
509
521
533
545
556
565
572
576
575
780
780
780
665
660
641
608
562
529
428
664
586
522
471
433
410
400
490
0
15
30
45
60
75
90
105
120
135
150
0.008 106
0.014 400
0.013 695
0.011 650
0.008 464
0.004 450
0
-0.004 450
-0.008 464
-0.011 650
-0.013 695
-0.014 400
0
0
0
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.014 400
-1.85
-3.59
-5.05
-6.08
-6.54
-6.35
-5.49
-4.04
-2.13
0
0
0
1.40
2.72
3.87
4.77
5.61
5.45
8.45
6.22
4.43
3.00
1.84
0.87
0
310
6.45
A plate-cam drives a reciprocating roller follower through the distance L = 1.25 in with
parabolic motion in 120 of cam rotation, dwells for 30 , and returns with cycloidal
motion in 120 , followed by dwells for the remaining cam angle. The external load on
the follower is F14  36 lb during the rise and zero during the dwells and the return. In
the notation of Fig. 6.41, R = 3 in, r = 1 in, lB  6 in, lC  8 in, and k  150 lb/in. The
spring is assembled with a preload of 37.5 lb when the follower is at the bottom of its
stroke. The weight of the follower is 1.8 lb, and the cam velocity is 140 rad/s. Assuming
no friction, plot the displacement, the torque exerted on the cam by the shaft, and the
radial component of the contact force exerted by the roller against the cam surface for one
complete cycle of motion.
For 0    60 , we use Eqs. (6.5a) (6.5c) with L  1.250 in and   120 .
y  2.500    in , y  2.387    in , y  1.140 in
2
For 60    120 , we use Eqs. (6.6a) – (6.6c) with L  1.250 in and   120 .
2
y  1.250 1  2 1      in , y  2.387 1     in , y  1.140 in


For 150    270 , we use Eqs. (6.13) with L  1.250 in and   120 .
Then we can use Eq. (6.50)
F23Y  F14  37.5  150 y  91.378 y lb
and Eqs. (6.48) and (6.52)
a tan   Y   Y 
T12  a tan  F23Y  Y F23Y
311
  t , deg
y, m
y , m/s
y , m/s2
F23Y , N
T12 , N·m
0
0
0
0.039 063
0.156 250
0.351 563
0.625 000
0.298 416
0.596 831
0.895 247
1.193 662
75
90
105
120
0.898 438
1.093 750
1.210 938
1.250 000
0.895 247
0.596 831
0.298 416
0
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
1.250 000
1.250 000
1.234 424
1.136 444
0.921 924
0.625 000
0.328 076
0.113 556
0.015 576
0
0
0
0
0
0
0
0
0
-0.174 808
-0.596 831
-1.018 854
-1.193 662
-1.018 854
-0.596 831
-0.174 808
0
0
0
0
0
0
0
37.5
177.7
183.5
201.1
230.4
271.4
63.1
104.1
133.4
151.0
156.8
225.0
225.0
225.0
107.0
44.4
60.1
131.3
202.4
218.1
155.5
37.5
37.5
37.5
37.5
37.5
37.5
37.5
177.7
0
15
30
45
60
0
1.139 863
1.139 863
1.139 863
1.139 863
1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
0
0
0
-1.266 070
-1.790 493
-1.266 070
0
1.266 070
1.790 493
1.266 070
0
0
0
0
0
0
0
1.139 863
360
54.8
120.0
206.3
324.0
75.3
93.2
79.6
45.1
0
0
0
-18.7
-26.5
-61.2
-156.7
-206.2
-130.2
-27.2
0
0
0
0
0
0
0
312
6.46
Repeat Problem 6.45 if friction exists with   0.04 and the cycloidal return takes place
in 180.
For 0    60 , we use Eqs. (6.6a) – (6.6c) with L  1.250 in and   120 .
y  2.500    in , y  2.387    in , y  1.140 in
2
For 60    120 , we use Eqs. (6.6a) – (6.6c) with L  1.250 in and   120 .
2
y  1.250 1  2 1      in , y  2.387 1     in , y  1.140 in


For 150    330 , we use Eqs. (6.13) with L  1.250 in and   180 .
Then we can use Eq. (6.50)
F  37.5  150 y  91.378 y lb
F23Y  14
1   5.6 y  16.8 tan  sgn y
and Eqs. (6.48) and (6.52)
a tan   Y   Y 
T12  a tan  F23Y  Y F23Y
313
  t , deg
y, m
y , m/s
y , m/s2
F23Y , N
T12 , N·m
0
0
0
0.039 063
0.156 250
0.351 563
0.625 000
0.298 416
0.596 831
0.895 247
1.193 662
75
90
105
120
0.898 438
1.093 750
1.210 938
1.250 000
0.895 247
0.596 831
0.298 416
0
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
1.250 000
1.250 000
1.245 305
1.213 957
1.136 444
1.005 624
0.828 639
0.625 000
0.421 361
0.244 376
0.113 556
0.036 043
0.004 695
0
0
0
0
0
-0.053 307
-0.198 944
-0.397 887
-0.596 831
-0.742 468
-0.795 775
-0.742 468
-0.596 831
-0.397 887
-0.198 944
-0.053 307
0
0
0
38
178
185
204
235
278
65
135
152
152
157
225
225
188
188
157
136
127
127
133
139
139
129
106
75
38
38
38
178
0
15
30
45
60
0
1.139 863
1.139 863
1.139 863
1.139 863
1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
0
0
0
-0.397 887
-0.689 161
-0.795 775
-0.689 161
-0.397 887
0
0.397 887
0.689 161
0.795 775
0.689 161
0.397 887
0
0
0
1.139 863
55
122
211
332
77
95
80
45
0
0
0
-10
-31
-54
-76
-94
-106
-104
-83
-51
-21
-4
0
0
0
314
Page intentionally blank.
315
Chapter 7
Spur Gears
7.1
Find the diametral pitch of a pair of gears having 32 and 84 teeth, respectively, whose
center distance is 3.625 in.
N 2 N3 32  84


 3.625 in
2P 2P
2P
116 teeth
P
 16 teeth/in
2  3.625 in 
R2  R3 
7.2
7.3
Ans.
Find the number of teeth and the circular pitch of a 6-in pitch diameter gear whose
diametral pitch is 9 teeth/in.
N  P(2R)   9 teeth/in  6.0 in   54 teeth
Ans.
p   P    9 teeth/in   0.349 1 in/tooth
Ans.
Determine the module of a pair of gears having 18 and 40 teeth, respectively, whose
center distance is 58 mm.
mN 2 mN3 m 18  40 


 58.0 mm
2
2
2
2  58.0 mm 
m
 2.0 mm/tooth
58 teeth
R2  R3 
7.4
7.5
Ans.
Find the number of teeth and the circular pitch of a gear whose pitch diameter is 200 mm
if the module is 8 mm/tooth.
N  (2R) m   200.0 mm  8 mm/tooth   25 teeth
Ans.
p   m   8 mm/tooth   25.13 mm/tooth
Ans.
Find the diametral pitch and the pitch diameter of a 40-tooth gear whose circular pitch is
3.50 in/tooth.
P   p    3.500 in/tooth   0.897 6 teeth/in
Ans.
D  2R  N P   40 teeth   0.897 6 teeth/in   44.563 in
Ans.
316
7.6
7.7
7.8
7.9
The pitch diameters of a pair of mating gears are 3.50 in and 8.25 in, respectively. If the
diametral pitch is 16 teeth/in how many teeth are there on each gear?
N2  2PR2  PD2  16 teeth/in  3.500 in   56 teeth
Ans.
N3  2PR3  PD3  16 teeth/in 8.250 in   132 teeth
Ans.
Find the module and the pitch diameter of a gear whose circular pitch is 40 mm/tooth if
the gear has 36 teeth.
m  p    40 mm/tooth    12.732 mm/tooth
Ans.
D  2R  mN  12.732 mm/tooth  36 teeth   458.4 mm
Ans.
The pitch diameters of a pair of gears are 60 mm and 100 mm, respectively. If their
module is 2.5 mm/tooth, how many teeth are there on each gear?
N2  2R2 m  D2 m   60 mm   2.5 mm/tooth   24 teeth
Ans.
N3  2R3 m  D3 m  100 mm   2.5 mm/tooth   40 teeth
Ans.
What is the pitch diameter of a 33-tooth gear if its circular pitch is 0.875 in/tooth?
D  2R  pN    0.875 in/tooth  33 teeth    9.191 in
7.10
Ans.
A shaft carries a 30-tooth, 3-teeth/in diametral pitch gear that drives another gear at a
speed of 480 rev/min. How fast does the 30-tooth gear rotate if the shaft center distance
is 9 in?
R2  N2 2P   30 teeth  2  3 teeth/in   5.000 in
R3   R2  R3   R2  9.000 in  5.000 in  4.000 in
2 
7.11
R3
4.000 in
3 
 480 rev/min   384 rev/min
R2
5.000 in
Ans.
Two gears having an angular velocity ratio of 3:1 are mounted on shafts whose centers
are 136 mm apart. If the module of the gears is 4 mm/tooth, how many teeth are there on
each gear?
  
3
 R2  R3   1  2  R2  1   R2  4 R2  136 mm 
 1
 3 
R2  34.0 mm
R3   R3  R2   R2  102.0 mm
2 R2 2  34.0 mm 

 17 teeth
m
4 mm/tooth
2 R 2 102.0 mm 
N3  3 
 51 teeth
m
4 mm/tooth
N2 
Ans.
Ans.
317
7.12
A gear having a module of 4 mm/tooth and 21 teeth drives another gear at a speed of 240
rev/min. How fast is the 21-tooth gear rotating if the shaft center distance is 156 mm?
R2  mN2 2   4 mm/tooth  21 teeth  2  42 mm
R3   R2  R3   R2  156 mm   42 mm  114 mm
2 
7.13
R3
114 mm
3 
 240 rev/min   651.4 rev/min
R2
42 mm
Ans.
A 4-tooth/in diametral pitch, 24-tooth pinion is to drive a 36-tooth gear. The gears are
cut on the 20° full-depth involute system. Find and tabulate the addendum, dedendum,
clearance, circular pitch, base pitch, tooth thickness, pitch circle radii, base circle radii,
lengths of paths of approach and recess, and contact ratio.
a  1 P  1  4 teeth/in   0.250 in
Ans.
d  1.25 P  1.25  4 teeth/in   0.312 5 in
Ans.
c  d  a   0.312 5 in    0.250 in   0.062 5 in
Ans.
p   P    4 teeth/in   0.785 4 in/tooth
Ans.
pb  p cos    0.785 4 in/tooth  cos 20  0.738 0 in/tooth
Ans.
t  p 2   0.785 4 in/tooth  2  0.392 7 in
Ans.
R2 
N2
24 teeth

 3.000 in
2 P 2  4 teeth/in 
R3 
N3
36 teeth

 4.500 in
2 P 2  4 teeth/in 
r2  R2 cos    3.0 in  cos 20  2.819 in ; r3  R3 cos    4.5 in  cos 20  4.229 in
CP  0.625 in [measured or by Eq. (7.10)]
PD  0.591 in [measured or by Eq. (7.11)]
CP  PD  0.625 in    0.591 in 
mc 

 1.647 teeth avg.
pb
0.738 0 in/tooth
Ans.
Ans.
Ans.
Ans.
Ans.
318
7.14
A 5-tooth/in diametral pitch, 15-tooth pinion is to mate with a 30-tooth internal gear. The
gears are 20° full-depth involute. Make a drawing of the gears showing several teeth on
each gear. Can these gears be assembled in a radial direction? If not, what remedy
should be used?
Since the addendum circle of internal gear 3 is of lesser radius (2.800 in) than its base
circle (2.819 in), contact is initiated to the left of point A before proper involute contact is
possible. This is similar to undercutting but on an internal gear it is called fouling.
With this condition the involute curves of the internal gear are extended radially to meet
the addendum circle and this results in converging radii; therefore the gears cannot be
assembled in the radial direction.
Ans.
One remedy is to reduce the internal gear addendum to match the base circle radius.
However, the internal gear is then non-standard. A better remedy is to reduce the
diametral pitch to 4 teeth/in so that the addendum circle of the internal gear is 2.750 in.
319
7.15
A 2½-teeth/in diametral pitch 17-tooth pinion and a 50-tooth gear are paired. The gears
are cut on the 20° full-depth involute system. Find the angles of approach and recess of
each gear and the contact ratio.
a  1 P  1  2.5 teeth/in   0.400 in
p   P    2.5 teeth/in   1.256 6 in/tooth
pb  p cos   1.256 6 in/tooth  cos 20  1.180 9 in/tooth
R2 
R3 
N3
50 teeth

 10.000 in
2 P 2  2.5 teeth/in 
r2  R2 cos    3.4 in  cos 20  3.195 in
r3  R3 cos   10.0 in  cos 20  9.397 in
CP  1.036 in [Eq. (7.10)]
CP 1.036 in
2 

 0.324 rad  18.58
r2
3.195 in
PD 0.894 in
2 

 0.280 rad  16.04
r2
3.195 in
PD  0.894 in [Eq. (7.11)]
CP 1.036 in
3 

 0.110 rad  6.32 Ans.
r3
9.397 in
PD 0.894 in
3 

 0.095 rad  5.45 Ans.
r3
9.397 in
mc 
7.16
N2
17 teeth

 3.400 in
2 P 2  2.5 teeth/in 
CP  PD 1.036 in    0.894 in 

 1.63 teeth avg.
pb
1.180 9 in/tooth
A gearset with a module of 5 mm/tooth has involute teeth with 22½° pressure angle, and
has 19 and 31 teeth, respectively. They have 1.0m for the addendum and 1.25m for the
dedendum.* Tabulate the addendum, dedendum, clearance, circular pitch, base pitch,
tooth thickness, base circle radii, and contact ratio.
a  1.0m  5.0 mm
d  1.35m  1.35  5 mm   6.75 mm
c  d  a  1.75 mm
p   m    5 mm/tooth   15.708 mm/tooth
pb  p cos   15.708 mm/tooth  cos 22.5  14.512 mm/tooth
t  p 2  7.854 mm
R2  N2 m 2  19 teeth  5 mm/tooth  2  47.500 mm
R3  N3m 2   31 teeth  5 mm/tooth  2  77.500 mm
r2  R2 cos    47.500 mm  cos 22.5  43.884 mm
r3  R3 cos    77.500 mm  cos 22.5  71.601 mm
CP  11.325 mm [Eq. (7.10)]
PD  10.640 mm [Eq. (7.11)]
CP  PD 11.325 mm   10.640 mm 
mc 

 1.51 teeth avg.
pb
14.512 mm/tooth
*
Ans.
In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
320
7.17
A gear with a module of 8 mm/tooth and 22 teeth is in mesh with a rack; the pressure
angle is 25°. The addendum and dedendum are 1.0m and 1.25m, respectively.* Find the
lengths of the paths of approach and recess and determine the contact ratio.
a  1.0m  8.0 mm
p   m   8 mm/tooth   25.133 mm/tooth
pb  p cos    25.133 mm/tooth  cos 25  22.778 mm/tooth
R2  N2 m 2   22 teeth 8 mm/tooth  2  88.0 mm
CP  a sin   18.930 mm [Fig. 7.10]
PD  16.243 mm [Eq. (7.11)]
CP  PD 18.930 mm   16.243 mm 
mc 

 1.54 teeth avg.
pb
22.778 mm/tooth
7.18
Ans.
Ans.
Ans.
Repeat Problem 7.15 using the 25° full-depth system.
a  1 P  1  2.5 teeth/in   0.400 in
p   P    2.5 teeth/in   1.256 6 in/tooth
pb  p cos   1.256 6 in/tooth  cos 25  1.138 9 in/tooth
R2 
R3 
N3
50 teeth

 10.000 in
2 P 2  2.5 teeth/in 
r2  R2 cos    3.4 in  cos 25  3.081 in
r3  R3 cos   10.0 in  cos 25  9.063 in
CP  0.875 in [Eq. (7.10)]
CP 0.875 in
2 

 0.284 rad  16.27
r2
3.081 in
PD 0.787 in
2 

 0.255 rad  14.63
r2
3.081 in
PD  0.787 in [Eq. (7.11)]
CP 0.875 in
3 

 0.097 rad  5.53 Ans.
r3
9.063 in
PD 0.787 in
3 

 0.087 rad  4.97 Ans.
r3
9.063 in
mc 
*
N2
17 teeth

 3.400 in
2 P 2  2.5 teeth/in 
CP  PD  0.875 in    0.787 in 

 1.46 teeth avg.
pb
1.138 9 in/tooth
In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.
Ans.
321
7.19
Draw a 2-tooth/in diametral pitch, 26-tooth, 20° full-depth involute gear in mesh with a
rack. (a) Find the lengths of the paths of approach and recess and the contact ratio. (b)
Draw a second rack in mesh with the same gear but offset ⅛ in further away from the
gear center. Determine the new contact ratio. Has the pressure angle changed?
(a)
(b)
CP  a sin   0.500 in sin 20 1.462 in [see figure]
PD  1.196 in [Eq. (7.11)]
CP  PD 1.462 in   1.196 in 
mc 

 1.80 teeth avg.
pb
1.476 in/tooth
Ans.
Ans.
Ans.
Since the pressure angle is a property that has determined the shapes of the teeth
on both the rack and the pinion, moving the rack by 0.125 in does not change the
tooth shapes or the pressure angle. The modified contact ratio is:
Ans.
CP  a sin    0.375 in sin 20 1.096 in [see figure]
Ans.
PD  1.196 in [Eq. (7.11)]
C P  PD 1.096 in   1.196 in 
mc 

 1.55 teeth avg.
Ans.
pb
1.476 in/tooth
322
7.20 to 7.24 Shaper gear cutters have the advantage that they can be used for either external or
internal gears and also that only a small amount of runout is necessary at the end of the
stroke. The generating action of a pinion shaper cutter can easily be simulated by
employing a sheet of clear plastic. The figure illustrates one tooth of a 16-tooth pinion
cutter with 20° pressure angle as it can be cut from a plastic sheet. To construct the
cutter, lay out the tooth on a sheet of drawing paper. Be sure to include the clearance at
the top of the tooth. Draw radial lines through the pitch circle spaced at distances equal
to one fourth of the tooth thickness as illustrated in the figure. Next, fasten the sheet of
plastic to the drawing and scribe the cutout, the pitch circle, and the radial lines onto the
sheet. Then remove the sheet and trim the tooth outline with a razor blade. Then use a
small piece of fine sandpaper to remove any burrs.
To generate a gear with the cutter, only the pitch circle and the addendum circle need
be drawn. Divide the pitch circle into spaces equal to those used on the template and
construct radial lines through them. The tooth outlines are then obtained by rolling the
template pitch circle upon that of the gear and drawing the cutter tooth lightly for each
position. The resulting generated tooth upon the gear will be evident. The following
problems all employ a standard 1-tooth/in diametral pitch 20 full-depth template
constructed as described above. In each case you should generate a few teeth and
estimate the amount of undercutting
Prob. No.
7.20
7.21
7.22
7.23
7.24
No. of Teeth
10
12
14
20
36
The diagram used to make the plastic template for Probs. 7.20 through 7.24 is shown
above.
The drawing generated for
Prob. 7.20 is shown at left.
Note how the tip(s) of the
shaper cutter slightly cut
away the material at the flank
of the tooth so that the tooth
is a small amount narrower
here than at its thickest
radius. This is the meaning
of the term undercut.
Probs. 7.21 - 7.24 are similar.
323
7.25
A 10-mm/tooth module gear has 17 teeth, a 20° pressure angle, an addendum of 1.0m,
and a dedendum of 1.25m.* Find the thickness of the teeth at the base circle and at the
addendum circle. What is the pressure angle corresponding to the addendum circle?
At the pitch circle:
rp  R  mN 2  10 mm/tooth 17 teeth  2  85.0 mm
t p   R N   85.0 mm  17  15.708 mm
inv   inv 20  0.014 904
At the base circle:
r  rb  R cos   85.0 mm  cos 20  79.874 mm
inv   inv0  0.0
From Eq. (7.16)
t

t  2r  p  inv   inv  
 2R

 15.708 mm

t  2  79.874 mm  
 0.014 904  0.0  17.142 mm
 2 85.0 mm 

At the addendum circle:
 15.708 mm

ta  2  95.0 mm  
 0.014 904  0.071 844  6.737 mm
 2  85.0 mm 

ra  R  a  85.0 mm  10.0 mm  95.0 mm
a  cos1  rb ra   cos1  79.874 mm 95.0 mm   32.78
*
In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.
Ans.
Ans.
Ans.
324
7.26
A 15-tooth pinion has 1½-tooth/in diametral pitch 20° full-depth involute teeth.
Calculate the thickness of the teeth at the base circle. What are the tooth thickness and
the pressure angle at the addendum circle?
At the pitch circle:
rp  R  N 2P  15 teeth  2 1.5 tooth/in   5.000 in
t p   R N    5.000 in  15 teeth   1.047 in
inv   inv 20  0.014 904
At the base circle:
r  rb  R cos    5.000 in  cos 20  4.698 in
inv   inv0  0.0
From Eq. (7.16)
t

t  2r  p  inv   inv  
 2R

 1.047 in

t  2  4.698 in  
 0.014 904  0.0   1.124 in
 2  5.000 in 

At the addendum circle:
ra  R  a  5.000 in  0.667 in  5.667 in
Ans.
  cos1  rb ra   cos1  4.698 in 5.667 mm   33.99
Ans.
 1.047 in

t  2  5.667 in  
 0.014 904  0.081 018  0.437 5 in
 2  5.000 in 

Ans.
7.27 A tooth is 0.785 in thick at a pitch circle radius of 8 in and has a pressure angle of 25°.
What is the thickness at the base circle?
At the base circle:
r  rb  R cos   8.000 in  cos 25  7.250 in
inv   inv0  0.0
From Eq. (7.16)
t

t  2r  p  inv   inv  
 2R

 0.785 in

t  2  7.250 in  
 0.029 975  0.0  1.146 in
 2  8.000 in 

Ans.
325
7.28
A tooth is 1.571 in thick at the pitch radius of 16 in and has a pressure angle of 20°. At
what radius does the tooth become pointed?
t

t  2r  p  inv   inv    0
 2R

inv   t p 2R  inv   1.571 in 2 16.000 in   0.014 904  0.063 991
r  rb cos   16.000 in  cos 20 cos31.647  17.661 in
7.29
Ans.
A 25˚ full-depth involute, 12-tooth/in diametral pitch pinion has 18 teeth. Calculate the
tooth thickness at the base circle. What are the tooth thickness and pressure angle at the
addendum circle?
At the pitch circle:
rp  R  N 2P  18 teeth  2 12.0 tooth/in   0.750 in
t p   R N    0.750 in  18 teeth   0.130 9 in
inv   inv 25  0.029 975
At the base circle:
r  rb  R cos    0.750 in  cos 25  0.680 in
inv   inv0  0.0
t

t  2r  p  inv   inv  
 2R

 0.130 9 in

t  2  0.680 in  
 0.029 975  0.0  0.159 4 in
 2  0.750 in 

At the addendum circle:
ra  R  a  0.750 in  0.083 in  0.833 in
Ans.
  cos1  rb ra   cos1  0.680 in 0.833 in   35.35
Ans.
 0.130 9 in

t  2  0.833 in  
 0.029 975  0.092 339  0.041 5 in
 2  0.750 in 

Ans.
326
7.30 A nonstandard 10-tooth 8-tooth/in diametral pitch involute pinion is to be cut with a 22½˚
pressure angle. What maximum addendum can be used before the teeth become pointed?
At the pitch circle:
rp  R  N 2P  10 teeth  2 8.0 tooth/in   0.625 in
t p   R N    0.625 in  10 teeth   0.196 3 in
At the addendum circle:
t

t  2r  p  inv   inv    0
 2R

inv   t p 2R  inv   0.196 3 in 2  0.625 in   0.021 514  0.178 594 ;   42.772
r  rb cos    0.625 in  cos 22.5 cos 42.772  0.786 6 in
a  r  R  0.786 6 in  0.625 in  0.161 6 in
7.31
Ans.
The accuracy of cutting gear teeth can be measured by fitting hardened and ground pins in
diametrically opposite tooth spaces and measuring the distance over the pins. For a 10tooth/in diametral pitch 20˚ full-depth involute system 96 tooth gear:
(a) Calculate the pin diameter that will contact the teeth at the pitch lines if there is to be
no backlash.
(b) What should be the distance measured over the pins if the gears are cut accurately?
(a)
R  N 2P   96 teeth  2 10.0 tooth/in   4.800 in
rb  R cos    4.800 in  cos 20  4.510 5 in
  rb tan    4.510 5 in  tan 20  1.641 7 in
   2 N   2  96 teeth   0.016362 rad  0.937 5
s  rb tan         4.510 5 in  tan  20.937 5  1.641 7 in   0.084 1 in
d  2s  2  0.084 1 in   0.168 2 in
(b)
Ans.
rs  rb cos       4.510 5 in  cos20.9375=4.829 4 in
distance over pins  2  rs  s   2  4.829 4 in  0.084 1 in   9.827 0 in
Ans.
327
7.32
7.33
A set of interchangeable gears with 4-tooth/in diametral pitch is cut on the 20° full-depth
involute system. The gears have tooth numbers of 24, 32, 48, and 96. For each gear,
calculate the radius of curvature of the tooth profile at the pitch circle and at the
addendum circle.
N , teeth
R  N 2P, in
rb  R cos  , in
 p  rb tan  , in
24
32
48
96
3.000
4.000
6.000
12.000
2.819
3.759
5.638
11.276
1.026
1.368
2.052
4.104
N , teeth
ra , in
rb ra
a  cos1  rb ra  , deg
a  rb tan a , in
24
32
48
96
3.250
4.250
6.250
12.250
0.86741
0.88442
0.90211
0.92052
29.84
27.82
25.56
23.00
1.617
1.983
2.697
4.786
Calculate the contact ratio of a 17-tooth pinion that drives a 73-tooth gear. The gears are
96-tooth/in diametral pitch and cut on the 20° full-depth involute system.
a  1 P  1 96 teeth/in  0.010 4 in
pb   P  cos    96 teeth/in  cos 20  0.030 75 in/tooth
N
N2
17 teeth
73 teeth

 0.088 55 in R3  3 
 0.380 21 in
2 P 2  96 teeth/in 
2 P 2  96 teeth/in 
CP  0.027 88 in [measured or by Eq. (7.10)]
PD  0.023 29 in [measured or by Eq. (7.11)]
CD  0.027 88 in    0.023 29 in 
mc 

 1.664 teeth avg.
pb
0.030 75 in/tooth
R2 
7.34
Ans.
A 25° pressure angle 11-tooth pinion is to drive a 23-tooth gear. The gears have a
diametral pitch of 8 teeth/in and have stub teeth. What is the contact ratio?
a  0.8 P  0.8 8 teeth/in  0.100 in (Notice the stub teeth.)
pb   P  cos    8 teeth/in  cos 25  0.355 9 in/tooth
N
N2
11 teeth
23 teeth

 0.687 5 in
R3  3 
 1.437 5 in
2 P 2 8 teeth/in 
2 P 2 8 teeth/in 
CP  0.208 9 in [measured or by Eq. (7.10)]
PD  0.191 0 in [measured or by Eq. (7.11)]
CD  0.208 9 in    0.191 0 in 
mc 

 1.124 teeth avg.
pb
0.355 9 in/tooth
R2 
Ans.
328
7.35
A 22-tooth pinion mates with a 42-tooth gear. The gears have full-depth involute teeth,
have a diametral pitch of 16 teeth/in, and are cut with a 17½° pressure angle.* Find the
contact ratio.
a  1 P  1 16 teeth/in  0.062 5 in
pb   P  cos    16 teeth/in  cos17.5  0.187 3 in/tooth
N
N
22 teeth
42 teeth
R2  2 
 0.687 5 in R3  3 
 1.312 5 in
2 P 2 16 teeth/in 
2 P 2 16 teeth/in 
CP  0.174 3 in [measured or by Eq. (7.10)]
PD  0.157 4 in [measured or by Eq. (7.11)]
CD  0.174 3 in    0.157 4 in 
mc 

 1.771 teeth avg.
pb
0.187 3 in/tooth
7.36
Ans.
The center distance of two 24-tooth, 20 pressure angle, full-depth involute spur gears
with diametral pitch of 2 teeth/in is increased by 0.125 in over the standard distance. At
what pressure angle do the gears operate?
The original two gears are identical.
N
2P
24 teeth

2  2 teeth/in 
R2  R3 
 6.00 in
When the gear centers are separated to
a non-standard distance, the base
circles do not change. The line of
contact adjusts to remain tangent to
the new locations of the two base
circles. The pressure angle changes.
Since the two base circles do not
change,
r2  r3   6.00 in  cos 20
 5.638 in
From the figure we can see that the
shaft center distance is related to the
new pressure angle as follows
r
r r
r2
 3  2 3
cos   cos   cos  
r r
5.638 in  5.638 in
   cos1 2 3  cos1
 21.56
R2  R3
12.125 in
R2  R3 
*
Such gears came from an older standard and are now obsolete.
Ans.
329
7.37
The center distance of two 18-tooth, 25 pressure angle, full-depth involute spur gears
with diametral pitch of 3 teeth/in is increased by 0.0625 in over the standard distance. At
what pressure angle do the gears operate?
Consult the figure and the discussion with the solution of Prob. 7.36.
N
18 teeth
R2  R3 

 3.00 in
2 P 2  3 teeth/in 
r2  r3   3.00 in  cos 25  2.719 in
   cos1
7.38
r2  r3
2.719 in  2.719 in
 cos1
 26.24
R2  R3
6.062 5 in
Ans.
A pair of mating gears have 24 teeth/in diametral pitch and are generated on the 20° fulldepth involute system. If the tooth numbers are 15 and 50, what maximum addendums
may they have if interference is not to occur?
R2 
N2
15 teeth

 0.312 5 in
2 P 2  24 teeth/in 
R3 
N3
50 teeth

 1.042 in
2 P 2  24 teeth/in 
r2  R2 cos    0.312 5 in  cos 20  0.293 7 in
r3  R3 cos   1.042 in  cos 20  0.978 8 in
From Eq. (7.12), using Eqs. (7.10) and (7.11),
a2  r22   R2  R3  sin 2   R2  0.235 9 in
Ans.
a3  r32   R2  R3  sin 2   R3  0.041 2 in
Ans.
2
2
7.39
A set of gears is cut with a 4½-in/tooth circular pitch and a 17½˚ pressure angle.* The
pinion has 20 full-depth teeth. If the gear has 240 teeth, what maximum addendum may it
have to avoid interference?
P   p    4.500 in/tooth   0.698 1 teeth/in
R2 
N2
20 teeth

 14.324 in
2 P 2  0.698 1 teeth/in 
R3 
N3
240 teeth

 171.887 in
2 P 2  0.698 1 teeth/in 
r3  R3 cos   171.887 in  cos17.5  163.932 in
a3  r32   R2  R3  sin 2   R3  1.344 in
2
*
Such gears came from an older standard and are now obsolete.
Ans.
330
7.40
Using the method described for Problems 7.20 to 7.24, cut a 1-tooth/in diametral pitch
20° pressure angle full-depth involute rack tooth from a sheet of clear plastic. Use a
nonstandard clearance of 0.35/P in order to obtain a stronger fillet. This template can be
used to simulate the generating action of a hob. Now, using the variable-center-distance
system, generate an 11-tooth pinion to mesh with a 25-tooth gear without interference.
Record the values found for center distance, pitch radii, pressure angle, gear blank
diameters, cutter offset, and contact ratio. Note that more than one satisfactory solution
exists.
One solution may be found by the procedure shown in the numeric example in Sec. 7.11
under the heading Center-Distance Modification. It proceeds as follows:
  20 , P  1 tooth/in , p   P  3.141 6 in/tooth ,
a  1 P  1.000 in , d  1.35 P  1.350 in , c  0.35 P  0.350 in ,
N2  11 teeth ,
N3  25 teeth ,
N
N
11 teeth
25 teeth
R2  2 
 5.500 in
R3  3 
 12.500 in
2 P 2 1 teeth/in 
2 P 2 1 teeth/in 
r2  R2 cos   5.168 in
r3  R3 cos   11.746 in
e  a  R2 sin 2   0.356 6 in
t2  2e tan   p 2  1.830 4 in
t3  p 2  1.570 8 in
N 2  t2  t3   2 R2
 inv   0.022 115 rad ,
   22.70
2 R2  N 2  N3 
R cos 
R cos 
R2  2
 5.602 2 in
R3  3
 12.732 4 in
cos  
cos  
R2  R3  18.334 7 in
Working depth  1.978 1 in
R2  a2  6.834 7 in
R3  a3  13.478 1 in
CP  1.696 1 in
PD  2.310 4 in


mc  CD pb  1.36 teeth avg.
inv   
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
331
7.41
Using the template cut in Problem 7.40 generate an 11-tooth pinion to mesh with a 44tooth gear with the long-and-short-addendum system. Determine and record suitable
values for gear and pinion addendum and dedendum and for the cutter offset and contact
ratio. Compare the contact ratio with that of standard gears.
  20 , P  1 tooth/in , p   P  3.141 6 in/tooth ,
a  1 P  1.000 in , d  1.35 P  1.350 in , c  0.35 P  0.350 in ,
N2  11 teeth ,
N3  44 teeth ,
N
N
11 teeth
44 teeth
R2  2 
 5.500 in
R3  3 
 22.000 in
2 P 2 1 teeth/in 
2 P 2 1 teeth/in 
r2  R2 cos   5.168 in
r3  R3 cos   20.673 in
Since, with standard gears, point C is to the left of point A, there is interference and
undercutting. This problem can be eliminated using the long-and-short-addendum system
as shown in the figure above. Since the interference is near point C, we reduce the
addendum of the gear until point C  is coincident with point A. Combining Eqs. (7.10)
and (7.12) we can show that
a3  r32   R2  R3  sin 2   R3  0.712 3 in
2
Ans.
Since the working depth of standard gears is retained, the new dedendum of the gear is
d3  1 P  1.35 P  a3  1.637 7 in
Ans.
Then, retaining the same clearance and pitch point P,
a2  d3  0.35 P  1.287 7 in ,
d2  1 P  1.35 P  a2  1.062 3 in
Ans.
Assuming a standard rack cutter with a  1 P  1 in , Fig. 7.26 shows the offset is
e2  a  d2  0.062 3 in ,
e3  a  d3  0.637 7 in
Ans.
From Eqs. (7.9), (7.10), and (7.11) the contact ratio is
CP  1.881 1 in
PD  2.519 0 in
mc  CD pb  1.49 teeth avg.
Ans.
It is not easy to find a “contact ratio” for standard gears since these would have
undercutting over the range CC  . The distance CD is slightly less than the distance
CD, but has eliminated the interference.
332
7.42
A pair of involute spur gears with 9 and 36 teeth are to be cut with a 20 full-depth cutter
with diametral pitch of 3 teeth/in.
(a) Determine the amount that the addendum of the gear must be decreased to avoid
interference.
(b) If the addendum of the pinion is increased by the same amount, determine the contact
ratio.
  20 , P  3 tooth/in , p   P  1.047 2 in/tooth ,
N2  9 teeth ,
N
9 teeth
R2  2 
 1.500 in
2 P 2  3 teeth/in 
N3  36 teeth ,
N
36 teeth
R3  3 
 6.000 in
2 P 2  3 teeth/in 
r2  R2 cos   1.410 in
r3  R3 cos   5.638 in
(a) From Eqs. (7.10) and (7.12)
a3  r32   R2  R3  sin 2   R3  0.194 3 in
2
  1 P  a3  0.139 1 in
Ans.
a3  1 P    0.194 3 in
(b) a2  1 P    0.472 4 in
From Eqs. (7.9), (7.10), and (7.11) the contact ratio is
CP  0.513 0 in
PD  0.866 7 in
mc  CD p cos   1.40 teeth avg.
7.43
Ans.
A standard 20° pressure angle full-depth involute 1-tooth/in diametral pitch 20-tooth
pinion drives a 48-tooth gear. The speed of the pinion is 500 rev/min. Using the position
of the point of contact along the line of action as the abscissa, plot a curve indicating the
sliding velocity at all points of contact. Note that the sliding velocity changes sign when
the point of contact passes through the pitch point.
  20 , P  1 tooth/in , a  1 P  1.000 in , 2  500 rev/min  52.360 rad/s ,
N3  48 teeth ,
R2  N2 2P  10.000 in
R3  N3 2P  24.000 in
r2  R2 cos   9.397 in
r3  R3 cos   22.553 in
Defining X to be the distance from the point of contact to the pitch point along the line of
action, then, since this is the distance to the instant center, the sliding velocity at the point
VX 3 / 2  X 3  2   X  R2 R3  1 2  74.176 X in/s
of contact is
N2  20 teeth ,
and, using Eqs. (7.9) and (7.10), X varies between
X final  PD  2.298 in .
X init  CP  2.579 in
and
333
7.44
Find the first-order kinematic coefficient of the gear train. What are the speed and
direction of gear 8?
 N 2   N 4   N5   N7 
18 15 33 16 5





44 33 36 48 88
 N 3   N5   N 6   N8 
7.45
82   
Ans.
8  82 2   5 881 200 rev/min ccw   68.18 rev/min ccw
Ans.
For the given the pitch diameters of a set of spur gears forming a train, compute the firstorder kinematic coefficient of the train. Determine the speeds and directions of rotation
of gears 5 and 7.
R R
R2 R4 7 in 9 in
7
7 30 in 9 in 21


72  52 5 6 

R3 R5 15 in 30 in 50
R6 R7 50 9 in 16 in 80
5  52 2   7 50120 rev/min cw   16.80 rev/min cw
52 
7  72 2   21 80120 rev/min cw   31.50 rev/min cw
Ans.
Ans.
Ans.
334
7.46
Use the truck transmission of Figure 7.30 and an input speed of 3 000 rev/min to find the
drive shaft speed for each forward gear and for the reverse gear.
First gear:
92 
N 2 N6 17 17 289


N3 N9 43 43 1 849
9  92 2   289 1 849 3 000 rev/min   468.9 rev/min
Second gear:
82 
N 2 N5 17 27 153


N3 N8 43 33 473
8  82 2  153 473 3 000 rev/min   970.4 rev/min
Third gear:
72 
Ans.
N 2 N 4 17 36 51


N3 N7 43 24 86
7  72 2   51 86 3 000 rev/min   1 779.1 rev/min
Fourth gear:
22  1.0
2  22 2  1.0 3 000 rev/min   3 000 rev/min
Reverse gear:
Ans.
92  
Ans.
Ans.
N 2 N6 N11
17 17 22
3 179


N3 N10 N9
43 18 43
16 641
9  92 2   3 179 16 641 3 000 rev/min   573.1 rev/min
Ans.
335
7.47
Consider the gears in a speed-change gearbox used in machine tool applications. By
sliding the cluster gears on shafts B and C, nine speed changes can be obtained. The
problem of the machine tool designer is to select tooth numbers for the various gears to
produce a reasonable distribution of speeds for the output shaft. The smallest and largest
gears are gears 2 and 9, respectively. Using 20 and 45 teeth for these gears, determine a
set of suitable tooth numbers for the remaining gears. What are the corresponding speeds
of the output shaft? Note that the problem has many solutions.
We also set N6  20 teeth (minimum).
Ans.
Since the largest speed reduction will be obtained with gears 2-5-6-9,
N N
20 20
C ,min  2 6  A 
450 rev/min  137 rev/min
N5 N9
N5 45
From this we get N5  29.197 , and we choose N5  30 teeth.
Ans.
Next, using distance units of circular pitch, the distance between shafts B and C is
BC  N5  N8  N6  N9  N7  N10  65 teeth . N8  BC  N5  35 teeth .
Ans.
Similarly the distance between shafts A and B is
AB  N2  N5  N3  N6  N4  N7  50 teeth . N3  AB  N6  30 teeth.
Ans.
Since the minimum is 20 teeth and since AB  N4  N7  50 teeth we see that
20  N4 , N7  30 and we choose
N4  N7  25 teeth
Ans.
N10  BC  N7  40 teeth .
and, finally,
Ans.
With all tooth numbers known, we can now find the output shaft speed for each gear
arrangement. These are:
Arrangement First-order kinematic coefficient, CA

Output shaft speed, C , rev/min
2-5-5-8
0.571
257.1
2-5-6-9
0.296
133.3
2-5-7-10
0.416
187.5
3-6-5-8
1.285
578.6
3-6-6-9
0.667
300.0
3-6-7-10
0.938
421.9
4-7-5-8
0.857
385.7
4-7-6-9
0.444
200.0
4-7-7-10
0.625
281.3
336
7.48
If internal gear 7 rotates at 60 rev/min ccw, determine the speed and direction of rotation
of arm 3?
N 2 N 4 N6 20 teeth 40 teeth 36 teeth 20


N 4 N5 N7 40 teeth 18 teeth 154 teeth 77
  3 60 rev/min  3 20
  7
72/3


; 3  81.1 rev/min ccw
2  3
0  3
77
72 
7.49
Ans.
If the arm in Figure P7.48 rotates at 300 rev/min ccw, find the speed and direction of
rotation of internal gear 7.
N 2 N 4 N6 20 teeth 40 teeth 36 teeth 20


N 4 N5 N7 40 teeth 18 teeth 154 teeth 77
  3 7  300 rev/min 20
  7
72/3


;   222.1 rev/min ccw
2  3 0  300 rev/min 77 7
72 
Ans.
337
7.50
If shaft C is stationary and gear 2 rotates at 800 rev/min ccw, what are the speed and
direction of rotation of shaft B?
N2
3
18 teeth
3
3  32 2   800 rev/min ccw   600 rev/min cw


4
N3
24 teeth
4
N N
18 teeth 20 teeth 3
85  5 7 

N6 N8 42 teeth 40 teeth 14
  3 0  600 rev/min
3
  8
Ans.
85/3

 ; 3  2 200 rev/min cw
5  3 5  600 rev/min 14
32  
7.51
In Figure P7.50, shaft B is stationary and shaft C is driven at 380 rev/min ccw. Determine
the speed and direction of rotation of shaft A?
N5 N7 18 teeth 20 teeth 3


N6 N8 42 teeth 40 teeth 14
  3 380 rev/min  3 3
  8
85/3

 ; 3  483.6 rev/min ccw
5  3
0  3
14
N
24 teeth
A  2   3 3  
483.6 rev/min ccw  644.8 rev/min cw
N2
18 teeth
85 
7.52
Ans.
In Figure P7.50, determine the speed and direction of rotation of shaft C if: (a) shafts A
and B both rotate at 360 rev/min ccw; and (b) shaft A rotates at 360 rev/min cw and shaft
B rotates at 360 rev/min ccw.
N5 N7 18 teeth 20 teeth 3


N6 N8 42 teeth 40 teeth 14
N
18 teeth
3   2  2  
360 rev/min ccw  270 rev/min cw
N3
24 teeth
85 
(a)
8  3
8  270 rev/min
3 ;
8  135 rev/min cw Ans.


5  3 360 rev/min  270 rev/min 14
N
18 teeth
3   2  2  
360 rev/min cw  270 rev/min ccw
N3
24 teeth
  3
8  270 rev/min
3 ;   289.3 rev/min ccw Ans.
  8
85/3


8
5  3 360 rev/min  270 rev/min 14
 
85/3
(b)
338
7.53
Ggear 2 is connected to the input shaft and arm 3 is connected to the output shaft.
Determine the speed reduction. What is the sense of rotation of the output shaft? What
changes could be made in the train to produce the opposite sense of rotation for the output
shaft?
N 2 N 4 N5 20 teeth 28 teeth 16 teeth
5


N 4 N5 N6 28 teeth 16 teeth 108 teeth 27
  3 0  3
5
5
  6
3    2
62/3


22
2  3 2  3 27
The speed reduction is 17 22  77.3
Ans.
The sense of the output rotation is opposite to the input sense.
Ans.
The opposite sense of rotation for the output shaft can be produced by replacing gears 4
and 5 by a single 44-tooth gear.
Ans.
62 
339
7.54
The Lévai type-L train illustrated in Figure 7.38 has N2 = 16T, N4 = 19T, N5 = 17T, N6 =
24T, and N7 = 95T. Internal gear 7 is fixed. Find the speed and direction of rotation of
the arm if gear 2 is driven at 100 rev/min cw.
N 2 N 4 N6 16 teeth 19 teeth 24 teeth 384


N 4 N5 N7 19 teeth 17 teeth 95 teeth 1 615
  3
0  3
384
  7
3  31.19 rev/min ccw
;
72/3


2  3 100 rev/min  3 1 615
72 
7.55
Ans.
The Lévai type-A train of Figure 7.38 has N2 = 20T and N4 = 32T.
(a) If the module is 6 mm/tooth, find the number of teeth on gear 5 and the crank arm
radius.
(b) If gear 2 is fixed and internal gear 5 rotates at 10 rev/min ccw, find the speed and
direction of rotation of the arm.
(a) N5  N2  2 N4  20 teeth  2  32 teeth   84 teeth
R3  m  N2  N4  2   6 mm/tooth  20 teeth  32 teeth  2  156 mm
N 2 N 4 20 teeth 32 teeth
5


N 4 N5 32 teeth 84 teeth
21
  3 10 rev/min  3
5
  5
52/3


2  3
0  3
21
Ans.
Ans.
(b) 52 
3  8.08 rev/min ccw
Ans.
340
7.56
The figure illustrates a possible arrangement of gears in a lathe headstock. Shaft A is
driven by a motor at a speed of 720 rev/min. The three pinions can slide along shaft A to
yield the meshes 2 with 5, 3 with 6, or 4 with 8. The gears on shaft C can also slide to
mesh either 7 with 9 or 8 with 10. Shaft C is the mandrel shaft.
(a) Make a table demonstrating all possible gear arrangements, beginning with the
slowest speed for shaft C and ending with the highest, and enter in this table the
speeds of shafts B and C.
(b) If the gears all have a module of m = 5 mm/tooth, what must be the shaft center
distances?
N2 = 16T, N3 = 36T, N4 = 25T, N5 = 64T, N6 = 66T, N7 = 17T, N8 = 55T, N9 = 79T, and
N10 = 41T
(a)
(b)
Gears
B , rev/min
C , rev/min
2-5-7-9
180.0
38.7
4-8-7-9
327.3
70.4
3-6-7-9
589.1
126.8
2-5-8-10
180.0
241.5
4-8-8-10
3-6-8-10
327.3
589.1
439.0
790.2
AB  m  N2  N5  2   5 mm/tooth 16 teeth  64 teeth  2  200 mm
Ans.
BC  m  N7  N9  2   5 mm/tooth 17 teeth  79 teeth  2  240 mm
Ans.
341
7.57
If shaft A is the output connected to the arm, and shaft B is the input driving gear 2,
determine the speed ratio. Can you identify the Lévai type for this train?
N2 = 16T, N3 = 18T, N4 = 16T, N5 = 18T, and N6 = 50T
N 2 N3 N5
16 teeth 18 teeth 18 teeth
9


N3 N 4 N 6
18 teeth 16 teeth 50 teeth
25
  A 0  A
9
 A 6
 A   9 34  2
62/

 ;
2   A 2   A
25
This train is Lévai type F.
62  
7.58
In Problem 7.57, shaft B rotates at 100 rev/min cw.
gears 3 and 4 about their own axes.
Ans.
Ans.
Find the speed of shaft A and of
Step
Arm
2
3
4
Locked
+1
+1
+1
+1
Arm fixed
0
+25/9
-200/81
+25/9
Total
+1
+34/9
-119/81
+34/9
A    9 34 2   9 34100 rev/min cw   26.47 rev/min cw
5
+1
+25/9
+34/9
6
+1
-1
0
Ans.
3   200 81 9 25 2   8 9 100 rev/min cw   88.89 rev/min ccw
Ans.
4   25 9 9 25 2  1.0100 rev/min cw   100.0 rev/min cw
Ans.
342
7.59
In the clock mechanism, a pendulum on shaft A drives an anchor (see Fig. 1.12c). The
pendulum period is such that one tooth of the 30T escapement wheel on shaft B is released
every 2 s, causing shaft B to rotate once every minute. Note that the second (to the right)
64T gear is pivoted loosely on shaft D and is connected by a tubular shaft to the hour hand.
(a) Show that the train values are such that the minute hand rotates once every hour and
that the hour hand rotates once every 12 hours. (b) How many turns does the drum on
shaft F make every day?
(a)
B  1.0 rev/min
N B NC
8 teeth 8 teeth
1


NC N D 60 teeth 64 teeth 60
 B  1 60 rev/min
D   DB
tD  60 min/rev
 
 DB
N D N E  1  28 teeth 8 teeth
1
 

N E N H  60  42 teeth 64 teeth 720
720 min/rev
 B  1 720 rev/min
H   HB
tH 
 12 hr/rev
60 min/hr
Ans.
   DB

 HB
(b)
   DB

 FB
Ans.
N D  1  8 teeth
1
 

N F  60  96 teeth 720
 B  1 720 rev/min  60 min/hr  24 hr/day   2 rev/day
F   FB
Ans.
343
Chapter 8
Helical Gears, Bevel Gears, Worms, and Worm Gears
8.1
A pair of parallel-axis helical gears has 14½° normal pressure angle, diametral pitch of 6
teeth/in, and 45° helix angle. The pinion has 15 teeth, and the gear has 24 teeth.
Calculate the transverse and normal circular pitches, the normal diametral pitch, the pitch
radii, and the equivalent tooth numbers.
N2  15 teeth ,
Pt  6 teeth/in ,
pt   Pt  0.523 6 in/tooth ,
R3  N3  2Pt   2.000 in ,
Ans.
Ne 2  N2 cos   42.43 teeth ,
Ne3  N3 cos   67.88 teeth
Ans.
3
A pair of parallel-axis helical gears are cut with a 20° normal pressure angle and a 30°
helix angle. They have diametral pitch of 16 teeth/in and have 16 and 40 teeth,
respectively. Find the transverse pressure angle, the normal circular pitch, the axial pitch,
and the pitch radii of the equivalent spur gears.
N2  16 teeth ,
t  tan 1  tan n cos   22.796 ,
8.3
Ans.
Ans.
R2  N2  2Pt   1.250 in ,
3
8.2
N3  24 teeth ,
Pn  Pt cos  8.485 teeth/in ,
pn  pt cos  0.370 2 in/tooth ,
N3  40 teeth ,
Ans.
Pt  16 teeth/in ,
pt   Pt  0.196 4 in/tooth ,
pn  pt cos  0.170 0 in/tooth ,
Ans.
R2  N2  2Pt   0.500 in ,
px  pt tan  0.340 1 in/tooth ,
R3  N3  2Pt   1.250 in ,
Re 2  R2 cos2   0.667 in ,
Re3  R3 cos2   1.667 in
Ans.
A parallel-axis helical gear set is made with a 20° transverse pressure angle and a 35°
helix angle. The gears have diametral pitch of 10 teeth/in and have 15 and 25 teeth,
respectively. If the face width is 0.75 in, calculate the base helix angle and the axial
contact ratio.
 b  tan 1  tan cos t   33.34 ,
Ans.
Pt  10 teeth/in ,
mx  F tan pt  1.67 teeth avg
pt   Pt  0.314 2 in/tooth ,
Ans.
344
8.4
A pair of helical gears is to be cut for parallel shafts whose center distance is to be about
3.5 in to give a velocity ratio of approximately 1.8. The gears are to be cut with a
standard 20° pressure angle hob whose diametral pitch is 8 teeth/in. Using a helix angle
of 30°, determine the transverse values of the diametral and circular pitches and the tooth
numbers, pitch radii, and center distance.
2 3  R3 R2  1.8 ,
R3  1.8R2 ,
R2  R3  R2  1.8R2  2.8R2  3.5 in ,
Pn  8.000 teeth/in ,
R2  1.25 in, R3  2.25 in ,
Pt  Pn cos  6.928 teeth/in ,
N2  2R2 Pt  17.3 teeth ,
pt   Pt  0.453 5 in/tooth
N3  2R3 Pt  31.2 teeth ,
Therefore we will use N2  17 teeth and N3  31 teeth
R2  N2  2Pt   1.227 in ,
R3  N3  2Pt   2.237 in ,
R2  R3  3.464 in
8.5
Ans.
Ans.
Ans.
Ans.
A 16-tooth helical pinion is to run at 1800 rev/min and drive a helical gear on a parallel
shaft at 400 rev/min. The centers of the shafts are to be spaced 11 in apart. Using a helix
angle of 23° and a pressure angle of 20°, determine the values for the tooth numbers,
pitch radii, normal circular pitch and diametral pitch, and the face width.
2 3  R3 R2  1800 400  4.5
R2  R3  R2  4.5R2  5.5R2  11.0 in ,
R3  4.5R2
R2  2.000 in, R3  9.000 in ,
Ans.
N2  16 teeth ,
N3   R3 R2  N2  4.5N2  72 teeth
Ans.
Pt  N2  2R2   4.000 teeth/in ,
Pn  Pt cos  4.345 teeth/in ,
Ans.
pt   Pt  0.785 4 in/tooth ,
pn   Pn  0.723 0 in/tooth ,
Ans.
px  pt tan  1.850 3 in/tooth
Therefore, we may choose F  3.750 in .
F   2 teeth  px  3.7 in
Ans.
345
8.6
The catalog description of a pair of helical gears is as follows: 14½° normal pressure
angle, 45° helix angle, diametral pitch of 8 teeth/in, 1.0-in face width, and normal
diametral pitch of 11.31 teeth/in. The pinion has 12 teeth and a 1.500-in pitch diameter,
and the gear has 32 teeth and a 4.000-in pitch diameter. Both gears have full-depth teeth,
and they may be purchased either right- or left-handed. If a right-hand pinion and lefthand gear are placed in mesh, find the transverse contact ratio, the normal contact ratio,
the axial contact ratio, and the total contact ratio.
R2  1.500 in  2  0.750 in ,
R3   4.000 in  2  2.000 in ,
a  1 Pt  0.125 in ,
pt   Pt  0.392 7 in/tooth ,
tan t  tan n cos  0.365 7 ,
t  tan 1  0.365 7   20.09  20.00 ,
pb  pt cos t  0.369 0 in/tooth ,
r2  R2 cos t  0.704 8 in ,
CP  0.307 7 in [Eq. (7.10)],
mt  CD pb  1.54 teeth avg ,
 b  tan 1  tan cos t   43.22 ,
mx  tan pt  2.55 teeth avg ,
8.7
r3  R3 cos t  1.879 4 in
PD  0.262 1 in [Eq. (7.11)],
Ans.
mn  mt cos2  b  2.91 teeth avg
Ans.
m  mx  mt  4.09 teeth avg
Ans.
In a medium-size truck transmission a 22-tooth clutch-stem gear meshes continuously
with a 41-tooth countershaft gear. The data indicate normal diametral pitch of 7.6
teeth/in, 18½° normal pressure angle, 23½° helix angle, and a 1.12-in face width. The
clutch-stem gear is cut with a left-hand helix, and the countershaft gear is cut with a
right-hand helix. Determine the normal and total contact ratios if the teeth are cut fulldepth with respect to the normal diametral pitch.
tan t  tan n cos  0.364 9 ,
t  tan 1  0.364 9  20.04  20.00
Pn  7.600 teeth/in ,
a  1 Pn  0.131 6 in ,
Pt  Pn cos  6.970 teeth/in ,
pt   Pt  0.450 8 in/tooth ,
R2  N2  2Pt   1.578 in ,
R3  N3  2Pt   2.941 in ,
r2  R2 cos t  1.483 in ,
CP  0.336 9 in [Eq. (7.10)],
pb  pt cos t  0.423 6 in/tooth ,
r3  R3 cos t  2.764 in
PD  0.262 1 in [Eq. (7.11)],
mt  CD pb  1.53 teeth avg ,
 b  tan 1  tan cos t   22.22 ,
mx  F tan pt  1.08 teeth avg ,
mn  mt cos2  b  1.79 teeth avg
Ans.
m  mx  mt  2.87 teeth avg
Ans.
346
8.8
A helical pinion is right-handed, has 12 teeth, has a 60° helix angle, and is to drive
another gear at a velocity ratio of 3.0. The shafts are at a 90° angle, and the normal
diametral pitch of the gears is 8 teeth/in. Find the helix angle and the number of teeth on
the mating gear. What is the shaft center distance?
 3    2  30 RH ,
N3  2 3  N2  36 teeth
R2  N2  2Pn cos 2   1.500 in ,
R3  N3  2Pn cos 3   2.598 in
R2  R3  4.098 in
8.9
8.10
Ans.
Ans.
A right-hand helical pinion is to drive a gear at a shaft angle of 90°. The pinion has 6
teeth and a 75° helix angle and is to drive the gear at a velocity ratio of 6.5. The normal
diametral pitch of the gear is 12 teeth/in. Calculate the helix angle and the number of
teeth on the mating gear. Also determine the pitch radius of each gear.
 3    2  15 RH ,
N3  2 3  N2  39 teeth
Ans.
R2  N2  2Pn cos 2   0.966 in ,
R3  N3  2Pn cos 3   1.682 in
Ans.
Gear 2 rotates clockwise and drives gear 3 counterclockwise at a velocity ratio of 2:1.
Use a normal diametral pitch of 5 teeth/in, a shaft angle of 50°, a shaft center distance of
about 10 in, and the same helix angle for both gears. Find the tooth numbers, the helix
angles, and the exact shaft center distance.
 2   3   2  25 ,
2 3  R3 R2  2.0 , R3  2.0R2 ,
Ans.
R2  R3  R2  2.0R2  3.0R2  10.0 in ,
R2  3.33 in, R3  6.66 in ,
N2  2Pn cos 2 R2  30.18 in ,
N3  2Pn cos 3 R3  60.36 in
Therefore we choose
N2  30 teeth ,
N3  60 teeth ,
Ans.
Ans.
R2  N2  2Pn cos 2   3.310 in ,
R3  N3  2Pn cos 3   6.620 in ,
R2  R3  9.930 in
Ans.
347
8.11
8.12
A pair of straight-tooth bevel gears are to be manufactured for a shaft angle of 90°. If the
driver is to have 18 teeth and the velocity ratio is to be 3:1, what are the pitch angles?
N2  18 teeth ,
N3  2 3  N2  3N2  54 teeth ,
 2  tan 1  N2 N3   18.43 ,
 3  90   2  71.57
A pair of straight-tooth bevel gears has a velocity ratio of 1.5 and a shaft angle of 75°.
What are the pitch angles?


sin 
  28.78 ,
  2 3   cos  
 2  tan 1 
8.13
 3  75   2  46.22
Ans.
A pair of straight-tooth bevel gears is to be mounted at a shaft angle of 120°. The pinion
and gear are to have 15 and 33 teeth, respectively. What are the pitch angles?


sin 
  27.00 ,  3  120   2  93.00
  N3 N 2   cos  
 2  tan 1 
8.14
Ans.
Ans.
A pair of straight-tooth bevel gears with diametral pitch of 2 teeth/in have 19 teeth and 28
teeth, respectively. The shaft angle is 90°. Determine the pitch diameters, pitch angles,
addendum, dedendum, face width, and pitch radii of the equivalent spur gears.
R2  N2  2P   4.750 in ,
R3  N3  2P   7.000 in ,
D2  2 R2  9.500 in,
D3  2 R3  14.000 in,
Ans.
 2  tan
 3  90   2  55.84 ,
Ans.
1
 N2 N3   34.16 ,
Using Table 8.2: m90  mG  N3 N2  1.474 , a3  0.376 in ,
d3  Whole depth  a3  0.718 in
Whole depth  2.188 2  1.094 in ,
Working depth  2.0 P  1.000 in ,
a2  Working depth  a3  0.624 in
c  0.188 P  0.002 in  0.096 in
d2  Whole depth  a2  0.470 in
Cone distance,  R2 sin  2  8.459 in
Re 2  R2 cos  2  5.740 in
Let F  0.3  2.538 in , say F = 2.5 in
Re3  R3 cos  3  12.467 in
Ans.
Ans.
Ans.
Ans.
Ans.
348
8.15
A pair of straight-tooth bevel gears with diametral pitch of 8 teeth/in have 17 teeth and 28
teeth, respectively, and a shaft angle of 105°. For each gear, calculate the pitch radius,
pitch angle, addendum, dedendum, face width, and equivalent number of teeth. Make a
sketch of the two gears in mesh. Use standard tooth proportions as for a 90° shaft angle.
R2  N2  2P   1.063 in ,
R3  N3  2P   1.750 in ,


sin 
  34.83 ,  3  120   2  70.17
N
N

cos



 3 2

8.16
Ans.
 2  tan 1 
Ans.
Using Table 8.2: mG  N3 N2  1.647, m90  1.996 , a3  0.0819 in ,
d3  Whole depth  a3  0.191 6 in
Whole depth  2.188 8  0.273 5 in ,
Working depth  2.0 P  0.250 0 in , c  0.188 P  0.002 in  0.025 5 in
a2  Working depth  a3  0.168 1 in
d2  Whole depth  a2  0.105 4 in
Cone distance,  R2 sin  2  1.860 in Let F  0.3  0.558 in , say F = 0.563 in
Re 2  R2 cos  2  1.294 in
Re3  R3 cos  3  5.159 in
Ne 2  2PRe 2  20.71 teeth ,
Ne3  2PRe3  82.54 teeth
Ans.
Ans.
Ans.
Ans.
Ans.
A worm having 4 teeth and a lead of 1.0 in drives a worm gear at a velocity ratio of 7.5.
Determine the pitch diameters of the worm and worm gear for a center distance of 1.75
in.
px 
N2  1 in 4 teeth  0.25 in/tooth
N3  2 3  N2  7.5  4 teeth   30 teeth
R3  N3 px 2  1.194 in
R2  1.75  R3  0.556 in
Ans.
349
8.17
8.18
Specify a suitable worm and worm gear combination for a velocity ratio of 60 and a
center distance of 6.50 in. Use an axial pitch of 0.500 in/tooth.
Use N2  1 tooth ,
N3  2 3  N2  60 1 teeth   60 teeth
R3  N3 px 2  4.775 in
R2  6.500  R3  1.725 in
A triple-threaded worm drives a worm gear having 40 teeth. The axial pitch is 1.25 in,
and the pitch diameter of the worm is 1.75 in. Calculate the lead and lead angle of the
worm. Find the helix angle and pitch diameter of the worm gear.
 N2 px  3 teeth 1.25 in/tooth   3.750 in
  tan 1  2 R2   tan 1  3.750 in  1.75 in   34.298
    34.298
R3  N3 px 2  40 teeth 1.25 in/tooth  2  7.957 in
8.19
Ans.
Ans.
Ans.
Ans.
Ans.
A triple-threaded worm with a lead angle of 20° and an axial pitch of 0.400 in/tooth
drives a worm gear with a velocity reduction of 15 to 1. Determine the following for the
worm gear: (a) the number of teeth, (b) the pitch radius, (c) the helix angle, (d) the pitch
radius of the worm, and (e) the center distance.
 N2 px  3 teeth  0.400 in/tooth   1.200 in
 2 tan    1.200 in 2 tan 20  0.524 7 in
R3  2 3  R2  15  0.524 7 in   7.871 in
R2 
N3  2 R3 px  123.6 teeth
    20.0
R2  R3  0.525 in  7.871 in  8.396 in
Ans.
Ans.
Ans.
Ans.
Ans.
350
8.20
The gear train consists of bevel gears, spur gears, and a worm and worm gear. The bevel
pinion is mounted on a shaft that is driven by a V-belt on pulleys. If pulley 2 rotates at
1200 rev/min in the direction indicated, find the speed and direction of rotation of gear 9.
R2 N 4 N6 N8 6 in 18 20 3
3


R3 N5 N7 N9 10 in 38 48 36 304
9  92 2   3 3041 200 rev/min   11.84 rev/min cw
92 
Ans.
351
8.21
The marine reduction differential has bevel gear 2 driven by the engine shaft A. Bevel
planets 3 mesh with fixed crown gear 4 and are pivoted on the spider (arm), which is
connected to propeller shaft B. Find the percentage speed reduction.
N2 = 36T, N3 = 21T, N4 = 52T; crown gear 4 is fixed
N 2 N3
36 teeth 21 teeth
9


N3 N 4
21 teeth 52 teeth
13
  B 0  B
9
 B 4
 B   9 22  2
42/

 ;
2  B 2  B
13
Speed reduction to 9/22 = 40.9% is speed reduction of 59.1%.
42  
Ans.
352
8.22
The tooth numbers for the automotive differential illustrated in Figure 8.28 are N2 = 17T,
N3 = 54T, N4 = 11T, and N5 = N6 = 16T. The drive shaft turns at 1200 rev/min. What is
the speed of the right wheel if it is jacked up and the left wheel is resting on the road
surface?
N2
17 teeth
2 
1 200 rev/min   377.8 rev/min
N3
54 teeth
N N
16 teeth 11 teeth
65   5 4  
 1
N4 N6
11 teeth 16 teeth
  3 6  377.8 rev/min
  6
65/3

 1 6  755.6 rev/min
5  3 0  377.8 rev/min
3 
Ans.
353
8.23
A vehicle using the differential illustrated in Figure 8.28 turns to the right at a speed of 30
mi/h on a curve of 80-ft radius. Use the same tooth numbers as in Problem 8.22. The
tire diameter is 15 in. Use 60 in as the distance between treads. Calculate the speed of
each rear wheel and the speed of the ring gear.
v
car  car 

 30 mi/h  5 280 ft/mi   60 min/h   33.0 rad/min
80 ft 
For the right and left wheels, respectively:
 33.0 rad/min 80 ft 12 in/ft  30 in   651.26 rev/min
v
6  R 
r
 2 rad/rev 15.0 2 in 
v
r
5  L 
 33.0 rad/min 80 ft 12 in/ft  30 in   693.27 rev/min
 2 rad/rev 15.0 2 in 
N5 N 4
16 teeth 11 teeth

 1
N4 N6
11 teeth 16 teeth
  3 651.3 rev/min  3
  6
65/3

 1
5  3 693.2 rev/min  3
Ans.
Ans.
65  
3  672.27 rev/min
Ans.
354
Page intentionally blank.
355
Chapter 9
Synthesis of Linkages
9.1
A function varies from 0 to 10. Find the Chebyshev spacing for two, three, four, five,
and six precision positions.
With x0  0.0 and xN 1  10.0 , Eq. (9.22) becomes:
x j  5.0  5.0cos
j\N
1
2
3
4
5
6
2
1.46447
8.53553
3
0.66987
5.00000
9.33013
(2 j  1)
2N
4
0.38060
3.08658
6.91342
9.61940
j  1, 2,..., N
5
0.24472
2.06107
5.00000
7.93893
9.75528
6
0.17037
1.46447
3.70591
6.29409
8.53553
9.82963
356
9.2
Determine the link lengths of a slider-crank linkage to have a stroke of 600 mm and an
advance to return ratio of 1.20.
After laying out the distance B1B2 = 600 mm, we see that the advance to return ratio is
Q  180    180     1.20
and, from this, our design must have   16.40. Therefore we construct the point C
such that the central angle B1CB2  2  32.80. Using this point C as the center of a
circle ensures that any point O2 on this circle will have the angle
B1O2 B2    16.40 and thus will be a possible solution point. One typical solution
uses the point O2 shown.
Choosing the point O2 shown we measure the distances r1  500.000 mm,
O2 B1  r3  r2  1 324.956 mm and O2 B2  r3  r2  801.946 mm and from these we find
r2  261.50 mm
r3  1 063.45 mm
Ans.
Ans.
357
9.3
Determine a set of link lengths for a slider-crank linkage such that the stroke is 16 in and
the advance to return ratio is 1.25.
After laying out the distance B1B2 = 16.00 in, we see that the advance to return ratio is
Q  180    180     1.25 and, from this, our design must have   20.00.
Therefore we construct the point C such that the central angle B1CB2  2  40.00.
Using this point C as the center of a circle ensures that any point O2 on this circle will
have the angle B1O2 B2    20.00 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the distances r1  10.00 in,
O2 B1  r3  r2  29.82 in and O2 B2  r3  r2  15.69 in and from these we find
r2  7.06 in
r3  22.75 in
Ans.
Ans.
358
9.4
The rocker of a crank-rocker linkage is to have a length of 500 mm and swing through a
total angle of 45 with an advance to return radio of 1.25. Determine a suitable set of
dimensions for r1 , r2 , and r3 .
After laying out the angle B1O4 B2    45 with BO4 = 500 mm, we see that the
advance to return ratio is Q  180    180     1.25 and, from this, we find that
  20.00.
Therefore we construct the point C such that the central angle B1CB2  2  40.00.
Then, using this point C as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    20.00 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4  556 mm ,
O2 B1  r3  r2  878 mm , and O2 B2  r3  r2  588 mm and from these we find
r1  556 mm
r2  145 mm
r3  733 mm
Ans.
Ans.
Ans.
359
9.5
A crank-rocker linkage is to have a rocker 6 ft in length and a rocking angle of 75 . If
the advance to return ratio is to be 1.32, what are a suitable set of link lengths for the
remaining three links?
After laying out the angle B1O4 B2    75 with BO4 = 6.00 ft = 72.00 in, we see that
the advance to return ratio is Q  180    180     1.32 and, from this, we find
that   24.83.
Therefore we construct the point C such that the central angle B1CB2  2  49.66.
Then, using this point C as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    24.83 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4  97.37 in ,
O2 B1  r3  r2  152.31 in , and O2 B2  r3  r2  76.29 in and from these we find
r1  97.37 in
r2  38.01 in
r3  114.30 in
Ans.
Ans.
Ans.
360
9.6
Design a crank and coupler to drive rocker 4 such that slider 6 will reciprocate through a
distance of 16 in with an advance to return ratio of 1.20. Use a  r4  16 in and r5  24
in with r4 vertical at midstroke. Record the location of O2 and dimensions r2 and r3 .
After laying out the angle
B1O4 B2    60 with BO4 = 16.00 in, we see that the
advance to return ratio is Q  180    180     1.20 and, from this, we find that
  16.36.
Therefore we construct the point D such that the central angle B1DB2  2  32.72.
Then, using this point D as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    16.36 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4  25.04 in ,
O2 B1  r3  r2  35.83 in , and O2 B2  r3  r2  21.96 in and from these we find
r1  25.04 in
r2  6.93 in
Ans.
Ans.
r3  28.90 in
Ans.
361
9.7
Design a crank and rocker for a six-bar linkage such that the slider in Figure P9.6
reciprocates a distance of 800 mm with an advance to return ratio of 1.12; use
a  r4  1 200 mm and r5  1 800 mm. Locate O4 such that rocker 4 is vertical when the
slider is at midstroke. Find suitable coordinates for O2 and lengths for r2 and r3 .
After laying out the angle
B1O4 B2    2sin 1  400 1200   38.94 with BO4 = 1 200
mm, we see that the advance to return ratio is Q  180    180     1.12 and, from
this, we find that   10.19.
Therefore we construct the point D such that the central angle B1DB2  2  20.38.
Then, using this point D as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    10.19 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4  1 979 mm ,
O2 B1  r3  r2  2 634 mm , and O2 B2  r3  r2  1 942 mm and from these we find
r1  1 979 mm
r2  346 mm
r3  2 288 mm
Ans.
Ans.
Ans.
362
9.8
Two postures of a folding seat used in the aisles of buses to accommodate extra
passengers are shown. Design a four-bar linkage to support the seat so that it will lock in
the open posture and fold to a stable closing posture along the side of the aisle.
The open position is a toggle position with no force tending to open or close the 3-4-5 triangle. Thus a small catch only allowing joint A to rotate very slightly past the 180°
position at A2 will keep the seat open.
9.9
Design a spring-operated four-bar linkage to support a heavy lid like the trunk lid of an
automobile. The lid is to swing through an angle of 80 from the closed to the open
posture. The springs are to be mounted so that the lid will be held closed against a stop,
and they should also hold the lid in a stable open posture without the use of a stop.
One
typical
solution
has
O2 A  AB  O4 B  O2O4 . A stop for the
closed posture may be provided with
point B slightly below the line O4 A .
The open posture is held stable by the
choice of the spring free length.
363
9.10
Ssynthesize a linkage to move AB from posture 1 to posture 2 and return.
9.11
Synthesize a linkage to move AB successively through postures 1, 2, and 3.
364
9.12 to 9.21* The figure illustrates a function-generator linkage in which the motion of rocker 2
corresponds to x and the motion of rocker 4 to the function y = f (x). Use four precision
points with Chebyshev spacing and synthesize a linkage to generate the functions in the
table. Plot a curve of the desired function and a curve of the actual function that the
linkage generates. Compute the maximum error between them in percent.
Prob. No.
Function,
y  f  x
Range
x0  0 ,
deg
y0 0 ,
deg
r2 r1
r3 r1
r4 r1
Max.
Error, deg
9.12, 9.22
log10 x
1 x  2
52.628
259.077
-3.352
0.845
3.485
0.0037
9.13, 9.23
sin x
0  x  2
-62.263
75.606
1.834
2.238
-0.693
0.1900
9.14, 9.24
tan x
0  x  4
269.709
124.189
-2.660
7.430
8.685
0.0380
9.15, 9.25
ex
0  x 1
241.644
40.422
-3.499
0.878
3.399
0.0258
9.16, 9.26
1x
1 x  2
33.804
120.213
-0.385
1.030
0.384
0.0161
9.17, 9.27
x1.5
0  x 1
-5.171
211.689
0.625
1.309
-0.401
0.1460
9.18, 9.28
x2
0  x 1
-29.321
233.836
2.523
3.329
-0.556
0.0673
9.19, 9.29
x 2.5
0  x 1
-88.313
44.492
-1.801
0.908
1.274
0.4120
9.20, 9.30
x3
0  x 1
-85.921
37.637
-1.606
0.925
1.107
0.5095
9.21, 9.31
x2
1  x  1
-21.180
-53.670
-0.610
0.565
0.380
2.3400
9.22 to 9.31 Repeat Probs. 9.12 through 9.21 using the overlay method.
The overlay method can be used to confirm the above solutions. Other nearby solutions
are also possible but are too numerous to display here.
*
Solutions for these problems were among the earliest computer work in kinematic synthesis and results are reported
in F. Freudenstein, 1958. “Four-bar Function Generators,” Machine Design, 30 (24) pp. 119-23.
365
9.32
The figure illustrates a coupler curve generated by a four-bar linkage (not shown). Link 5
is to be attached to the coupler point, and link 6 is to be a rotating member with O6 as the
frame connection. In this problem we wish to find a coupler curve from the Hrones and
Nelson atlas [14] or by precision postures such that, for an appreciable distance, point C
moves through an arc of a circle. Link 5 is then proportioned so that D lies at the center
of curvature of this arc. The result is then called a hesitation motion because link 6
hesitates in its rotation for the period during which point C traverses the approximate
circular arc. Make a drawing of the complete linkage and plot the first-order kinematic
coefficient of link 6 for 360 of displacement of the input link.
This coupler curve for point C was found from the Hrones and Nelson atlas, page 150.
The hesitation is shown by the following plot of the first-order kinematic coefficient 6 .
366
9.33
Synthesize a four-bar linkage to obtain a coupler curve having an approximate straightline segment. Then, using the suggestion included in Figure 9.42b or Figure 9.44b,
synthesize a dwell motion. Using an input crank angular velocity of unity, plot the firstorder kinematic coefficient 6 of rocker 6 versus the input crank displacement.
The Hrones and Nelson atlas [14] contains a wide variety of coupler curves similar to the
one shown; this one is from page 93.
The dwell in the rotation of link 6 is shown by the following plot of the first-order
kinematic coefficient 6 .
367
9.34
Synthesize a dwell mechanism using the idea suggested in Fig. 9.42a and the Hrones and
Nelson atlas [14]. Rocker 6 is to have a total angular displacement of 60 . Using this
displacement as the abscissa, plot the first-order kinematic coefficient 6 of the motion of
the rocker to illustrate the dwell motion.
The Hrones and Nelson atlas contains a wide variety of coupler curves similar to the one
shown here; this one is from page 93.
The dwell in the rotation of link 6 is shown by the following plot of the first-order
kinematic coefficient 6 .
368
Page intentionally blank.
369
Chapter 10
Spatial Mechanisms and Robotics
10.1
Use the Kutzbach criterion to determine the mobility of the SSC linkage. Identify any
idle freedoms and state how they can be removed. What is the nature of the path
described by point B?
RBA  RO3O  75 mm, RBO3  150 mm, and 2 = 30
n  3 , j2  3 , j3  1 , j1  j4  j5  0
m  6  3  1  4 1  3  2   2
Ans.
There is one idle freedom, the rotation of link 3 about its own axis. This idle freedom may
be eliminated by employing a two-freedom pair, such as a universal joint, in place of one of
the two spheric pairs, either at B or at O3.
Ans.
The path described by point B is the curve of intersection of a cylinder of radius BA about
the y axis and a sphere of radius BO3 centered at O3.
Ans.
370
10.2
For the SSC linkage illustrated in Fig. P10.1 express the posture of each link in vector
form.
RO3O  75ˆi mm , R BA  75cos 2ˆi  75sin  2kˆ  64.952ˆi  37.500kˆ mm ,
R  R ˆj mm , R  150 mm .
AO
Ans.
BO3
AO
Substituting these into RO3O  R BO3  R AO  R BA gives
R AO  11 250  cos 2  1ˆj  144.889ˆj mm ,
R BO3  75  cos  2  1 ˆi  11 250  cos  2  1ˆj  75sin  2kˆ
 10.048ˆi  144.889ˆj  37.500kˆ mm
10.3
Ans.
Ans.
For the linkage of Figure P10.1 with VA  50ˆj mm/s , use vector analysis to find the
angular velocities of links 2 and 3 and the velocity of point B at the posture specified.
The velocity of point B is given by VB  VBO3  VA  VBA , or
V  ω × R  50ˆj  ω × R
B
3
BO3
2
BA
Assuming that the idle freedom is not active we can set ω3 R BO3  0 , where
ω   x ˆi   y ˆj   z kˆ and ω   ˆj . Expanding these and using the position data from
3
3
3
3
2
2
Prob. 10.2 gives four simultaneous equations:
10.048 144.889 37.500   2   0 
 0
 37.500
0
37.500 144.889 3x   0 


 0
37.500
0
10.048  3y   50.000

 z 

0
 64.952 144.889 10.048
 3   0 
Solving these gives ω  2.570ˆj rad/s
2
3  1.327 rad/s
ω3  1.158ˆi  0.086ˆj  0.643kˆ rad/s ,
VB  96.361ˆi  50.000ˆj  166.903kˆ mm/s , VB  0.199 m/s
Ans.
Ans.
Ans.
371
10.4
Solve Problem 10.3 using graphic techniques.
The position and velocity solutions are shown in the figure below. After the top and front
views are drawn to scale, first and second auxiliary views are drawn to view rod O3B in
true length and end views, respectively.
Next a velocity polygon is drawn with origin at point B. The velocity VA is drawn true
length, downward in the front view. The direction of VBA is added horizontal in the front
view and perpendicular to link 2 in the top view. This direction is projected to the first
auxiliary view where it intersects the line of VB, which is perpendicular to rod 3 in this
view. This completes the velocity polygon for the equation VB  VA  VBA . Projecting
this to all other views, we can measure the true lengths of VB from the second auxiliary
view, and VBA from the top view.
The angular velocities are then found from
2 
VBA 305 mm/s

 4.07 rad/s
RBA
75 mm
3 
VBO3
RBO3

222 mm/s
 1.48 rad/s
150 mm
VB  222 mm/s
Ans.
Ans.
Ans.
The results show typical graphical error when compared with the analytic solution in
Prob. 10.3.
372
10.5
For the spheric RRRR linkage in the posture illustrated, use vector algebra to make
complete velocity and acceleration analyses at the posture indicated.
RO2O  7kˆ in, RO4O  2ˆi in, R AO2  3ˆi in, R BO4  9ˆj in, R BA  5ˆi  9ˆj  7kˆ in, and ω2  60kˆ rad/s.
R AO2  3ˆi in ,
RO4O2  2ˆi  7kˆ in ,
R BO4  9ˆj in .
Substituting these into R AO2  R BA  RO4O2  R BO4 gives
R  5ˆi  9ˆj  7kˆ in , R  12.450 in
BA
BA
The velocity analysis proceeds as follows:
VA  VAO2  ω2 × R AO2  60kˆ rad/s × 3ˆi in  180ˆj in/s
VB  VBO4  ω4 × R BO4

  
  ˆi  ×  9ˆj in   9 kˆ
4
Ans.
4
Since the two revolutes at A and B have axes that intersect at O, this is a spheric linkage;
therefore triangle AOB rotates about O as a rigid link with point O stationary. From this
we see that the axis of rotation of link 3 passes through O and is perpendicular to both
VA and VB . Therefore ω3  3ˆi and
V  ω × R   ˆi × 5ˆi  9ˆj  7kˆ  7 ˆj  9 kˆ
BA
3
BA
3


3
3
Substituting these into VB  VA  VBA or 94kˆ  180ˆj  73ˆj  93kˆ , and equating
components gives
ω3  ω4  180 7  25.714ˆi rad/s ,
Ans.
V  180ˆj  231.4kˆ in/s , and V  231.4kˆ in/s .
Ans.
BA
B
To find accelerations we first calculate
AnAO2  22 R AO2  10 800ˆi in/s 2 ,
An   2 R  5 951ˆj in/s2 ,
BO4
4
BO4
AtAO2  α 2 × R AO2  0
AtBO4  α 4 × R BO4  9 4kˆ
Ans.
373
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3 × R AO   4 629kˆ in/s2 ,
At  α × R  7 y ˆi   7 x  3 z  ˆj  3 y kˆ
AO
3
AO
3
3
3
3
A  3 ×  3 × R BO   5 951ˆj in/s2 ,
AtBO  α3 × R BO  93z ˆi  23z ˆj   9 3x  23y  kˆ
n
BO
Substituting these into A A  AnAO2  AnAO  AtAO
equating components gives
10 800 
73y
0
0  4 629
73x
,
33z ,
33y
and A B  AnBO4  AnBO  AtBO , and
0
93z ,
5 951  5 951
23z ,
9 4 
, and
93x
From these we solve for α 3  1 543ˆj rad/s and α 4  343ˆi rad/s .
A  5 951ˆj  3 087kˆ in/s2
2
B
2
23y
.
Ans.
Ans.
374
10.6
Solve Problem 10.5 using graphic techniques.
To avoid confusion, the position and velocity solutions are shown in a separate figure
above. The acceleration solution is shown below. The results agree (within graphic
error) with those of the previous analytic solution of Prob. 10.5.
375
10.7
Solve Problem 10.5 using transformation matrix techniques.
Following the conventions of Sec. 10.6 and Fig. 10.11, the Denavit-Hartenberg parameter
values are:
i,j
ai , j
i, j
i, j
si , j
1,2
0
23.20°
1  0
0
2,3
0
94.90°
0
3,4
0
77.47°
4,1
0
90.00°
2  101.54º
3  67.30º
4  90.00º
0
0
Using Eqs. (10.13) and (10.16) we find
0
0
0
1
0 0.91914 0.39394 0 

T12  
0 0.39394 0.91914 0 


0
0
1
0
 0.20005 0.08369 0.97620 0 
 0.20005 0.08369 0.97620 0 
 0.97979 0.01709 0.19932 0 


 , T13   0.90056 0.37679 0.21685 0  ,
T23  

 0.38598 0.92252
0
0.99635 0.08542 0 
0
0




0
0
0
1
0
0
0
1


 0.38591 0.20015 0.90057 0 
0 1 0 0 
 0.92254 0.08372 0.37671 0 


 , T14  0 0 1 0  .
T34  

1 0 0 0 
0
0.97618 0.21695 0 




0
0
0
1

0 0 0 1 
Next, from Eqs. (10.23) and (10.27),
0
0.91914 0.39394 0 
0 1 0 0 

1 0 0 0 
 0.91914
0
0
0 



D1 
, D2 
,
0 0 0 0 
 0.39394
0
0
0




0
0
0
0
0 0 0 0 

0
0.21685 0 
 0
0 0 1 0 
 0

0 0 0 0 
0
0.97620 0 

.
D3 
, D4  
0.21685 0.97620
1 0 0 0 
0
0




0
0
0
 0
0 0 0 0 
Now, from Eq. (10.28), D11  D22  D33  D44  0 , we get the following set of
equations:
0.97620 0  2   0 
 0
0
0.39394 0.21685 1     0     0  rad/s

 3   1  
0.91914
60
0
0  4   1
376
From these we find 2  65.275 rad/s , 3  0 , and 4  25.714 rad/s . Then, from Eq.
(10.29) we find the velocity matrices
0 25.714 0
0 25.714 0
 0 60 0 0 
 0
 0
 60 0 0 0 
 0


0
0
0
0
0 0 0 




2 
rad/s, 3 
rad/s,  4 
rad/s.
 0 0 0 0
 25.714 0
 25.714 0 0 0
0
0






0
0
0
0 0 0
 0 0 0 0
 0
 0
These can be used with Eq. (10.30) to find the velocities of all moving points.
Acceleration analysis follows similar steps. From Eq. (10.32) we get the following set of
equations:
0.97620
0  2   1 543
 0
0.39394 0.21685 1     0  rad/s 2

 3 

0.91914
0
0  4   0 
From these we find 2  0 , 3  1 580 rad/s 2 , and 4  343 rad/s2 . From these and Eq.
(10.33) we find the acceleration matrices
0
0
0 0 0 0
0 0
 0 0 343 0 
0 0 0 0
0 0 1 543 0 
 0 0 0 0
2




 rad/s2.
2 
, 3 
rad/s ,  4  
0 0 0 0
0 1 543
 343 0 0 0 
0
0






0
0
0 0 0 0
0 0
 0 0 0 0
These can be used with Eq. (10.34) to find the accelerations of all moving points.
377
10.8
Solve Problem 10.5 except with   90 .
The position vectors for this new posture are:
x ˆ
y ˆ
z ˆ
i  RBA
j  RBA
k , RO4O2  2ˆi  7kˆ in , R BO4  9sin  4ˆj  9cos 4kˆ .
R AO2  3ˆj in , R BA  RBA
Substituting these into R AO2  R BA  RO4O2  R BO4 and separating components gives
x
y
z
 9cos 4  7 , and squaring and adding these gives
RBA
 2 , RBA
 9sin  4  3 , RBA
2
2
2
2
If we now define Z  tan 4 2 
2  9  126cos 4  7  54sin 4  32  RBA
 155.
2
2
and use the identities cos4  1  Z  1  Z  and sin4  2Z 1  Z 2  , then the
above equation can be reduced to 114Z 2  108Z  138  0 . The root of interest here is
Z  0.72419 , which corresponds to  4  71.823 . With this the four position vectors are
R AO2  3ˆj in , R BA  2ˆi  11.551ˆj  4.192kˆ in , RO4O2  2ˆi  7kˆ in , R BO4  8.551ˆj  2.808kˆ in
The velocity analysis proceeds as in Prob. 10.5:

  
 

VA  VAO2  ω2 × R AO2  60kˆ rad/s × 3ˆj in  180ˆi in/s
V V
 ω ×R
  ˆi × 8.551ˆj  2.808kˆ in  2.808 ˆj  8.551 kˆ
B
BO4
4
BO4
4

4
4
Since the two revolutes at A and B have axes which intersect at O, this is a spheric
linkage; therefore triangle AOB rotates about O as a rigid link with point O stationary.
From this we see that the axis of rotation of link 3 passes through O and is perpendicular
to both VA and VB . Therefore ω3  0.950103ˆj  0.311953kˆ and
VBA  ω3 × R BA  7.5863ˆi  0.6243ˆj  1.9003kˆ
Substituting these into VB  VA  VBA and equating components gives
3  23.726 rad/s and 4  5.273 rad/s , and from these we get
ω3  22.542ˆj  7.401kˆ rad/s , ω4  5.273ˆi rad/s ,
Ans.
ˆ
ˆ
ˆ
ˆ
ˆ
VBA  180i  14.803j  45.085k in/s , and VB  14.803j  45.085k in/s .
Ans.
To find accelerations we first calculate
AtAO2  α 2 × R AO2  0 ,
AnAO2  22 R AO2  10 800ˆj in/s2 ,
An   2 R  237.709ˆj  78.047kˆ in/s2 , At  α × R  2.808 ˆj  8.551 kˆ
BO4
4
BO4
BO4
4
BO4
4
4
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3R AO   1 332ˆj  4 058kˆ in/s2 ,
At  α × R   7 y  3 z  ˆi  7 x ˆj  3 xkˆ ,
AO
3
AO
3
3
3
3
AnBO  3 ×  3 × R BO   1 126ˆi in/s2 ,
AtBO  α3 × R BO   2.808 3y  8.5513z  ˆi   2.808 3x  23z  ˆj  8.551 3x  23y  kˆ
Substituting these into A A  AnAO2  AnAO  AtAO and A B  AnBO4  AnBO  AtBO , and equating
components gives
0  73y  33z ,
0  1126  2.8083y  8.5513z ,
10 800  1 332  73x ,
237.709  2.808 4  2.8083x  23z , 0  4 058  33x , 78.047  8.551 4  8.5513x  23y .
From these we solve for α3  1 302ˆi  49ˆj  115kˆ rad/s 2 and α 4  1 304ˆi rad/s 2 .
ˆ in/s2
A A  10 800ˆj in/s 2 , AtBO  3 430ˆj  11 230k
Ans.
Ans.
378
10.9
Determine the advance-to-return ratio for Problem 10.5. What is the total angle of
oscillation of link 4?
Advance-to-return ratio = 181.1/178.9 = 1.012,
 4  46.5
Ans.
379
10.10 For the spheric RRRR linkage determine whether the crank is free to turn through a
complete revolution. If so, find the angle of oscillation of link 4 and the advance-toreturn ratio.
RO2O  150 mm, RO4O  225 mm, RAO2  37.5 mm, RBO4  262 mm, RBA  412 mm,
  120, and ω  30kˆ rad/s.
2

Advance-to-return ratio = Q = 187/173 = 1.081,
 4  38
Ans.
380
10.11 Use vector algebra to make complete velocity and acceleration analyses of the linkage of
Figure P10.10 at the posture specified.
The position vectors were found from a graphic analysis done on a CAD software system
(see Prob. 10.12). For the posture with  2  120 they are as follows:
R
 225ˆi  150kˆ mm ,
R  236.975ˆj  111.744kˆ mm ,
O4O2
BO4
R AO2  18.750ˆi  32.476ˆj mm ,
R BA  243.750ˆi  204.499ˆj  261.744kˆ mm .
The velocity analysis for this posture, with ω2  36kˆ rad/s , proceeds as follows:
V  ω × R  1.169ˆi  0.675ˆj m/s , V   ˆi × R  0.111744 ˆj  0.236975 kˆ .
A
2
AO2
B
4
BO4
4
4
Since all revolute axes intersect at O, this is a spheric linkage and triangle AOB (link 3)
rotates about O. Thus the axis of rotation of link 3 passes through O and is perpendicular
to VA and VB . Calling this axis uˆ ,  3  3uˆ  ,
uˆ   V × V  V × V  0.462932ˆi  0.801731ˆj  0.378051kˆ
B
A
B
A
VBA  3 × R BA  0.1325373ˆi  0.2133203ˆj  0.2900913kˆ
Substituting these into VB  VA  VBA , equating components, and solving gives
3  8.820 rad/s and 4  10.797 rad/s . From these
3  4.083ˆi  7.071ˆj  3.334kˆ rad/s , and 4  10.797ˆi rad/s .
Ans.
For acceleration analysis we first calculate
AnAO2  2 × 2 × R AO2  24.300ˆi  42.089ˆj m/s2 , AtAO2  2 × R AO2  0 ,
A
n
BO4


  ×   × R   27.626ˆj  13.027kˆ m/s ,
2
4
4
BO4
AtBO4   4ˆi × R BO4  0.111744 4ˆj  0.236975 4kˆ
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3 × R AO   2.251ˆi  3.898ˆj  11.023kˆ m/s2 ,
An   ×   × R   22.117ˆi  10.447ˆj  4.926kˆ m/s2 ,
BO
A
t
AO
3
3
BO
 3 × R AO   0.1503y  0.0323z  ˆi   0.0193z  0.1503x  ˆj   0.0323x  0.0193y  kˆ ,
AtBO  3 × R BO    0.1123y  0.2373z  ˆi   0.2253z  0.1123x  ˆj   0.2373x  0.2253y  kˆ
Next, from A A  AnAO2  AtAO2  AnAO  AtAO and A B  AnBO4  AtBO4  AnBO  AtBO , we
separate components and obtain
24.300  2.251  0.1503y  0.0323z ;
0  22.117  0.1123y  0.2373z ;
42.089  3.898  0.1503x  0.0193z ; 27.626  0.112 4  10.447  0.1123x  0.2253z ;
0  11.023  0.0323x  0.0193y ; 13.027  0.237 4  4.926  0.2373x  0.2253y ;
From these α  273ˆi  115ˆj  148kˆ rad/s 2 and α  130ˆi rad/s2 .
Ans.
3
4
381
10.12 Solve Problem 10.11 using graphic techniques.
The position and velocity solutions are shown first with the acceleration solution on the
next diagram. The results verify those of the analytic solution in Prob. 10.11.
382
10.13 Solve Problem 10.11 using transformation matrix techniques.
The Denavit-Hartenberg parameters are:
a12  0, 12  14.04, 12  1  60.00, s12  0,
a23  0,  23  104.41,  23  2  69.52, s23  0,
a34  0,  34  49.34,  34  3  93.18, s34  0,
a41  0,  41  90.00  41  4  64.75 s41  0.
From Eqs. (10.13) and (10.16) the transformation matrices are:
 0.50000 0.84015 0.21005 0 
 0.86603 0.48506 0.12127 0 

T12  

0
0.24260 0.97014 0 


0
0
0
1

 0.61206 0.39312 0.68618 0 
 0.75747 0.04214 0.65151 0 

T13  
 0.22720 0.91852 0.32356 0 


0
0
0
1

0.42650 0.90449 0 0 
 0
0
1 0 
T14  
0.90449 0.42650 0 0 


0
0 1
 0
Next, from Eqs. (10.24) and (10.27),
0
0.97015 0.12127 0 
0 1 0 0 

1 0 0 0 
 0.97015
0
0.21005 0



D1 
D2 
0 0 0 0 
 0.12127 0.21005
0
0




0
0
0
0
0 0 0 0 

0
0.32357 0.65151 0 

0 0 1 0 
 0.32357

0 0 0 0 
0
0.68618 0 


D3 
D4  
 0.65151 0.68618
1 0 0 0 
0
0




0
0
0
0

0 0 0 0 
and from Eq. (10.28) we get the following set of equations
 0.21005 0.68618 0  2 
0
 0 
0.12127 0.65151 1     0    0  rad/s

 3
  1 

 0.97015 0.32357 0  4 
1 
 36
from which we find 2  33.670 rad/s , 3  10.307 rad/s , and 4  10.798 rad/s .
383
With these values and Eqs. (10.29) we find the velocity matrices
0 10.798 0 
 0 36 0 0 
 0 3.335 4.083 0 
 0
36 0 0 0 
3.335


0
7.072 0 
0
0
0
0 




2 
, 
, 4 
.Ans.
 0 0 0 0  3  4.083 7.072
10.798 0
0
0
0
0






0
0
0
0
0
0
 0 0 0 0
 0
 0
These can be used with Eq. (10.30) to find the velocities of all moving points.
Ans.
The acceleration analysis follows parallel steps using Eqs. (10.32) and (10.33)
2  152 rad/s 2
 0.21005 0.68618 0  2   183.0
0.12127 0.65151 1     254.6  m/s 2 ;
3  220 rad/s 2

 3 

 0.97015 0.32357 0  4   76.4 
4  130 rad/s 2
1
0 0 0 0
 0 148 273 0 
 0 0 130 0 
0 0 0 0
 148 0 115 0 
 0 0 0 0




.
2 
3 
,
, 4  
Ans.
0 0 0 0
 273 115 0 0 
130 0 0 0 






0
0 0
0 0 0 0
 0
 0 0 0 0
Although the global axes have changed because of the Denavit-Hartenberg conventions,
these results correlate with and verify those of Probs. 10.11 and 10.12.
Ans.
384
10.14 The figure illustrates the top, front, and auxiliary views of a spatial slider-crank RSSP
linkage. In the construction of many such linkages, a provision is made to vary the angle
; thus, the stroke of slider 4 becomes adjustable from zero, when  = 0, to twice the
crank length, when  = 90. With  = 30, use vector algebra to make a complete
velocity analysis of the linkage at the given posture.
RAO  2 in, RBA  6 in,    240 , ω  24ˆi rad/s.
The position vectors were found from a graphic analysis done on a CAD software system
(see Prob. 10.15). For the posture where  2  240 they are as follows:
R  0.866 025ˆi  1.500 000ˆj  1.000 000kˆ in ,
R  6.588 787ˆi in ,
AO
BO
R BA  5.722 762ˆi  1.500 000ˆj  1.000 000kˆ in .
The velocity analysis for this posture, with 2  24 rad/s , proceeds as follows:
ω  20.784610ˆi 12.0ˆj rad/s , V  ω × R  12.0ˆi  20.784 610ˆj  41.569 219kˆ in/s ,
A
2
2
AO
VBA  ω3 × R BA    1.5  ˆi   5.722 762  3x  ˆj   1.53x  5.722 7623y  kˆ , VB  VB ˆi .
y
3
z
3
z
3
Substituting these into VB  VA  VBA and separating into components,
VB  12.0
3y  1.5

0.0  20.784 610 
3x
3z
 5.722 7623z
0.0  41.569 219  1.53x  5.7227623y
However, this is a set of only three equations and there are four unknown variables. This
results from the fact that the linkage has two degrees of freedom and the connecting rod is
free to rotate about the axis AB. If we assume that this second “idle freedom” is inactive,
then we can set 3 R BA  0 to get a fourth equation:
0.0  5.722 7623x 1.53y  1.03z
The four equations can now be solved to give
ω3  2.309 401ˆi  6.658 519ˆj  3.228 373kˆ rad/s and VB  13.815 960ˆi in/s
Ans.
385
10.15 Solve Problem 10.14 using graphic techniques.
The graphic solution is shown in the figure above. The results verify those found in
Prob. 10.14. They are
3 
VBA 46.528 in/s

 7.766 rad/s , and VB  14.197 in/s
RBA
6 in
Ans.
386
10.16 Solve Problem 10.14 using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
12  90 ,
12  1  30 ,
a12  0 ,
s12  0 ,
s14  4 .
a14  0 ,
14    30 ,
14  0 ,
From Eqs. (10.13) and (10.12) we obtain
cos 1 0 sin 1 0 
 0   2sin 1 
 sin  0  cos  0 
 0   2 cos  
1
1
1


T12 
RA  T12    
,
;
 0 1
2  0 
0
0


  

0
1
 0 0
1   1 
0
0
0
1 0

0 

0 cos   sin   sin  
0    sin  
4
,
.
T14  
RB  T14     4
0 sin  cos  4 cos  
0   4 cos  


  

0
1
1
0 0

1  

From these and the length of the connecting rod we write
t
2
2
2
2
RBA
  RB  RA   RB  RA    2sin 1    2cos 1  4 sin    4 cos    62
This reduces to 42  44 sin  cos 1  32  0 , which has a solution
4  2sin  cos 1  4sin 2  cos 2 1  32
By differentiating the above equation with respect to time we obtain
244  44 sin  cos 1  414 sin  sin 1  0
which has for a solution
24 sin  sin 1
4 

 2sin  cos 1  4  1
and, from Eqs. (10.23), (10.24), (10.27), and (10.29),
0 
0 1 0 0 
0 0 0
1 0 0 0 
0 0 0 sin  
,
;
D1  
D4  
0 0 0 0 
0 0 0  cos  




0 
0 0 0 0 
0 0 0
 0 1 0 0 
0 0 0  sin 4 




1 0 0 0 
0 0 0 cos 4 


2 
4 
,
.
 0 0 0 0
0 0 0





0
 0 0 0 0 
0 0 0

Now, with   30 , 1  30 , and 1  24 rad/s , the above formulae give
4  6.588 in , 4  13.816 in/s , and
Ans.
387
0
0
 1.000 in 


 41.569 in/s 


 1.732 in 
 3.294 in 
 24.000 in/s 
 6.908 in/s 
 , RB  
 , RA  
 , RB  
 . Ans.
RA  


 5.706 in 


11.965 in/s 
0
0








1
1
0
0








These results agree with those of Probs. 10.14 and 10.15 once the Denavit-Hartenberg
coordinate directions are considered. Note, however, that the loop-closure equation was
never used. No coordinate system was fixed to link 3 and no velocity of link 3 was
found. This is because of our lack of information about the degree of freedom
representing the spin of link 3 about the line AB.
388
10.17 Solve Problem 10.14 with  = 60 using vector algebra.
The position vectors were found from a graphic analysis done on a CAD software system
(see Prob. 10.18). For the posture with  2  240 they are as follows:
R  1.500 000ˆi  0.866 025ˆj  1.000 000kˆ in , R  7.352 350ˆi in ,
AO
BO
R BA  5.852 350ˆi  0.866 025ˆj  1.000 000kˆ in .
The velocity analysis for this posture, with 2  24 rad/s , proceeds as follows:
ω  12.0ˆi  20.784 610ˆj rad/s , V  ω ×R  20.784 610ˆi  12.0ˆj  41.569 219kˆ in/s ,Ans.
2
A
2
AO
VBA  ω3 ×R BA    0.866 025  ˆi  5.852 3503z  3x  ˆj   0.866 0253x  5.852 3503y  kˆ ,
y
3
z
3
VB  VB ˆi .
Substituting these into VB  VA  VBA and separating into components we get
3y  0.866 0253z
VB  20.784 610 
0.0  12.0

3x
 5.852 3503z
0.0  41.569 219  0.866 0253x  5.852 3503y
However, this is a set of only three equations and there are four unknown variables. This
results from the fact that the linkage has two degrees of freedom and the connecting rod is
free to rotate about the axis AB. If we assume that this second “idle freedom” is inactive,
then ω3 R BA  0 .
0.0  5.852 3503x  0.866 0253y  3z
The four equations can now be solved to give
ω3  1.333 333ˆi  6.905 692ˆj  1.822 629kˆ rad/s and VB  26.112 859ˆi in/s
Ans.
389
10.18 Solve Problem 10.14 with  = 60 using graphic techniques.
The graphic solution is shown in the figure above. The results verify those found in
Prob. 10.17. They are
V
45.542 in/s
Ans.
3  BA 
 7.257 rad/s , and VB  26.350 in/s
RBA
6 in
390
10.19 Solve Problem 10.14 with  = 60 using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
12  90 ,
12  1  30 ,
a12  0 ,
s12  0 ,
s14  4 .
a14  0 ,
14    60 ,
14  0 ,
From Eqs. (10.13) and (10.12) we obtain
cos 1 0 sin 1 0 
 0   2sin 1 
 sin  0  cos  0 
 0   2 cos  
1
1
1


T12 
RA  T12    
,
;
 0 1
2  0 
0
0


  

0
1
 0 0
1   1 
0
0
0
1 0

0 

0 cos   sin   sin  
0    sin  
4
,
.
T14  
RB  T14     4
0 sin  cos  4 cos  
0   4 cos  


  

0
1
1
0 0

1  

From these and the length of the connecting rod we write
t
2
2
2
2
RBA
  RB  RA   RB  RA    2sin 1    2cos 1  4 sin    4 cos    62
This reduces to 42  44 sin  cos 1  32  0 , which has a solution
4  2sin  cos 1  4sin 2  cos 2 1  32
By differentiating the above equation with respect to time we obtain
244  44 sin  cos 1  414 sin  sin 1  0
which has for a solution
24 sin  sin 1
4 

 2sin  cos 1  4  1
and, from Eqs. (10.23), (10.24), (10.27), and (10.29),
0 
0 1 0 0 
0 0 0
1 0 0 0 
0 0 0 sin  
,
;
D1  
D4  
0 0 0 0 
0 0 0  cos  




0 
0 0 0 0 
0 0 0
 0 1 0 0 
0 0 0  sin 4 




1 0 0 0 
0 0 0 cos 4 


2 
4 
,
.
 0 0 0 0
0 0 0





0
 0 0 0 0 
0 0 0

Now, with   60 , 1  30 , and 1  24 rad/s , the above formulae give
4  7.352 in , 4  26.112 in/s , and
391
0
0
 1.000 in 


 41.569 in/s 


 1.732 in 
 6.367 in 
 24.000 in/s 
 22.614 in/s 
,
,
 , RB  
.
RA  
RB  
RA  


 3.676 in 


 13.056 in/s 
0
0








1
1
0
0








These results agree with those of Probs. 10.17 and 10.18 once the Denavit-Hartenberg
coordinate directions are considered.
392
10.20 The figure illustrates the top, front, and profile views of an RSRC crank and oscillatingslider linkage. Link 4, the oscillating slider, is rigidly attached to a round rod that rotates
and slides in the two bearings. (a) Use the Kutzbach criterion to find the mobility of this
linkage. (b) With crank 2 as the driver, find the total angular and linear travel of link 4.
(c) Write the loop-closure equation for this linkage and use vector algebra to solve it for
all unknown position data.
RAO  4 in, RBA  12 in, 2  40 , and ω  48ˆi rad/s.
393
(a)
The RSRC linkage has n = 4, j1 = 2, j2 = 1, j3 = 1. The Kutzbach criterion gives
m  6  n  1  5 j1  4 j2  3 j3  6(4  1)  5(2)  4(1)  3(1)  1
Ans.
(b)
Since vectors do not show the rotation , matrix methods were necessary and are
shown in Prob. 11.23. See (c) for the vector solution. Together, they show:
Ans.
135   4  45 ;
 4  90 .
Ans.
7.314 in  yB  15.314 in ;
yB  8.000 in .
(c)
R B  R AB  RQ  R AQ
y ˆj  x ˆi  y ˆj  z kˆ  4ˆi  4sin  ˆj  4cos kˆ
B
AB
AB
AB
2
2
Separating components,
yB  yAB  4sin  2 , z AB  4cos 2
xAB  4 ,
and, from the length of link 3,
2
2
2
xAB
 y 2AB  z AB
 RAB
 4   4sin  2  yB    4cos 2   122
2
2
2
yB2  8sin  2 yB  112  0
yB  4sin  2  4 7  sin 2  2
Ans.
R AB  4ˆi  4 7  sin 2  2 ˆj  4cos 2kˆ
For   40 , R  y ˆj  8.320ˆj in , R
Ans.
2
B
B
AB
 4.000ˆi  10.891ˆj  3.064kˆ in .
Ans.
394
10.21 Use vector algebra to find VB, 3, and 4 for Problem 10.20.
RQ  4ˆi in ,
R B  8.320ˆj in , and R AB  4.000ˆi  10.891ˆj  3.064kˆ in .
First
we
identify,
for
 2  40 , that
R AQ  2.571ˆj  3.064kˆ in ,
Next we note that the rotation axis of the revolute at B is j× R AB  3.064ˆi  4.000kˆ and,
normalizing this, we can express the apparent angular velocity ω3/ 4 axis as
ˆ  0.608ˆi  0.794kˆ . Then the angular velocity of link 3 can be written as
ω
3/ 4
ω3  ω4  ω3/ 4  0.6083/ 4ˆi  4 ˆj  0.7943/ 4kˆ .
V  ω × R  147.081ˆj  123.415kˆ in/s ,
A
2
With
AQ
this
done,
V  V ˆj ,
B
we
can
B
VAB  ω3 × R AB  (3.0644  8.6463/ 4 )ˆi  5.0393/ 4ˆj   4.0004  6.6233/ 4  kˆ .
find
and
Now,
setting VB  VAB  VA and equating components, we get the following equations:
3.064 4  8.6463/ 4  0
VB
 5.0393/ 4  147.081
4.000 4  6.6233/ 4  123.415
which can be solved to give 4  19.444 rad/s , 3/ 4  6.891 rad/s , VB  112.358 in/s .
Ans.
ω3  4.190ˆi  19.444ˆj  5.471kˆ rad/s , ω4  19.444ˆj rad/s , VB  112.358ˆj in/s .
Note that if ω were written as ω   x ˆi   y ˆj   z kˆ , then the above set of simultaneous
3
3
3
3
3
equations would have four unknowns and could not be solved.
395
10.22 Solve Problem 10.21 using graphic techniques.
The graphic solution is shown in the figure above. The results verify those found in Prob.
10.21. 3 and 4 are not apparent in the graphic method, but VB  112.3 in/s .
Ans.
396
10.23 Solve Problem 10.21 using transformation matrix techniques.
The Denavit-Hartenberg parameters from the global coordinate system to joint A are:
a12  0 ,
12  0 ,
12   2  40 ,
s12  4 in ,
and, proceeding in the other direction around the loop, to joint A, they are:
16  90 ,
a16  0 ,
16  0 ,
s16  0 ,
a65  0 ,
 65  0 ,
s65  yB ,
 65  0 ,
a54  0 ,
 54  0 ,
54   4 ,
s54  0 ,
a43  0 ,
 43  90 ,
 43  0 ,
s43  0 ,
a3 B  0 ,
3B  3 ,
s3 B  0 ,
3B  0 ,
aBA  12 in ,
sBA  0 .
 BA  0 ,
 BA  90 ,
From Eqs. (10.12) we obtain for a first path to A:
cos  2  sin  2 0 0 
 sin 
cos  2 0 0 
2

T12 
,
 0
0
1 4 


0
0 1
 0
397
and along the other path to joint A, from Eqs. (10.13) and (10.16), we get:
1 0 0 0 
0 0 1 0 
,
T16  
0 1 0 0 


0 0 0 1 
1
0
T65  
0

0
0
0 
,
yB 

1
0 0
1 0
0 1
0 0
cos  4
 sin 
4
T54  
 0

 0
 sin  4
cos  4
1
0
T43  
0

0
0
0
 cos  4
 0
T14  
  sin  4

 0
0 0
0 0 
,
1 0

0 1
 cos  4
 0
T13  
  sin  4

 0
0
0 1 0 
,
1 0 0

0 0 1
0
cos  3
 sin 
3
T3 B  
 0

 0
0
 1
TBA  
0

0
0
1 0
0 0
T15  
0 1

0 0
 sin  3
cos  3
0
0
0 0
0 0 
,
1 0

0 1
 cos 3 cos 4

sin  3
T1B  
  cos 3 sin  4

0

0 
 sin  3 cos 4

  cos
0 0 12 
3
, T1 A  


0 1 0
 sin  3 sin  4


0 0 1 
0

1 0
0
1
0
0
 sin  4
0
0
1
 cos 4
0
0
0
0 sin  4
1
0
0 cos  4
0
0
0
yB 
,
0

1
0
yB 
,
0

1
0
yB 
,
0

1
 sin  3 cos  4
sin  4
cos  3
0
sin  3 sin  4
cos  4
0
0
cos 3 cos 4
sin  4
sin  3
0
 cos 3 sin  4
cos 4
0
0
12sin  3 cos 4 
yB  12 cos 3 
.
12sin  3 sin  4 

1

0
yB 
,
0

1
At the terminations of these two paths, the position of point A must agree. Therefore,
cos 3 cos 4 sin  4 12sin  3 cos  4  0 
cos 2  sin  2 0 0   4   sin  3 cos  4
 sin 




cos  2 0 0   0    cos  3
sin  3
0
yB  12 cos  3  0 
2


 0
0
1 4   0    sin  3 sin  4  cos  3 sin  4 cos  4 12sin  3 sin  4  0 

  
 
0
0 1  1  
0
0
0
1
 0
 1 
 4 cos  2   12sin  3 cos  4 
 4sin    y  12 cos  
2
3 

 B
 4   12sin  3 sin  4 

 

1
1

 

398
Noting from the figure that 90  3  0 and 180   4  0 , we can solve these three
equations for the position results
3   sin 1
 1  cos  3
 4   tan
1 cos2 
1
2
2
yB  4 7  sin 2  2  4sin  2
and, at  2  40 , these give 3  24.83 ,  4  52.55 , and yB  8.320 in .
For velocity analysis we begin by using Eqs. (10.23) and (10.27) to find
0 1 0 0 
0 1 0 0 
1 0 0 0 
1 0 0 0 
,
,
Q1  
D1  Q1  
0 0 0 0 
0 0 0 0 




0 0 0 0 
0 0 0 0 
0
yB cos  4 
0 1 0 0 
 0  cos  4
1 0 0 0 
cos

0
 sin  4
0
4
,
,
Q3  
D3  T13Q3T131  
0 0 0 0 
 0
sin  4
0
 yB sin  4 




0
0
0
0 0 0 0 
 0

0 1 0 0 
1 0 0 0 
,
Q4  
0 0 0 0 


0 0 0 0 
 0 0 1 0
 0 0 0 0
1
,
D4  T14Q4T14  
 1 0 0 0 


 0 0 0 0
0 6 0 0
0 0 0 0
,
Q6  
0 0 0 1 


0 0 0 0
0 0 0 0 
0 0 0 1 
.
D6  T16Q6T161  
0 0 0 0 


0 0 0 0 
Next we write from Eq. (10.29)
 0  2 0 0 


 0 0 0
 2  D2 2   2
 0 0 0 0


 0 0 0 0 
and, along the other path,
 0
 cos 4 3
4
yB cos 4 3 


cos 4 3
0
 sin  4 3
yB

3  D6 yB  D4 4  D3 3  
  4
sin  4 3
0
 yB sin  4 3 


0
0
0
 0

399
 0 0 4

0 0 0
 4  D6 yB  D4 4  
  4 0 0

 0 0 0
0

yB 
0

0 
Since the velocity of point A must agree along the two paths
 4 cos  2 
 4 cos  2 
 4sin  
 4sin  
2
2

2
 3 
 4 
 4 




1
1




 4sin  2 2   4sin  2 cos  4 3  yB cos 4 3  4 4 

 

 4 cos  2 2    4 cos 2 cos 4 3  4sin  4 3  yB 

  4 cos  2 4  4sin  2 sin  4 3  yB sin  4 3 
0

 

0
0

 

At the posture where  2  40 , 3  24.83 ,  4  52.55 ,
yB  8.320 in , and
 2  48.0 rad/s , these equations can be solved for 3  6.891 rad/s ,  4  19.444 rad/s ,
and yB  112.357 in/s . With these values we can evaluate
4.191 19.444 34.865 
0 19.444
0 
 0
 0
 4.191


0
5.471 112.357 
0
0
0
112.357 


3 
and  4 
 19.444 5.471
 19.444 0
0
45.513 
0
0 




0
0
0 
0
0
0 
 0
 0
from which we write the vector forms of the results:
VB  112.357ˆj in/s , ω3  5.471ˆi  19.444ˆj  4.191kˆ rad/s , and ω4  19.444ˆj rad/s . Ans.
Note that the global x1 and z1 axis orientations in this solution differ from those of Prob.
10.21 because of the conventions of the Denavit-Hartenberg parameters. This is also the
reason that the components of 3 seem switched.
400
10.24 For the SCARA robot find the transformation matrix T15 relating the posture of the tool
coordinate system to the ground coordinate system when the joint actuators are set to the
values 1  30 , 2  60 , 3  2 in, and 4  0 . Also find the absolute position of the
tool point that has coordinates x5 = y5 = 0, z5 = 1.5 in.
a12  a23  10 in , a34  a45  0 , 12  34  45  0 ,  23  180 , 12  1 ,  23  2 , 34  0 ,  45  4 , s12  12 in ,
s23  0 , s34  3 , and s45  2 in .
See the solution to Prob. 10.25 for the formulae before numeric evaluation.
 0.866 0.500 0 17.321 in   0  17.321 in 
 0.500 0.866 0
 0  

0
0






R1  T15 R5 

 0
0
1 8.000 in  1.5 in   6.500 in 


 

0
0
1
1
 0
 1  

Ans.
401
10.25 Solve Problem 10.24 using arbitrary (symbolic) values for the joint variables.
a12  a23  10 in , a34  a45  0 , 12  34   45  0 ,  23  180 , 12  1 ,  23  2 ,
34  0 ,  45  4 , s12  12 in , s23  0 , s34  3 , s45  2 in
cos 1  sin 1 0 10 cos 1 in 
 sin  cos  0 10sin  in 
1
1
1

T12  
 0
0
1
12 in 


0
0
1
 0

1 0 0 0 
0 1 0 0 

T34  
0 0 1 3 


0 0 0 1 
cos 2 sin 2 0 10 cos 1 in 
 sin   cos  0 10sin  in 
2
2
1

T23  
 0

0
1
0


0
0
1
 0

cos 4  sin 4 0 0 
 sin  cos  0 0 
4
4

T45  
 0
0
1 2 in 


0
0 1 
 0
cos 1  2  sin 1  2  0 10 cos 1  10 cos 1  2  in 


sin 1  2   cos 1  2  0 10sin 1  10sin 1  2  in 

T13  T12T23 


0
0
1
12 in


0
0
0
1


cos 1  2  sin 1  2  0 10 cos 1  10 cos 1  2  in 


sin 1  2   cos 1  2  0 10sin 1  10sin 1  2  in 
T14  T13T34  


0
0
1
12  3 in


0
0
0
1


cos 1  2  4  sin 1  2  4  0 10cos 1  10cos 1  2  in 


sin 1  2  4   cos 1  2  4  0 10sin 1  10sin 1  2  in 

T15  T14T45 


0
0
1
10  3 in


0
0
0
1


 0 
 0 

R5  
1.500 in 


 1 
10 cos 1  10 cos 1  2  in 


10sin 1  10sin 1  2  in 
R1  T15 R5  


8.500  3 in


1


Ans.
Ans.
402
10.26 For the gantry robot shown, find the transformation matrix T15 relating the posture of the
tool coordinate system to the ground coordinate system when the joint actuators are set to
the values 1  450 mm , 2  180 mm , 3  50 mm , and 4  0 . Also find the absolute
position of the tool point that has coordinates x5 = y5 = 0, z5 = 45 mm.
a12  a23  a34  a45  0 , 12  90 ,  23  90 , 34   45  0 , 12   23  90 , 34  0 ,  45  4 ,
s12  1 , s23  2 , s34  3 , and s45  50 mm .
See the solution to Prob. 10.27 for the formulae before numerical evaluation.
0 1 0 180 mm 
 180 mm 
0 0 1 100 mm 
 145 mm 



T15 
R1  
1 0 0 450 mm 
 450 mm 




1
1
0 0 0



Ans.
403
10.27 Repeat Problem 10.26 using arbitrary (symbolic) values for the joint variables.
a12  a23  a34  a45  0 12  90  23  90 34   45  0 12   23  90 34  0
,
,
,
,
,
,
 45  4 s12  1 s23  2 s34  3 s45  50 mm
,
,
,
,
0 0 1 0 
1 0 0 0 

T12  
0 1 0 1 


0 0 0 1 
1 0 0 0 
0 1 0 0 

T34  
0 0 1 3 


0 0 0 1 
0 0 1 0 
1 0 0 0 

T23  
0 1 0 2 


0 0 0 1 
0 
cos 4  sin 4 0
 sin  cos  0
0 
4
4

T45 
 0
0
1 50 mm 


0
0
1 
 0
0 1 0 2 
0 1 0 2 
0 0 1 0 
0 0 1  
3


T13  T12T23 
T14  T13T34  
1 0 0 1 
1 0 0 1 




0 0 0 1 
0 0 0 1 
2
  sin 4  cos 4 0

 0
0
1 3  50 mm 

T15  T14T45 
 cos 4  sin 4 0

1


0
0
1
 0

 0 
 0 

R5  
 45 mm 


 1 
2


   95 mm 

R1  T15 R5   3


1


1


Ans.
Ans.
404
10.28 For the SCARA robot of Problem 10.24 in the posture described, find the instantaneous
velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 1.5 in, if the actuators
have (constant) velocities of 1  0.20 rad/s , 2  0.35 rad/s , and 3  4  0 .
0 1 0 0 
1 0 0 0 

D1  Q1  
0 0 0 0 


0 0 0 0 
0 0 0 0 
0 0 0 0 
1

D3  T13Q3T13  
0 0 0 1


0 0 0 0 
0 1 0 5.000 in/rad 
1 0 0 8.660 in/rad 
1

D2  T12Q2T12  
0 0 0

0


0
0 0 0

0
0 1 0

 1 0 0 17.321 in/rad 

D4  T14Q4T141  
0 0 0

0


0
0 0 0

0.150 0 1.750 in/s 
 0 0.200 0 0 
 0
0.200
 0.150 0 0 3.031 in/s 
0
0 0 

 2  1  D11  
3  2  D22  
 0
 0

0
0 0
0 0
0




0
0 0
0 0
0
 0
 0

0
0.150
0

1.750
in/s
0
0.150
0

1.750
in/s




 0.150 0 0 3.031 in/s 
 0.150 0 0 3.031 in/s 
 5  4  D44  

4  3  D33  
 0

 0

0 0
0
0 0
0




0 0
0
0 0
0
 0

 0

 2  1  D11  1 D1  D11  1
0 0 0 0
0 0 0 0


0 0 0 0


0 0 0 0
 4   3  D33  3 D3  D33  3
0

0

0

0
0
0
0
0
0 0.606 in/s2 

0 0.350 in/s2 

0
0

0
0

 1.750 in/s 
 0.433 in/s 

R5  5 R1  


0


0


 3   2  D22  2 D2  D22  2
0 0 0 0.606 in/s 2 


0 0 0 0.350 in/s 2 


0 0 0

0


0
0 0 0

 5   4  D44  4 D4  D44  4
0 0 0 0.606 in/s 2 


0 0 0 0.350 in/s 2 


0 0 0

0


0
0 0 0

 0.541 in/s 2 


0.088 in/s 2 
R5   5  55  R1  


0


0


Ans.
405
10.29 For the gantry robot of Problem 10.26 in the posture described, find the instantaneous
velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 45 mm, if the actuators
have (constant) velocities of 1  2  0 , 3  40 mm/s , and 4  20 rad/s .
0
0 0 0

0 0 0

0

D1  Q1  
0 0 0 1 mm/mm 


0
0 0 0

0
0
0
0


0 0 0 1 mm/mm 

D3  T13Q3T131  
0 0 0

0


0
0 0 0

0 0 0 1 mm/mm 
0 0 0

0

D2  T12Q2T121  
0 0 0

0


0
0 0 0

0
0

1
450
mm/rad


0 0 0

0

D4  T14Q4T141  
1 0 0 180 mm/rad 


0
0 0 0

2  1  D11  0
3  2  D22  0
0
0 0 0

0 0 0 40 mm/s 

4  3  D33  
0 0 0

0


0
0 0 0

 0 0 20 9 000 mm/s 
0 0 0
40 mm/s 
5  4  D44  
 20 0 0 3 600 mm/s 


0
0 0 0

 2  1  D11  1D1  D11 1  0
3   2  D22  2 D2  D22 2  0
 4  3  D33  3 D3  D33  3  0
5   4  D44  4 D4  D44 4  0
0


 40 mm/s 

R5  5 R1  


0


0


0 
0 
R5   5  55  R1   
0 
 
0 
Ans.
406
10.30 The SCARA robot of Problem 10.24 is to be guided along a path for which the origin of
the end effector O5 follows the straight line given by
RO (t )  1.6t  4.0  ˆi1  1.2t  3.0  ˆj1  2.0kˆ 1 in
5
with t varying from 0.0 to 5.0 s; the orientation of the end effector is to remain constant
with kˆ 5  kˆ 1 (vertically downward) and î 5 radially outward from the base of the robot.
Find expressions for how each of the actuators must be driven, as functions of time, to
achieve this motion.
From the problem statement we construct the figure shown for the path described.
From this we write the transformation T15.
0.800 0.600 0 4.000  1.600t in 
0.600 0.800 0 3.000  1.200t in 

T15  
 0

0
1
2.000 in


0
0
1
 0

However, as shown in the solution for Prob. 10.25,
407
cos 1  2  4  sin 1  2  4  0 10 cos 1  10 cos 1  2  in 


sin 1  2  4   cos 1  2  4  0 10sin 1  10sin 1  2  in 

T15  T14T45 
.


0
0
1
10  3 in


0
0
0
1


Equating these gives
10 cos 1  10 cos 1  2  in  4.000  1.600t in  4 1.000  0.400t  in
1
10sin 1  10sin 1  2  in  3.000  1.200t in  3 1.000  0.400t  in
 2
10  3  2 in
1  2  4    36.87
 3
 4
Squaring and adding Eqs. (1) and (2) gives
200  200cos 2 in 2  25 1.000  0.800t  0.040t 2  in 2
2  cos1  0.020t 2  0.050t  0.875
180  2  0
Ans.
where the quadrant was found from the figure of the robot.
Expanding the trigonometric functions and recognizing that 2 is now known, Eqs. (1)
and (2) become
10 1  cos 2  cos 1  10  sin 2  sin 1 in  4 1.000  0.400t  in
10  sin 2  cos 1  10 1  cos 2  sin 1 in  3 1.000  0.400t  in
which can be solved for sin 1 and cos 1
sin 1  3 1  cos 2   4sin 2  
cos 1  4 1  cos 2   3sin 2  
where   20 1  cos 2  1.000  0.400t 
From the ratio of these we get
 3 1  cos 2   4sin 2 
1  tan 1 

 4 1  cos 2   3sin 2 
Ans.
where the quadrant of 1 is found by considering the signs of the numerator and
denominator separately.
With both 1 and 2 known, Eqs. (3) and (4) give
3  8.000 in
Ans.
4  1  2  36.87
Ans.
408
10.31 The gantry robot of Problem 10.26 is to travel a path for which the origin of the end
effector O5 follows the straight line given by
RO (t )  120t  300  ˆi1  150ˆj1   90t  225 kˆ 1 mm
5
with t varying from 0.0 to 4.0 s; the orientation of the end effector is to remain constant
with kˆ 5  ˆj1 (vertically downward) and ˆi5  ˆi1 . Find expressions for the positions of
each of the actuators, as functions of time, for this motion.
From Prob. 10.27 and the problem statement we can write
2
  sin 4  cos 4 0
 1 0 0 120t  300 mm 
 0
 0 0 1
0

1



50
mm
150 mm 
3

T15  
 cos 4  sin 4 0
 0 1 0 90t  225 mm 
1

 

0
0
1
1
 0
 0 0 0

Equating individual elements and solving, we get
1  90t  225 mm
2  120t  300 mm
3  100 mm
4  90
Ans.
Ans.
Ans.
Ans.
409
10.32 The end effector of the SCARA robot of Problem 10.24 is working against a force
loading of 10ˆi1  5tkˆ 1 lb and a constant torque loading of 25kˆ 1 in  lb as it follows the
trajectory described in Prob. 10.30. Find the torques required at the actuators, as
functions of time, to achieve the motion described.
Using the 1 , 2 , 3 , and 4 values from Prob. 10.30 and formulae from Probs. 10.25
and 10.28,
0 1 0 0 
0 1 0 10sin 1 
1 0 0 0 
1 0 0 10 cos  
1

D1  Q1  
D2  T12Q2T121  
0 0 0 0 
0 0 0

0




0
0 0 0 0 
0 0 0

 0 1 0 10sin 1  10sin 1  2 
0 0 0 0 


0 0 0 0 
1 0 0 10 cos 1  10 cos 1  2  
1
1



D3  T13Q3T13 
D4  T14Q4T14 
0 0 0

0 0 0 1
0




0
0 0 0 0 
 0 0 0

From these the Jacobian and loads are
0
0
0
0

 0 
0

 0 
0
0
0




1

 25 in  lb 
1
0
1
J 
F 


0 10sin 1 0 10sin 1  10sin 1  2  
 10 lb 
0 10 cos 1 0 10 cos 1  10 cos 1  2  
 0 




0
1
0
0

 5 lb 
Now, from Eq. (10.54), we get
1  25 in  lb
Ans.
 2  25  100sin 1 in  lb
Ans.
 3  5 lb
Ans.
 4  25  100sin 1  100sin 1  2  in  lb
Ans.
410
10.33 The end effector of the gantry robot of Problem 10.26 is working against a force loading
of 20ˆi1  10tˆj1 N and a constant torque loading of 5ˆj1 N  m as it follows the trajectory
described in Prob. 10.31. Find the torques required at the actuators, as functions of time,
to achieve the motion described.
Using the 1 , 2 , 3 , and 4 values from Prob. 10.31 and formulae from Probs. 10.27
and 10.29,
0 0 0 0 
0 0 0 1 
0 0 0 0 
0 0 0 0 


D1  Q1  
D2  T12Q2T121  
0 0 0 1 
0 0 0 0 




0 0 0 0 
0 0 0 0 
0 0 0 0 
0 0 1 0.225  0.090t 
0 0 0 1
0 0 0 0.300  0.120t 


D3  T13Q3T131  
D4  T14Q4T141  
0 0 0 0 
1 0 0

0




0
0 0 0 0 
0 0 0

From these, the Jacobian and loads are
0
0 0 0

 0 
0 0 0

5 N  m 
1




0 0 0

 0 
0
J 
F 


0 1 0 0.225  0.090t 
 20 N 
0 0 1 0.300  0.120t 
 10t N 




0
1 0 0

 0 
Now, from Eq. (10.54), we get
1  0
Ans.
 2  20 N
Ans.
 3  10t N
Ans.
 4  0.500  1.200t  1.200t 2 N  m
Ans.
411
PART 3
DYNAMICS OF MACHINES
412
Page intentionally blank.
413
Chapter 11
Static Force Analysis
11.1
The figure illustrates four linkages and the external forces and torques exerted on or by
the linkages. Sketch the free-body diagram of each part of each linkage. Do not attempt
to show the magnitudes of the forces, except roughly, but do sketch them in their proper
locations and orientations.
414
415
11.2
If the force P  0.9 kN, determine the torque T12 that must be applied to crank 2 to
maintain the linkage in static equilibrium.
RAO2  75 mm, RBA  350 mm.
Kinematic analysis:
  sin 1  r sin     sin 1  75 mmsin105 350 mm  11.95
Force analysis:
P  900ˆi N
 F  P  F ˆj  F
14
34
cosˆi  sinˆj  900 Nˆi  F ˆj  F 0.978ˆi  0.207ˆj  0
14
900 N  0.978F34  0
F14  0.207 F34  0
34
F34  900 N 0.978  920 N


F14  0.207  920 N   190 N
F32  F34  920 N 0.978ˆi  0.207ˆj  900ˆi  190ˆj N
 M  T  r  F  T  75 mm cos105ˆi  sin105ˆj ×  900ˆi  190ˆj N   0
12
2
32
T12  61.5kˆ N  m  0
12
T12  61.5kˆ N  m
Ans.
416
11.3
If T12  100 N  m cw for the linkage illustrated in Figure P11.2, determine the force P to
maintain static equilibrium.
RAO2  r  75 mm, RBA   350 mm
Kinematic analysis:
  sin 1  r sin     sin 1  75 mmsin105 350 mm  11.95
x  r cos  cos  75 mmcos105  350 mmcos11.95  323 mm
Force analysis:
 MOz  xF14  T12  0
P  F14 tan   310 N tan11.95  1 463 N
F14  T12 x  100N  m 0.323 m  310 N
Ans.
417
11.4
Dtermine the forces acting on the ground and the torque T12 to maintain static
equilibrium for the four-bar linkage illustrated in Figure P11.4a.
RAO2  3.5 in; RBA  RBO4  6 in; RCO4  4 in; RDO4  7 in; RO2O4  2 in.
Kinematic analysis:
R AO2  3.5 in210  3.031ˆi  1.750ˆj in
R  6 in82.83  0.749ˆi  5.953ˆj in
BA
R BO4  6 in135.53  4.282ˆi  4.203ˆj in
R  4 in135.53  2.855ˆi  2.802ˆj in
CO4
Force analysis:
M  R
O4
× P  R BO4 × F34  0
CO4
 2.855ˆi  2.802ˆj in  ×  70.050ˆi  71.365ˆj lb 
  4.282ˆi  4.203ˆj in  ×  cos82.83ˆi  sin 82.83ˆj F  0
34
 400.027 in  lb  4.773 inF34  kˆ  0
F34  83.807 in  lb
F  P  F  F  0
F  F  F  0
M  T  R × F  0
i4
34
i2
32
O2

14
12
12
AO2
32
 
F34  83.807 lb82.83  10.465ˆi  83.151ˆj lb
F  59.585ˆi  11.786ˆj lb  60.739 lb168.81 Ans.
14
F12  10.465ˆi  83.151ˆj lb  83.807 lb82.83 Ans.

T12kˆ  3.031ˆi  1.750ˆj in × 10.465ˆi  83.151ˆj lb  0
T12  233.7 in  lb
T12  233.7kˆ in  lb
Ans.
418
11.5
What torque must be applied to link 2 of the linkage illustrated in Figure P11.4b to
maintain static equilibrium?
RAO2  3.5 in; RBA  RBO4  6 in; RCO4  4 in; RDO4  7 in; RO2O4  2 in.
Kinematic analysis:
R AO2  3.5 in240  1.750ˆi  3.031ˆj in
R  6 in105.26  1.579ˆi  5.788ˆj in
BA
R BO4  6 in152.64  5.329ˆi  2.757ˆj in
R  7 in152.64  6.217ˆi  3.217ˆj in
DO4
Force analysis:
M  R
O4
DO4
 P  R BO4  F34  0
 6.217ˆi  3.217ˆj in  × 50ˆi lb    5.329ˆi  2.757ˆj in  × cos105.26ˆi  sin105.26ˆj F  0
34
 160.850 in  lb  4.415 inF34  kˆ  0
F34  36.429 lb105.26  9.588ˆi  35.144ˆj lb
F34  36.429 in  lb
M  T  R
O2

12
AO2
× F32  0
 

T12kˆ  1.750ˆi  3.031ˆj in × 9.588ˆi  35.144ˆj lb  0
T12  90.56 in  lb
T12  90.56kˆ in  lb
Ans.
419
11.6
Sketch a complete free-body diagram of each link and determine the force P to maintain
static equilibrium.
RAO2  100 mm; RBA  150 mm; RBO4  125 mm; RCO4  200 mm; RCD  400 mm; RO2O4  60 mm.
Kinematic analysis:
R AO2  100 mm90  100ˆj mm
R  150 mm  4.86  149ˆi  13ˆj mm
BA
R BO4  125 mm44.3  89ˆi  87ˆj mm
R  200 mm44.3  143ˆi  140ˆj mm
CO4
R DC  400 mm  20.44  375ˆi  140ˆj mm
Force analysis:
M  T  R
O2
12

AO2
× F32  0
 

90kˆ N  m  100ˆj mm × cos 4.86ˆi  sin  4.86ˆj F32  0
 90 N  m  99.640 mmF32  kˆ  0
M  R
O4
BO4
F32  903 N  4.86
× F34  R CO4 × F54  0
89ˆi  87ˆj mm  ×  cos175.14ˆi  sin175.14ˆj 903 N
 143ˆi  140ˆj mm  ×  cos  20.44ˆi  sin  20.44ˆj F  0
54
86 N  m  181 mmF54  kˆ  0

F54  472 N  20.44=44ˆi  165ˆj N
F  Pˆi  443 Nˆi  165 Nˆj  F ˆj=0
P  443ˆi N
16
Ans.
420
11.7
Determine the torque T12 required to drive slider 6 of Figure P11.7 against a load of
P  100 lb at a crank angle of   30 , or as specified by your instructor.
y
RAO2  2.5 in; RO2O4  6 in; RCO
 10 in; RBO4  16 in; and RBC  8 in.
2
Kinematic analysis:
R AO2  2.5 in30  2.165ˆi  1.250ˆj in
R  16 in73.37  4.578ˆi  15.331ˆj in
R AO4  7.466 in73.37  2.136ˆi  7.154ˆj in
R  8 in175.20  7.972ˆi  0.669ˆj in
BO4
CB
Force analysis:
 F  Pˆi  F16ˆj  cos175.20ˆi  sin175.20ˆj F56  0


F56   P cos175.20  100 lb 0.996 100.351 lb
F  100.351 lb175.20  100ˆi  8.392ˆj lb
56
M  R
O4
BO4
×F54  R AO4 ×F34  0
 4.578ˆi  15.331ˆj in  × 100ˆi  8.392ˆj lb  +  2.136ˆi  7.154ˆj in  × cos163.37ˆi  sin163.37ˆj F  0
34
 1 571.519 in  lb  7.466 inF34  kˆ  0
M  T  R
O2

12
AO2
 F32  0
 
F34  210.488 lb163.37=  201.684ˆi  60.240ˆj lb

T12kˆ  2.165ˆi  1.250ˆj in  201.684ˆi  60.240ˆj lb  0
T12  382.52 in  lb
T12  382.52kˆ in  lb
Ans.
421
11.8
Sketch complete free-body diagrams of each link and determine the torque T12 that must
be applied to link 2 to maintain static equilibrium at the posture shown.
RAO2  200 mm; RBA  400 mm; RCA  RO4O2  700 mm; and RCO4  350 mm.
Kinematic analysis:
R AO2  200 mm60  100ˆi  173ˆj mm
RCO4  350 mm  109.05  114ˆi  331ˆj mm
R  400 mm  46.06  278ˆi  288ˆj mm , R  700 mm  46.06  486ˆi  504ˆj mm
BA
CA
Force analysis:
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,

. Then
F34r and F34
M  T + R
O4
14
CO4

 
45kˆ N  m + 350kˆ mmF34  0
 M  R  P  R × F + R × F  0
A
BA

× F34  45kˆ N  m + 114ˆi  331ˆj mm × cos 19.05ˆi  sin  19.05ˆj F34  0
CA
43

F34
 129 N  19.05=122ˆi  42ˆj N
r
43
CA
 278ˆi  288ˆj mm  ×  350ˆi N  +  486ˆi  504ˆj mm  ×  122ˆi  42ˆj N 
+  486ˆi  504ˆj mm  ×  cos 70.95ˆi  sin 70.95ˆj F  0
r
43
101kˆ N  m  41kˆ N  m+624kˆ mmF43r  0 , F43r  228 N70.95=74ˆi  215ˆj N
F  Fr  F  48ˆi  257ˆj N  261 N100.58
43
43
43
Now the lines of action for other forces may be found as shown.
 F  F43r  F43  P  F23  0
74ˆi  215ˆj N  122ˆi  42ˆj N  350ˆi N  F  0 , F =398ˆi  257ˆj N  474 N  32.85

23
23
 

 MO2  T12  R AO2 × F32  T12kˆ  100ˆi  173ˆj mm × 398ˆi  257ˆj N  0
T12  94.55 N  m
T12  94.55kˆ N  m
Ans.
422
11.9
Sketch free-body diagrams of each link and show all the forces acting. Find the
magnitude and direction of the torque that must be applied to link 2 at the posture
illustrated to drive the linkage against the forces shown.
Kinematic analysis:
R AO2  4 in30  3.464ˆi  2.000ˆj in
R  14 in67.81  5.288ˆi  12.963ˆj in
RCO4  10 in84.34  0.987ˆi  9.951ˆj in
R  14 in34.61  11.523ˆi  7.951ˆj in
BA
CA
R DO4  7 in84.34  0.691ˆi  6.966ˆj in
Force analysis:
RAO2  4 in; RCA  14 in; RO4O2  14 in; RCO4  10 in; RDO4  7 in; RBA  14 in; and RBC  8 in.
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,

. Then
F34r and F34
M  R
O4
DO4
× PD  R CO4 ×F34  0
 0.691ˆi  6.966ˆj in  ×  193ˆi  52ˆj lb  + 0.987ˆi  9.951ˆj in  × cos 5.66ˆi  sin 5.66ˆj F  0

34
1 380kˆ in  lb  9.999kˆ inF34  0
 M  R × P  R × F  R × F  0
A
BA
B
CA
43
CA

F34
 138 lb  5.66=137ˆi  14ˆj lb
r
43
5.288ˆi  12.963ˆj in  ×  100ˆi lb  + 11.523ˆi  7.951ˆj in  ×  137ˆi 14ˆj lb 
+ 11.523ˆi  7.951ˆj in  ×  cos  95.66ˆi  sin  95.66ˆj F  0
r
43
1 296kˆ in  lb  1 251kˆ in  lb 10.683kˆ inF43r  0 , F43r  238 lb  95.66=  23ˆi  237ˆj lb
F  Fr  F  160ˆi  223ˆj lb  274 lb  125.66
43
43
43
Now the lines of action for other forces may be found as shown.
 F  F43r  F43  PB  F23  0
23ˆi  237ˆj lb  137ˆi  14ˆj lb 100ˆi lb  F  0 , F =260ˆi  223ˆj lb  343 lb40.62
M  T  R
O2
12
T12  252 in  lb
AO2
× F32  0;

23
23
 

T12kˆ  3.464ˆi  2.000ˆj in × 260ˆi  223ˆj lb  0
T12  252kˆ in  lb
Ans.
423
11.10 The figure illustrates a four-bar linkage with external forces applied at points B and C.
Draw a free-body diagram of each link and show all the forces acting on each. Find the
torque that must be applied to link 2 to maintain static equilibrium at the posture
illustrated.
RAO2  75 mm; RCA  300 mm; RO4O2  400 mm; RCO4  RBA  200 mm; RBC  150 mm.
Kinematic analysis:
R AO2  75 mm  30  65ˆi  38ˆj mm
R  200 mm16.00  192ˆi  55ˆj mm
BA
RCO4  200 mm124.56  113ˆi  165ˆj mm
R  300 mm42.38  222ˆi  202ˆj mm
CA
Force analysis:
Ans.
 M  R ×P  R ×P  R ×F  0
A
BA
B
CA
C
CA
43
192ˆi  55ˆj mm ×  354ˆi  354ˆj N  + 222ˆi  202ˆj mm × 1 800ˆi N  + 222ˆi  202ˆj mm× cos124.56ˆi  sin124.56ˆj F  0
43
87.438kˆ N  m  363.000kˆ N  m  297kˆ mmF  0 ,
F  927124.56 N=  526ˆi  763ˆj N
r
43
43
F  F  P  P  F  0
43
B
C
23
526ˆi  763ˆj N  354ˆi  354ˆj N+1 800ˆi N  F23  0 ,
F =  920ˆi  1 117ˆj N  1 447 N  129.48
23
M  T  R
O2
12
AO2
T12  107.57 N  m
× F32  0;

 

T12kˆ  65ˆi  38ˆj mm × 920ˆi  1 117ˆj N  0
T12  107.57kˆ N  m
Ans.
424
11.11 Draw a free-body diagram of each member of the linkage and find the magnitudes and the
directions of all forces and moments. Compute the magnitude and direction of the torque
that must be applied to link 2 to maintain static equilibrium at the posture indicated.
RAO2  4 in; RCA  10 in; RO4O2  RCO4  8 in; RDO4  6 in; RDC  4 in; RBA  14 in; and RBC  5 in.
Kinematic analysis:
R AO2  4 in180  4.000ˆi in
R  14 in55.98  7.834ˆi  11.603ˆj in
BA
RCO4  8 in124.23  4.500ˆi  6.614ˆj in
R  10 in41.41  7.500ˆi  6.614ˆj in
CA
R DO4  6 in95.27  0.551ˆi  5.975ˆj in
Force analysis:
Ans.
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,

F34r and F34
. Then
M  R
O4
DO4
× PD  R CO4 ×F34  0
 0.551ˆi  5.975ˆj in  ×  156ˆi  90ˆj lb  +  4.500ˆi  6.614ˆj in  × cos34.23ˆi  sin 34.23ˆj F = 0

34
883kˆ in  lb  8.000kˆ inF34  0

F34
 110 lb34.23=91ˆi  62ˆj lb
425
 M  R × P + R × F  R × F  0
A
BA
B
CA
43
CA
r
43
 7.834ˆi  11.603ˆj in  ×  120ˆi lb  +  7.500ˆi  6.614ˆj in  ×  91ˆi  62ˆj lb 
+  7.500ˆi  6.614ˆj in  ×  cos 55.77ˆi  sin  55.77ˆj F  0
r
43
F43r  154 lb  55.77=87ˆi  127ˆj lb
F34  F43  4ˆi  189ˆj lb  189 lb88.79 Ans.
43
43
43
Now the lines of action for other forces may be found as shown.
4ˆi  189ˆj lb  120ˆi lb  F23  0 ,
 F  F43  PB  F23  0 ,
F23 =124ˆi  189ˆj lb  226 lb56.73 ,
F32  F23  124ˆi  189ˆj lb  226 lb123.27 Ans.
F  F P F  0,
4ˆi  189ˆj lb  156ˆi  90ˆj lb  F  0 ,
1 392kˆ in  lb  137kˆ in  lb  9.921kˆ inF43r  0 ,
F  Fr  F  4ˆi  189ˆj lb  189 lb  91.21 ,

34
D
14
14
F14 =152ˆi  279ˆj lb  318 lb  61.42 ,
M  T  R
O2
12
T12  756 in  lb
AO2
× F32  0;
F12  F32  124ˆi  189ˆj lb  226 lb56.73 Ans.
T kˆ  4.000ˆi in × 124ˆi  189ˆj lb = 0
12

 
T12  756kˆ in  lb

Ans.
426
11.12 Determine the magnitude and direction of the torque that must be applied to link 2 to
maintain static equilibrium at the posture illustrated.
RAO2  3 in; RCA  14 in; RBA  7 in; and RBC  8 in.
Kinematic analysis:
R AO2  3 in90  3.000ˆj in
RCA  14 in  12.37  13.675ˆi  3.000ˆj in
R BA  7 in  34.93  5.739ˆi  4.009ˆj in
Force analysis:
287kˆ in  lb  300kˆ in  lb  14kˆ inF14  0
F  P  P  F  F  0
B
C
14
23
50ˆj lb  100ˆi lb  1ˆj lb  F23  0 ,
M  T  R
O2
12
T12  300 in  lb
AO2
F14  1 lb90=1ˆj lb
 F32  0
F23 =100ˆi  51ˆj lb  112 lb  27.02
T kˆ + 3.000ˆj in × 100ˆi  51ˆj lb  0
12

 
T12  300kˆ in  lb

Ans.
427
11.13 Figure P11.13a shows the Figee floating crane with lemniscate boom configuration, and
Figure 11.13b shows is a schematic diagram of the crane with dimensions given in the
legend. The lifting capacity is 16 T (where 1 T = 1 metric ton =1 000 kg) including the
grab which is about 10 T. The maximum outreach is 30 m, which corresponds to the
position  2  49 . Minimum outreach is 10.5 m at  2  132. For the maximum
outreach posture and a grab load of 10 T (under standard gravity), find the bearing
reactions at A, B, O2 , and O4 , as well as the torque T12 at O2. Notice that the
photograph shows a counterweight on link 2; neglect this weight and also the weights of
the members.
(a) Photograph and (b) RAO2  14.7 m; RBA  6.5 m; RBO4  19.3 m; RCA  22.3 m; RCB  16 m. and
RO2O4  6.4ˆi  5.3ˆj m.
(Courtesy of B.V. Machinefabriek Figee, Haarlem, Holland).
428
Kinematic analysis:
R AO2  14.700 m49.00  9.644ˆi  11.094ˆj m , R BO4  19.300 m59.70  9.739ˆi  16.663ˆj m
R BA  6.500 m2.37  6.494ˆi  0.269ˆj m , RCA  22.300 m14.39  21.600ˆi  5.543ˆj m
Force analysis:
Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane
would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is
W  mg  1 000 kg  9.81 m/s2   9.810 kN . Therefore, the stated load on the crane is F  98.100 kN .
 M  R ×F  R ×F  0
A
BA
43
CA
 6.494ˆi  0.269ˆj m  × cos59.70ˆi  sin 59.70ˆj F +  21.600ˆi  5.543ˆj m ×  98.100ˆj kN   0
43
5.471kˆ mF43  2 119kˆ kN  m  0 ,
F  38759.70 kN=195ˆi  334ˆj kN ,
43
F  F  F  F  0
43
Ans.
23
195ˆi  334ˆj kN  98.1ˆj kN  F23  0 ,
F =  195ˆi  236ˆj kN  307 kN 129.59 ,
23
M  M  R
O2
F14  F34  387 kN59.70=195ˆi  334ˆj kN ,
12
AO2
M12  113 kN  m
×F32  0 ;
F12 =  F32 = 195ˆi  236ˆj kN  307 kN129.59 ,
Ans.
M kˆ  9.644ˆi  11.094ˆj m × 195ˆi  236ˆj kN  0
12

M12  113kˆ kN  m


Ans.
429
11.14 Repeat Problem 11.13 for the minimum outreach posture.
Kinematic analysis:
R AO2  14.700 m132.00  9.836ˆi  10.924ˆj m , R BO4  19.300 m120.35  9.751ˆi  16.656ˆj m
R  6.500 m3.81  6.486ˆi  0.432ˆj m , R  22.300 m15.83  21.454ˆi  6.083ˆj m
BA
CA
Force analysis:
Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane
would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is
W  mg  1 000 kg  9.81 m/s2   9.810 kN . Therefore, the stated load on the crane is F  98.1 kN .
 M  R ×F  R ×F  0
A
BA
43
CA
 6.486ˆi  0.432ˆj m  × cos120.35ˆi  sin120.35ˆj F +  21.454ˆi  6.083ˆj m  ×  98.1ˆj kN  = 0
43
5.815kˆ mF43  2 105kˆ kN  m  0 ,
F  362 kN120.35=  183ˆi  312ˆj kN ,
43
F  F  F  F  0
43
23
F23 =183ˆi  214ˆj kN  282 kN  49.51 ,
M  M  R
O2
12
AO2
M12  109 kN  m
×F32  0,
F14  F34  362 kN120.35= 183ˆi  312ˆj kN
183ˆi  312ˆj kN  98.1ˆj kN  F  0
Ans.
23
F12 =  F32 =183ˆi  214ˆj kN  282 kN 49.51 , Ans.
M kˆ  9.836ˆi  10.924ˆj m × 183ˆi  214ˆj kN  0
12

M12  109kˆ kN  m


Ans.
430
11.15 Repeat Problem 11.7 assuming coefficients of Coulomb friction c  0.20 between links
1 and 6 and c  0.10 between links 3 and 4. Determine the torque T12 necessary to
drive the system, including friction, against the load P.
y
RAO2  2.5 in; RO2O4  6 in; RCO
 10 in; RBO4  16 in; and RBC  8 in.
2
See the figure and solution for Prob. 11.7 for the kinematic and frictionless solutions. For
friction between links 1 and 6, the friction angle is   tan 1  0.20  11.31 . Because the
impending motion VC6 /1 is to the left the friction force f16  c F16n is toward the right.
Also, since the non-friction normal force F16n is downward (from the solution for Prob.
11.7), the total force F16 acts at the angle 90  11.31  78.69 . Therefore,
F  Pˆi  cos 78.69ˆi  sin  78.69ˆj F ˆj  cos175.20ˆi  sin175.20ˆj F  0



100 lb  cos 78.69F16  cos175.20F56  0 ,
F16  8.710 lb ,
F56  102.066 lb ,

16

56
sin  78.69F16  sin175.20F56  0
F  102.066 lb175.20  101.708ˆi  8.541ˆj lb . For
56
friction between links 3 and 4, the friction angle is   tan 1  0.10   5.71 . Because the
impending motion VA3 / 4 is upward, the friction force f 43 is downward, and the reaction
force f34  c F34n is upward. Also, since the non-friction normal force F34n is toward the
left (from the solution for Prob. 11.7), the total force F34 acts at the angle
163.37  5.71  157.66 . Therefore,
 MO4  R BO4 ×F54  R AO4 ×F34  0
 4.578ˆi 15.331ˆj in  × 101.708ˆi  8.541ˆj lb   2.136ˆi  7.154ˆj in  × cos157.66ˆi  sin157.66ˆj F  0
34
 1 598.386 in  lb  7.429 inF34  kˆ  0 ,
F34  215.157 lb157.66= 199.007ˆi  81.784ˆj lb
M  T  R
T12kˆ  2.165ˆi  1.250ˆj in × 199.007ˆi  81.784ˆj lb  0
O2
12
AO2
T12  425.82 in  lb
× F32  0 ,


T12  425.82kˆ in  lb

Ans.
431
11.16 Repeat Problem 11.12 assuming a coefficient of static friction   0.15 between links 1
and 4. Determine the torque T12 necessary to overcome friction.
RAO2  3 in; RCA  14 in; RBA  7 in; and RBC  8 in.
See the figure and solution for Prob. 11.12 for the kinematic and frictionless solution. For
friction between links 1 and 4, the friction angle is   tan 1  0.15  8.53 . Since the
impending motion VC4 /1 is to the right the friction force f14  c F14n is toward the left.
Also, since the non-friction normal force F14n is upward (from the solution of Prob. 11.12),
the total force F14 acts at the angle 90  8.53  98.53 . Therefore,
M  R P  R ×P  R ×F  0
A
BA
B
CA
C
CA
14
5.739ˆi  4.009ˆj in  × 50ˆj lb  + 13.675ˆi  3.000ˆj in  ×  100ˆi lb 
+ 13.675ˆi  3.000ˆj in  ×  cos 98.53ˆi  sin 98.53ˆj F  0
14
287kˆ in  lb  300kˆ in  lb  13.078kˆ inF14  0 , F14  0.994 lb98.53=  0.147ˆi  0.983ˆj lb
F  P  P  F  F  0
B
C
14
23
50ˆj lb  100ˆi lb  0.147ˆi  0.983ˆj lb  F23  0 , F23 =100.1ˆi  51.0ˆj lb  112.4 lb  26.98
T kˆ  3.000ˆj in  100.1ˆi  51.0ˆj lb  0
M  T  R  F  0
O2
12
AO2
T12  300.4 in  lb
32
12

 
T12  300.4kˆ in  lb

Ans.
432
11.17 In each case shown, pinion 2 is the driver, gear 3 is an idler, the gears have diametral pitch
of 6 and 20 pressure angle. For each case, sketch the free-body diagram of gear 3 and
show all forces acting. For (a) pinion 2 rotates at 600 rev/min and transmits 18 hp to the
gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits 25 hp to the gearset.
a) R2 
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
R3 
 600 rev/min  2   62.832 rad/s cw ,
N3
34 teeth

 2.833 in
2 P 2  6 teeth/in 
R
1.500 in
3  2  2 
62.832 rad/s  33.268 rad/s ccw
60 s/min
R3
2.833 in
18 hp  550 ft  lb/s  hp 12 in/ft   1 260 lb
P
F23t 

R33
 2.833 in  33.268 rad/s 
2 
F23  F23t cos   1 260 lb cos20  1 341 lb ,
F  F  F  F  0
23
b) R2 
2 
43
13
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
 900 rev/min  2   94.248 rad/s ccw ,
F43  F23  1 341 lb
Ans.
F13  F  F  1 260 lb  1 260 lb  2 520 lb
Ans.
t
23
R3 
3 
t
43
N3
36 teeth

 3.000 in
2 P 2  6 teeth/in 
R2
1.500 in
2 
94.248 rad/s  47.124 rad/s cw
R3
3.000 in
60 s/min
 25 hp  550 ft  lb/s  hp 12 in/ft   1 167 lb
P
F23t 

R33
 3.000 in  47.124 rad/s 
F23  F23t cos   1 167 lb cos20  1 242 lb ,
F43  F23  1 242 lb
Ans.
433
F  F  F  F  0
23
43
13
F13    F23  F43 
  1 242 lb  20  1 242 lb110 
 1 050 lb225
c) R2 
2 
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
 900 rev/min  2   94.248 rad/s ccw ,
F23t 
60 s/min
P
R33

3 
N3
36 teeth

 3.000 in
2 P 2  6 teeth/in 
R2
1.500 in
2 
94.248 rad/s  47.124 rad/s cw
R3
3.000 in
 25 hp  550 ft  lb/s/hp 12 in/ft   1 167 lb
 3.000 in  47.124 rad/s 
F23  F23t cos   1 167 lb cos20  1 242 lb ,
F  F  F  F  0
23
R3 
Ans.
43
13
F43  F23  1 242 lb
Ans.
F13    F23  F43 
  1 242 lb  20  1 242 lb  70 
 2 250 lb135
Ans.
434
11.18 A 15-tooth spur pinion has a diametral pitch of 5 and 20 pressure angle, rotates at 600
rev/min, and drives a 60-tooth gear. The drive transmits 25 hp. Construct a free-body
diagram of each gear showing upon it the tangential and radial components of the forces
and their proper directions.
R2 
2 
N2
15 teeth

 1.500 in
2 P 2  5 teeth/in 
 600 rev/min  2   62.832 rad/s
F32t 
60 s/min
R3 
N3
60 teeth

 6.000 in
2 P 2  5 teeth/in 
3 
R2
1.500 in
2 
62.832 rad/s  15.708 rad/s
R3
6.000 in
 25 hp  550 ft  lb/s  hp 12 in/ft   1 751 lb , F r  F t tan   637 lb
P

32
32
R33
1.500 in  62.832 rad/s 
F23t  F32t  1 751 lb
F23r  F32r  637 lb
Note that both free-body diagrams are incomplete. The driving torque T12 is not shown on
body 2, and the loading (either a force or a torque) is not shown on body 3; also the
directions of the transmitted forces may not be correct since the direction of rotation of
each gear is not known.
435
11.19 A 16-tooth pinion on shaft 2 rotates at 1 720 rev/min and transmits 5 hp to the doublereduction gear train. All gears have 20 pressure angle. Find the magnitude and direction
of the radial force that each bearing exerts against the shaft.
R2 
N2
16 teeth

 1.000 in
2 P 2  8 teeth/in 
RA 
NA
64 teeth

 4.000 in
2P
2  8 teeth/in 
RB 
NB
24 teeth

 2.000 in
2 P 2  6 teeth/in 
R4 
NA
36 teeth

 3.000 in
2 P 2  6 teeth/in 
2 
1 720 rev/min  2   180.118 rad/s
3 
R2
1.000 in
2 
180.118 rad/s  45.029 rad/s
RA
4.000 in
60 s/min
R
2.000 in
 4  B 3 
45.029 rad/s  30.020 rad/s
R4
3.000 in
F23t 
F43t 
P
RA3

 5 hp  550 ft  lb/s/hp 12 in/ft   183 lb , F  F t cos   195 lb
23
23
 4.000 in  45.029 rad/s 
RA t 4.000 in
F23 
183 lb  366 lb
RB
2.000 in
F43  F43t cos   390 lb
436
Choosing a coordinate system with origin at C as shown we have
FA  F23  195 lb20  183ˆi  67ˆj lb
R A  4.000ˆj  2.000kˆ in
F  F  390 lb  20  366ˆi  133ˆj lb
R  2.000ˆj  10.000kˆ in
B
43
B
FC  F ˆi  FCy ˆj
F  F x ˆi  F y ˆj
RC  0
x
C
D
D
D
 M  R ×F  R ×F  R ×F  0
C
A
A
B
B
D
R D  12.000kˆ in
D
 4.000ˆj  2.000kˆ in  × 183ˆi  67ˆj lb  2.000ˆj 10.000kˆ in  × 366ˆi 133ˆj lb  12.000kˆ in ×  F ˆi  F ˆj  0
 133ˆi  366ˆj  732kˆ in  lb   1 334ˆi  3 664ˆj  732kˆ in  lb    12 F ˆi 12 F ˆj in   0
x
D
y
D
F  F  F  F  F  0
B
C
x
D
FD  350 lb163.42  336ˆi  100ˆj lb Ans.
FDx  336 lb, FDy  100 lb
A
y
D
D
183ˆi  67ˆj lb  366ˆi 133ˆj lb  F  336ˆi 100ˆj lb  0 , F  216 lb188.86  214ˆi  33ˆj lb Ans.
C
C
437
11.20 Solve Problem 11.17 if each pinion has right-hand helical teeth with a 30 helix angle
and a 20 pressure angle. All gears in the train are helical, and the normal diametral
pitch is 6 teeth/in for each case.
Since the pressure angles and the helix angle are related by cos  tan n tan t ,
t  tan 1  tan n cos   tan 1  tan 20 cos30  22.80
a) R2 
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
R3 
 600 rev/min  2   62.832 rad/s cw ,
N3
34 teeth

 2.833 in
2 P 2  6 teeth/in 
R
1.500 in
3  2  2 
62.832 rad/s  33.268 rad/s ccw
60 s/min
R3
2.833 in
18 hp  550 ft  lb/s/hp 12 in/ft   1 260 lb
P
F23t 

R33
 2.833 in  33.268 rad/s 
2 
F23r  F23t tan t  1 260 lb  tan 22.80  530 lb ,
F  1 260ˆi  530ˆj  727kˆ lb
F23a  F23t tan  1 260 lb  tan 30  727 lb
F  1 260ˆi  530ˆj  727kˆ lb
F  F  F  F  0
 M R ˆj×F  R ˆj F  M  0
F13  2 521ˆi lb
23
23
43
13
3
23
3
43
43
Ans.
13
 2.833ˆj in  × 1 260ˆi  530ˆj  727kˆ lb   2.833ˆj in  × 1 260ˆi  530ˆj  727kˆ lb  M  0
 2.833ˆj in  × 1 260ˆi  530ˆj  727kˆ lb   2.833ˆj in  × 1 260ˆi  530ˆj  727kˆ lb  M  0
13
13
M13  4 119ˆi in  lb This moment must be supplied by the shaft bearings.
Ans.
438
b) R2 
2 
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
 900 rev/min  2   94.248 rad/s ccw ,
R3 
3 
N3
36 teeth

 3.000 in
2 P 2  6 teeth/in 
R2
1.500 in
2 
94.248 rad/s  47.124 rad/s cw
R3
3.000 in
60 s/min
 25 hp  550 ft  lb/s/hp 12 in/ft   1 167 lb
P
F23t 

R33
 3.000 in  47.124 rad/s 
F23r  F23t tan t  1 167 lb  tan 22.80  491 lb , F23a  F23t tan  1 167 lb  tan 30  674 lb
F  1 167ˆi  491ˆj  674kˆ lb
F  491ˆi  1 167ˆj  674kˆ lb
23
F  F  F  F  0
 M R ˆj×F  R ˆi ×F  M  0
23
43
13
3
23
3
43
43
F13  677ˆi  677ˆj lb
Ans.
13
3.000ˆj in  × 1 167ˆi  491ˆj  674kˆ lb   3.000ˆi in  ×  491ˆi 1 167ˆj  674kˆ lb  M  0
13
M13  2 022ˆi  2 022ˆj in  lb This moment must be supplied by the shaft bearings.
c) R2 
2 
N2
18 teeth

 1.500 in
2 P 2  6 teeth/in 
 900 rev/min  2   94.248 rad/s ccw ,
R3 
3 
Ans.
N3
36 teeth

 3.000 in
2 P 2  6 teeth/in 
R2
1.500 in
2 
94.248 rad/s  47.124 rad/s cw
R3
3.000 in
60 s/min
 25 hp  550 ft  lb/s/hp 12 in/ft   1 167 lb
P
F23t 

R33
 3.000 in  47.124 rad/s 
F23r  F23t tan t  1 167 lb  tan 22.80  491 lb ,
F  1 167ˆi  491ˆj  674kˆ lb
F23a  F23t tan  1 260 lb  tan 30  674 lb
F  491ˆi  1 167ˆj  674kˆ lb
F  F  F  F  0
 M R ˆj×F  R ˆi ×F  M  0
F13  1 658ˆi  1 658ˆj lb
23
23
43
13
3
23
3
43
43
Ans.
13
3.000ˆj in  × 1 167ˆi  491ˆj  674kˆ lb    3.000ˆi in  × 491ˆi 1 167ˆj  674kˆ lb  M  0
13
M13  2 022ˆi  2 022ˆj in  lb This moment must be supplied by the shaft bearings. Ans.
439
11.21 Analyze the gear shaft of Example 11.8 and find the bearing reactions FC and FD .
The solution is shown in Fig. 11.20c.
FC  118ˆi  140ˆj  251kˆ lb
F  71ˆi  1551kˆ lb
D
Ans.
440
11.22 In each of the bevel gear drives shown, bearing A takes both thrust load and radial load,
whereas bearing B takes only radial load. The teeth are cut with a 20 pressure angle. For
(a) T2  180ˆi in  lb and for (b) T2  240kˆ in  lb . Compute the bearing loads for each
case.
a)   tan 1  32 teeth 16 teeth   63.43 
F32t  T2 R2  180 in  lb 0.69 in  261 lb
F32r  F32t tan  cos   42.5 lb
F  42.5ˆi  261ˆj  84.9kˆ lb
F32a  F32t tan  sin   84.9 lb
23
 M  R ×F  R ×F  T  0
A
BA
B
PA
23
3
 2.000kˆ in  ×  F ˆi  F ˆj  1.380ˆi  2.360kˆ in  ×  42.5ˆi  261ˆj  84.9kˆ lb   T kˆ  0
 2.000 inF ˆi  2.000 inF ˆj   615.65ˆi 16.98ˆj  360kˆ in  lb  T kˆ  0
x
B
y
B
y
B
3
x
B
3
F  F  F  F  0
FB  8.5ˆi  308ˆj lb
F  51ˆi  47ˆj  85kˆ lb
b)   tan 1 18 teeth 24 teeth   36.87 
F32t  T2 R2  240 in  lb 1.28 in  188 lb
F32r  F32t tan  cos   54.6 lb
F  54.6ˆi  188ˆj  40.9kˆ lb
F32a  F32t tan  sin   40.9 lb
T3  360kˆ in  lb
A
B
23
A
Ans.
Ans.
32
 M  R ×F  R ×F  T  0
A
BA
B
PA
32
2
 2.000kˆ in  ×  F ˆi  F ˆj   1.280ˆi  0.800kˆ in  × 54.6ˆi 188ˆj  40.9kˆ lb    240kˆ in  lb  0
 2.000 inF ˆi  2.000 inF ˆj   150ˆi  8ˆj  240kˆ in  lb    240kˆ in  lb   0
x
B
y
B
y
B
x
B
F  F  F  F  0
A
B
23
FB  4ˆi  75ˆj lb
F  50ˆi  263ˆj  41kˆ lb
A
Ans.
Ans.
441
11.23 The figure shows a gear train composed of a pair of helical gears and a pair of straighttooth bevel gears. Shaft 4 is the output of the train and delivers 6 hp to the load at a speed
of 370 rev/min. All gears have pressure angles of 20 . If bearing E is to take both thrust
load and radial load, whereas bearing F is to take only radial load, determine the force that
each bearing exerts against shaft 4.
The diameters of the bevel gears at their large ends are
R4  N4 2P   40 teeth   2  8 teeth/in   2.500 in
R3  N3 2P   20 teeth   2  8 teeth/in   1.250 in
  tan 1  R4 R3   63.43
  tan 1  R3 R4   26.57
The average pitch radii are
R4,avg  R4  0.500sin   2.053 in
R3,avg  R3  0.500sin   1.026 in
4 
 370 rev/min  2   38.746 rad/s
60 s/min
 6 hp  550 ft  lb/s/hp 12 in/ft   498 lb
P
F34t 

R4,avg4
 2.053 in  38.746 rad/s 
F34r  F34t tan  cos   81 lb
F  162ˆi  81ˆj  498kˆ lb
F34a  F34t tan  sin   162 lb
34
 M  R ×F  R ×F  T  0
E
FE
F
PE
34
4
 2.500ˆi in  ×  F ˆj  F kˆ   0.723ˆi  2.053ˆj in  ×  162ˆi  81ˆj  498kˆ lb    1 022ˆi in  lb  0
 2.500 inF ˆj  2.500 inF kˆ   1 022ˆi  360ˆj  274kˆ in  lb    1 022ˆi in  lb  0
y
F
z
F
z
F
y
F
F  F  F  F  0
E
F
34
FF  110ˆj  144kˆ lb
F  162ˆi  191ˆj  354kˆ lb
E
Ans.
Ans.
442
11.24 Using the data of Problem 11.23, find the forces exerted by bearings C and D onto shaft 3.
Which of these bearings should take the thrust load if the shaft is to be loaded in
compression?
The pitch radius of the helical gear S is
RS  NS 2P  35 teeth  2 12 teeth/in   1.458 in
F23t  F43t R3,avg RS  498 lb 1.026 in 1.458 in   350 lb F23  F23 tan    350 lb  tan 20  128 lb
r
t
F23a  F23t tan   350 lb  tan 30  202 lb
F  128ˆi  202ˆj  350kˆ lb
23
M  R
C
DC
×FD  R PC ×F43  R RC ×F23  0
1.750ˆj in  ×  F ˆi  F kˆ   1.026ˆi  2.697ˆj in × 162ˆi  81ˆj  498kˆ lb  1.458ˆi  0.875ˆj in × 128ˆi  202ˆj  350kˆ lb  0
1.750 inF ˆi 1.750 inF kˆ   1 343ˆi  511ˆj  354kˆ in  lb   306ˆi  511ˆj  407kˆ in  lb  0
x
D
z
D
z
D
x
D
F  F  F  F  F  0
C
D
23
43
FD  30ˆi  942kˆ lb
F  64ˆi  121ˆj  94kˆ lb
C
Since the thrust force is in the jˆ direction, C should be a thrust bearing.
Ans.
Ans.
Ans.
443
11.25 Use the method of virtual work to solve the slider-crank linkage of Problem 11.2.
RAO2  75 mm, RBA  350 mm.
  sin 1  r sin     sin 1  75 mmsin105 350 mm  11.95
x  r cos  cos  75 mmcos105  350 mmcos11.95  323 mm
The first-order kinematic coefficient is
x  dx d  RI24 I12  x tan   323 mm tan11.95  68.36 mm
T12  Px  900 N  68.36 mm  61.5 N  m cw
Ans.
444
11.26 Use the method of virtual-work to solve the four-bar linkage of Problem 11.5.
RI24 I12  2.577 in , RI24 I14  4.577 in , RDO4  7 in152.64
The first-order kinematic coefficient is
4  d4 d2  RI24 I12 RI24 I14  2.577 in 4.577 in  0.563
T14  RDO4 P sin152.64  7 in  50 lb  sin152.64  161 in  lb cw
T12  T14 d4 d2  T144   161 in  lb cw  0.563  90.56 in  lb ccw
Ans.
445
11.27 Use the method of virtual work to analyze the crank-shaper linkage of Problem 11.7.
Given that the load remains constant at P  100ˆi lb, find and plot a graph of the crank
torque T12 for all postures in the cycle using increments of 30 for the input crank.
y
RAO2  2.5 in; RO2O4  6 in; RCO
 10 in; RBO4  16 in; and RBC  8 in.
2
xAO4  RAO4 cos 4  2.5cos 2
 6  2.5sin  2 

 2.5cos 2 
yAO4  RAO4 sin  4  6  2.5sin  2
 4  tan 1 
RAO4  42.25  30sin  2
yI24 I14  RAO4 sin 4
 4 
yBC  8sin 5  16  16sin  4
5  sin 1  2  2sin  4 
yI46C  yBC  xC xCB 
 8sin 5  8cos 5  16 cos  4   8cos 5  
 8  sin 5  2 cos  4 tan 5 
dxC d4  yI46 I14  16  8sin 5  16cos 4 tan 5
T12   dxC d4  d4 d2  P
d 4 yI24 I14  6

 1  6sin  4 RAO4
d 2
yI24 I14
446
Values for one cycle are shown in the following table.
 2 (deg.)
 4 (deg.)
RAO4 (in)
d 4 d 2
5 (deg.)
dxC d 4 (in)
T12 (in  lb)
0
30
60
90
120
150
180
210
240
270
300
330
360
67.38
73.37
81.30
90.00
98.70
106.63
112.62
114.50
108.05
90.00
71.95
65.50
67.38
6.500
7.566
8.260
8.500
8.260
7.566
6.500
5.220
4.034
3.500
4.034
5.220
6.500
0.147 93
0.240 15
0.281 97
0.294 12
0.281 97
0.240 15
0.147 93
0.045 93
0.414 16
0.714 29
0.414 16
0.045 93
0.147 93
8.85
4.80
1.32
0.00
1.32
4.80
8.85
10.37
5.65
0.00
5.65
10.37
8.85
15.727
15.715
15.871
16.000
15.760
14.946
13.811
13.346
14.722
16.000
15.703
15.774
15.727
232.66
377.40
447.53
470.59
444.38
358.93
204.31
61.30
609.72
1 142.86
650.35
72.45
232.66
The values of T12 from this table are graphed as follows:
447
11.28 Use the method of virtual work to solve the four-bar linkage of Problem 11.10.
3  d3 d2  RI I
RI23I13  75 mm 688 mm  0.1089
RB  dR B d2  3kˆ  R BI13  0.1089kˆ   568 mm135.33  61.881 mm45.33
R  dR d   kˆ  R  0.1089kˆ   662 mm124.56   72.153 mm34.56
23 12
C
C
2
3
CI13
T12  PB RB  PC RC  0
T12  PB RB  PC RC
   500 N135   61.881 mm45.33   1 800 N0  72.153 mm34.56 
 30.940cos89.67 N  m  129.876cos34.56 N  m  107 N  m
T12  107 N  m cw
Ans.
448
11.29 A car (link 2) which weighs 2 000 lb is slowly backing a 1 000 lb trailer (link 3) up a 30
inclined ramp. The car wheels are of 13-in radius, and the trailer wheels have 10-in
radius; the center of the hitch ball is also 13 in above the roadway. The centers of mass of
the car and trailer are located at G2 and G3 , respectively, and gravity acts vertically
downward. The weights of the wheels and friction in the bearings are considered
negligible. Assume that there are no brakes applied on the car or on the trailer, and that
the car has front-wheel drive. Determine the loads on each of the wheels and the
minimum coefficient of static friction between the driving wheels and the road to avoid
slipping.
For the trailer:
RG3B  37.801ˆi  34.526ˆj in  51.196 in42.41
F  756 lb120  378ˆi  655ˆj lb
 M  50.000 inF kˆ  R  W  0
B
13
G3 B
F  F  F  W  0
13
23
3
13
Ans.
F23  512 lb42.41  378ˆi  345ˆj lb
3
For the car:
 MP  80 inF12Rkˆ  32 in  2 000 lb  kˆ  116ˆi  13ˆj in  F32  0

 F  F  f ˆi  F  F  0
F
12
12
R
12
32
  f12 F12F  378 lb 1 106 lb

F12R  1 239ˆj lb
F F  1 106ˆj lb
Ans.
f12  378ˆi lb
Ans.
  0.34
Ans.
12
Ans.
449
11.30 Repeat Problem 11.29 assuming that the car has rear-wheel drive rather than front-wheel
drive.
The entire solution is identical with that of Prob. 13.29 except that friction force f12 acts
on the rear wheel of the car instead of on the front wheel. The solution process and all
values are the same until the final step. Then
Ans.
  f12 F12R  378 lb 1 239 lb
  0.31
450
11.31 The low-speed disk cam with oscillating flat-faced follower is driven at a constant shaft
speed. The displacement curve for the cam has a full-rise cycloidal motion, defined by Eq.
(6.13) with parameters L  30 ,   150 , and a prime circle radius Ro  30 mm ; the
instant pictured is at 2  112.5. A force of FC  8 N is applied at point C and continues
at 45 from the face of the follower as illustrated. Use the virtual work approach to
determine the torque T12 required on the camshaft at the instant shown to produce this
motion.
RO2O3  50 mm; RB  42 mm; and RC  150 mm.
The moment on link 3 caused by the output load is
M13  RCO3 ×FC  150 mm 8 N  sin  135 kˆ  0.849kˆ N  m
From Eq. (6.13b),
L
2  30 
112.5 
y  1  cos

1  cos 2
  0.200

  150 
150 
From virtual work
T12   d3 d2 M13   yM13  0.200 0.849kˆ N  m  0.170kˆ N  m


Ans.
11.32 Repeat Problem 11.31 for the entire lift portion of the cycle, finding T12 as a function of
2.
From Prob. 11.31
M13  RCO3  FC  150 mm 8 N  sin  135 kˆ  0.849kˆ N  m
360 2 
L
2  30 
1  cos

1  cos
  0.200 1  cos 2.4 2 

  150 
150 
T12   yM13  0.200 1  cos2.42  0.849kˆ N  m  0.170 1  cos2.42  kˆ N  m Ans.
y 


451
11.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an applied
torque T. Assume a coefficient of static friction of  between the disk and ground and
that all other joints are frictionless. A force F is acting vertically downward on the bar at a
distance d from the pivot O2 . Assume that the weights of the links are negligible in
comparison to F. Find an equation for the torque T required as a function of the distance
x  RCO2 , and an equation for the final distance x that is reached when friction no longer
allows further movement.
d cos 
d
 FB
x cos 
x
Rd
T  F23 R sin   FB
sin 
 MC  0
x
But, for geometric compatibility, R  x sin  . Therefore,
2
T  FB d sin 2   FB d  R x 
M  0
O2
F32  FB
Ans.
Also,
R
sec
1  tan 2 
1
x
R
R
R
1
sin 
tan 
tan 
tan 2 
Motion is still possible as long as tan    , or as long as
x  R 1  2 1
Ans.
452
11.34 For the linkage in the posture illustrated, an external torque T14  50kˆ in  lb is acting
on link 4 about O4 and a horizontal force FC is acting at point C on link 3 to hold the
linkage in static equilibrium. Assume that gravity is acting into the page and the effects
of friction can be neglected. (a) Draw free body diagrams of links 2, 3, and 4. (b)
Determine the magnitudes, directions, and locations of all internal reaction forces. (c)
Determine the magnitude and direction of the external force acting at point C.
RO4O2  2ˆi  0.864ˆj in, RAO2  2.6 in, RCA  6 in.
Since link 2 is a two-force member with no external moment, its free-body diagram must
be as illustrated in the figure below. Also, this establishes the line of action of F23 on the
free-body diagram of link 3. Since link 3 is a three-force member with no external
moment and since, without friction, the line of action of F43 must be perpendicular to the
surface of link 4, then the free body diagram must be as illustrated in the figure below.
Therefore, the three free-body diagrams appear as shown.
453
Because the measured distance d = 0.50 in shows that F34 falls within the length of link 4,
block 4 does not tumble. The sense of the applied torque T14 establishes the sense of each
force of the couple F34 and F14. From F34 the sense of F43 becomes known. The sum of
moments on link 3 about points A and C then establishes the senses of the forces F23 and
FC. From F23, the senses of the forces on link 2 become known.
Starting with link 4, the sum of moments about point O4, we find
T
50 in  lb
F34  14 
 100 lb
Ans.
d
0.50 in
and from this we know
F14  F43  100 lb
Ans.
Taking moments on link 3 about point C gives us
RO4C  d F43 1.50 in 100 lb
F23 

 50 lb
Ans.
RBC
3.00 in
and from this
F32  F12  50 lb
Ans.


Finally, taking moments about point A on link 3, we find
RO4 A  d F43  4.50 in 100 lb
FC 

 75 lb
RCA
6 in
The directions of all forces are as shown in the free-body diagrams.


Ans.
454
11.35 A horizontal force FC  25 N is acting at point C on link 4 and an external torque T2 is
acting on link 2. The coefficient of friction between link 4 and the ground link is   0.3
and the coefficient of friction between link 2 and the ground link is not specified, but is
large enough that there is no slip at E. Assume that gravity is acting into the plane of the
figure. Block 4 is 60 mm wide by 40 mm high with pin B centrally located. Point C is 5
mm above the centerline. (a) Determine the magnitude and direction of the external
torque T12 necessary to overcome the force FC . (b) Determine the magnitudes, the
directions, and the locations of the internal reaction forces. (c) Determine whether link 4
slipping or tipping on the ground link 1.
R BA  180 mm 30, REA  30 mm.
Since link 3 is a two-force member in compression, the free-body diagram of link 4
without friction must be as shown in the figure on the left below. Note that the force F14
must act downward for the forces to sum to zero. Also note that the forces are concurrent
at the point marked D. Now, since the problem statement says “overcome the force FC,”,
the impending motion of link 4, with friction, must be to the right. Also, since the
coefficient of friction is   0.3 , the friction angle is   tan 1  0.3  16.7 . Therefore,
the free-body diagram of link 4 with friction must be as shown in the figure on the right.
455
The force polygon for link 4 is as shown next, and we note that F14n is larger than F14
without friction. Thus, F14f can not be added to the frictionless solution by superposition.
From this force polygon we measure
F14  18.23 N   106.7 and F34  34.92 N  30
Then, from action and reaction, noting again that link 3 is a two-force member,
F43  34.92 N   150, F23  34.92 N  30, F32  34.92 N   150
From this, we see that the free-body diagram of link 2 must be as shown here
where we have already found
F32  34.92 N   150 and, therefore, F12  34.92 N  30
Taking moments about point A, we find the torque T12
T12  F12 d   34.92 N  26 mm   0.908 N  m cw
Ans.
Ans.
Ans.
Ans.
The friction angle between the frame and link 2 is   60 . Therefore, the coefficient of
friction at that location must be
Ans.
  tan 60  1.73
Since F14 acts down on the top surface of link 4, link 4 is slipping (not tipping) on the
ground link.
Ans.
456
11.36 For the linkage in the posture illustrated, a constant external torque T12  180kˆ in  lb is
acting on link 2 about the shaft O2 and a horizontal external force P is acting on link 5 to
hold the linkage in static equilibrium. Assume that gravity is acting into the page and
effects of friction in the linkage can be neglected. (a) Draw free body diagrams for links
2, 3, 4, 5, and 6. (b) Determine the magnitudes, directions, and locations of the internal
reaction forces. (c) Determine the magnitude and direction of the external force P.
RO2  7.0ˆi  5.0ˆj in, RAO2  5.0 in, RBA  2.75 in, RCD  6.0 in, and RBD  2.818 in.
Blocks 5 and 6 are each 2 in by 1 in with the pins centrally located.
The free body diagrams of links 2 and 3 are as shown in the figures below. Starting with
link 3, we recognize this as a two-force member with no external moments. This gives the
lines of action of F23 and F43 and, therefore, of F32 onto link 2 and F34 onto link 4.
Next, on link 2, we know the length of the link and the magnitude of the applied torque.
Thus,
F32  T12 RAO2  180 in  lb 5.0 in  36 lb
Ans.
F12  F32  F23  F43  F34  36 lb
Ans.
Now we we draw the free-body diagrams of links 6, 4, and 5 as shown next. Recognizing
that link 6 is a two-force member, we find the lines of action of F16, F46, and F64. Then we
see that both link 4 and link 5 are three-force members with no applied moments.
Therefore, each has a concurrency point which establishes the remaining lines of action.
457
Now, summing moments on link 4 about point C, using measured distances, we find
2.25 in
Ans.
F64 
F34  19.06 lb
4.25 in
F16  F46  F64  19.06 lb
Ans.
Then, summing forces on link 4 gives
F54    F64  F34   19.06ˆi  36ˆj lb  40.73 lb117.9
Ans.
and summing forces on link 5 gives
F45  F54  19.06ˆi  36ˆj lb  40.73 lb  62.1
F  36ˆj lb
P  19.06ˆi lb
15
Ans.
Ans.
458
11.37 For the linkage in the posture illustrated, a torque T12  0.24kˆ N  m is acting on link 2 at
the crankshaft O2. A force P is applied to point D on link 4 at an angle of 45o to hold the
linkage in static equilibrium. Gravity is acting into the plane and friction in the linkage
can be neglected. Determine the magnitude and the direction of the force P. Determine the
magnitudes, directions, and locations of the internal reaction forces in the linkage. Is link
4 slipping or tipping on the ground link?
R AO2  40ˆi  30ˆj mm, RBO2  70 mm, and R DO2  40ˆi  50ˆj mm, and RCB  30 mm.
The free-body diagrams of links 2 and 3 are as shown in the figures below. Starting with
link 3, we recognize this as a two-force member with no external moments. This gives the
lines of action of F23 and F43 and, therefore, of F32 onto link 2 and F34 onto link 4.
Next, on link 2, we know the length of the link and the magnitude of the applied torque.
Thus,
F32  T12 d  0.24 N  m 30 mm  8 N
Ans.
F12  F32  F23  F43  F34  8 N
Ans.
The free-body diagram of link 4 appears as in the figure on the left below. Treating link 4
as a three-force member with no external moment we would find that force F14 must act
vertically through concurrency point s; however, this would fall outside of the physical
support surface of link 4. Therefore, we recognize that link 4 is tipping rather than
slipping, (Ans.) and it is acting as a four-force member with no external moment.
459
Therefore we group these into two pairs of forces; the pair  F34  F14C  must act through the
concurrency point q, and must be equal and opposite to the pair
 P  F  acting through
B
14
the concurrency point p. Therefore, the forces on member 4 must agree with the force
polygon on the right below.
Measuring from the force polygon, we find
F14B  5.33 N 90 , F14C  13.33 N   90
and
P  11.31 N 45
Ans.
Ans.
460
11.38 For the linkage in the posture illustrated, a force P  10 lb is applied on link 4. A torque
T12 is acting on link 2 at the crankshaft O2 to hold the linkage in static equilibrium.
Gravity is acting into the plane and friction in the linkage can be neglected. Is the
impending motion of link 3 slipping or tipping in link 4? Is the impending motion of link
4 slipping or tipping on the ground link? Determine the magnitudes, directions, and
locations of the internal reaction forces in the linkage. Determine the magnitude and the
direction of the torque T12 .
y
y
 6.6 in; and REO
 9.2 in. .
R AO2  5.0 in  60; RCO
2
2
The free-body diagrams of links 2 and 3 are as shown in the figures below. Starting with
link 3, we recognize this as a two-force member with no external moments. This gives the
lines of action of F23 and F43 and, therefore, of F32 onto link 2 and F34 onto link 4. Note
461
that F43 acts through the center of block 3; therefore, the impending motion of block 3 is
slipping, not tipping in link 4, (Ans.).
The free-body diagram of link 4 appears as in the figure below. If we consider link 4 we
find that the two forces P and F34 are parallel; therefore, they do not cross and link 4
cannot be a three-force member with no external moment. F14 cannot be a single force,
but must form a couple, F14C and F14D . Therefore, we recognize that the impending motion
of link 4 is tipping rather than slipping on the ground link 1, (Ans.) and link 4 acts as a
four-force member with no external moment.
The sum of vertical forces on link 4 gives F34  P  10 lb and, therefore
F43  F32  10 lb90 and F34  F23  F12  10 lb  90
The moment formed by the couple P and F34 is of magnitude
Pd  P RAO2 sin 30  10 lb  5 in  sin 30  25 in  lb

Ans.

The magnitude of the two forces in the opposing couple are
F14C  F14D  25 in  lb 2.6 in  9.96 lb
F14C  9.62 lb180 and F14D  9.62 lb0
Therefore,
Finally, from summing moments on link 2, we find
T12  F32 RAO2 sin 30  10 lb 5 in  sin 30  25 in  lb cw


Ans.
Ans.
462
11.39 Links 2 and 3 are pinned together at B and a constant vertical load P  800 kN is applied
at B. Link 2 is fixed in the ground at A and link 3 is pinned to the ground at C. The length
of link 2 is 8 m and it has a 150-mm solid square cross-section. The length of link 3 is 0.5
m and it has a solid circular cross-section with diameter D. Both links are made from a
steel with a modulus of elasticity E  207 GPa and a compressive yield strength
Syc  200 MPa. Using the theoretic values for the end condition constants of each link,
determine: (a) the slenderness ratio, the critical load, and the factor of safety guarding
against buckling of link 2; and (b) the minimum diameter Dmin of link 3 if the static factor
of safety guarding against buckling is to be N  2 .
(a) area moment of inertia and the radius of gyration of link 2, respectively, are
4
bh3 150 mm  150 mm 
I2 

 0.421 88 104 m
12
12
4
I
b /12
b
0.150 m
k2  2 


 0.043 30 m
2
A2
b
12
12
Therefore, the slenderness ratio of link 2 is
L
8m
Sr2  2 
 184.75
Ans.
k2 0.043 30 m
The end-condition constant for link 2 (with fixed-pinned ends) is C2  2. Therefore, the
slenderness ratio at the point of tangency is
3
463
2  2   207 109 Pa 
2C2 E

 202.14
 Sr D2  
S yc
200 106 Pa
Comparing these gives Sr2   Sr  D . Therefore, the Johnson parabolic equation must be
2
used to determine the critical load of link 2.
The critical unit load of link 2 is
2
Pcr
 S yc Sr2  1
2
 S yc  

A2
 2  C2 E
Substituting the given information into this equation, the critical load of link 2 is
Pcr2  2.62 106 N
Ans.
The axial compressive load F2 is the component of the vertical load P acting along the
central axis of the link 2 as shown in this figure
30o
F2
P
60o
F3
Therefore, the axial compressive load in link 2 is
F2  P cos30o  692.8 kN
Similarly, the axial compressive load in link 3 is
F3  P sin 30o  400 kN
The factor of safety guarding against buckling for link 2 is
Pcr
2.62 106 N
N2  2 
 3.782
F2 692.8 103 N
Therefore, link 2 is safe against this axial compressive load.
Ans.
(b) The end condition constant for link 3 (with pinned-pinned ends) is C3  1 . The
slenderness ratio at the point of tangency for link 3 is
 Sr  D  
3
2C3 E
2 1 207 109 Pa

 142.93
S yc
200 106 Pa
(1)
464
The cross-sectional area and the area moment of inertia of link 3 are
 D2
 D4
A3 
and I 3 
4
64
Therefore, the radius of gyration for the link 3, in terms of the diameter D, is
I
D
k3  3 
A3 4
Therefore, the slenderness ratio for link 3, in terms of diameter D, is
L 0.5 m 2 m
Sr  3 

k3 D 4
D
To determine the minimum diameter of link 3 to prevent buckling, the critical load must
be derived in terms of the diameter D for both the Euler column formula and the Johnson
parabolic equation. The critical load can be written as
Pcr3  N3 F3
3
Substituting N3  2 and F3 from above, the critical load is
Pcr3  2  400 kN  800 kN
CASE (1): The critical load in terms of the diameter D from the Euler column formula is
 C  2E  
N 
Pcr3  A3  3 2    4.011 44 1011 4  D 4
m 
 Sr 3  
Equating with the load F3 gives
Pcr3   4.011 44 1011 N m4  D4  800 kN
Therefore, the minimum diameter from the Euler column formula is
DEuler  0.038 m
(2)
CASE (2). The critical load can be written in terms of the diameter D from the Johnson
parabolic equation as
2

S S  1 

Pcr  A3  S yc   yc r 

 2   C3 E 
3
3


or as
Pcr3  1.570 8 108 Pa  D2  15 377.3 N
Equating with the load F3 gives
Pcr3  1.570 8 108 Pa  D2  15 377.3 N  800 kN
Therefore, the minimum diameter of the link 3 from the Johnson parabolic equation is
DJohnson  0.072 m
(3)
Note that the diameter given by the Euler column formula,Eq.(2) is smaller than the
diameter from the Johnson parabolic equation, Eq. (3). i.e.,( DEuler  DJohnson ).
In certain cases, the claim that the bigger of the two diameters is the correct answer
may not be true. Therefore, to determine the minimum diameter of the column we need to
decide which of the two criteria is valid. The slenderness ratio must be compared with the
slenderness ratio at the point of tangency for both diameters.
Tthe slenderness ratio of link 3 is
465
2m
 52.63
DEuler
Comparing with Eq. (1), the conclusion is
Sr3E   Sr D3
Sr3E 
Therefore, the Euler column formula is not appropriate. So Dmin  0.038 m. From Eq. (3),
the slenderness ratio of link 3 can be written as
2m
Sr3J 
 27.78
DJohnson
The conclusion is
Sr3J   Sr D
3
Therefore, the Johnson parabolic equation is the valid equation. The minimum diameter of
the column 3 from the Johnson parabolic equation is
Dmin  DJohnson  0.072 m  72 mm.
Ans.
466
11.40 The horizontal link 2 is subjected to the load F  150 kN at C as illustrated. The link is
supported by the solid circular aluminum link 3. The lengths of the links are
L2  RCA  5 m, RBA  3 m, and L3  RBD  3 m. The end D of link 3 is fixed in the
ground and the opposite end B is pinned to link 2; (that is, the effective length of the link
is LEFF  0.5 L2 ). For aluminum, the yield strength is Syc = 370 MPa and the modulus of
elasticity is E = 207 GPa. Determine the diameter d of the solid circular cross-section of
link 3 to ensure that the static factor of safety is N = 2.5.
The cross-sectional area and the area moment of inertia of link 3, respectively, are
A = π d2 / 4
and
I = π d4 / 64
Therefore, the radius of gyration of the link is
I
 d 4 64 d


A
d2 4 4
Using the effective length Leff  0.5 L2 , the slenderness ratio of the link is
k
Sr  Leff k  0.5  5 m  d / 4  10 m d
(1)
The slenderness ratio at the point of tangency is
 Sr D   2E Sr   2  207 109 Pa 370 106 Pa  105.09
(2)
Taking moments about A gives
3 m P  5 m F cos60
where P is the compressive load acting at B on link 3. Solving this equation, the
compressive load is
 5 m 150 000 N 0.5  125 000 N
P
3m
The factor of safety guarding against buckling of link 3 is defined as
N  Pcr P
Substituting N = 2.5, the critical unit load is
Pcr  2.5 125 000 N   312 500 N
467
Using the Euler column formula, the critical unit load can be written as
Pcr   2 E
A Sr 2
Which, with the available data and Eq. (1), can be written as
2
9
312 500 N   207 10 Pa 

2
d2 4
10 m d 
Rearranging this equation gives
d 4 194.76 109 m4
Therefore, the diameter of link 3 is
d  0.0664 m  66.4 mm
Using the Johnson parabolic equation, the critical unit load can be written as
2
Pcr  S  1  Sy Sr 
y
A
E  2  
(3)
which can be written as
 370 106 10 m d 
312 500 N
1
6
 370 10 Pa 


d2 4
207 109 Pa 
2

2
Rearranging this equation gives
d 2  5.60 103 m2
Therefore, the diameter of link 3 is
d  0.074 8 m  74.8 mm
(4)
To check which answer is valid, that is, Eq. (3) or Eq. (4), recall that the slenderness ratio,
from Eq. (1), is
Sr = 10 m/ d
To check the Euler column formula, the slenderness ratio is
(Sr)Euler = 10 m / (0.0664 m) = 150.6
Comparing this answer with Eq. (2) indicates that
(Sr)Euler > (Sr)D that is 150.6 > 105.09
Therefore, the Euler column formula is a valid equation. The correct diameter of link 3 is
d = 66.4 mm
Ans.
Next we check of the Johnson parabolic equation. The slenderness ratio is
(Sr)Johnson = 10 m / (0.0748 m) = 133.69
Therefore
(Sr)Johnson< (Sr)D that is 133.69 < 105.09
which is not possible. Therefore, the Johnson parabolic equation is not valid.
468
11.41 The horizontal link 2 is subjected to the inclined load F  8 000 N at C as illustrated. The
link is supported by a solid circular cross-section link 3 whose length L3  BD  5 m . The
end D of the vertical link 3 is fixed in the ground link and the end B supports link 2 (that
is, the effective length of the link is Leff  0.5 L3 ). Link 3 is a steel with a compressive
yield strength Syc  370 MPa and a modulus of elasticity E  207 GPa . Determine the
diameter d of link 3 to ensure that the factor of safety guarding against buckling is
N  2.5. Also, answer the following statements true or false and briefly give your
reasons. (a) The slenderness ratio at the point of tangency between Euler’s column formula
and Johnson’s parabolic equation does not depend on the geometry of the column. (b)
Under the same loading conditions, a link with pinned-pinned ends gives a higher factor of
safety against buckling than an identical link with fixed-fixed ends. (c) If the slenderness
ratio Sr  (Sr ) D at the point of tangency then the critical unit load does not depend on the
yield strength of the column material.
The cross-sectional area and the area moment of inertia of link 3, respectively, are
A = π d2 / 4
and I = π d4 / 64
Therefore, the radius of gyration of the link is
k
I

A
 d 4 64 d

 d2 4 4
Taking moments about A gives
 4 m P  5 m  F cos 60
469
where P is the compressive load acting at B on link 3 . Rearranging this equation, the
compressive load P is
P   5 m 8 000 N  0.5  4 m   5 000 N
Using the effective length Leff  0.5L3 , the slenderness ratio of the link is
Sr  Leff / k  0.5  5 m   d 4   10 m d
(1)
The slenderness ratio at the point of tangency is
 Sr D    2E S yc 
12
   2  207 109 Pa 370 106 Pa 
12
 105.087
The factor of safety guarding against buckling can be written as
N = Pcr / P
Therefore, the critical unit load is
Pcr  NP  2.5  5 000 N  12 500 N
and the critical unit load for this factor of safety is
Pcr 12 500 N

A
 d2 4
From the Euler column formula, the critical unit load can be written as
Pcr  2 E
 2
A
Sr
Equating Eqs. (2) and (3) gives
12 500 N  2  207 109 Pa

2
 d2 4
10 m d 
Rearranging and solving, the diameter of the link using the Euler formula is
d  0.0297 m  29.7 mm
(2)
(3)
(4)
From the Johnson parabolic equation, the critical unit load can be written as
Pcr
1S S 
 Sy   y r 
A
E  2 
2
Substituting the known data gives
 370 106 Pa 10 m d 
12 500 N
1
6

370

10
Pa



 d2 4
207 109 Pa 
2

2
Rearranging and solving, the diameter of the link using the Johnson parabolic equation is
d  0.0676 m  67.6 mm
(5)
To determine the correct diameter, that is, Eq. (4) or Eq. (5), the slenderness ratio from Eq.
(1) is
Sr  10 m d
Using Eq. (4), the slenderness ratio, from the Euler column formula, is
(Sr)Euler = 10 m / 0.0297 m = 336.7
Since 336.7 > 105.087, (Sr)EULER > (Sr)D is valid. Therefore, the diameter of link 3 is
d = 29.7 mm
Ans.
To check the Johnson parabolic equation. Using Eq. (5), the slenderness ratio, from the
Johnson parabolic equation, is
470
(Sr)Johnson = 10 m / 0.0676 m = 147.93
Since 147.93 > 105.087, (Sr)Johnson > (Sr)D which is not possible. Therefore, the Johnson
parabolic equation is not valid.
Statement (a) is true.
The reason is that the slenderness ratio at the point of tangency is defined as
Ans.
(Sr ) D    2E / S yc 
12
which is a function of only the material properties of the link (and does not depend on the
geometry of the link).
Statement (b) is false.
Ans.
The factor of safety is defined as N = Pcr / P. Therefore, a higher value for the critical load
Pcr will give a higher value for the factor of safety. The critical load Pcr is greater for fixedfixed ends (C = 4) than for pinned-pinned ends (C = 1) from both the Euler column
formula and the Johnson parabolic equation.
Statement (c) is true.
Ans.
2
P  E
When Sr > (Sr)D, then we must use the Euler column formula; that is, cr  2 , which
A
Sr
does not depend on the yield strength of the material.
471
11.42 A load PA is acting at A and a load PB is acting at B of the horizontal link 3. Link 3 is
pinned to the vertical link 2 at O and link 2 is fixed in the ground link 1 at D. The lengths
are AO  4 ft, OB  2 ft, and DO  6 ft. Both links have solid circular cross-sections
with diameter D  2 in and are made from a steel alloy with a compressive yield strength
Syc  85 000 psi, a tensile yield strength Syt  75 000 psi, and modulus of elasticity
E  30 106 psi. Assuming that links 2 and 3 are in static equilibrium and using the
theoretic value for the end-condition constant for link 2, determine: (a) the magnitude of
the force PB that is acting as shown at B if PA  30 000 lb , (b) the critical load, the critical
unit load, and the factor of safety to guard against buckling for link 2, and (c) the diameter
of a solid circular cross-section for link 2 that will ensure the factor of safety guarding
against buckling of the link is N = 4.
(a) The free body diagram of link 3 is as shown in the figure below.
472
Taking moments about O gives
 M O   4 ft  PAy   2 ft  PBy  0
which can be written as
 4 ft  PA cos30   2 ft  PB cos30
Therefore, the reaction force at B is
PB  2 PA
Substituting the given data into this equation, the reaction force at B is
PB  2  30 000 lb  60 000 lb
Taking the sum of the forces in the y-direction on link 3 gives
 F y  PAy  PBy  F23y  0
which can be written as
F23y  PA cos30  PB cos30
Substituting Eq. (1) gives
F23y  3PA cos30  3  30 000 lb  cos30  77 942 lb
(1)
Ans.
(2)
Taking the sum of the forces in the x-direction gives
 F x  PAx  PBx  F23x  0
which can be written as
F12x  PA sin 30  PB sin 30
Substituting Eq. (1), the x-component of the reaction force at O is
F23x  45 000 lb
(b)
A

The cross-sectional area and the area moment of inertia of link 2, respectively, are
D2 

(2 in)2  3.141 6 in 2
4
4
The radius of gyration of link 2 is
and
I

64
I
0.785 4 in 4

 0.5 in
A
3.141 6 in 2
The slenderness ratio of link 2 can be definded as
k
D4 

64
(2 in)4  0.785 4 in 4
473
L 6 ft 12 in/ft

 144
k
0.5 in
and the slenderness ratio of link 2 at the point of tangency can be written as
2CE
 Sr  D  
S yc
Sr 
(3)
where the end condition constant for fixed-pinned end conditions (using the theoretical
value) is C = 2 that is, the effective length for fixed-pinned ends is Leff  0.5L . Therefore,
the slenderness ratio of link 2 at the point of tangency is
2  2  30 106 psi
(4)
 118.04
85 103 psi
In order to determine the critical load on link 2, we must first determine if this link is an
Euler column or a Johnson column. The criterion for using the Johnson parabolic equation
is
Sr   Sr  D
 Sr  D  
From Eqs. (3) and (4) this implies that 144  118.04 which is a contradiction. Therefore,
link 2 is an Euler column. The critical load on link 2 from the Euler column equation can
be written as
 C 2 E 
(5)
Pcr  A  2 
 Sr 
Substituting the known values into this equation, the critical load on link 2 is
 2 2 30 106 psi 
Ans.
Pcr   3.141 6 in 2  
  89 717 lb
1442


The critical unit load on link 2 is
Pcr
89 717 lb

 28 558 lb/in 2
Ans.
2
A 3.141 6 in
Link 2 is in compression as shown in the figure.
474
The definition of the factor of safety for link 2 is
P
N  cry
F32
where, from Eq. (2), F32y  77 942 lb is the compressive load at point O on link 2.
Therefore, the factor of safety for link 2 is
89 717 lb
N
 1.15
Ans.
77 942 lb
(c)
For the circular cross-section of link 2 and a factor of safety N = 4, the critical load
for buckling can be written as
Pcr
N
4
77 942 lb
Therefore, the required critical load for the buckling is
Pcr  4  77 942 lb  311 769 lb
(6)
The cross-section area and the area moment of inertia of the solid circular link 2,
respectively, are
A   D2 4 and I   D4 64
Therefore, the radius of gyration of the link is
I
 D 4 64 D


A
 D2 4
4
The slenderness ratio of link 2 is
L 6 ft 12 in/ft 288 in
(7)
Sr  

k
D4
D
First consider the Euler column formula. Substituting Eqs. (6) and (7) into Eq. (5), the
critical load is
k
475
 C 2 E   D 2  2 2  30 106 psi 
Pcr  A  2  

  311 769 lb
4   288 in D 2 
 Sr 


Rearranging this equation gives
4  311 769 lb(288 in) 2
D4 
 55.6 in 4
3
6
2
2 30 10 lb/in
Therefore, the diameter of the cross-section of the link 2 is D  2.73 in ,
Substituting this into Eq. (7), the slenderness ratio of link 2 is
288 in 288 in
Sr 

 105.5
(8)
D
2.73 in
Compaing Eq. (8) with Eq. (4) gives that Sr   Sr D , that is, 105.5  118.04 . Therefore,
the condition for link 2 to be an Euler column is not satisfied, that is, the assumption that
the link is an Euler column is not correct.
Now we assume that link 2 is a Johnson column, the Johnson parabolic equation is
2

1  S y Sr  
Pcr  A  S y 

 
CE  2  


Substituting Eqs. (7) into this equation gives
2
 S yc  288 in D  
 2
1
Pcr  D  S yc 

 
4
2  30 106 psi 
2

 
Rearranging this equation gives
2
 4P
 S yc  288 in   1
1
2
cr
D 


 
2  30 106 psi 
2
 
  S yc
Substituting the known values into this equation gives
2
 4  311,769 lb
 85 103 psi  288 in  
1
1
2
D 

 7.65 in 2

 
6
3

2

30

10
psi
2

85

10
psi


 
Therefore, the diameter of link 2 from the Johnson parabolic equation is D  2.77 in .
Substituting this into Eq. (7), the slenderness ratio of link 2 is
288 in 288 in
Sr 

 104.1
D
2.77 in
Comparing with Eq. (4) now shows that Sr   Sr D , that is, that 104.1  118.04 .
Therefore, the assertion that the column is a Johnson column is valid. The diameter of link
2 to guard against buckling is D  2.77 in .
Ans.
476
11.43 The horizontal link 2 is subjected to the load P = 5 000 N and is supported by the vertical
link 3 which has a constant circular cross-section. The lengths are AC  5 m, AB  4 m,
and DB  L3  5 m. For the vertical link 3, the end D is fixed in the ground link and the
end B supports link 2 (that is, the effective length of link 3 is LEFF  0.5 L3 ). The yield
strength and the modulus of elasticity for the aluminum link 3 are Sy = 370 MPa and E =
207 GPa, respectively. Determine the diameter d of link 3 to ensure that the static factor
of safety guarding against buckling is N = 2.5.
The cross-sectional area and the second moment of area of link 3 can be written as
A = π d2 / 4 and I = πd4 /64
Therefore, the radius of gyration of the link is
k
I

A
 d 4 64 d

d2 4 4
Using the effective length Leff  0.5 L3 , the slenderness ratio of the link is
Sr  Leff k  0.5  5 m  d 4   10 m d
The slenderness ratio at the point of tangency is
 Sr  D    2 E S y 
12
   2  207 109 Pa 370 106 Pa 
12
Taking moments about A of all forces on link 2 we find
M A  4 m  F32y  5 m  P sin 53.13  0
And therefore the vertical load on link 3 is
F32y  P sin 53.13 5 m 4 m  5 000 N
 105.09
477
The factor of safety guarding against buckling of link 3 can be written as
N  Pcr P  2.5
Therefore, the critical unit load is
Pcr  NP  2.5  5 000 N=12 500 N
(i) Using the Euler column formula, the critical unit load can be written as
Pcr  2 E
 2
A
Sr
which can be written as
2
9
12 500 N   207 10 Pa 

2
d2 4
10 m d 
Rearranging this equation gives
d 4  7.79 107 m4
Therefore, the diameter of link 3 is
d  0.0297 m  29.7 mm
(ii) Using the Johnson parabolic equation, the critical unit load can be written as
2
Pcr
1  S y Sr 
 Sy  

A
E  2 
which can be written as
 370 106 Pa 10 m d 
12 500 N
1
6

370

10
Pa



 d2 4
207 109 Pa 
2

(1)
2
Rearranging this equation gives
d 2  4.57 103 m2
Therefore, the diameter of link 3 is
d  0.0676 m  67.6 mm
(2)
To check which answer is valid, that is, Eq. (1) or Eq. (2), recall that the slenderness ratio
is defined as
Sr = 10 m/ d
To check the Euler column formula. The slenderness ratio is
(Sr)Euler = 10 m/ (0.0297 m) = 336.7
or (Sr)Euler > (Sr)D since the values are 336.7 > 105.09. Therefore, the Euler column
formula is the valid equation. The correct diameter of the link is
d = 29.7 mm
Ans.
Using the Johnson parabolic equation, the slenderness ratio is (Sr)Johnson = 10 / 0.0676 =
147.93 or (Sr)Johnson > (Sr)D. Since 147.93 > 105.09, which is not possible, therefore the
Johnson parabolic equation is not valid.
478
11.44 The horizontal link 2 is pinned to the vertical wall at A and pinned to link 3 at B. The
opposite end of link 3 is pinned to the wall at C. A vertical force P  25 kN is acting on
link 2 at B. Link 2 has a 20 mm by 30 mm solid rectangular cross-section and link 3 has
a 40 mm by 40 mm solid square cross-section. The length of link 3 is BC  1.2 m and
the angle ABC  30 . The two links are made from a steel alloy with a tensile yield
strength Syt  190 MPa , a compressive yield strength Syc  205 MPa and a modulus of
elasticity E  207 GPa . Using the theoretic value for the end-condition constant for link
3, determine: (a) the value of the slenderness ratio at the point of tangency between the
Euler column formula and the Johnson parabolic formula; (b) the critical load and the
factor of safety guarding against buckling of link 3; and (c) the minimum width of the
square cross-section of link 3 for the factor of safety to guard against buckling to be N = 1.
The free body diagram of link 2 is as shown below.
Taking moments about A gives
479
M  R F  R P  0
A
BA
y
32
BA
Therefore, the y-component of the reaction force at B is
F32y  P  25 kN
The free body diagram of link 3 is as shown in the next figure.
Taking moments about C gives
 M C  RBA F23y  RAC F23x  0
or
RBA y
F23  tan 60F23y
RAC
Therefore, the x- and y-components of the reaction force at B are
F23x   tan 60 25 kN  43.3 kN
and
F23y   F32y  25 kN
The magnitude of the tensile load T on link 2 at B is
T  F32x   F23x  43.3 kN
F23x 
(1)
The magnitude of the compressive load Papp on link 3 at B is
Papp  F23 
 F    F   50 kN
x 2
23
y 2
23
(2)
Link 2 is subjected to the tensile load T which creates a tensile stress  in the link. The
factor of safety guarding against yielding for the link is defined as
480
N  Syt 
(3)
where the tensile stress can be written as
(4)
 T A
and the cross-sectional area of the link is
(5)
A  (0.02 m)(0.03 m)  6 104 m2
Substituting Eqs. (1) and (5) into Eq. (4), the tensile stress is
43.3 103 N
(6)

 72.17 MPa
6 104 m2
Substituting Eq. (6) and the tensile yield strength into Eq. (3), the factor of safety guarding
against yielding for link 2 is
190 MPa
N
 2.63
72.17 MPa
(a) The slenderness ratio of link 3, at the point of tangency, can be written as
2CE
 Sr  D  
S yc
Using the theoretical value (for pinned-pinned ends), the end-condition constant for the
link is
(7)
C 1
Substituting E  207 GPa , Syc  205 MPa and Eq. (7) into Eq. (6), the slenderness ratio,
at the point of tangency, is
 Sr  D  
2 1 207 109 Pa
 141.18
205 106 Pa
(b) The cross-sectional area of link 3 is
A  b2  (0.040 m)2  1.6 103 m2
The second moment of area of the link is
b4 (0.040 m)4
I

 2.13 107 m4
12
12
and the radius of gyration of the link is
Ans.
(8)
(9)
(10)
I
2.13 107 m4

 1.155 102 m
3
2
A
1.6 10 m
Therefore, the slenderness ratio is
L
1.2 m
Sr  
 103.92
(11)
k 1.155 102 m
In order to determine the critical load for link 3, we must first determine if this column is
an Euler column or a Johnson column. The criterion for using the Johnson parabolic
equation is
Sr   Sr  D
k
481
From Eqs. (8) and (11) we have Sr   Sr  D that is, 103.92  141.18 . Therefore, link 3 is a
Johnson column. The critical load for link 3 (that is, the Johnson parabolic equation) can
be written as
2

1  S yc Sr  
(12)
Pcr  A  S yc 

 
CE  2  


Substituting the known values into this equation, the critical load is
2

 205 106 Pa 103.92  
1
3
2
6
Pcr  1.6 10 m  205 10 Pa 

 
1 207 109 Pa 
2

 
Therefore, the critical load is
Ans.
(13)
Pcr  239.14 kN
The definition of the factor of safety guarding against buckling is
P
(14)
N  cr
Papp
where Papp is the compressive load on column 3 at B and is given by Eq. (2). Substituting
Eqs. (2) and (13) into Eq. (14), the factor of safety guarding against buckling is
239.14 kN
N
 4.78
Ans.
50 kN
(c) The critical load for a factor of safety N  1 is Pcr  NPapp  1 50 kN  50 kN
If we assume that link 3 is an Euler column, then the critical load on link 3 (that is, the
Euler column equation) can be written as
C 2 EA
Pcr 
(15)
Sr2
From Eqs. (9) and (10), the radius of gyration of link 3 is
I
b 4 /12
b


2
A
b
12
and from Eq. (11), the slenderness ratio of link 3 is
L
Sr   12 L / b
(16)
k
Substituting Eqs. (9) and (16) into Eq.(15), the critical load on link 3 can be written as
C 2 Eb 2
Pcr 
12 L2 / b 2
Rearranging this equation, the minimum width of the link can be written as
k
1/4
 12L2 Pcr 
b

2
 C E 
Substituting the known values into this equation gives
482
1/4
 12  1.2 m 2  50 103 N 
b
  0.025 5 m  25.5 mm
 1  2  207 109 Pa 


To check whether the assumption of an Euler column is correct, from Eq. (16), the new
slenderness ratio of link 3 is
Sr  12L / b  12 1.2 m  / (25.5 103 m)  163.02
Comparing this result with Eq. (8) we have Sr   Sr  D
that is,
163.01  141.18 . So
this verifies that link 3 is indeed an Euler column.
If we assume that link 3 is a Johnson column, then substituting Eqs. (9) and (16) into
Eq. (12), the critical load can be written as
2


1  S yc ( 12 L / b)  
2 
Pcr  b S yc 


 

CE 
2
 

or as
2
12 L2  S yc 
2
Pcr  S ycb 


CE  2 
Rearranging this equation, the new width of the link can be written as
2

12 L2  S yc  
b   Pcr 

  S yc
CE  2  


Substituting the known values into this equation, the width is
b
2

12 1.2 m   205 106 Pa 2 
3
50 10 N 
 
2
1 207 109 Pa 

 

(205 106 Pa )  26 mm
Now we must check if the assertion that the link is a Johnson column is correct. From Eq.
(16), the new slenderness ratio of link 3 is
Sr  12L / b  12 1.2 m  / (2.60 102 m)  159.88
Comparing this result with Eq. (8) we have Sr   Sr  D , that is, 159.88  141.18 . This
means that the assumption that link 3 is a Johnson column is, invalid. Link 3 is an Euler
column and the minimum width of the square cross-section (in order for the factor of
safety to guard against buckling to be N = 1) is b = 25.5 mm.
Ans.
483
11.45 The link BC = 1.2 m and 25 mm square cross-section is fixed in the vertical wall at C and
pinned at B to a circular steel cable AB with diameter d = 20 mm. The distance AC = 0.7
m. The mass m of a container, suspended from pin B, produces a gravitational load at B
which results in the moment at point C in the wall M C  8 000 Nm ccw. The yield
strength and modulus of elasticity of the steel cable AB and the steel link BC are
Sy  370 MPa and E  207 GPa, respectively. Given that m = 2 000 kg, determine: (a)
the tension in the cable AB and the factor of safety guarding against tensile failure; (b) the
compressive load acting in link BC; (c) the factor of safety guarding against buckling of
link BC (Use the theoretical value of the end-condition constant assuming that the link has
fixed-pinned ends). If M C  10 000 N  m ccw, then determine the maximum mass of a
container that can be suspended from pin B before buckling of link BC begins (that is, the
factor of safety guarding against buckling failure is N  1 ).
(a) The angle between the cable and the link is
 0.7 
  tan 1 
  30.256
 1.2 
The cable AB and the link BC are in static equilibrium. The free body diagram of link BC
is as shown in the following figure.
484
Summing moments about C at the wall gives
 M C M C  L sin 30.256P  Lmg  0
(1)
where P is the tension in the cable AB. Rearranging Eq. (1), the tension can be written as
mg  M C L
P
sin 30.256
Substituting the given data into this equation, the tension is
2 000 kg  9.81 m/s 2  8 000 N  m 1.2 m
Ans.
P
 25 708 N
sin 30.256
The axial stress in the cable can be written as
P
P
A  
A  d2 4
Substituting the given data into this equation, the axial stress in the cable is
25 708 N
A 
 81.83 MPa
2
  0.020 m  4
The factor of safety guarding against tensile failure in the cable can be written as
Sy
370 MPa
N

 4.52
Ans.
 A 81.83 MPa
(2)
(b) The compressive load in link BC can be obtained by summing forces in the x-direction;
that is,
x
 F  Pc  P cos30.256  0
where Pc is the compressive load in the link. Therefore, the compressive load is
Pc  P cos30.256  22 206 N
Ans.
(c) To determine whether the link is an Euler column or a Johnson column, we first find
the slenderness ratio at the point of tangency between the Euler column formula and the
Johnson parabolic equation
2 EC
 Sr  D  
Sy
where the theoretical value of the end-condition constant corresponding to fixed-pinned
end conditions is C  2 . Therefore, the slenderness ratio at the point of tangency D is
485
 Sr  D  
2  207 109 Pa  2
 148.62
370 106 Pa
The radius of gyration of the link can be written as
I
bh3 12
h
k


A
bh
12
Which, for the given data, is
0.025 m
k
 0.007 22 m
12
The slenderness ratio of the link can be written as
L
1.2 m
Sr  
 166.2
k 0.007 22 m
Since Sr   Sr  D , therefore the link is an Euler column. The critical load using the Euler
column formula can be written as
C 2 AE 2 2 (0.025 m)2 (207 109 Pa)
Pcr 

 92 452 N
Sr 2
(166.2)2
Therefore, the factor of safety guarding against buckling of the link is
P
92 367 N
Ans.
N  cr 
 4.16
Pc 22 206 N
Combining Eqs. (1) and (2), the compressive load exerted on the link, as a function of
the mass of the container, is
mg  M C L
mg  M C L
Pc  P cos30.256 
cos30.256 
(3)
sin 30.256
tan 30.256
For a factor of safety guarding against buckling N = 1, the critical load must be equal to
the compressive load; that is,
mg  M C L
Pcr  Pc 
tan 30.256
Rearranging this equation, the mass of the container can be written as
tan 30.256Pcr  M C L
m
g
Substituting the given data into this equation, the mass of the container is
tan 30.256(92 367 N)  10 000 Nm 1.2 m
m
 6 342 kg
Ans.
(4)
9.81 m/s 2
Substituting Eq. (4) into Eq. (3), the compressive load is
mg  M C L (6 342 kg)(9.81 m/s 2 )  10 000 Nm 1.2 m
Pc 

 92 370 N
tan 30.256
tan 30.256
Note that the compressive load in the link is equal to the critical load. Also, note that it is
important to show that this answer cannot be obtained using the factor of safety found in
part (iii) because the moment has changed; that is, mnew  2 000 kg  4.16  8 320 kg.
486
11.46 A vertically upward force F is applied at C of the horizontal link 4 which is pinned to the
ground at B and pinned to the vertical links 2 and 3 at A.
The lengths
AB  2 ft and AC  5 ft and the three links are made from a steel alloy with a compressive
yield strength Syc  60 000 psi and a modulus of elasticity E  30 106 psi. Link 2 has a
hollow circular cross-section with an outside diameter D  2 in, wall thickness
t  0.25 in, and length L  5 ft. Using the theoretic value for the end-condition constant
for link 2, determine: (a) the value of the slenderness ratio at the point of tangency
between the Euler-column formula and the Johnson parabolic formula; (b) the critical load
acting on link 2; and (c) the force F in order for the factor of safety of link 2 to guard
against buckling to be N  1 . (d) If link 2 has a solid circular cross-section with diameter
D  3 in and F  20 000 lb then determine the maximum length of link 2 in order for the
factor of safety to guard against buckling to be N  2 .
(a) The cross-sectional area and the second moment of area for the hollow circular link 2,
respectively, are
487

 4    2 in   1.5 in   4  1.374 4 in
and
I   D4  d 4
 64    2 in 1.5 in  64  0.536 9 in
A   D2  d 2

2
4
2
4
2
4
Therefore, the radius of gyration for the link is
I
0.536 9 in 4

 0.625 in
A
1.374 4 in 2
and the slenderness ratio of the link is
L
60 in
Sr  
 96
k 0.625 in
The slenderness ratio at the point of tangency can be written as
2CE
 Sr  D  
Sy
k
where the theoretical value for the end-condition constant for link 2 is C = 2 . Therefore,
the slenderness ratio at the point of tangency is
 Sr  D  
2  2  30 106 psi
 140.50
60 103 psi
Ans.
(b) The critical load is determined by first finding if the link is to be considered an Euler
column or a Johnson column. The criterion for using the Johnson parabolic equation is
Sr   Sr D . In this example, we have 96 < 140.50. Therefore, the link is regarded as a
Johnson column. The critical load from the Johnson parabolic equation is
2

1  S y Sr  
Pcr  A  S y 

 
CE  2  


Substituting the known values into this equation gives the critical load as
2

 60 103 psi  96  
1
2
3
Ans.
Pcr  1.374 4 in  60 10 psi 

   63 213 lb
2  30 106 psi 
2

 
(c) The definition of the factor of safety of the link is
N  Pcr Papp
From the given factor of safety for the link, this equation can be written as
N  Pcr Papp  1
Therefore, the applied force at point A can be written as
Papp  Pcr 1  63 213 lb
To determine the force F acting at C in link 4, we take moments about B which gives
 M B  3 ft  F  2 ft  Papp  0
Therefore, the force F acting at point C is
2 ft  Papp 2 ft  63 213 lb
F

 42 142 lb
3 ft
3 ft
Ans.
488
(d) Link 2 is a solid circular column with factor of safety of N = 2. To determine the
applied force we take moments about B; that is,
MB  0
Therefore, the applied force is
3 ft  F 3 ft  20 000 lb
Papp 

 30 000 lb
2 ft
2 ft
From the definition of the factor of safety
N  Pcr Papp  2
Therefore, the critical load is
Pcr  2  Papp  2  30 103 lb  60 000 lb
(1)
The cross-sectional area is
2
A   D2 4    3 in  4  7.07 in 2
The second moment of area is
4
I   D4 64    3 in  64  3.98 in 4
The radius of gyration is
I
3.98 in 4

 0.75 in
A
7.06 in 2
First, the critical unit load from the Johnson parabolic equation is
2
Pcr 
1  S y Sr  
  Sy 

 
A 
CE  2  


With known values, this equation can be written as
2
Sr 2
60 000 lb 
 60 000 psi  

60
000
psi



 
7.07 in 2
2  30 106 psi 
2
 

Solving this equation gives the slenderness ratio from the Johnson parabolic formula as
Sr  184.10
Second, the critical unit load from the Euler column formula is
Pcr C 2 E

A
Sr 2
Substituting the known values into this equation gives
60 000 lb 2   2 30 106 psi

7.07 in 2
Sr 2
Therefore, the slenderness ratio from the Euler formula is
Sr  264.14
A comparison of the two slenderness ratios shows that
264.14  184.10
In other words
(Sr)Euler > (Sr)Johnson
The length of link 2 can be written as
k
489
L  k  Sr
where the slenderness ratio that is used in this equation is for the Euler column formula.
Therefore, the length of the link is
Ans.
L  0.75 in  264.14  198.18 in  16.5 ft
As an alternative method to determine the critical load, consider the effective length of the
link; that is, LEFF. From the end condition constant
C = (1/α)2
This defines α = 0.707. Therefore, the effective length of the link can be written as
Leff   L  0.707  5 ft 12 in/ft  42.43 in
The slenderness ratio of the link is
L
42.43 in
Sr  eff 
 67.88
k
0.625 in
The slenderness ratio at the point of tangency is
 Sr  D  
2E
2  30 106 psi

 99.32
Sy
60 103 psi
The critical load is determined by first finding if the link is an Euler column or a Johnson
column. The criterion for the Johnson column is
Sr   Sr  D
In this example, we have
67.88 < 99.32
Therefore, the Johnson parabolic equation must be applied. The equation can be written as
2

1  S y Sr  
Pcr  A  S y  
 
E  2  


Substituting the known values into this equation, the critical load is
2

 60 103 psi  67.88  
1
2
3
Pcr  1.3744 in 60 10 psi 

   63.214 lb
30 106 psi 
2

 
Note that this answer is in good agreement with Eq. (1).
490
11.47 A force F is acting at C perpendicular to link 2 and the end A is pinned to the ground and
the supporting link 3 is pinned to link 2 at B and pinned to the ground at D. The lengths
are AC  200 mm and AD  150 mm. Link 3 has a circular cross-section with diameter
D  5 mm. Both links are made of a steel alloy with a compressive yield strength
Syc  420 MPa and modulus of elasticity E  206 GPa. Using the theoretic value for the
end-condition constant for link 3, determine: (a) the slenderness ratio at the point of
tangency between the Euler column formula and the Johnson parabolic formula; (b) the
critical load and the critical unit load acting on the link; and (c) the force F for the factor of
safety to guard against buckling to be N  1 . If the force F  3 000 N then for link 3
determine: (1) the critical load for the factor of safety to guard against buckling to be
N  1; and (2) the slenderness ratio.
(a) From the pinned-pinned end conditions, the end condition constant for link 3 is
C=1
Therefore, the slenderness ratio of link 3 at the point of tangency is
 Sr  D  
2CE
2 1 206 109 Pa

 98.4
S yc
420 106 Pa
Ans.
(1)
(2)
(b) In order to determine the critical load on link 3, we first determine if this link is an
Euler column or a Johnson column. The Euler and Johnson criterion, respectively, are
(3)
Sr   Sr  D and Sr   Sr D
The cross-sectional area of link 3 is
491
A   D2 4   (0.005 m)2 4  1.963 105 m2
and the second moment of area of link 3 is
I   D4 64   (0.005 m)4 64  3.068 1011 m4
The radius of gyration of link 3 is
I
3.068 1011 m4

 1.25 103 m
5
2
A
1.963 10 m
The slenderness ratio of link 3 is
L
0.15 m
Sr  
 120
k 1.25 103 m
Substituting Eqs. (2) and (5) into Eq. (3) gives
Sr   Sr  D that is, 120  98.4
(4)
k
(5)
Therefore, link 3 is an Euler column. The critical load on link 3 from the Euler column
equation can be written as
 C 2 E 
(6)
Pcr  A  2 
 Sr 
Substituting the known values into this equation, the critical load on link 3 is
1  2  206 109 Pa 
Ans.
Pcr  1.963 105 m2 
  2 772 N
1202


Therefore, the critical unit load on link 3 is
Pcr
2 772 N

 141.2 106 Pa
Ans.
5
2
A 1.963 10 m
(c) The free body diagram of link 2 is shown in the figure. Taking moments about pin A
can be written as
 M A  (RCAx F y  RCAy F x )  (RBAx F32y  RBAy F32x )  0
or as
( RCA cos2 )( F sin F )  ( RCA sin 2 )( F cos F )  ( RBA cos 2 )( F32 sin 3 )  ( RBA sin 2 )( F32 cos 3 )  0
492
or as
RCA F sin( F  2 )  RBA F32 sin(3  2 )  0
Rearranging this equation, the force is
R F sin( 2  3 )
F  BA 32
(7)
RCA sin( F   2 )
The free body diagram of link 3 is shown in the following figure. Note that link 3 is in
compression.
The reaction force FB  F23   F32 ; therefore, the magnitude of the reaction force FB is
equal to the magnitude of the internal force F32 ; that is, Eq. (7) can be written as
R F sin ( 2  3 )
F  BA B
(8)
RCA sin( F   2 )
The factor of safety guarding against buckling for link 3 can be written as
P
N  cr
(9)
FB
Rearranging Eq. (9), the compressive load on link 3 can be written as
P
2 772 N
FB  cr 
 2 772 N
(10)
N
1
Substituting 2  120, 3  60,  F  210, RCA  0.2 m, RBA  0.15 m, and putting
Eq.(10) into Eq. (8), the force is
0.15 m  2 772 N  sin(120  60)
F
 1 800.6 N
Ans.
 0.2 m  sin(210  120)
(1) Rearranging Eq. (7) gives
R F sin( F   2 )
F32  CA
RBA sin( 2  3 )
493
Substituting F  3 000 N and the given data into this equation gives
0.2 m  3 000 N  sin(210  120)
FB  F32 
 4 618.8 N
(11)
 0.15 m  sin(120  60)
Rearranging Eq. (9) and substituting Eq. (11) and the factor of safety N = 1 into the
resulting equation, the critical load applied on link 3 is
Ans.
(12)
Pcr  NFB  1 4 618.8 N  4 618.8 N
(2) If we assume that the link is an Euler column. Rearranging Eq. (6), the slenderness
ratio of link 3 can be written as
C 2 EA
Pcr
Substituting Eqs. (1), (4) and (12), into Eq. (13), the slenderness ratio of link 3 is
 Sr Euler 
 Sr Euler 
(13)
1  2  206 109 Pa 1.963 105 m2
 92.97
4 618.8 N
Next, if we assume that the link is a Johnson column. The Johnson parabolic equation is
2
Pcr 
1  S y Sr  
  Sy 

 
A 
CE  2  


Rearranging this equation, the slenderness ratio can be written as
 Sr Johnson 
P 
2 
S y  cr  CE

Sy 
A
or as
2
4 618.8 N 

6
(14)
1 206 109 Pa  92.3
 420 10 Pa 
6
5
2 
420 10 Pa 
1.963 10 m 
From Eq. (2), the slenderness ratio of link 3 at the point of tangency is  Sr D  98.4 .
 Sr Johnson 
Note that
 Sr Johnson   Sr D
that is,
92.3  98.4
which is the correct answer. Therefore, the link must be a Johnson column. The correct
value for the slenderness ratio is given by Eq. (14); that is,
Ans.
Sr   Sr Johnson  92.3
As a check, we note that
 Sr Euler   Sr D that is,
92.97  98.4 which is not possible. Therefore, the link must
be a Johnson column. So again the correct value for the slenderness ratio is
Sr   Sr Johnson  92.3
494
11.48 For the four-bar linkage in the posture illustrated, the torque acting on link 2 at the
crankshaft O2 is T12   6 700kˆ ft  lb. There is also a torque T14 acting on link 4 at the
crankshaft O4 to hold the linkage in static equilibrium. The length of the coupler link 3 is
AB = 6 in and the cross-section is rectangular with width 3t and thickness 4t (into the
plane). The coupler link is a steel alloy with compressive yield strength S yc  60 kpsi and a
modulus of elasticity E = 30 Mpsi. If the critical load for the coupler link is
Pcr  150 000 lb, and the effects of gravity are ignored, then determine: (a) the factor of
safety guarding against buckling; (b) the slenderness ratio at the point of tangency between
the Euler column formula and the Johnson parabolic equation; and (c) the numeric value
of the parameter t. (d) If the rectangular cross-section is replaced by a circular crosssection of the same material and diameter d = 0.200 in, then determine the factor of safety
guarding against buckling of the coupler link.
The free body diagrams of links 2 and 3 are shown in the figures below:
Since link 3 is perpendicular to link 2 at A, the sum of the moments on link 2 shows that
495
F23  F32 
 6 700 ft  lb 12in/ft   23 210 lb
T12

RAO2
 6 in  tan 30
The factor of safety guarding against buckling of the coupler link is
P
150 000 lb
Ans.
N  cr 
 6.46
F23 23 210 lb
Since the width 3t is less than the thickness 4t then link 3 will buckle in the xy plane
(referred to as in-plane buckling) as shown in this figure
The slenderness ratio at the point of tangency for link 3 can be written as
2  1  30 10 psi
2C E

 99.346
 Sr  D 3  
3
Syc
60 10 psi
6
Ans.
The cross-sectional area of link 3 is
A  3t  4t  12 t 2
The area moment of inertia of the rectangular cross-section can be written as
I 3  b h3 12
Substituting the thickness b = 4t and the width h = 3t, the area moment of inertia of link 3
is
(4t ) (3t )3
I3 
 9 t4
12
However, if the thickness b = 3t and the width h = 4t then the area moment of inertia of
link 3 is
(3t ) (4t )3
I3 
 16 t 4
12
The smaller of the two values must be used, see Example 11.14 (the member will begin to
buckle in the weakest plane), that is, the value 9t4 must indeed be used to determine the
slenderness ratio of link 3 and the critical unit load. Also, note that if link 3 is assumed to
be buckling out of the x-y plane then the end-conditions of link 3 are not known. It is
common to assume that for out of plane buckling that the ends of link 3 are fixed-fixed. In
such a case, the end-condition constant for a fixed-fixed link (see Sec. 11.15) is C  4 .
The radius of gyration of link 3 is defined as
496
k3 
I3
9t 4

 0.866t
A3
12t 2
Substituting and the length L3  AB  6.0 in into Eq. (11), the slenderness ratio of link 3
can be written in terms of the unknown parameter t (expressed in inches) as
L 6.000 in 6.928 in
Sr  3 

k3
0.866 t
t
If we assume that link 3 is an Euler column, according to the Euler column formula, the
critical load for link 3 can be written as
C 2 E 
Pcr  A 

2
 Sr3 
Substituting the known information and the given data into this equation gives
6
2


2 1    30 10  psi
150 000 lb  12t 

2
  6.928 in 


 

t

or
6
150 000 lb  74.026 10 lb in 4  t 4


Rearranging this equation, the value of the parameter t is
t  0.212 17 in
Substituting, the slenderness ratio of the coupler link 3 is
6.928 in
Sr 
 32.654
0.212 17 in
However, we recall that the Euler column formula is only valid when
Sr   Sr  D
Comparing these values shows that the coupler link 3 cannot be an Euler column, that is,
link 3 must be a Johnson column. Therefore, the value of the parameter t must be obtained
as follows.
Using the Johnson parabolic equation, the critical load for link 3 can be written as
2

1  Syc Sr  
Pcr  A  Syc 

CE  2  

Substituting the data gives
2

6.928 in  

3
60(10)
psi


 
1
3
t
150 000 lb  12t 2 60 10  psi 

 
6
2
1  30 10 psi 

 


 
or
145.894 lb 
3

150 000 lb  12t 2 60 10 psi 

t2

Rearranging this equation, the parameter is
497
t  0.459 in
The area moment of inertia for a circular cross-section is
I3 
 d4

  0.200 in 
64
The cross-sectional area of link 3 is
A
 d2
4
64

4
 0.000 078 54 in 4
  0.200 in 
4
Ans.
2
 0.031 42 in 2
The radius of gyration of link 3 is
0.000 078 54 in 4
k3 
 0.050 in
0.031 42 in 2
The slenderness ratio is
6.0 in
 120
0.050 in
Comparing values, the conclusion is that the slenderness ratio
 Sr  D  Sr
Sr 
Therefore, the circular cross-section link 3 is an Euler column.
Substituting the given data, the critical load for link 3 can be written as
Pcr 
 2 EA
 2 30 10 psi  0.031 42 in 2 
6

 6 460 lb
Sr2
1202
Therefore, the factor of safety guarding against buckling is
6 460 lb
N
 0.28
Ans.
23 210 lb
Since the factor of safety is less than one then buckling is predicted to occur in the coupler
link.
498
11.49 The single-cylinder engine is in static equilibrium due to the external force F   15ˆi kN
acting at point D. The cross-section of the connecting rod (link 3) ) of length L3 = 200 mm
is rectangular with width 4 t and thickness 2 t and the material is a steel alloy with
compressive yield strength Syc  205 MPa and a modulus of elasticity E  207 GPa.
Assume for practical purposes that the thickness (2t) of link 3 must be greater than 5 mm
and that buckling will occur in the x-y plane. The effects of gravity and friction in the
engine can be ignored. Determine: (a) The slenderness ratio of link 3 in terms of the
unknown thickness parameter t. (b) The slenderness ratio of link 3 at the point of tangency
between the Euler column formula and the Johnson parabolic equation. (c) The thickness
parameter t to ensure that the factor of safety guarding against buckling for link 3 is N = 5.
Since link 3 is a two-force member with no external moments, the free body diagram of
links 3 must appear as in the diagram on the left below. Since friction is neglected, and
since link 4 is a three-force member with no external moment, the lines of action of the
three forces F14 , F34 , and F must intersect at the concurrency point e as shown in the
free body diagram of link 4 in the center below. From this, the force polygon for the
forces on link 4 is drawn as shown in the diagram on the right below.
499
Either by measuring from the above force polygon, or by the following calculation, we can
find the compressive force in link 3:
F
15 kN
F23  F43  F34 

 19 580 N
cos 40 cos 40
The area moment of inertia of link 3 (which has rectangular cross-section) can be written
as
bh 3 (4t ) (2t )3 8 t 4
I3 


12
12
3
The radius of gyration of link 3 can be written as
I3
(8 3) t 4
t2
1


 t
A3
(2t )(4t )
3
3
The slenderness ratio of link 3 in terms of the unknown thickness parameter t (expressed
in mm) can be written as
L 200 mm 200 mm 3 346.4 mm
Sr3  3 


Ans.
k3
t
t
t 13
Since link 3 is bending in the x-y plane the end-conditions for this link are pinned-pinned.
Recall that the end-condition constant for a pinned-pinned link with in-plane bending is
C3 = 1 (refer to Sec.11.16). Substituting the given compressive yield strength and modulus
of elasticity, the slenderness ratio at the point of tangency for link 3 is
k3 
 Sr  D 3  
2 C3 E
2(1)(207  109 Pa)

 141.18
S yc
205  106 Pa
Ans.
The Euler column formula is valid when
Sr   Sr  D
Substituting the above values for link 3 gives
346.4 mm
 141.18
t
Therefore, for an Euler column, the thickness parameter t for link 3 is t  2.45 mm .
However, this violates the given condition that the thickness of link 3 (2t) must be greater
than 5 mm. Therefore, the Euler column formula is not valid and the Johnson parabolic
equation must be used.
500
The criterion for using the Johnson parabolic equation is Sr   Sr  D ,that is, the thickness
of link 3 is t  2.45 mm . Using the Johnson parabolic equation, the critical load acting on
link 3 can be written as
2

1  Syc Sr3  
Pcr  A  S yc 

C3 E  2  

Substituting the given data, the critical load acting on link 3 can be written as
2

0.3464 m  

6

 (205  10 Pa)
 
1
t
Pcr  8t 2  205  106 Pa 

 
9
(1.0)(207  10 Pa) 
2

 


 
Therefore, the critical load acting on link 3 is
617.07 N 

Pcr  8t 2 205  106 Pa 

t2

To ensure that the buckling factor of safety for link 3 is N = 5, the critical load can be
written as
Pcr  NF43  5  19.58 kN  97.90 kN
Then substituting Equation (11b) into Equation (10b) gives
617.07 N 

9.79  104 N  8t 2 205  106 

t2

Rearranging this equation, the thickness parameter t for link 3 is
t  7.92 mm
Note that this parameter value satisfies the criterion that the thickness (2t = 15.8 mm) of
link 3 must be greater than 5 mm.
501
11.50 Tthe vertical link 2 is rigidly fixed to the ground (at A) and pinned to the horizontal link 3
at B. Link 4 is pinned to link 3 at H and the mass of link 4 is 200 kg. The vertical link 5
is pinned to link 3 at D and to the ground at O5. Link 2 has a solid circular cross-section
with diameter d2 = 25 mm, a compressive yield strength Syc = 370 MPa, and a modulus of
elasticity E = 205 GPa. The factor of safety guarding against buckling, for link 2, is N =
2.5 and the end-condition constant C = 2. Assume that gravity is acting vertically
downward and links 2, 3, and 5 are massless compared to link 4. (a) For link 2
determine: (1) the slenderness ratio; and (2) the slenderness ratio at the point of tangency
between the Euler column formula and the Johnson parabolic formula. (b) Determine the
critical unit load acting on link 2. (c) If the diameter of link 2 is increased to d2 = 90 mm,
then is link 2 an Euler column or a Johnson column?
AB = 2.5 m and DB = BH = 1.5 m.
The second moment of area for the solid circular cross-section of link 2 is
4
 d 24   0.025 m 
8
I

 1.9175 10 m4
64
64
and the cross-sectional area of link 2 is
2
 d 22   0.025 m 
4
A

 4.9087 10 m2
4
4
The radius of gyration of link 2 is
1.9175 10 m4
I

 0.006 25 m
4
A
4.9087 10 m2
8
k
502
Check: The radius of gyration of a solid circular cross-section, see Table 4, Appendix A, is
k  d 4  0.025 m 4  0.006 25 m
The slenderness ratio of link 2 is
L
2.5 m
Sr  
 400
k 0.006 25 m
The slenderness ratio at the point of tangency between the Euler column formula and the
Johnson parabolic formula, where the end condition constant for link 2 is given as C = 2
can be written as
 Sr  D  
Comparing values indicates that
2  2  205 GPa 
2CE

 147.90
Syc
370MPa
Sr   Sr  D
Therefore, link 2 is an Euler column.
The free body diagrams for links 3 and 4 are shown in the figures below.
Taking the sum of the vertical forces on link 4 we find
 F4y  F34y  m4g  0
From this we obtain
F34y  m4 g   200 kg   9.807 m/s2   1 961 N
Taking the sum of moments on link 3 about point B gives
 M B3  RDB F53y  RHB F43y  0
However, since RDB = RHB, this can be written as
F53y  F43y  1 961 N
and the sum of the vertical forces on link 3 gives
F23y  F43y  F53y  3 922 N
Therefore, the reaction force acting from link 3 onto link 2 at pin B is
F23  3 922ˆj N
The critical load on link 2 can be written as
Pcr  NF23y  2.5  3 922 N   9 805 N
Ans.
503
The critical unit load acting on link 2 is
Pcr
9 805 N

 19.97 MPa
A 4.9087 10 4 m2
Ans.
If the diameter of link 2 is d 2  90 mm, then the new cross-sectional area of link 2 is
A
 d 22

  0.090 m 
2
 0.006 362 m2
4
4
The new radius of gyration of the solid circular cross-section is
d
0.090 m
k2  2 
 0.022 50 m
4
4
Therefore, the new slenderness ratio is
L
2.500 m
Sr  2 
 111.111
k2 0.02250 m
The slenderness ratio at the point of tangency is still
2CE
 147.90
 Sr  D  
Syc
Comparing values indicates that now
Sr   Sr  D
Therefore, link 2 is now a Johnson column.
Ans.
504
Page intentionally blank.
505
Chapter 12
Dynamic Force Analysis*
12.1
The steel bell crank is used as an oscillating cam follower. Using 0.282 lb/ in 3 for the
density of steel, find the mass moment of inertia of the lever about an axis through O.
For the vertical arm, using Appendix A, Table 5,
m  wht    0.750 in  3.750 in  0.375 in   0.282 lb/in3   0.297 lb
2
2
IG  m  a 2  c 2  12   0.297 lb 386 in/s 2   0.750 in   3.750 in   12  0.000 939 in  lb  s 2
IO  IG  md 2  0.000 939 in  lb  s 2   0.297 lb 386 in/s 2  1.500 in   0.002 672 in  lb  s 2
2
For the horizontal arm, using Appendix A, Table 5,
m  wht    0.750 in  6.000 in  0.375 in   0.282 lb/in 3   0.476 lb
IG  m  a 2  c 2  12   0.476 lb 386 in/s 2   0.750 in    6.000 in   12  0.003 755 in  lb  s 2
2
2
IO  IG  md 2  0.003 755 in  lb  s 2   0.476 lb 386 in/s 2  3.375 in   0.017 798 in  lb  s 2
2
For the roller, using Appendix A, Table 5,
2
m   r 2t     0.500 in   0.500 in   0.282 lb/in 3   0.111 lb
*
Unless instructed otherwise, solve all problems without friction and without gravitational loads.
506
IG  mr 2 2   0.111 lb 386 in/s2   0.500 in  2  0.000 036 in  lb  s2
2
IO  IG  md 2  0.000 036 in  lb  s 2   0.111 lb 386 in/s 2   6.000 in   0.010 364 in  lb  s 2
2
For the composite lever
m   0.297 lb    0.476 lb    0.111 lb   0.884 lb
IO   0.002 672 in  lb  s 2    0.017 798 in  lb  s 2    0.010 364 in  lb  s 2   0.030 834 in  lb  s 2
Ans.
507
12.2
A 5- by 50- by 300-mm steel bar has two round steel disks, each 50 mm in diameter and
20 mm long, welded to one end. A small hole is drilled 25 mm from the other end.
Using 7.80 Mg/ m3 for the density of steel, find the mass moment of inertia of this bart
about an axis through the hole.
Dimensions are in millimeters.
For the rectangular bar, using Appendix A, Table 5,
m  wht    0.050 m  0.300 m  0.005 m   7.80 Mg/m3   0.585 kg
2
2
IG  m  a 2  c 2  12   0.585 kg   0.050 m    0.300 m   12  0.004 509 kg  m 2


IO  IG  md 2  0.004 509 kg  m2   0.585 kg  0.125 m   0.013 650 kg  m2
2
For the two circular disks, using Appendix A, Table 5,
2
m  2 r 2t   2  0.025 m   0.020 m   7.80 Mg/m3   0.613 kg
IG  mr 2 2   0.613 kg  0.025 m  2  0.000 191 kg  m2
2
IO  IG  md 2  0.000 191 kg  m2   0.613 kg  0.250 m   0.038 480 kg  m2
2
Assume that the mass and inertia of the small drilled hole are negligible.
For the composite lever
m   0.585 kg    0.613 kg   1.198 kg
IO   0.013 650 kg  m2    0.038 480 kg  m2   0.052 130 kg  m2
Ans.
508
12.3
Determine the reaction forces at the joints and the external torque applied to the input link
2 of the four-bar linkage in the posture illustrated. For the constant angular velocity
ω2  180kˆ rad/s, the known kinematics are: α3  4 950kˆ rad/s2 , α 4  8 900kˆ rad/s2 ,
A  6 320ˆi  750ˆj ft/s2 , and A  2 280ˆi  750ˆj ft/s2 . The weights and mass moments of
G3
G4
inertia of the links are: w3  0.708 lb, w4  0.780 lb, IG2  0.0258 in  lb  s2 , IG3  0.0154 in  lb  s2 ,
and IG4  0.0112 in  lb  s2 .
RAO2  3 in, RO4O2  7 in, RBA  8 in, RBO4  6 in, RG3 A  4 in, and RG4O4  3 in.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 139ˆi  16ˆj lb  140 lb186.77
t 3   I G3 α 3
 55ˆi  18ˆj lb  58 lb198.21
t 4   I G4 α 4
   0.708 lb 32.2 ft/s 2   6 320ˆi +750ˆj ft/s 2 

   0.015 4 in  lb  s 2  4 950kˆ rad/s 2

   0.780 lb 32.2 ft/s 2   2 280ˆi +750ˆj ft/s 2 

   0.011 2 in  lb  s 2  8 900kˆ rad/s 2

 76kˆ in  lb
 100kˆ in  lb
h3  t3 f3  128 in  lb  140 lb   0.545 in , h4  t4 f 4  100 in  lb   58 lb   1.714 in
Next, the free-body diagrams are drawn with the inertia forces applied. Since the lines of
action for the forces on the free-body diagrams cannot be discovered from two- and threeforce member concepts, the force F34 is divided into radial and transverse components.
509
(Note that it is totally coincidental that the reaction components are also exactly aligned
with the radial and transverse axes of link 3. This results from the perpendicularity of
links 3 and 4.)
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 1.800ˆi  2.400ˆj in  ×  55ˆi  18ˆj lb   100kˆ in  lb    3.600ˆi  4.800ˆj in  ×  0.800ˆi  0.600ˆj F  0
t
164.400kˆ in  lb  100kˆ in  lb   6.000kˆ in  F  0
34
t
34
F34t  44 lb
 M  R ×f  t  R ×F  0
3.200ˆi  2.400ˆj in  ×  139ˆi 16ˆj lb    76kˆ in  lb   6.400ˆi  4.800ˆj in  ×  0.600ˆi  0.800ˆj F  0
A
G3 A
3
3
BA
r
43
 282.400kˆ in  lb   76kˆ in  lb   8.000kˆ in  F  0
r
43
r
43
F43r  26 lb
F  F  f  F  0
M  R × F  T  0
43
O2
3
23
AO2
32
12
F34  20ˆi  47ˆj lb  51 lb67.26
F  159ˆi  64ˆj lb  171 lb21.82
23
T12  192kˆ in  lb
Ans.
510
12.4
Determine the reaction forces at the joints and the external torque applied to the input link
2 of the four-bar linkage in the posture illustrated. For the constant angular velocity
ω2  200kˆ rad/s, the known kinematics are: α3  6 500kˆ rad/s 2 , α 4  240kˆ rad/s2 ,
A  3 160ˆi  262ˆj ft/s2 , and A  800ˆi  2 110ˆj ft/s2 . The weights and mass
G3
G4
moments of inertia of the links are: w3  2.65 lb, w4  6.72 lb, IG2  0.023 9 in  lb  s2 ,
IG3  0.060 6 in  lb  s2 , and IG4  0.531 in  lb  s2 .
RAO2  2 in, RO4O2  13 in, RBA  17 in, RBO4  8 in, RG3 A  8.5 in, and RG4O4  4 in.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 260ˆi  22ˆj lb  261 lb  4.74
t 3   I G3 α 3
 167ˆi  440ˆj lb  471 lb69.24
t 4   I G4 α 4
   2.650 lb 32.2 ft/s 2   3 160ˆi +262ˆj ft/s 2 

   0.060 6 in  lb  s 2  6 500kˆ rad/s 2

   6.720 lb 32.2 ft/s 2   800ˆi  2 110ˆj ft/s 2 

   0.531 in  lb  s 2  240kˆ rad/s 2

 394kˆ in  lb
 127kˆ in  lb
h3  t3 f3   394 in  lb   261 lb   1.509 in , h4  t4 f 4  127 in  lb   471 lb   0.271 in
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
511
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
1.461ˆi  3.724ˆj in  × 167ˆi  440ˆj lb   127kˆ in  lb    2.922ˆi  7.447ˆj in  ×  0.931ˆi  0.365ˆj F  0
t
34
 21.641kˆ in  lb  127kˆ in  lb   8.000kˆ in  F  0
t
34
F34t  19 lb
M  R
A
G3 A
×f3  t 3  R BA ×F43t  R BA ×F43r  0
 7.254ˆi  4.431ˆj in  ×  260ˆi  22ˆj lb   394kˆ in  lb   14.508ˆi  8.862ˆj in  ×  17ˆi  7ˆj lb 
 14.508ˆi  8.862ˆj in  ×  0.365ˆi  0.931ˆj F  0
r
43
 1 311.583kˆ in  lb  394kˆ in  lb  165.162kˆ in  lb    11.887kˆ in  F  0
r
43
F43r  63 lb
F  F  f  F  0
F  F  f  F  0
F  F  F  0
M  R  F  T  0
O2
34
4
14
43
3
23
32
12
AO2
32
12
F34  6ˆi  66ˆj lb  66 lb  95.06
Ans.
F  162ˆi  375ˆj lb  408 lb  113.27 Ans.
14
F23  266ˆi  44ˆj lb  270 lb  170.57 Ans.
F12  266ˆi  44ˆj lb  270 lb9.43
Ans.
T  439kˆ in  lb
Ans.
12
512
12.5
Determine the reaction forces at the joints and the crank torque of the slider-crank linkage
in the posture illustrated. For the constant angular velocity ω2  210kˆ rad/s, the known
kinematics are: α  7 670kˆ rad/s2 , A  7 820ˆi  4 876ˆj ft/s2 , and A  7 850ˆi ft/s2 . The
3
G3
G4
weights and mass moments of inertia of the links are: w3  3.40 lb, w4  2.86 lb,
IG2  0.352 in  lb  s 2 , and IG3  0.108 in  lb  s2 .
RAO2  3 in, RBA  12 in, and RG3 A  4.5 in.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f 4  m4 AG4
f3  m3 AG3
   3.40 lb 32.2 ft/s 2  7 820ˆi  4 876ˆj ft/s 2 
   2.860 lb 32.2 ft/s 2   7 850ˆi ft/s 2 
 826ˆi  515ˆj lb  973 lb31.95
 697ˆi lb  697 lb0.00
t 3   I G3 α 3

ˆ rad/s 2
   0.108 in  lb  s 2  7 670k
t 4   IG4 α 4  0

ˆ in  lb
 828k
h3  t3 f3  828 in  lb   973 lb   0.851 in , h4  t4 f 4  0
Next, the free-body diagrams with inertia forces are drawn and the solution proceeds.
M  R
A
G3 A
×f3  t 3  R BA ×f4  R BA ×F14  0
 4.429ˆi  0.795ˆj in  × 826ˆi  515ˆj lb   828kˆ in  lb   11.811ˆi  2.121ˆj in  ×  697ˆi lb 
 11.811ˆi  2.121ˆj in  × F ˆj  0
 2 938kˆ in  lb  828kˆ in  lb  1 478kˆ in  lb   11.811kˆ in  F  0
14
14
F14  444 lb
F  f  F  F  0
F  F  f  F  0
F  F  F  0
M  R × F  T  0
O2
4
14
34
43
3
23
32
12
AO2
32
12
F14  444ˆj lb  444 lb  90
F  697ˆi  444ˆj lb  826 lb147.50
34
Ans.
Ans.
F23  1 523ˆi  71ˆj lb  1 525 lb  177.33 Ans.
F  1 523ˆi  71ˆj lb  1 525 lb  177.33 Ans.
12
T12  3 080kˆ in  lb
Ans.
513
12.6
Determine the reaction forces at the joints and the crank torque of the slider-crank linkage
in the posture illustrated if the external force acting through pin B of the piston is
FB  800î lb. For the constant angular velocity ω2  160kˆ rad/s, the known kinematics
A  2 640 ft/s2150,
A  6 130 ft/s2158.3,
α  3 090kˆ rad / s2 ,
are:
and
G2
3
AG4  6 280 ft/s 180.
2
w2  0.95 lb,
G3
The weights and mass moments of inertia of the links are:
w3  3.50 lb,
w4  2.50 lb,
IG2  0.003 69 in  lb  s2 ,
IG3  0.108 in  lb  s2 .
RAO2  3 in, RBA  12 in, RG2O2  1.25 in, and RG3 A  3.5 in.
The d’Alembert inertia forces and offsets are:
f 2  m2 AG2
   0.95 lb 32.2 ft/s 2  2 286ˆi  1 320ˆj ft/s 2  t 2   IG2 α 2  0
 67ˆi  39ˆj lb  78 lb  30.00
h2  t2 f 2  0
f 4  m4 AG4
f3  m3 AG3
   3.50 lb 32.2 ft/s 2  5 696ˆi  2 267 ˆj ft/s 2 
 619ˆi  246ˆj lb  666 lb  21.70
t 3   I G3 α 3

   0.110 in  lb  s 2  3 090kˆ rad/s 2
   2.500 lb 32.2 ft/s 2   6 280ˆi ft/s 2 
 488ˆi lb  488 lb0.00

t 4   IG4 α 4  0
 340kˆ in  lb
h3  t3 f3   340 in  lb   666 lb   0.510 in , h4  t4 f 4  0
Next, the free-body diagrams are drawn with inertia forces and the solution proceeds.
and
514
M  R
A
G3 A
×f3  t 3  R BA ×f4  R BA ×FB  R BA ×F14  0
3.473ˆi  0.438ˆj in  × 619ˆi  246ˆj lb   340kˆ in  lb   11.906ˆi 1.500ˆj in  ×  488ˆi lb 
 11.906ˆi  1.500ˆj in  ×  800ˆi lb   11.906ˆi  1.500ˆj in  × F ˆj  0
14
 1 125kˆ in  lb  340kˆ in  lb   732kˆ in  lb   1 200kˆ in  lb   11.906kˆ in  F  0
14
F14  27 lb
F  f  F  F  F  0
F  F  f  F  0
F  F  F  0
M  R  F  T  0
O2
4
B
14
43
3
23
32
12
AO2
32
34
12
F14  27ˆj lb  27 lb90
F  312ˆi  27ˆj lb  313 lb  4.95
Ans.
F23  307ˆi  219ˆj lb  377 lb144.50
Ans.
F12  374ˆi  258ˆj lb  454 lb145.40
Ans.
T12  108kˆ in  lb
Ans.
34
Ans.
515
12.7
Determine the reaction forces at the joints and the torque applied to the input link 2 of the
four-bar linkage in the posture when 2  53. For the constant angular velocity
ω  12kˆ rad/s ccw,
the
known
kinematics
are:
  0.7,
  20.4,
2
3
4
3  85.6 rad/s cw,
 4  172 rad/s cw,
AG4  97.8 m/s2270.
The masses and mass moments of inertia of the links are:
2
m2  5.2 kg,
2
m3  65.8 kg,
m4  21.8 kg,
AG3  96.4 m/s 259,
2
IG2  2.3 kg  m2 ,
and
IG3  4.2 kg  m2 , and
IG4  0.51 kg  m2 .
RAO2  0.3 m, RO4O2  0.9 m, RBA  1.5 m, RBO4  0.8 m, C  33, RCA  0.85 m,
 D  53, RDO4  0.4 m,   16, RG3 A  0.65 m,   17, and RG4O4  0.45 m.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 1 210ˆi  6 227ˆj N  6 343 N79
t 3   I G3 α 3
 2 132ˆj N  2 132 N90
t 4   I G4 α 4
   65.8 kg   18.394ˆi  94.629ˆj m/s 2 

   4.200 kg  m 2  85.6kˆ rad/s 2
 360kˆ N  m

   21.8 kg   97.800ˆj m/s 2 

   0.51 kg  m 2  172kˆ rad/s 2
 88kˆ N  m

516
h3  t3 f3   360 N  m   6 343 N   0.057 m , h4  t4 f 4  88 N  m   2 132 N   0.041 m
Next, the free-body diagrams are drawn with inertia forces. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 0.390ˆi  0.225ˆj m  ×  2 132ˆj N   88kˆ N  m    0.750ˆi  0.279ˆj m  ×  0.348ˆi  0.937ˆj F  0
t
34
831kˆ N  m  88kˆ N  m   0.800kˆ m F  0
t
34
F34t  1 149 N
M  R
A
G3 A
×f3  t 3  R BA ×F43t  R BA ×F43r  0
 0.631ˆi  0.154ˆj m  × 1 210ˆi  6 227ˆj N   360kˆ N  m   1.499ˆi  0.019ˆj m  ×  400ˆi  1 077ˆj N 
 1.499ˆi  0.019ˆj m  ×  0.937ˆi  0.348ˆj F  0
r
43
3 743kˆ N  m  360kˆ N  m  1 622kˆ N  m   0.515kˆ in  F  0
r
43
34
4
14
F34  10 416ˆi  2 792ˆj N  11 171 N14.5
F  10 416ˆi  4 924ˆj N  11 521 N154.7
14
Ans.
43
3
23
F23  9 207ˆi  3 435ˆj N  9 827 N  20.5
Ans.
32
12
F12  9 207ˆi  3 435ˆj N  9 827 N  20.5
Ans.
T12  2 907kˆ N  m
Ans.
F43r  11 117 N
F  F  f  F  0
F  F  f  F  0
F  F  F  0
M  R × F  T  0
O2
AO2
32
12
Ans.
517
12.8
Solve Problem 12.7 with an external force FD  12ˆi kN acting at point D.
Here, the method of superposition is used for the solution. The force components of
Prob. 12.7 are denoted with primes and the additional force increments with double
primes. The figures here show the incremental forces only.
 M  R
O4
DO4
×FD  R BO4 ×F34  0
 0.114ˆi  0.383ˆj m  × 12 000ˆi N    0.750ˆi  0.279ˆj m  ×  0.999ˆi  0.013ˆj F34  0
 4 596kˆ N  m  0.269kˆ m F   0
F34  17 087 N
F34  17 085ˆi  215ˆj N  17 087 N  179.3
F  5 086ˆi  215ˆj N  5 091 N2.4
F34  6 670ˆi  2 577ˆj N  7 150 N158.9
F  5 330ˆi  2 577ˆj N  5 920 N  154.2
F23  17 085ˆi  215ˆj N  17 087 N  179.3
F23  7 879ˆi  3 650ˆj N  8 684 N  155.1 Ans.
F12  17 085ˆi  215ˆj N  17 087 N  179.3
F12  7 879ˆi  3 650ˆj N  8 684 N  155.1 Ans.
T12  1 499kˆ N  m
Ans.
34
14
T12  4 406kˆ N  m
14
Ans.
Ans.
518
12.9
Make a complete kinematic and dynamic analysis of the four-bar linkage of Problem 12.7
using the data in the figure caption but in the posture when 2  170. The constant
angular velocity of the input link 2 is 2  12 rad/s ccw, and the external force at point D
is FD  8.94 kN64.3 .
Kinematic Analysis:

 
VA  ω 2 × R AO2  12kˆ rad/s × 0.295ˆi  0.052ˆj m

 0.625ˆi  3.545ˆj m/s  3.600 m/s  100
VB  VA  ω3 × R BA  ω 4 × R BO4
  0.625ˆi  3.545ˆj m/s   3kˆ rad/s  × 1.304ˆi  0.740ˆj m   4kˆ rad/s  ×  0.109ˆi  0.793ˆj m 

 
 
 0.625ˆi  3.545ˆj m/s  0.7403ˆi  1.3043ˆj m  0.7934 ˆi  0.1094 ˆj m
ω3  3.020kˆ rad/s
ω4  3.607kˆ rad/s
V  2.859ˆi  0.393ˆj m/s  2.886 m/s172.16
B

2
A A   22 R AO2  α 2 × R AO2   12 rad/s  × 0.295ˆi  0.052ˆj m
 42.543ˆi  7.502ˆj m/s2  43.200 m/s2  10

Ans.

519
A B  A A  32 R BA  α3 × R BA  42 R BO4  α 4 × R BO4


 
 
  1.419ˆi  10.311ˆj m/s    kˆ rad/s  ×  0.109ˆi  0.793ˆj m 
  32.069ˆi  3.940ˆj m/s    0.740 ˆi  1.304 ˆj m    0.793 ˆi  0.109 ˆj m 
 42.543ˆi  7.502ˆj m/s 2  11.893ˆi  6.749ˆj m/s 2   3kˆ rad/s 2 × 1.304ˆi  0.740ˆj m
2

2
4
2
3
α3  0.390kˆ rad/s
3
AG  A A   R G A  α 3 × R G A
3
2
3

4
4
α 4  40.816kˆ rad/s
2
3
3
 
 
2
Ans.

 42.543ˆi  7.502ˆj m/s2  4.149ˆi  4.234ˆj m/s 2  0.390kˆ rad/s 2 × 0.455ˆi  0.464ˆj m
AG3  38.575ˆi  11.913ˆj m/s2  40.373 m/s2  17.16
AG   R G O  α 4 × RG O
4

2
4
4 4
4 4
 

 0.932ˆi  5.780ˆj m/s 2  40.816kˆ rad/s 2 × 0.072ˆi  0.444ˆj m

Ans.

AG4  19.054ˆi  2.841ˆj m/s2  19.265 m/s2  8.48
Ans.
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 2 538ˆi  784ˆj N  2 657 N162.84
t 3   I G3 α 3
 415ˆi  62ˆj N  420 N171.52
t 4   I G4 α 4
   65.8 kg   38.575ˆi  11.913ˆj m/s 2 

   4.200 kg  m 2  0.390kˆ rad/s 2

   21.8 kg  19.054ˆi  2.841ˆj m/s 2 

   0.51 kg  m 2  40.816kˆ rad/s 2

 1.638kˆ N  m
 21kˆ N  m
h3  t3 f3  1.638 N  m   2 657 N   0.001 m , h4  t4 f 4   21 N  m   420 N   0.050 m
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
520
M  R
O4
G4O4
×f4  t 4  R DO4 ×FD  R BO4 ×F34t  0
 0.072ˆi  0.444ˆj m  ×  415ˆi  62ˆj N    21kˆ N  m    0.284ˆi  0.282ˆj m  × 3 877ˆi  8 056ˆj N 
  0.109ˆi  0.793ˆj m  ×  0.991ˆi  0.136ˆj F  0
t
34
180kˆ N  m   21kˆ N  m   3 379kˆ N  m   0.800kˆ m F34t  0
F34t  3 972 N
 M  R ×f  t  R ×F  R ×F  0
 0.455ˆi  0.464ˆj m  ×  2 538ˆi  784ˆj N   1.638kˆ N  m   1.304ˆi  0.740ˆj m  ×  3 935ˆi  542ˆj N 
 1.304ˆi  0.740ˆj m  ×  0.136ˆi  0.991ˆj F  0
A
G3 A
3
3
BA
t
43
r
43
BA
r
43
1 534kˆ N  m   2kˆ N  m   3 618kˆ N  m   1.192kˆ in  F43r  0
F43r  1 747 N
F34  4 173ˆi  1 189ˆj N  4 339 N  164.10
F  F  f  F  F  0
F  F  f  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
D
43
3
23
32
12
AO2
32
14
12
Ans.
F14  711ˆi  6 929ˆj N  6 966 N  84.14 Ans.
F23  1 635ˆi  1 972 N  2 562 N  129.65
Ans.
F12  1 635ˆi  1 972 N  2 562 N  129.65
Ans.
T12  668kˆ N  m
Ans.
521
12.10 Repeat Problem 12.9 in the posture when 2  200. The constant angular velocity of the
input link 2 is 2  12 rad/s ccw, the external force at point C is FC  8.49 kN45 , and
there is no external force at point D.
Kinematic Analysis:
 

VA  ω 2 × R AO2  12kˆ rad/s × 0.282ˆi  0.103ˆj m

 1.231ˆi  3.383ˆj m/s  3.600 m/s  70
VB  VA  ω3 × R BA  ω 4 × R BO4
 1.231ˆi  3.383ˆj m/s   3kˆ rad/s  × 1.198ˆi  0.902ˆj m   4kˆ rad/s  ×  0.016ˆi  0.799ˆj m 

 
 
 1.231ˆi  3.383ˆj m/s  0.9023ˆi  1.1983ˆj m  0.7994 ˆi  0.0164 ˆj m
ω3  2.846kˆ rad/s
ω4  1.671kˆ rad/s
V  1.336ˆi  0.027ˆj m/s  1.337 m/s178.84
B

Ans.

2
A A   22 R AO2  α 2 × R AO2   12 rad/s  × 0.282ˆi  0.103ˆj m

 40.595ˆi  14.775ˆj m/s2  43.200 m/s220
A B  A A  32 R BA  α 3 × R BA  42 R BO4  α 4 × R BO4
 

 
 
ˆ
ˆ
ˆ
ˆ
ˆ
  0.045i  2.233 j m/s    k rad/s  ×  0.016i  0.799 j m 
  30.937ˆi  9.702ˆj m/s    0.902 ˆi  1.198 ˆj m    0.799 ˆi  0.016 ˆj m 
ˆ rad/s 2 × 1.198ˆi  0.902ˆj m
 40.595ˆi  14.775ˆj m/s 2  9.703ˆi  7.306ˆj m/s 2   3k
2
2
4
2
3
3
4
4

522
α3  8.760kˆ rad/s2
α 4  48.558kˆ rad/s2
AG3  A A   R G3 A  α3 × R G3 A

2
3
 
 
 
 40.595ˆi  14.775ˆj m/s 2  3.169ˆi  4.204ˆj m/s 2  8.760kˆ rad/s 2 × 0.391ˆi  0.519ˆj m
Ans.

AG3  41.972ˆi  7.146ˆj m/s  42.576 m/s 9.66
2
AG4  42 R G4O4  α 4 × R G4O4

2
 
 
 0.343ˆi  1.209ˆj m/s 2  48.558kˆ rad/s 2 × 0.123ˆi  0.433ˆj m
Ans.

AG4  21.365ˆi  4.754ˆj m/s2  21.887 m/s212.54
Ans.
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
   65.8 kg   41.972ˆi  7.146ˆj m/s 2 
   21.8 kg   21.365ˆi  4.754ˆj m/s 2 
 2 762ˆi  470ˆj N  2 802 N  170.34  466ˆi  104ˆj N  477 N  167.46
t 3   I G3 α 3
t 4   I G4 α 4

   4.200 kg  m 2  8.760kˆ rad/s 2


   0.51 kg  m 2  48.558kˆ rad/s 2

 37kˆ N  m
 25kˆ N  m
h3  t3 f3   37 N  m   2 802 N   0.013 m , h4  t4 f 4   25 N  m   477 N   0.052 m
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
523
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 0.123ˆi  0.433ˆj m    466ˆi  104ˆj N    25kˆ N  m    0.016ˆi  0.799ˆj m   0.999ˆi  0.020ˆj F  0
t
34
 214kˆ N  m   25kˆ N  m   0.800kˆ m F34t  0
F34t  299 N
 M  R ×f  t  R ×F  R ×F  R ×F  0
 0.391ˆi  0.519ˆj m  ×  2 762ˆi  470ˆj N   37kˆ N  m   0.291ˆi  0.799ˆj m  × 6 003ˆi  6 003ˆj N 
 1.198ˆi  0.902ˆj m  ×  299ˆi  6ˆj N   1.198ˆi  0.902ˆj m  ×  0.020ˆi  0.999ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
1 250kˆ N  m  37kˆ N  m   3 050kˆ N  m  277kˆ N  m   1.179kˆ in  F43r  0
F43r  1 260 N
F  F  f  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
14
43
3
C
32
12
AO2
32
23
12
F34  274ˆi  1 266ˆj N  1 295 N  77.80 Ans.
F  192ˆi  1 370ˆj N  1 383 N82.01 Ans.
14
F23  2 968ˆi  6 799 N  7 419 N  113.58
Ans.
F12  2 968ˆi  6 799 N  7 419 N  113.58
Ans.
T12  1 612kˆ N  m
Ans.
524
12.11 Make a complete dynamic analysis of the four-bar linkage of Problem 12.7, but in the
posture when 2  270. The constant angular velocity of the input link 2 is
2  18 rad/s ccw and the external force at point D is FD  8.94 kN64.3 . The known
kinematics
are:
3  46.6 ,
4  80.5 ,
3  178 rad/s2 cw ,
 4  256 rad/s2 cw ,
AG3  112 m/s222.7 , and AG4  119 m/s2352.5 .
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   65.8 kg  103.324ˆi  43.221ˆj m/s 2 
 6 799ˆi  2 844ˆj N  7 370 N  157.30
t 3   I G3 α 3

   4.200 kg  m 2  178kˆ rad/s 2

f 4  m4 AG4
   21.8 kg  117.982ˆi  15.533ˆj m/s 2 
 2 572ˆi  339ˆj N  2 594 N172.50
t 4   I G4 α 4

   0.51 kg  m 2  256kˆ rad/s 2

 748kˆ N  m
 131kˆ N  m
h3  t3 f3   748 N  m   7 370 N   0.101 m , h4  t4 f 4  131 N  m   2 594 N   0.050 m
525
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  R
O4
G4O4
×f4  t 4  R DO4 ×FD  R BO4 ×F34t  0
 0.059ˆi  0.446ˆj m  ×  2 572ˆi  339ˆj N   131kˆ N  m    0.276ˆi  0.290ˆj m  ×  3 877ˆi  8 056ˆj N 
  0.131ˆi  0.789ˆj m  ×  0.986ˆi  0.164ˆj F  0
t
34
1 127kˆ N  m  131kˆ N  m   3 348kˆ N  m   0.800kˆ m F34t  0
F34t  2 613 N
 M  R ×f  t  R ×F  R ×F  0
 0.300ˆi  0.577ˆj m  ×  6 799ˆi  2 844ˆj N    748kˆ N  m   1.031ˆi  1.089ˆj m  ×  2 578ˆi  429ˆj N 
 1.031ˆi  1.089ˆj m  ×  0.164ˆi  0.986ˆj F  0
A
G3 A
3
3
BA
t
43
r
43
BA
r
43
3 070kˆ N  m  748kˆ N  m   3 250kˆ N  m   0.839kˆ in  F43r  0
F43r  677 N
F  F  f  F  F  0
F  F  f  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
D
43
3
23
32
12
AO2
32
14
12
F34  2 466ˆi  1 097ˆj N  2 699 N156.01 Ans.
F14  1 411ˆi  9 153ˆj N  9 261 N  98.76 Ans.
F23  9 265ˆi  1 747 lb  9 428 N10.68
Ans.
F12  9 265ˆi  1 747 lb  9 428 N10.68
Ans.
T12  2 780kˆ N  m
Ans.
526
12.12 Make a complete dynamic analysis of the four-bar linkage in the posture when 2  90 ,
3  23.9, and 4  91.7. For the constant angular velocity 2  32 rad/s ccw , the known
kinematics
are:
3  221 rad/s2 ccw,
 4  122 rad/s2 ccw,
AG3  88.6 m/s2255,
and
AG4  32.6 m/s2244. There is an external force FC  632 N342 acting at point C. The
masses and mass moments of inertia of the links are: m2  0.5 kg, m3  4 kg,
m4  1.5 kg, IG2  0.005 N  m  s2 , IG3  0.011 N  m  s2 , and IG4  0.002 3 N  m  s2 .
RAO2  120 mm, RO4O2  300 mm, RBA  320 mm, RBO4  250 mm, C  15, RCA  360 mm,
 D  0, RDO  0, RG O  0,   8, RG3 A  200 mm,   0, and RG4O4  125 mm.
4
2 2
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   4.0 kg   22.931ˆi  85.581ˆj m/s 2 
 92ˆi  342ˆj N  354 N75
t 3   I G3 α 3

   0.011 kg  m 2  221kˆ rad/s 2
 2.431kˆ N  m

f 4  m4 AG4
  1.5 kg   14.291ˆi  29.301ˆj m/s 2 
 21ˆi  44ˆj N  49 N64
t 4   I G4 α 4

   0.002 3 kg  m 2  122kˆ rad/s 2
 0.281kˆ N  m

527
h3  t3 f3   2.431 N  m   354 N   0.007 m , h4  t4 f 4   0.281 N  m   49 N   0.006 m
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 0.004ˆi  0.125ˆj m  ×  21ˆi  44ˆj N    0.281kˆ N  m    0.008ˆi  0.249ˆj m  ×  0.999ˆi  0.030ˆj F  0
t
34
 2.844kˆ N  m   0.281kˆ N  m   0.250kˆ m F34t  0
F34t  12.5 N
 M  R ×f  t  R ×F  R ×F  R ×F  0
 0.170ˆi  0.106ˆj m  × 92ˆi  342ˆj N    2.431kˆ N  m   0.280ˆi  0.226ˆj m  × 601ˆi  195ˆj N 
  0.292ˆi  0.130ˆj m  × 12.5ˆi N    0.292ˆi  0.130ˆj m  ×  0.030ˆi  0.999ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
 48kˆ N  m   2.431kˆ N  m   191kˆ N  m   1.623kˆ N  m    0.296kˆ in  F  0
r
43
F43r  496 N
F  F  f  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
14
43
3
C
32
12
AO2
32
23
12
F34  2ˆi  496ˆj N  496 N  89.71
Ans.
F14  24ˆi  452ˆj N  453 N93.02
Ans.
F23  690ˆi  643 lb  944 N  137
Ans.
F12  690ˆi  643 lb  944 N  137
Ans.
T12  82.83kˆ N  m
Ans.
528
12.13 Make a complete kinematic and dynamic analysis of the four-bar linkage in Problem.
12.12 but in the posture when 2  260 .
Kinematic Analysis:
 

VA  ω 2 × R AO2  32kˆ rad/s × 0.021ˆi  0.118ˆj m

 3.782ˆi  0.667ˆj m/s  3.840 m/s  10
VB  VA  ω3 × R BA  ω 4 × R BO4
  3.782ˆi  0.667ˆj m/s   3kˆ rad/s  ×  0.138ˆi  0.289ˆj m   4kˆ rad/s  ×  0.183ˆi  0.171ˆj m 

 
 
 3.782ˆi  0.667ˆj m/s  0.2893ˆi  0.1383ˆj m  0.1714 ˆi  0.1834 ˆj m

ω3  10.554kˆ rad/s
ω4  4.314kˆ rad/s
V  0.736ˆi  0.789ˆj m/s  1.079 m/s46.99
B

2
A A   22 R AO2  α 2 × R AO2    32 rad/s  × 0.021ˆi  0.118ˆj m
Ans.

 21.338ˆi  121.013ˆj m/s2  122.880 m/s280
A B  A A  32 R BA  α 3 × R BA  42 R BO4  α 4 × R BO4

 
 
  3.402ˆi  3.174ˆj m/s    kˆ rad/s  ×  0.183ˆi  0.171ˆj m 
 
 21.338ˆi  121.013ˆj m/s 2  15.374ˆi  32.158ˆj m/s 2   3kˆ rad/s 2 × 0.138ˆi  0.289ˆj m
2

2
4
 2.562ˆi  92.029ˆj m/s    0.289 ˆi  0.138 ˆj m    0.171 ˆi  0.183 ˆj m 
2
3
α3  199.622kˆ rad/s
4
4
α 4  352.356kˆ rad/s2
AG3  A A   R G3 A  α 3 × R G3 A

3
2
2
3
 
 
 
 21.338ˆi  121.013ˆj m/s 2  6.718ˆi  21.240ˆj m/s 2  199.622kˆ rad/s 2 × 0.060ˆi  0.191ˆj m
AG3  52.748ˆi  87.796ˆj m/s  102.423 m/s 59.00
2
AG4  42 R G4O4  α 4 × RG4O4

 
2

 1.701ˆi  1.587ˆj m/s 2  352.356kˆ rad/s 2 × 0.091ˆi  0.085ˆj m
AG4  31.651ˆi  30.477ˆj m/s2  43.939 m/s243.92
Ans.

Ans.

Ans.
529
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
  1.5 kg   31.651ˆi  30.477ˆj m/s 2 
   4.0 kg   52.748ˆi  87.796ˆj m/s 2 
 47ˆi  46ˆj N  66 N  136.08
 211ˆi  351ˆj N  410 N  121.00
t 3   I G3 α 3

ˆ rad/s 2
   0.011 N  m  s 2  199.622k

ˆ Nm
 2.196k
t 4   I G4 α 4

ˆ rad/s 2
   0.002 3 N  m  s 2  352.356k

ˆ Nm
 0.810k
h3  t3 f3   2.196 N  m   410 N   0.005 m , h4  t4 f 4   0.810 N  m   66 N   0.012 m
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 0.091ˆi  0.085ˆj m  ×  47ˆi  46ˆj N    0.810kˆ N  m    0.183ˆi  0.170ˆj m  × 0.682ˆi  0.731ˆj F  0
t
34
8.212kˆ N  m  0.810kˆ N  m   0.250kˆ m F  0
t
34
F34t  36 N
 M  R ×f  t  R ×F  R ×F  R ×F  0
 0.060ˆi  0.191ˆj m  ×  211ˆi  351ˆj N    2.196kˆ N  m   0.066ˆi  0.354ˆj m  ×  601ˆi  195ˆj N 
  0.138ˆi  0.289ˆj m  ×  25ˆi  26ˆj N    0.138ˆi  0.289ˆj m  ×  0.731ˆi  0.682 ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
19kˆ N  m   2.196kˆ N  m   226kˆ N  m  3.629kˆ N  m    0.305kˆ in  F  0
r
43
F43r  658 N
F  F  f  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
14
43
3
C
32
12
AO2
32
23
12
F34  505ˆi  422ˆj N  659 N  39.88
Ans.
F14  458ˆi  468ˆj N  655 N134.39
Ans.
F23  115ˆi  124 N  169 N46.97
Ans.
F12  115ˆi  124 N  169 N46.97
Ans.
T12  11.03kˆ N  m
Ans.
530
12.14 Make a complete kinematic and dynamic analysis of the four-bar linkage in Problem.
12.12 but in the posture when 2  300 .
Kinematic Analysis:
 

ˆ rad/s × 0.060ˆi  0.104ˆj m
VA  ω 2 × R AO2  32k

 3.326ˆi  1.920ˆj m/s  3.840 m/s30
VB  VA  ω3 × R BA  ω 4 × R BO4
  3.326ˆi  1.920ˆj m/s   3kˆ rad/s  ×  0.093ˆi  0.306ˆj m   4kˆ rad/s  ×  0.147ˆi  0.202ˆj m 

 
 
 3.326ˆi  1.920ˆj m/s  0.3063ˆi  0.0933 ˆj m  0.2024 ˆi  0.1474 ˆj m

ω3  1.597kˆ rad/s
ω4  14.071kˆ rad/s
V  2.844ˆi  2.066ˆj m/s  3.515 m/s36.04
B

A A   22 R AO2  α 2 × R AO2    32 rad/s  × 0.060ˆi  0.104ˆj m
2
Ans.

 61.440ˆi  106.417ˆj m/s2  122.880 m/s2120
A B  A A  32 R BA  α 3 × R BA  42 R BO4  α 4 × R BO4

 
 
  29.105ˆi  39.995ˆj m/s    kˆ rad/s  ×  0.147ˆi  0.202ˆj m 
 
 61.440ˆi  106.417ˆj m/s 2  0.237ˆi  0.780ˆj m/s 2   3kˆ rad/s 2 × 0.093ˆi  0.306ˆj m
2

2
4

 
 
90.782ˆi  145.632ˆj m/s 2  0.3063ˆi  0.0933ˆj m  0.202 4ˆi  0.147 4ˆj m
α3  671.302kˆ rad/s2
AG3  A A   R G3 A  α 3 × R G3 A
2
3


α 4  567.507kˆ rad/s2
 
 
 
 61.440ˆi  106.417ˆj m/s 2  0.079ˆi  0.504ˆj m/s 2  671.302kˆ rad/s 2 × 0.031ˆi  0.198ˆj m
AG3  71.399ˆi  85.103ˆj m/s  111.087 m/s 50.00
2
AG4   R G4O4  α 4 × R G4O4

2
4
 
2
 
 14.547ˆi  20.022ˆj m/s 2  567.507kˆ rad/s 2 × 0.073ˆi  0.101ˆj m
AG4  71.865ˆi  21.406ˆj m/s2  74.986 m/s216.59
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
Ans.

Ans.

Ans.
531
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 286ˆi  340ˆj N  444 N  130.00
t 3   I G3 α 3
 108ˆi  32ˆj N  112 N  163.41
t 4   I G4 α 4
   4.0 kg   71.399ˆi  85.103ˆj m/s 2 

   0.011 N  m  s 2  671.302kˆ rad/s 2
  1.5 kg   71.865ˆi  21.406ˆj m/s 2 

 7.384kˆ N  m

   0.002 3 N  m  s 2  567.507kˆ rad/s 2

 1.305kˆ N  m
h3  t3 f3   7.384 N  m   444 N   0.017 m , h4  t4 f 4  1.305 N  m  112 N   0.012 m
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  R
O4
G4O4
×f4  t 4  R BO4 ×F34t  0
 0.073ˆi  0.101ˆj m  ×  108ˆi  32ˆj N   1.305kˆ N  m    0.147ˆi  0.202ˆj m  × 0.809ˆi  0.588ˆj F  0
t
34
13.273kˆ N  m  1.305kˆ N  m   0.250kˆ m F  0
t
34
F34t  58 N
 M  R ×f  t  R ×F  R ×F  R ×F  0
 0.032ˆi  0.197ˆj m  ×  286ˆi  340ˆj N    7.384kˆ N  m   0.013ˆi  0.360ˆj m  ×  601ˆi  195ˆj N 
  0.094ˆi  0.306ˆj m  ×  47ˆi  34ˆj N    0.094ˆi  0.306ˆj m  ×  0.588ˆi  0.809ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
 46kˆ N  m  7.384kˆ N  m   219kˆ N  m  11.170kˆ N  m    0.256kˆ in  F  0
r
43
F43r  603 N
F  F  f  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
34
4
14
43
3
C
32
12
AO2
32
23
12
F34  401ˆi  453ˆj N  606 N  48.50
Ans.
F14  293ˆi  486ˆj N  567 N121.13
Ans.
F23  86ˆi  81 N  119 N43.26
Ans.
F12  86ˆi  81 N  119 N43.26
Ans.
T12  13.85kˆ N  m
Ans.
532
12.15 Make a complete kinematic and dynamic analysis of the offset slider-crank linkage in the
posture when 2  120, and the constant angular velocity 2  18 rad/s cw . The masses
and mass moments of inertia of the links are: m2  2.5 kg, m3  7.4 kg, m4  2.5 kg,
IG2  0.005 N  m  s2 , and IG3  0.013 6 N  m  s 2 . The external forces acting at points B
and C are FB  2 000ˆi N and FC  1 000ˆi N, respectively.
a  0.06 m, RAO2  0.1 m, RBA  0.38 m, C  32, RCA  0.4 m,   22, and RG3 A  0.26 m.
Kinematic Analysis:
 

VA  ω 2 × R AO2  18kˆ rad/s × 0.050ˆi  0.087ˆj m

 1.559ˆi  0.900ˆj m/s  1.800 m/s30
VB  VA  ω3 × R BA
V ˆi  1.559ˆi  0.900ˆj m/s    kˆ rad/s  ×  0.351ˆi  0.147ˆj m 
B
3

 
 1.559ˆi  0.900ˆj m/s  0.1473ˆi  0.3513ˆj m
ω3  2.567kˆ rad/s


VB  1.182ˆi m/s
2
A A   22 R AO2  α 2 × R AO2   18 rad/s  × 0.050ˆi  0.087ˆj m
Ans.

 16.200ˆi  28.059ˆj m/s2  32.400 m/s2  60
A B  A A  32 R BA  α3 × R BA

 
 
A ˆi  13.890ˆi  27.093ˆj m/s    0.147 ˆi  0.351 ˆj m 

AB ˆi  16.200ˆi  28.059ˆj m/s2  2.310ˆi  0.966ˆj m/s 2  3kˆ rad/s 2 × 0.351ˆi  0.147ˆj m

2
3
B
A B  25.156ˆi m/s 2
α3  77.188kˆ rad/s 2
AG3  A A   R G3 A  α 3 × R G3 A

2
3
3
 
 
 
 16.200ˆi  28.059ˆj m/s 2  1.713ˆi  0.021ˆj m/s 2  77.188kˆ rad/s 2  0.260ˆi  0.003ˆj m
AG3  14.466ˆi  7.969ˆj m/s2  16.516 m/s2  28.85
Ans.

Ans.
533
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
   7.4 kg  14.466ˆi  7.969ˆj m/s 2 
   2.5 kg   25.156ˆi m/s 2 
 107ˆi  59ˆj N  122 N151.15
t 3   I G3 α 3

 63ˆi N  63 N180
   0.013 6 N  m  s 2  77.188kˆ rad/s 2

t 4   IG4 α 4  0
 1.050kˆ N  m
h3  t3 f3  1.050 N  m  122 N   0.009 m , h4  t4 f 4  0
Next, the free-body diagrams with inertia forces are drawn. Here the lines of action for the
forces on the free-body diagrams can all be discovered from two- and three-force member
concepts.
M  R
A
G3 A
×f3  t 3  RCA ×FC  R BA ×f4  R BA ×FB  R BA ×F14  0
 0.260ˆi  0.003ˆj m  ×  107ˆi  59ˆj N    1.050kˆ N  m    0.395ˆi  0.065ˆj m  ×  1 000ˆi N 
  0.351ˆi  0.147 ˆj m  ×  63ˆi N    0.351ˆi  0.147 ˆj m  ×  2 000ˆi N    0.351ˆi  0.147ˆj m  × 1.000ˆj  F  0
14

 
 
 
 
 

15kˆ N  m  1.050kˆ N  m  65kˆ N  m  9.261kˆ N  m  294kˆ N  m  0.351kˆ in F14  0
F14  639 N
F  F  f  F  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
F14  639ˆj N  639 N90.00
Ans.
14
4
B
34
F34  2 063ˆi  639ˆj N  2 160 N  17.21 Ans.
43
3
C
23
F23  3 170ˆi  698ˆj N  3 246 N  12.42
Ans.
32
12
F12  3 170ˆi  698ˆj N  3 246 N  12.42
Ans.
T12  241kˆ N  m
Ans.
AO2
32
12
534
12.16 Perform a kinematic and dynamic analysis of the offset slider-crank linkage of Problem
14.15 for a complete rotation of the crank. The forces FB  1 000ˆi N and FC  0 when
the velocity of link 4 is to the right; and the forces FB  FC  0 when the velocity of link
4 is to the left. Plot the crank torque T12 versus the crank angle  2 .
Kinematic Analysis:
jR1  R2e j2  R3e j3  R4
j 0.060  0.100e j2  0.380e j3  RB
0.060  0.100sin  2  0.380sin 3  0
3   sin 1  0.158  0.263sin  2 
RB  0.100cos 2  0.380cos3
RG3  jR1  R2e j2  RG3 Ae  3
RG3  j 0.060  0.100e j2  0.260e  3
j   
j   22
RG3  0.100cos 2  0.260cos 3  22  j 0.060  0.100sin  2  0.260sin 3  22 
The first-order kinematic coefficients are found as follows:
j 0.100e j2  j 0.380e j33  R4
0.100cos 2  0.380cos33  0
0.100sin  2  0.380sin 33  R4
3  0.263cos 2 cos3
RB  0.100  sin  2  cos 2 tan 3 
j   22
RG3  j 0.100e j2  j 0.260e  3 3
RG3  0.100sin  2  0.260sin 3  223   j 0.100cos  2  0.260cos 3  22 3 
Similarly, the second-order kinematic coefficients are as follows:
0.100e j2  j 0.380e j33  0.380e j332  RB
0.100sin  2  0.380cos33  0.380sin 332  0
0.100cos 2  0.380sin 33  0.380cos332  RB
3  0.263sin  2 cos3  tan 332 ,
RB   0.100cos 3   2   0.38032  cos3
j   22
j   22
RG3  0.100e j2  j 0.260e  3 3  0.260e  3 32
RG3   0.100cos 2  0.260sin  3  22  3  0.260cos  3  22  32 
 j  0.100sin  2  0.260cos  3  22  3  0.260sin  3  22  32 
Dynamic Analysis:
By virtual work we can formulate the dynamic input torque requirement as:
T12  f3 RG3  t 3 3kˆ  f4 RB  FB RB
The individual elements of this equation are:
f3  m3 AG3  m3RG322    2 397.6 kg/s 2  RG3
f3   240cos 2  623sin  3  22  3  623cos  3  22  32 
 j  240sin  2  623cos  3  22  3  623sin  3  22  32 
535
2
f3 R G   240 cos  2  623sin  3  22   3  623 cos  3  22   3   0.100 sin  2  0.260 sin  3  22   3 
3
  240 sin  2  623 cos  3  22   3  623sin  3  22   3  0.100 cos  2  0.260 cos  3  22  3 
2
f3 RG3  62sin 3  22   2 3 1  3   62cos 3  22   2 3  16233
t   I α   I   2kˆ    4.406 N  m  kˆ
3
G3
3
G3 3
2
3
t 3 3kˆ    4.406 N  m 33
f4  m4 A B  m4 RB22   810 kg/s2  RB
f4  81cos 3   2   30832  ˆi N cos3
f4 RB  sin 3   2  8.1cos 3   2   30.832  N  m cos2 3


FB  500 1  sgn  cos 2 tan 3  sin 2  ˆi N=500 1+sgn sin 3  2  cos 3  ˆi N


FB RB  50 1+sgn sin 3   2  cos3  sin 3   2  cos3 N  m
Finally, putting these pieces together, we obtain:
T12  62sin 3  22   2  3 1  3   62 cos 3  22   2  3  15833
 sin 3   2  8.1cos 3   2   30.832  cos2  3


 50 1+sgn sin 3   2  cos 3  sin 3   2  cos 3 N  m
The plot of this torque requirement is shown below. The sinusoidal curve in the first half
of the cycle is caused primarily by the mass of the connecting rod; the mass of the piston
is included also. The applied force FB causes the rise in the second half of the cycle.
Note that the mass of link 3 causes dynamic torque, which helps to overcome up to one
third of the applied force effect.
536
12.17 Make a complete dynamic analysis of the slider-crank linkage in the posture when
2  120. For the constant angular velocity 2  24 rad/s cw , the known kinematics
  89.3 rad/s2 ccw,
A  40.6ˆi m/s2 ,
are:
and
R  0.374 m,
  9,
3
B
3
B
AG3  40.6ˆi  22.6ˆj m / s2 . The constant crank torque is T12  60 N  m. The masses and
mass moments of inertia of the links are m2  1.5 kg,
m3  3.5 kg,
m4  1.2 kg,
IG2  0.010 N  m  s , and IG3  0.060 N  m  s .
2
2
a  0, RAO2  0.1 m, RBA  0.45 m, C  0, RCB  0,   0, and RG3 A  0.2 m.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
   3.5 kg   40.6ˆi  22.6ˆj m / s 2 
  1.2 kg   40.6ˆi m/s 2 
 142ˆi  79ˆj N  163 N150.90
t 3   I G3 α 3

   0.060 N  m  s 2  89.3kˆ rad/s 2

 49ˆi N  49 N180
t 4   IG4 α 4  0
 5.358kˆ N  m
h3  t3 f3   5.358 N  m  163 N   0.033 m , h4  t4 f 4  0
M  R
A
G3 A
×f3  t 3  R BA ×f4  R BA ×F14  0
 0.196ˆi  0.038ˆj m  ×  142ˆi  79ˆj N    5.358kˆ N  m    0.442ˆi  0.087 ˆj m  ×  49ˆi N 
  0.442ˆi  0.087 ˆj m  × 1.000ˆj  F  0
14

 
 
 

10.039kˆ N  m  5.358kˆ N  m  4.243kˆ N  m  0.442kˆ in F14  0
F14  1 N
F  F  f  f  F  0
14
4
3
23
F14  1ˆj N  1 N  90.00
F  191ˆi  78ˆj N  206 N-22.21,
23
Ans.
Ans.
537
12.18 Repeat Problem 12.17 in the posture where 2  240. The known kinematics are: 3  11.1,
R  0.392 m,   112 rad/s2 cw, A  35.2ˆi m/s2 , and A  31.6ˆi  27.7ˆj m / s2 .
B
G3
B
3
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
   3.5 kg   31.6ˆi  27.7ˆj m / s 
  1.2 kg   35.2ˆi m/s 2 
2
 111ˆi  97ˆj N  147 N  138.76
t 3   I G3 α 3

   0.060 N  m  s 2  112kˆ rad/s 2

 42ˆi N  42 N180
t 4   IG4 α 4  0
 6.720kˆ N  m
h3  t3 f3   6.720 N  m  147 N   0.046 m , h4  t4 f 4  0
M  R
A
G3 A
×f3  t 3  R BA ×f4  R BA ×F14  0
 0.196ˆi  0.038ˆj m  ×  111ˆi  97ˆj N    6.720kˆ N  m    0.442ˆi  0.087 ˆj m  ×  42ˆi N 
  0.442ˆi  0.087 ˆj m  × 1.000ˆj F  0
14
 22.961kˆ N  m   6.720kˆ N  m   3.637kˆ N  m    0.442kˆ in  F  0
14
F14  29 N
F  F  f  f  F  0
14
4
3
23
F14  29ˆj N  29 N90.00
Ans.
F23  153ˆi  68 N  168 N24.11
Ans.
538
12.19 Make a complete kinematic and dynamic analysis of the offset slider-crank linkage in the
posture when 2  120 and the constant angular velocity 2  6 rad/s ccw . The external
forces at points B and C are F  50ˆi kN and F  80 kN  60, respectively The
B
C
masses and mass moments of inertia of the links are: m2  10 kg, m3  140 kg,
m4  50 kg, IG2  2.0 N  m  s 2 , and IG3  8.42 N  m  s 2 .
a  0.008 m, RAO2  0.25 m, RBA  1.25 m, RCA  1.0 m, C  38,   18, and
RG3 A  0.75 m.
Kinematic Analysis:
 

VA  ω2 × R AO2  6kˆ rad/s × 0.125ˆi  0.217ˆj m

 1.299ˆi  0.750ˆj m/s  1.500 m/s  150
VB  VA  ω3 × R BA
V ˆi   1.299ˆi  0.750ˆj m/s    kˆ rad/s  × 1.227ˆi  0.225ˆj m 
B
3

 
 1.299ˆi  0.750ˆj m/s  0.2253ˆi  1.2273 ˆj m

VB  1.162ˆi m/s
ω3  0.611kˆ rad/s

2
A A   22 R AO2  α 2 × R AO2    6 rad/s  × 0.125ˆi  0.217ˆj m
Ans.

 4.500ˆi  7.794ˆj m/s 2  9.000 m/s 2  60
A B  A A  32 R BA  α3 × R BA

 
 
A ˆi   4.041ˆi  7.710ˆj m/s    0.225 ˆi  1.227 ˆj m 

AB ˆi  4.500ˆi  7.794ˆj m/s2  0.459ˆi  0.084ˆj m/s 2  3kˆ rad/s 2 × 1.227ˆi  0.225ˆj m

2
3
B
A B  5.455ˆi m/s2
α3  6.284kˆ rad/s 2
AG3  A A   R G3 A  α 3 × R G3 A

2
3
3
 
 
 
 4.500ˆi  7.794ˆj m/s 2  0.247ˆi  0.133ˆj m/s 2  6.284kˆ rad/s 2 × 0.660ˆi  0.357ˆj m
Ans.

539
AG3  6.496ˆi  3.514ˆj m/s2  16.516 m/s2  28.41
Ans.
Dynamic Analysis:
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
  140 kg   6.496ˆi  3.514ˆj m/s 
   50 kg   5.455ˆi m/s 2 
2
 909ˆi  492ˆj N  1 034 N151.59
t 3   I G3 α 3

   8.42 N  m  s 2  6.284kˆ rad/s 2

 273ˆi N  273 N180
t 4   IG4 α 4  0
 52.911kˆ N  m
h3  t3 f3   52.911 N  m  1 034 N   0.051 m , h4  t4 f 4  0
M  R
A
G3 A
×f3  t 3  RCA ×FC  R BA ×f4  R BA ×FB  R BA ×F14  0
 0.660ˆi  0.357ˆj m  ×  909ˆi  492ˆj N    52.911kˆ N  m    0.263ˆi  0.301ˆj m  ×  40 000ˆi  69 282ˆj N 
 1.227ˆi  0.225ˆj m  ×  273ˆi N   1.227ˆi  0.225ˆj m  ×  50 000ˆi N   1.227ˆi  0.225ˆj m  × 1.000ˆj  F  0
 0.207kˆ N  m    52.911kˆ N  m    6 157kˆ N  m    61.425kˆ N  m    11 250kˆ N  m   1.227kˆ in  F  0
14
14
F14  14 280 N
F  F  f  F  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
F14  14 280ˆj N  14 280 N90.00
Ans.
14
4
B
34
F34  50 273ˆi  14280ˆj N  52 262 N  15.86
Ans.
43
3
C
23
F23  11 182ˆi  54 510ˆj N  55 645 N78.41
Ans.
32
12
F12  11 182ˆi  54 510ˆj N  55 645 N78.41
Ans.
T12  9 240kˆ N  m
Ans.
AO2
32
12
540
12.20 Find the driving torque and the reaction forces at the joints for the crossed-linkage in the
posture illustrated. For the constant angular velocity 2  10 rad/s ccw , the known
kinematics are: 3  1.43 rad/s cw, 4  11.43 rad/s cw, 3   4  84.8 rad/s2 ccw, and
A  25.92ˆi  24.58ˆj ft/s2 . The external force at point C is F  30ˆj lb. The weights and
C
G3
mass moments of inertia of the links are: w3  4 lb, IG2  IG4  0.063in  lb  s2 , and
IG3  0.497 in  lb  s2 .
RAO2  6 in, RO4O2  18 in, RBA  18 in, RBO4  6 in, RCA  24 in, and RG3 A  12 in.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3

   4 lb  25.92ˆi  24.58ˆj ft/s 2
 32.2 ft/s  f  m A  0
2
4
4
G4
 3.220ˆi  3.053ˆj lb  4.437 lb  136.52
t 3   I G3 α 3

   0.497 in  lb  s 2  84.8kˆ rad/s 2
 42.146kˆ in  lb

t 4   I G4 α 4

   0.063 in  lb  s 2  84.8kˆ rad/s 2
 5.342kˆ in  lb
h3  t3 f3   42.146 in  lb   4.437 lb   9.499 in , h4  0

541
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M  t  R
O4
4
BO4
×F34t  0
 5.342kˆ in  lb    0.857ˆi  5.938ˆj in  ×  0.990ˆi  0.143ˆj F  0
t
34
F34t  1.123 lb
 M  R ×f  t  R ×F  R ×F  R ×F  0
 9.429ˆi  7.423ˆj in  ×  3.217ˆi  3.056ˆj lb    42.163kˆ in  lb   18.857ˆi  14.846ˆj in  ×  30ˆj lb 
 14.143ˆi  11.135ˆj in  ×  1.011ˆi  0.161ˆj lb   14.143ˆi  11.135ˆj in  ×  0.143ˆi  0.990ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
 52.69kˆ in  lb   42.16kˆ in  lb   565.71kˆ in  lb   8.98kˆ in  lb   15.59kˆ in  F  0
r
43
F43r  42.95 lb
F  F  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  T  0
O2
34
14
43
3
32
12
AO2
C
32
23
12
F34  5.03ˆi  42.68ˆj lb  42.98 lb  96.7 Ans.
F  5.03ˆi  42.68ˆj lb  42.98 lb83.3
Ans.
14
F23  1.81ˆi  9.62ˆj lb  9.79 lb  100.67 Ans.
F  1.81ˆi  9.62ˆj lb  9.79 lb  100.67 Ans.
12
T12  19.45kˆ in  lb
Ans.
542
12.21 Find the driving torque and the reaction forces at the joints for Problem 12.20 under the
same dynamic conditions, but with crank 4 as the driver of the linkage.
Given the same dynamic conditions, the d’Alembert forces and torques are the same as in
Prob. 12.20. However, crank 2 is now a two-force member with no applied moment.
Therefore the free-body diagrams appear as:
The solution now proceeds as follows:
 MB  RG3B ×f3  t3  RCB ×FC  R AB ×F23  0
 4.714ˆi  3.712ˆj in  ×  3.217ˆi  3.056ˆj lb    42.163kˆ in  lb    4.714ˆi  3.712ˆj in  ×  30ˆj lb 
  14.143ˆi  11.135ˆj in  ×  0.500ˆi  0.866ˆj F  0
23
 26.348kˆ in  lb   42.163kˆ in  lb    141.429kˆ in  lb   17.815kˆ in  F  0
23
43
F23  4.413ˆi  7.644ˆj lb  8.826 lb 120 Ans.
F12  4.413ˆi  7.644ˆj lb  8.826 lb 120 Ans.
F  7.631ˆi  40.700ˆj lb  41.409 lb79.38 Ans.
4
F14  7.631ˆi  40.700ˆj lb  41.409 lb79.38 Ans.
T  15.77kˆ in  lb
Ans.
F23  8.826 lb
F  F  F  0
F  F  f  F  F  0
F  F  F  0
M  R × F  t  T  0
O4
32
12
23
3
34
14
BO4
C
34
14
43
14
543
12.22 Using the same force FC as in Problem 12.20, compute the crank torque and the reaction
forces at the joints in the posture when 2  210 . For the constant angular velocity
the
known
kinematics
are:
2  10 rad/s ccw ,
3  14.7,
4  164.7
3  4.73 rad/s ccw, 4  5.27 rad/s cw, 3   4  10.39 rad/s cw, and AG3  26 ft / s220.85.
The d’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3

   4 lb  24.3ˆi  9.25ˆj ft/s 2
 32.2 ft/s 
2
f4  m4 AG4  0
 3.018ˆi  1.150ˆj lb  3.230 lb  159.15
t 3   I G3 α 3

   0.063 in  lb  s 2  10.39kˆ rad/s 2

t 4   I G4 α 4

   0.063 in  lb  s 2  10.39kˆ rad/s 2

 0.655kˆ in  lb
 0.655kˆ in  lb
h3  t3 f3   0.655 in  lb   3.230 lb   0.203 in , h4  0
Next, the free-body diagrams are drawn. Since the lines of action for the forces on the
free-body diagrams cannot be discovered from two- and three-force member concepts, the
force F34 is divided into radial and transverse components.
M  t  R
O4
4
BO4
×F34t  0
 0.655kˆ in  lb    5.788ˆi  1.579ˆj in  ×  0.263ˆi  0.965ˆj F  0
t
34
F34t  0.109 lb (neglegible)
M  R
A
G3 A
t
r
×f3  t 3  RCA × FC  R BA × F43
 R BA × F43
0
11.605ˆi  3.053ˆj in  ×  3.018ˆi  1.150ˆj lb    0.655kˆ in  lb    23.210ˆi  6.106ˆj in  ×  30ˆj lb 
 17.408ˆi  4.579ˆj in  ×  0.029ˆi  0.105ˆj lb   17.408ˆi  4.579ˆj in  ×  0.965ˆi  0.263ˆj  F  0
r
43
 4.132kˆ in  lb   0.655kˆ in  lb    696kˆ in  lb    1.695kˆ in  lb   8.997kˆ in  F  0
r
43
3
C
23
F34  75ˆi  20ˆj lb  78 lb  15.20
F  78ˆi  11ˆj lb  79 lb8.02
AO2
32
12
T12  177kˆ in  lb
F43r  78 lb
F  F  f  F  F  0
M  R × F  T  0
43
O2
23
Ans.
Ans.
Ans.
544
12.23 Make a kinematic and dynamic analysis of the linkage for a complete rotation of the
crank with the constant angular velocity 2  10 rad/s ccw. The external force at point C
is F  500ˆi  886ˆj lb for 90    300, and F  0 otherwise. The weights and the
C
2
C
mass moments of inertias of the links are: w3  222 lb, w4  208 lb, IG3  226 in  lb  s2 ,
and IG4  264 in  lb  s 2 .
RAO2  16 in, RO4O2  RBA  40 in, RBO4  56 in, RG3 A  32 in, and RG4O4  20 in.
Kinematic Analysis
R2e j2  R3e j3  R1  R4e j4
16e j2  40e j3  40  56e j4
16cos 2  40cos3  40  56cos 4
16sin  2  40sin 3  56sin  4
Eliminating  3 we find  4 from the roots of the quadratic
17  8cos2  tan 2 4 2  56sin 2  tan 4 2   48cos2 123  0
Then 3  tan 1  7sin 4  2sin 2   7 cos 4  5  2cos 2 
RG3  16e j2  32e j3
RG4  40  20e j4

RC  16e j2  56.143e  3
The first-order kinematic coefficients are:
j   4.086
Ans.
Ans.
Ans.
Ans.
545
j16e j2  j 40e j33  j56e j4 4
16sin  2  40sin 33  56sin  4 4
3  14sin 4  2  35sin 4  3 
16cos 2  40cos33  56cos 4 4
4  10sin 2  3  35sin 4  3 
Ans.
RG3  j16e 2  j32e 3 3  16  sin 2  2sin 33   j16  cos 2  2cos 33 
Ans.
RG4  j 20e j4 4  20sin  4 4  j 20cos 4 4
Ans.
j
j
j   4.086
RC  j16e j2  j56.143e  3
3
  16sin  2  56.143sin 3  4.086  3   j 16 cos  2  56.143cos 3  4.086  3 
The second-order kinematic coefficients are:
16e j2  j 40e j33  40e j332  j56e j4 4  56e j4 42
Ans.
16cos 2  40sin 33  40cos332  56sin  4 4  56cos 4 42
16sin  2  40cos33  40sin 332  56cos 4 4  56sin  4 42
3  14cos  4   2   35cos  4  3 32  49 42  35sin  4  3 
 4  10cos  4   2   2532  35cos  4  3  42  35sin  4  3 
j
j
Ans.
Ans.
j
RG3  16e 2  j32e 3 3  32e 3 32
  16 cos  2  32sin 33  32 cos 3332  j  16sin  2  32 cos 33  32sin 332 
RG4  j 20e j  4  20e j  42   20sin  4 4  20 cos  4 42   j  20 cos  4 4  20sin  4 42 
4
4
Ans.
Ans.
Dynamic Analysis
By virtual work we can formulate the dynamic input torque requirement as:
M12  f3 RG3  t 3 3kˆ  f4 RG4  t 4  4kˆ  FC RC
The individual elements of this equation are:
2
f3  m3 AG3  m3RG3 22    222 lb 386 in/s 2  10 rad/s  RG3  57.513 lb/inRG3
  920 cos  2  1 840 sin 33  1 840 cos 3332  j  920 sin  2  1 840 cos 33  1 840 sin 332  lb
f3 RG3   920cos  2  1 840sin 33  1 840cos 3332  16sin  2  32sin 33  in  lb
  920sin  2  1 840cos 33  1 840sin 332  16cos  2  32cos 33  in  lb
f3 RG3   29 447 in  lb   sin 3   2 3 1  3   cos 3   2 3  233
t 3   IG3 α3   IG3322kˆ    22 600 in  lb 3kˆ
t  kˆ    22 600 in  lb   
3
3
3 3
f4  m4 AG4  m4 RG4     208 lb 386 in/s 2  10 rad/s  RG4  53.886 lb/inRG4
2
2
2
 1 078sin  4 4  1 078cos  4 42   j  1 078cos  4 4  1 078sin  4 42  lb
f4 RG4  1 078sin  4 4  1 078cos 4 42   20sin  4 4  in  lb
  1 078cos  4 4  1 078sin  4 42   20cos  4 4  in  lb
f4 RG4  21 560 4 4 in  lb
546
t 4   IG4 α 4   IG4 422kˆ    26 400 in  lb  4kˆ
t  kˆ  26 400   in  lb
4
4
4 4
FC RC  1 000 lb  cos120  16sin  2  56.143sin 3  4.086 3 in 
 1 000 lb  sin120 16cos  2  56.143cos 3  4.086 3 in 
FC RC  16 000sin  2  120  56 143sin 3  124.0863 in  lb
Reassembling the elements we must remember that force FC is nonzero for only a portion
of the cycle. Therefore,
T12   29 447    sin       1      cos        2     22 600      21 560    26 400   in  lb
3
2
3
3
3
2
3
3
3
3
3
4
4
4
4
=29 447 cos 3  2  3  29 447sin 3   2  3 1  3   81 4943 3  47 960 4 4 in  lb
for the entire cycle and, for 90   2  300 , an additional increment is added:
Ans.
T12  16 000sin 2  120  56 143sin 3  124.086 3 in  lb
This input torque requirement is shown in the following plot. Notice the small
discontinuities in the curve when force FC begins and ends its effect.
547
12.24 The motor is geared to a shaft on which a flywheel is mounted. The mass moments of
inertia of the parts are: flywheel, I  2.73 in  lb  s2 ; flywheel shaft, I  0.015 5 in  lb  s2 ;
gear, I  0.172 in  lb  s2 ; pinion, I  0.003 49 in  lb  s2 ; and motor, I  0.086 4 in  lb  s2 .
If the motor has a starting torque of 75 in  lb, determine the angular acceleration of the
flywheel shaft at the instant the motor is started.
If we identify the motor shaft as 2 and the flywheel shaft as 3 then
I 2  0.086 4 in  lb  s2  0.003 49 in  lb  s 2  0.089 89 in  lb  s2
I3  2.73 in  lb  s2  0.015 5 in  lb  s 2  0.172 in  lb  s 2  2.917 5 in  lb  s2
3    R2 R3  2
3    R2 R3  2
M  R F  I 
M  T  R F  I 
3
t
3 23
2
12
3 3
2
t
32
2 2
F23t    R2 R32  I 32
T12  I 22  R2 F32t   I 2   R2 R3  I 3  2


2
Now, substituting the numeric values,
2
75 in  lb  0.089 89 in  lb  s 2  1.0 in 4.5 in  2.917 5 in  lb  s 2   2   0.233 96 in  lb  s 2   2
2  75 in  lb 0.233 96 in  lb  s2  320.56 rad/s 2
3    R2 R3 2   1.0 in 4.5 in  320.56 rad/s2  71.24 rad/s2
Ans.
548
12.25 The disk cam of Problem 11.31 is driven at the constant input shaft speed
2  20 rad/s ccw. Both the cam and the follower have been balanced so that the centers
of mass of each are located at their respective fixed pivots. The mass and radius of
gyration of the cam are 0.075 kg and 30 mm, respectively, and the mass and radius of
gyration of the follower are 0.030 kg and 35 mm, respectively. At the instant illustrated,
determine the torque T12 required on the camshaft to produce this motion.
IG3  m3k32   0.030 kg  0.035 m   0.000 036 75 kg  m2
2
For full-rise cycloidal cam motion, Eq. (6.13),
L
2  30 
112.5 
y  1  cos

1  cos 2
  0.200

  150 
150 
 360 30 sin 2 112.5  0.480

150
2
1502
2
3  y22  0.480  20 rad/s   192 rad/s2
y 
2 L
sin
2

t3   IG33    0.000 036 75 kg  m2  192 rad/s2   0.007 056 N  m
By virtual work,
T12   y t3  R CO3 × FC kˆ 
 0.200 0.007 056 N  m   0.150 m 8 N  sin  45    0.168 N  m ccw


Ans.
549
12.26 Repeat Problem 12.25 with the constant shaft speed 2  40 rad/s ccw.
IG3  m3k32   0.030 kg  0.035 m   0.000 036 75 kg  m2
2
For full-rise cycloidal cam motion, Eq. (6.13),
L
2  30 
112.5 
y  1  cos

1  cos 2
  0.200

  150 
150 
 360 30 sin 2 112.5  0.480

150

1502
2
3  y22  0.480  40 rad/s   768 rad/s2
y 
2 L
2
sin
2

t3   IG33    0.000 036 75 kg  m2  768 rad/s2   0.028 224 N  m
By virtual work,
T12   y t3  R CO3 × FC kˆ 
 0.200 0.028 224 N  m   0.150 m 8 N  sin  45    0.164 N  m ccw


Ans.
550
12.27 A rotating drum is pivoted at O2 and is decelerated by the double-shoe brake mechanism.
The weight and radius of gyration of the drum are 230 lb and 5.66 in, respectively. The
brake is actuated by the force P  100ˆj lb . Assume that the contact points between the
two shoes and the drum are C and D, where the coefficients of Coulomb friction are
  0.300. Determine the angular deceleration of the drum and the reaction force F12 at
the fixed pivot.
 M   12ˆi in  ×  Pˆj  3ˆj in  ×  F ˆi   0
F  P  F  F  0
5
F65  4P  400 lb
65
5
65
F35  400ˆi  100ˆj lb  412 lb14.04
35
The friction angle is   tan 1  0.300   16.70 .
 M   22ˆj in  ×F   5ˆi  9ˆj in  ×   cos ˆi  sin ˆj F  0
3
53
23
F32  733ˆi  220ˆj lb  766 lb16.70
F23  766 lb
 M   25ˆj in  ×F  5ˆi  9ˆj in  ×  cos ˆi  sin ˆj F  0
4
64
24
F42  1 333ˆi  400ˆj lb  1 392 lb  163.30
F24  1392 lb
IG2  m2 k22   230 lb 386 in/s2   5.66 in   19.084 in  lb  s 2
2
 M   8ˆi in  ×  733ˆi  220ˆj lb   8ˆi in  ×  1 333ˆi  400ˆj lb   I α
O2
G2
F  F  F  F  0
2
12
32
42
2
α 2  261kˆ rad/s
F  600ˆi  180ˆj lb  626 lb16.70
2
12
Ans.
Ans.
Note that gravitational effects are not yet included. If gravity acts in the jˆ direction then
the ĵ component is 410 lb. Since the main bearing at O2 supports this weight, it does not
affect the friction forces and can be added by superposition. If weights of the other parts
were known, however, these weights might have some small effect on the friction forces
and the braking forces, and would have to be included simultaneously. Superposition
could not be applied.
551
12.28 The length of link 4 is 0.20 m, symmetric about O4, and the ground bearing is midway
between E and G2. Link 2 is in translation with a velocity VG2  0.114 8ˆj m/s and an
acceleration A  0.35ˆj m/s2 ; and the acceleration of the mass center of link 3 is
G2
AG 3  1.053ˆi  0.432ˆj m/s2 .
The kinematic coefficients
of the linkage are
  40 m/m2 (where R 43 is the
  2 m/m , and R43
3  11.5 m1 , 3  380 m2 , R43
vector from G3 to G4 ). A moment M12 acts on the input link 2 and a torque T4 acts on
link 4. The masses and second moments of mass of the links are m2  m4  0.5 kg,
m3  1 kg, IG  IG  2 kg  m2 , and IG  5 kg  m2 . Gravity is in the negative zdirection. Determine the internal reaction forces, the moment M12, and the torque T4 .
2
4
3
RG2O4  0.15 m, and REG2  0.20 m.
The free-body diagram for link 2 is shown below. Recall that gravity acts in the negative
z direction and the effects of friction are neglected.
552
Since the center of mass G2 is translating in the y direction, the sum of the external forces
in the x direction acting on link 2 shows
F12x  F32x  P x  m2 AGx2  0
(1)
Since friction is neglected, the sum of the external forces in the y direction acting on link
2 gives
F32y  P y  m2 AGy2
Substituting the given information, the y component of the reaction force between links 2
and 3 is
Ans. (2)
F32y   0.5 kg  ( 0.35 m/s2 )   40 N  sin120  34.816 N
Since link 2 is not rotating the angular acceleration  2  0 . Therefore, the sum of the
external moments on link 2 acting about G2 can be written as
 R7 F12x  R9 P x  M12  0
(3)
x
12
x
32
y
32
Therefore, there are still 4 unknowns for link 2, namely the forces F , F , and F and
the moment M 12 .
The free-body diagram for link 3 is shown in this figure.
The sum of the external forces in the x direction acting on link 3 can be written as
F23x  F43x  m3 AGx3
(4)
The sum of the external forces in the y direction acting on link 3 can be written as
F23y  F43y  m3 AGy3
(5)
The sum of the external moments acting about the center of mass G3 can be written as
R43F43  RG2G3 F23y  I G33
(6)
The vector R 43 points from the center of mass of link 3 to the location of the reaction
force F43, and the vector R G2G3 points from the center of mass of link 3 to the center of
mass of link 2.
Equations (4), (5), and (6) contain two new unknown variables, namely the internal
reaction force F43 and the location of this force (i.e., R43). Note that the force F43 is
553
perpendicular to the slot since friction is neglected. Therefore, F43x and F43y are not
independent unknowns (the angle is known). Therefore, there are 6 equations and 6
unknown variables, namely the forces F12x , F32x , F32y , F43 , the moment M12, and the
distance R43.
These six unknowns can now be solved for by inspection. Substituting Eq. (2) and
the given acceleration of the center of mass G2 into Eq. (5), the y component of the
internal reaction force between links 3 and 4 is
F43y  F43 sin 60  1 kg  ( 0.432 m/s2 )  (34.816 N)  34.384 N
Ans.
Therefore, the force between links 3 and 4 is
34.384 N
Ans.
F43 
 39.70 N
sin 60
Substituting known values into Eq. (4), the internal reaction force between links 2 and
3 is
Ans.
F23x  (  39.70 N)cos60  1 kg  1.053 m/s2   20.91 N
Substituting known values into Eq. (6) gives
R34 (39.70 N)  (0.259 81 m)(34.816 N)   5 kg  m2  (0.983 rad/s2 )
Rearranging this equation, the unknown distance is
13.961 N  m
R43 
 0.351 66 m
39.70 N
Therefore, the distance from the ground pin O4 to the line of action of the internal
reaction force F43 is
s  R43  0.300 m  0.351 66  0.300 m  51.66 mm
Since the distance s is less than the length of link 4 then link 3 is sliding along link 4 (i.e.,
there is sliding contact and not tipping). The internal reaction force between links 3 and 4
acts within the physical limits of link 4.
Substituting known values into Eq. (1) gives
F12x  20.91 N   40 N  cos60  0
Ans.
Rearranging this equation, the unknown force is
F12x  40.91 N
Substituting known values into Eq. (3) gives
 0.1 m (40.91 N)  0.2 m (40 N)cos120  M12  0
Rearranging this equation, the unknown moment acting on link 2 is
M12  0.091 Nm
The free-body diagram for link 4 is shown in this figure.
Ans.
Ans.
554
Since G4 is coincident with the ground pivot O4 , the sum of the external forces in the x
direction acting on link 4 can be written as
F34 cos60  F14x  m4 AGx4  0
(7)
The sum of the external forces in the y direction acting on link 4 can be written as
(8)
F34 sin 60  F14y  0
The sum of the external moments acting about the center of mass of link 4 can be written
as
ZF34  T4  IG4  4
(9)
Equations (7), (8), and (9) contain three new unknown variables, namely the internal
reaction forces F14x , F14y , and the moment T4. These three unknown variables can now be
solved as follows. Substituting known values into Eq. (7), the x component of the force
between links 1 and 4 is
F14x  19.85 N
Ans.
Substituting known values into Eq. (8), the y component of the internal reaction force
between links 1 and 4 is
F14y  34.38 N
Ans.
Substituting known values into Eq. (9), the moment acting on link 4 is
Ans.
T4   2 kg  m2  ( 0.983 rad/s2 )  (0.051 66 m)(39.70 N)  4.017 Nm
The negative sign indicates that the moment acting on link 4 is clockwise.
555
12.29 The kinematic coefficients for the elliptic trammel linkage are 3  2 rad/m,
3  6.928 rad/m2 , R4  1.732 m/m, and R4  8 m/m2 . A linear spring is attached
between O and A with a free length L  0.5 m and spring constant K  2 500 N/m. A
viscous damper with damping coefficient C  45 N  s/m is connected between the ground
and link 4. The masses and mass moments of inertia of the links are: m2  0.75 kg,
m3  2.0 kg, m4  1.5 kg, IG2  0.25 N  m  s 2 , IG3  1.0 N  m  s 2 and IG4  0.35 N  m  s2 .
The velocity, acceleration, and force acting on the input link 2 are V  5ˆj m/s ,
A2
A A2  20ˆj m/s , and F  200 ĵ N, respectively.
2
If gravity acts in the negative y-
direction, determine: (a) the first-order kinematic coefficients of the spring and the
viscous damper; (b) the equivalent mass of the linkage; and (c) the horizontal force P
acting on link 4.
.
RBA  1 m and RG3G2  0.5 m.
556
(a) The vectors for the linear spring are shown in the following figure.
Vectors for the linear spring.
The vector loop for the linear spring can be written as
?
I
R S  R2  0
From which, the magnitudes can be written as the scalar equation
RS  R2
Differentiating this with respect to the input position R2 , the first-order kinematic
coefficient of the spring is
RS  1 m/m
Ans. (1)
Note that the sign is positive because, for a negative input, the length of the linear spring
is decreasing. Also, note that the first-order kinematic coefficient of the mass center of
input link 2 is
yG 2  RS  1 m/m
The vectors for the viscous damper are shown in the following figure.
Vectors for the viscous damper.
The vector loop for the damper can be written as

?

R9  R C  R 4  0
Since all components of this equation are horizontal, this gives
R9  RC  R4  0
Differentiating with respect to the input position R2 gives
 RC  R4  0
Rearranging and substituting the given data, the first-order kinematic coefficient of the
viscous damper is
RC   R4  1.732 m/m
Ans. (2)
The positive sign agrees with our intuition since, for a positive input, the length of the
viscous damper is increasing.
557
(b) The equivalent mass of the mechanism can be written as
4
mEQ   Aj
(3)
j 2
A2  m2  xG2  yG2   IG 22
For link 2:
2
2
(4)
2
The vector loop for the center of mass of the input link can be written as
I
??
RG  R 2  0
2
The x and y components of this equation give
yG 2  R2
xG2  0 and
Differentiating these with respect to the input position R2 give
yG 2  R2  1 m/m
xG 2  0 and
Substituting these values into Eq. (4) gives
A2   0.75 kg   02  12    0.25 kg  m2   0   0.75 kg
2
(5)
A3  m3  xG 32  yG 32   I G 332
For link 3:
(6)
The vector loop for the center of mass of link 3 can be written as
I
??

RG3  R 2  R G3G2
The x and y components are
xG3   RG3G2 cos3  0.25 m
yG3  R2  RG3G2 sin 3  0.433 m
and
Differentiating these with respect to the input position R2 give
xG 3  RG3G2 sin 33  0.866 m/m
yG 3  1  RG3G2 cos33  0.5 m/m
and
Substituting these and other known data into Eq. (6) gives
2
A3   2.0 kg  (0.866 m/m) 2   0.5 m/m    1.0 kg  m 2  ( 2 rad/m) 2  6 kg


2
2
A4  m4  xG 4  yG 4   I G 442
For link 4:
(7)
(8)
Note from given data that X G 4  R4  1.732 m/m; therefore, Eq. (8) can be written as
A4  1.5 kg  [ 1.732 m/m   02 ]   0.35 kg  m2   0   4.5 kg
2
2
(9)
Therefore, substituting Eqs. (5), (7), and (9) into Eq. (3), the equivalent mass of the
mechanism is
Ans. (10)
mEQ  0.75 kg  6 kg  4.5 kg  11.25 kg
(c) The power equation for the mechanism can be written as
dT dU dW f
F  VA2  P  VB4 


dt
dt
dt
Substituting the time rate of change of energy terms into the right-hand side gives
4
4
4
j 2
j 2
j 2
F  VA2  P  VB4   Aj R2 R2   B j R23   m j gyG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
2
558
The linear velocity of link 2 and the force acting on link 2 are both in the same direction
(that is, both downward). Assuming that the force P is in the same direction as the
velocity of link 4 (that is, to the right), then the above equation can be written as
4
4
4
j 2
j 2
j 2
FVA2  PVB4   Aj R2 R2   B j R23   m j gyG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
2
The velocity of the input link 2 is VA2  R2 and the velocity of link 4 is VB4  R4 ;
therefore, this equation can be written as
4
4
4
j 2
j 2
j 2
FR2  PR4   Aj R2 R2   B j R23   m j gyG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
2
Dividing by the input velocity R2 throughout gives the equation of motion for the
mechanism, that is
4
4
4
(11)
F  PR4   Aj R2   B j R22   m j gYG  K s  RS  RS 0  RS  CRC2 R2
j 2
j 2
j
j 2
where the first-order kinematic coefficient of link 4 is given as R4  1.732 m/m . The
sum of the Bj terms can be written as
4
1 4 dA
(12)
 Bj   j
2 j  2 d 2
j 2
B2  m2 ( xG 2 xG2  yG 2 yG2 )  I G222
For link 2:
and this has a value of
B2   0.75 kg  0  0   0.25 kg  m2  0   0




(13)
B3  m3 xG 3 xG3  yG 3 yG3  I G333
For link 3:
which has a value of
B3   2 kg [(0.866 m/m)(4 m/m2 )  (0.5 m/m)(0)]  1 kg  m2  (2 rad/m)(0.928 rad/m2 )  20.78 kg/m (14)


B4  m4 xG 4 xG4  yG 4 yG4  I G444
For link 4:
which has a value of
B4  1.5 kg  [(1.732 m/m)(8 m/m2 )  (0)(0)]  0.35 kg  m 2  0  0   20.78kg/m (15)


Substituting Eqs. (13), (14), and (15) into Eq. (12) gives
4
 B  B  B  B  0  20.78 kg/m  20.78 kg/m  41.56 kg/m
j 2
j
2
3
4
(16)
The change in potential energy due to gravity is
4
 m gy
j 2
j
Gj
For link 2:
m2 gyG 2  (0.75 kg)(9.81 m/s2 ) 1 m/m  7.36 N
For link 3:
m3 gyG 3  (2 kg)(9.81 m/s2 )(0.5 m/m)  9.81 N
For link 4:
m4 gyG 4  (1.5 kg)(9.81 m/s2 )(0)  0
Summing these three values gives
4
 m gy  7.36 N  9.81 N  0  17.17 N
j 2
j
Gj
(17)
559
Then substituting Eqs. (1), (2), (10), (16), and (17) into Eq. (11) gives
F  PR4  11.25 kg  R2   41.56 kg/m R22  17.17 N  KS  RS  RS 0  (1 m/m)  C(1.732 m/m)2 R2
Rearranging this equation, the force acting on link 4 can be written as
1
  F  11.25 kg  R2   41.56 kg/m  R22  17.17 N  K S  RS  RS 0   3CR2 
P
R4 
The input velocity is R2  5 m/s , the input acceleration is R2  20 m/s2 , and the force
is F  200 N. Substituting these values and the known data into this equation, the force
acting on link 4 can be written as
P
 200 N  11.25 kg  ( 20 m/s2 )   41.56 kg/m  ( 5 m/s) 2  17.17 N

1


1.732 m/m 
  2500 N/m  0.866 m  0.5 m   3  45 Ns/m  ( 5 m/s) 
or as
P
1
200 N  225 N  1 039 N  17.17 N  915.0 N  675.0 N 
1.732
Therefore, the force acting on link 4 is
Ans.
P  733.93 N
The negative sign indicates that the force P (acting on link 4) is acting to the left; that is,
in the opposite direction to the velocity of the output link 4. Recall that the force P was
originally assumed to be acting to the right.
560
12.30 The input crank of the four-bar linkage is rotating with the constant angular velocity ω2 =
10 rad/s ccw. The angular acceleration of link 3 and the acceleration of the mass center of
link 3 are 3  84.8 rad/s2 ccw and AG3  310ˆi  295ˆj in/s2 , respectively. The masses
and second moments of mass of the links are specified in Prob. 12.20 with the exception
that the weight of link 3 is w3 = 10 lb. The spring has stiffness k  12 lb/in and free
length R0  4.5 in. The viscous damper has a damping coefficient C  0.25 lb  s/in. The
external force acting at point C is FC  125 ĵ lb and gravity is in the negative y-direction.
Determine the equivalent mass moment of inertia of the linkage; and the driving torque
T2 .
RAO2  6 in, RO4O2  18 in, RBA  18 in, RBO4  6 in, RCA  24 in, and RG3 A  12 in.
From Prob. 12.20, the given data are
w3  10 lb, IG3  0.497 in  lb  s 2 , IG2  IG4  0.063 in  lb  s2 , 3  1.43 rad/s cw,
4  11.43 rad/s cw,  4  84.8 rad/s2 ccw.
Figure P12.30 shows vectors that are used throughout the solution.
The first-order kinematic coefficient of link 3 can be written as
 1.43 rad/s
3  33  3 
 0.143 rad/rad
2
10 rad/s
561
The first-order kinematic coefficient of link 4 can be written as
 11.43 rad/s
4  4 
 1.143 rad/rad
2
10 rad/s
The angular acceleration of link 3 can be written as
3  322  3 2
Rearranging this, the second-order kinematic coefficient for link 3 can be written as
 3  3 2 84.8 rad/s 2  (0.143 rad/rad)(0)


3  33 

 0.84 rad/rad 2
2
2
2
10 rad/s 
Similarly, the second-order kinematic coefficient of link 4 is
    84.8 rad/s 2  (0.143 rad/rad)(0)
4  4 2 4 2 
 0.84 rad/rad 2
2
2
10 rad/s 
Since the mass centers G2 and G4 are located at the fixed pivots O2 and O4, respectively,
the first- and second-order kinematic coefficients of these mass centers are
xG2  0
xG 2  0
xG2  0
yG2  0
yG 2  0
yG2  0
xG4  18 in
xG 4  0
xG4  0
yG4  0
yG 4  0
yG4  0
To find the first- and second-order kinematic coefficients for the center of mass G3, the
vector loop for the center of mass of link 3 can be written as
RG3  R 2  R33
where R2 = 6 in, θ2 = 60º, R33 = 12 in, and θ33 = θ3 = 321.8º. The x and y components of
the vector equation for the center of mass of link 3 are
xG3  R2 cos 2  R33 cos 3   6 in  cos 60  12 in  cos321.8  12.4303 in
yG3  R2 sin 2  R33 sin 3   6 in  sin 60  12 in  sin 321.8  2.2247 in
The first-order kinematic coefficients for the center of mass of link 3 are
xG 3   R2 sin 2  R33 sin 33    6 in  sin 60  12 in  sin321.8(0.143 rad/rad)  6.2573 in/rad
yG 3  R2 cos 2  R33 cos33   6 in  cos60  12 in  cos321.8(0.143 rad/rad)  1.6515 in/rad
The second-order kinematic coefficients for the center of mass of link 3 can be written as
xG3   R2 cos 2  R33 cos332  R33 sin 33
yG3   R2 sin 2  R33 sin 332  R33 cos33
Therefore,
xG3    6 in  cos60  12 in  cos321.8(0.143 rad/rad) 2  12 in  sin321.8(0.848 rad/rad 2 )  3.100 in/rad 2
yG3    6 in  sin 60  12 in  sin321.8(0.143 rad/rad) 2  12 in  cos321.8(0.848 rad/rad)  2.952 in/rad 2
4
4
j 2
j 2
To determine  A j , note that  Aj  I EQ (that is, the equivalent mass moment of
inertia). Therefore, the units must be in  lb  s2 .
For link 2:
562
A2  m2 ( xG22  yG22 )  IG222  m2 (0  0)  0.063 in  lb  s 2 (1 rad/rad) 2  0.063 in  lb  s 2
For link 3:
A3  m3 ( xG23  yG23 )  I G 332
10 lb
2
[(6.2573 in/rad)2  1.6515 in/rad  ]   0.497 in  lb  s2  (0.143 rad/rad)2  1.094 in  lb  s 2
2
386.1 in/s
For link 4:
A4  m4 ( xG24  yG24 )  IG442  m4 (0  0)   0.063 in  lb  s 2  (1.143 rad/rad) 2  0.0823 in  lb  s2

Therefore, the equivalent mass moment of inertia of the mechanism is
4
I EQ   Aj  A2  A3  A4  0.063 in  lb  s2  1.094 in  lb  s 2  0.0823 in  lb  s 2  1.239 in  lb  s2
Ans.
j 2
1 4 dAj
To determine  B j , note that  B j  
; therefore, the units must be in  lb  s2 .
2 j 2 d 2
j 2
j 2
For link 2:
B2  m2 ( xG 2 xG 2  yG 2 yG 2 )  IG 222  m2 (0  0)   0.063 in  lb  s 2  (1 rad/rad)(0)  0
4
4
For link 3:
B3  m3 ( xG 3 xG 3  yG 3 yG 3 )  I G 333

10 lb
[(6.2573 in/rad)(3.100 in/rad 2 )  (1.6515 in/rad)(2.952in/rad 2 )]
2
386.4 in/s
  0.497 in  lb  s 2  ( 0.143 rad/rad)(0.848 rad/rad 2 )
 0.436 in  lb  s 2
For link 4:
B4  m4 ( xG 4 xG 4  yG 4 yG 4 )  I G 444
 m4 (0  0)   0.063 in  lb  s 2  (1.143 rad/rad)(0.848 rad/rad 2 )  0.061 in  lb  s 2
Therefore, the sum of these coefficients is
4
 B  B  B  B  0  0.436 in  lb  s  0.061 in  lb  s  0.497 in  lb  s
2
j 2
j
2
3
2
2
4
The power equation can be written as
dT dU dW f
P


(1)
dt
dt
dt
The left-hand side of the power equation can be written
P  T22  FC  VC
(2)
The unknown torque T2 is taken to be positive in the same direction as the input angular
velocity (that is, counterclockwise). The velocity of point C can be written as
VC  ( xC ˆi  yC ˆj)2
(3)
The first-order kinematic coefficients for the path of point C can be obtained from the
vector equation
RC  R 2  R3
563
The X and Y components of this vector equation are
xC  R2 cos2  R3 cos3
yC  R2 cos2  R3 sin 3
Therefore, the first-order kinematic coefficients for point C are
xC   R2 sin 2  R3 sin 33    6 in  sin 60   24 in  sin321.8(0.143 rad/rad)  7.318 53 in/rad
(4a)
yC  R2 cos 2  R3 cos33   6 in  cos60   24 in  cos321.8(0.143 rad/rad)  0.302 94 in/rad
(4b)
Substituting Eqs. (4) into Eq. (3), the velocity of point C can be written as
VC   7.318 53 in/rad  ˆi   0.302 94 in/rad  ˆj 2
Therefore, the power due to the vertically downward force at point C is
FC  VC   125 lb  ˆj  ( xC ˆi  yC ˆj)2   125 lb  yC 2
  125 lb  (0.302 94 in/rad)2    37.87 in  lb/rad  2
(5)
The negative sign indicates that the vertical force and the vertical component of the
velocity of point C are in opposite directions (that is, that the vertical component of the
velocity of point C is upwards).
Substituting Eq. (5) into Eq. (2), the net power is
P  T22   37.87 in  lb/rad  2
(6)
Now consider the right-hand side of the power equation, see Eq. (1). In general, the time
rate of change of kinetic energy can be written as
4
4
dT
  Aj   B j 3
dt j 2
j 2
However, the generalized inputs for this problem are   2 ,   2  2 , and
  2   2 . Therefore, this equation can be written as
4
4
dT
  Aj2 2   B j 23
dt j 2
j 2
The constant angular velocity of the input link is 2  10 rad/s ccw. Therefore, the time
rate of change of the kinetic energy is
dT
 [1.239 in  lb  s 2  (0)   0.497 in  lb  s 2  (10 rad/s) 2 ]2    49.7 in  lb  2
(7)
dt
The time rate of change of the potential energy due to gravity is
4
dU g
  m j gyG j  m3 gyG 3 2  10 lb 1.6515 in/rad  2  16.515 in  lb  2
(8)
dt
j 2
The vector loop equation for the spring can be written as
R 2  R S  R10  0
The x and y components are
R2 cos 2  RS cos S  0
(9a)
R2 sin  2  RS sin S  R10  0
(9b)
From Eq. (9a) the stretched length of the spring (for this position of the mechanism) is
Rs  xA  R2 cos 2   6 in  cos 60  3 in
564
Differentiating Eqs. (9) with respect to the input position gives
 R2 sin  2  RS cos S  RS sin SS  0
R2 cos  2  RS sin S  RS cos SS  0
Substituting the known information gives
  6 in  sin 60o  RS  0
Therefore, the first-order kinematic coefficient for the spring is
Rs    6 in  sin 60  5.19 in/rad
The time rate of change of the potential energy in the spring is
dU s
 k ( Rs  Rso ) Rs2  12 lb/in  (3 in  4.5 in)(5.19 in/rad)2  93.42 in  lb/rad  2
dt
The vector loop equation for the viscous damper can be written as
R 4  RC  R11  0
The x and y components of this equation are
R4 cos4  RC cosC  R11 cos 11  0
(10)
R4 sin 4  RC sin C  R11 sin 11  0
Differentiating these with respect to the input position gives
 R4 sin 44  RC cos C  RC sin CC  0
R4 cos 44  RC sin C  RC cos CC  0
Substituting the known information (with the angle 4  261.8, and the first-order
kinematic coefficient 4  1.143 rad/rad ) gives
  6 in  sin 261.8(1.143 rad/rad)  RC cos180  0
Therefore, the first-order kinematic coefficient for the damper is
RC   6 in  sin 261.8(1.143 rad/rad)  6.79 in/rad
The positive sign indicates that the length of the vector R C is increasing for positive
input. The velocity of point B at the end of the damper is
VB  RC2   6.79 in/rad  (10 rad/s)  67.9 in/s
The time rate of change of the dissipative effect of the damper is
dW f
 CRc22 2   0.25 lb  s/in  (6.79 in/rad) 2 10 rad/s  2  114.92 in  lb/rad  2 (11)
dt
Therefore, from Eqs. (7), (8), (10), and (11), the right hand side of the power equation,
Eq. (1), can be written as
dT dU dW f
 
   49.7 in  lb/rad  2  16.515 in  lb/rad  2  93.42 in  lb/rad  2  114.92 in  lb/rad  2
(12)
dt dt dt
 175.16 in  lb/rad  2
Substituting Eqs. (6) and (12) into Eq. (1) gives
T22   37.87 in  lb/rad  2  175.16 in  lb/rad  2
(13)
The equation of motion is obtained by dividing both sides of Eq. (13) by the input ngular
velocity ω2. Therefore, the equation of motion for this problem can be written as
T2  37.87 in  lb  175.16 in  lb
Ans.
565
The driving torque acting on the input crank is
T2  213.03 in  lb
The positive sense indicates that the driving torque acting on the input crank is in the
same direction as the input angular velocity. Therefore, the driving torque acting on the
input crank is
Ans.
T2  213.03 in  lb ccw
566
12.31 For the Scotch-yoke linkage in the posture illustrated, the angle   30 , and the angular
velocity and acceleration of the input link 2 are ω  15kˆ rad/s and α  2 kˆ rad/s2 ,
2
2
respectively.
The accelerations of the centers of mass of the links are
ˆ
AG2  5.4i  11.3ˆj m/s2 , AG3  10.8ˆi  22.6ˆj m/s2 , and AG4  22.6jˆ m/s2 . The masses
and mass moments of inertia of the links are: m2  5 kg , m3  5 kg , m4  15 kg ,
IG2  0.02 N  m  s2 , I G3  0.12N  m  s2 , and IG4  0.08 N  m  s2 . Gravity is in the
negative y-direction, an external force P  125ˆj N is acting on link 4, and an unknown
torque T2 is acting on link 2. Determine the internal reaction forces and the torque T2 .
Indicate the point(s) of contact between link 4 and the ground link.
R2  1 m .
567
The free-body diagram for link 2 is shown in the following figure.
The sum of the forces acting on link 2 in the x direction can be written as
 F x  F12x  F32x  m2 AGx
(1)
The sum of the forces acting on link 2 in the y direction can be written as
 F y  F12y  F32y  W2  m2 AGy2 .
(2)
2
The sum of the moments acting on link 2 about point O2 can be written as
 M R F  R F  R W  T  I   m (R A  R A )
O2
x
2
y
32
y
2
x
32
x
5
2
2
G2
2
2
x
5
y
G2
y
5
x
G2
(3)
Therefore, there are three equations and five unknowns for the free-body diagram of link
2. The unknowns are the four reaction forces F12x , F12y , F32x , F32y and the crank torque T2 .
The free-body diagram for link 3 is shown in the next figure.
The sum of the forces acting on link 3 in the x direction can be written as
 F x F23x  m3 AGx3
(4)
The sum of the forces acting on link 3 in the y direction can be written as
 F y F23y  m3 AGy3
(5)
The sum of the moments acting on link 3 about the center of mass G3 can be written as
 M R F  I 
G3
7
43
G3
3
Since link 3 cannot rotate, the angular acceleration is  3  0 . Therefore, this equation can
be written as
568
R7 F43  0
(6)
Equations (4), (5), and (6) contain two new unknowns, F43 and R7 . Therefore, there are
now a total of six equations and seven unknowns.
If we assume that link 4 is only sliding on ground link 1, then the free-body diagram for
link 4 is as shown in the following figure.
G4
R8
R9
F34
W4
F14
P
The sum of the forces acting on link 4 in the x direction can be written as
 F x F14  m4 AGx4
Since link 4 can only accelerate in the y direction, that is, since AGx4  0 , this becomes
F14  0
The sum of the forces acting on link 4 in the y direction can be written as
 F y F34  P  W4  m4 AGy4
(7)
(8)
The sum of the moments acting on link 4 about the center of mass G 4 can be written as
M
G4
R8 F34  R9 F14  I G 4 4
Since link 4 can not rotate, that is, since  4  0 , this becomes
R8 F34  R9 F14  0
(9)
From Equations (7) and (9), the distance R9   , which means that link 4 attempts to tip.
For tipping, the free body diagram of link 4 is modified as shown in this figure.
569
G4
R8 F34
R10
F14T
W4 R11
F14B
P
The sum of the forces acting on link 4 in the x direction can be written as
 F x F14T  F14 B  m4 AGx4  0
(10)
The sum of the forces acting on link 4 in the y direction can be written as
 F y F34  P  W4  m4 AGy4
(11)
Note that Eq. (11) is the same as Eq. (8).
The sum of the moments acting on link 4 about the center of mas G 4 can be written as
R8 F34  R10 F14T  R11F14B  0
(12)
Equations (10), (11) and (12) contain two new unknowns F14T and F14B . Therefore, there
are a total of nine equations and nine unknowns.
Substituting m3  5 kg and AGx3  10.8 m/s2 into Eq.(4), we have
F23x  (5 kg)(10.8 m/s2 )  54 N
Substituting m2  5 kg , AGx2  5.4 m/s2 , and F23x  54 N into Eq.(1), we have
F12x  (5kg)(5.4 m/s2 )  ( 54 N)  81 N
Equation (6) implies that either R7  0 or F43  0 . Since there is contact between links 3
and 4, the internal reaction force F43 cannot be zero. Therefore
R7  0
Substituting m4  15 kg , AGy4  22.6 m/s2 , W4  m4 g  (15 kg)(9.81 m/s2 )  147 N , and
P  125 N into Eq. (11) we have
F34  (15 kg)(22.6 m/s2 )  147 N 125 N  361 N
Substituting m3  5 kg ,
AGy3  22.6 m/s2 , W3  m3 g  (5)(9.8) kg  m/s 2  49 N , and
F34  361 N into Eq. (5), we have
F23y  (5 kg)(22.6 m/s2 )  49 N  361 N  523 N
570
Substituting m2  5 kg ,
AGy2  11.3 m/s2 , W2  m2 g  (5 kg)(9.81 m/s2 )  49 N , and
F23y  523 N into Eq. (2), we have
F12y  (5 kg)(11.3 m/s2 )  49 N  523 N  628.5 N
From Eq. (12), we have
R10 F14T  R11F14B  R8 F34
Using Eqs. (10) and (13), we have
RF
F14T   8 34 ,
R11  R10
and
F14B 
Ans.
(13)
R8 F34
R11  R10
R8  R2x  R7  R2x  R2 sin 30  0.5 m , R10  0.4 m ,
F34  361 N into Eq. (14), we have
(0.5 m)(361 N)
F14T  
 361 N
0.9 m  0.4 m
and
(0.5 m)(361 N)
F14 B 
 361 N
0.9 m  0.4 m
From Eq. (3), we have
T2  I G22  m2 ( R5x AGy2  R5y AGx2 )  R5xW2  R2x F23y  R2y F23x
Substituting
(14)
R11  0.9 m
and
Ans.
Ans.
1
Substituting W2  49 N, IG2  0.02 N  m  s2 ,  2  2 rad/s 2 , m2  5 kg , R5x  R2 sin30  0.25 m ,
2
1
R5y   R2 cos30  0.433 m , AGx2  5.4 m/s2 , AGy2  11.3 m/s2 , R2x  R2 sin 30  0.5 m ,
2
y
R2   R2 cos30  0.866 m , F23x  54 N , and F23y  523 N into Eq. (22a), we have
T2  (0.02 N  m  s 2 )(2 rad/s 2 )  (5 kg) (0.25 m)(11.3 m/s 2 )  (0.433 m)(5.4 m/s 2 ) 
(0.25 m)(49 N)  (0.5 m)(523 N)  ( 0.866 m)( 54 N) N  m
Ans.
 229.46 N.
Since F14T  361 N , therefore F41T  361 N ; this means that link 4 is pushing to the
right on ground link 1 at the upper-right corner of the slot. Also, since F14B  361 N ,
therefore F41B  361 N ; this means that link 4 is pushing to the left on ground link 1 at
the lower-left corner of the slot. That is, link 4 is attempting to tip clockwise.
Ans.
571
12.32 For the parallelogram four-bar linkage in the posture illustrated, the angular velocity and
acceleration of the input link 2 are 2  2 rad / s ccw and  2  1 rad / s2 ccw, respectively.
The first and second-order kinematic coefficients of the center of mass of link 3 are
xG 3  0.141 m/rad, yG 3  0.141 m/rad, xG3  0.141 m/rad 2 , and yG3  0.141 m/rad 2 .
The masses and second moments of mass of the links are: m2  m4  0.5 kg,
IG2  IG4  2 kg  m2 , m3  1 kg, and IG3  5 kg  m2 . Gravity is in the negative z-direction,
the force F  100ˆi N acts at point C, and the torque T  10kˆ N.m acts on link 4.
C
4
Determine: (a) the acceleration of the mass center of link 3; (b) the internal reaction
forces F23 and F43 , and (c) the torque T2 .
RBO2  RAO4  0.2 m, RBA  RO2O4  0.3 m, and RCB  0.1 m.
The acceleration of the mass center G3 can be written as
2
AGx  xG 22  xG 2  0.141 m   2 rad/s     0.141 m  1 rad/s2   0.423 m/s2


2
y
2
AG  yG 2  yG 2   0.141 m   2 rad/s     0.141 m  1 rad/s2   0.705 m/s2


The free-body diagram of link 2 is shown in the following figure.
3
3
3
3
3
3
Ans.
Ans.
572
The sum of the forces in the x direction acting on link 2 can be written as
 F x F12x  F32x  0
(1)
The sum of the forces in the y direction can be written as
 F y F12y  F32y  0
(2)
The sum of the moments acting about the mass center G2 can be written as
 M G2 RB cos B  F32y  RB sin B  F32x  T2  IG22
(3)
The unknown variables in Eqs. (1), (2), and (3) are F12x , F12y , F32x , F32y , and T2 . Therefore,
the total number of unknown variables is five and there are only three equations.
The free body diagram of Link 3 is shown in this figure.
The sum of the forces in the x direction acting on link 3 can be written as
 F x F23x  F43x  FC  m3 AGx3
(4)
The sum of the forces in the y direction acting on link 3 can be written as
 F y F23y  F43y  m3 AGy3
(5)
The sum of the external moments acting about the mass center G3 can be written as
 M R cos  F   R sin   F   0
G3
BA
BA
y
23
BA
BA
x
23
(6a)
The new unknown variables in Eqs. (4), (5), and (6a) are F43x and F43y . Therefore, the total
number of equations is six and the total number of unknown variables is seven; that is,
F12x , F12y , F32x , F32y , T2 , F43x and F43y . Note that  BA  0; therefore, Eq. (6a) can be written
as
RBA  F23y   0
Therefore, either
RBA  0 or F23y  0
The free body diagram of link 4 is shown in the next figure.
.
(6b)
573
The sum of the forces in the x direction acting on link 4 can be written as
 F x  F14x  F34x  0
(7)
The sum of the forces in the y direction acting on link 4 can be written as
 F y F34y  F14y  0
(8)
The sum of the moments acting on link 4 about the mass center G4 can be written as
 M R cos  F   R sin   F   T  I 
G4
A
4
y
34
A
4
x
34
4
G4
(9)
4
The new unknown variables in Eqs. (7), (8), and (9) are F14x and F14y . Therefore, there are a
total of nine equations and nine unknown variables; that is, F12x , F12y , F32x , F32y , T2 ,F43x ,F43y ,
Ans.
F14x , and F14y .
The solution procedure will be the method of inspection. From Eq. (6b), the reaction
force
(10)
F23y  0
Substituting Eq. (10) into Eq. (5), the reaction force
F43y  m3 AGy3  F23y  1 kg(0.705 m/s2 )  0  0.705 N
Rearranging Eq. (9), the reaction force F34x can be written as
F 
x
34

I G4 4  T4  RA cos  4  F34y 
 RA sin  4
2kg  m 2 (1 rad/s2 )  10 N  m  0.2 m  0.707  0.705 N 
0.2 m  0.707 
Therefore, the total reaction force is
F43  55.87 N 180.72
 55.87 N
Ans.
x
23
Solving Eq. (7), the reaction force F can be written as
F23x  m3 AGx3  F43x  FC
 1 kg  0.423 m/s2    55.87 N   100 N  43.71 N
Therefore, the total reaction force is
F23  43.71 N180
Rearranging Eq. (3), the input torque T2 can be written as
Ans.
T2  I G2 2  RB sin  2  F32x   RB cos  2  F32y 
 2 kg  m 2 1 rad/s2   0.2 m  0.707  43.71 N  0.2 m  0.707  0  8.18 N  m
The positive sign indicates that the input torque is counterclockwise.
Ans.
Ans.
574
12.33 For the mechanism in the posture illustrated, the distance ROG3  2.5 m and the velocity
and acceleration of the input link 2 are V  10ˆi m/s and A  10ˆi m/s2 , respectively.
2
2
The first- and second-order kinematic coefficients of link 3 are 3  0.20 m1 , and
3  0.1386 m2 , respectively. The masses and second moments of mass of links 2
and 3 are m2  3 kg, m3  5 kg, I G2  1.5 kg  m2 , and I G3  7.5 kg  m2 . Gravity is in the
negative y-direction, the force FC  10 ĵ N acts at point C, and the line of action of the
unknown force P acting on link 2 is parallel to the x-axis. Determine: (a) the internal
reaction forces; (b) the force P; and (c) indicate the point(s) of contact of link 2 with the
ground link.
Link 2 is 1.5 m by 0.75 m, RG2G3  0.5 m, and RCG3  3.5 m .
The free-body diagram of link 2 is shown in the following figure.
The sum of the forces acting on link 2 in the x direction can be written as
 F x F32X  P  m2 AGx2
(1)
Note that the direction of the external force P is assumed to be in the positive x direction.
The sum of the forces acting on link 2 in the y direction can be written as
(2)
 F y F12y  F32y W2  m2 AGy
2
Note that AGx2  10 m/s2 and AGy2  0 .
The sum of the moments acting on link 2 about the mass center G2 can be written as
M
G2
R12x F12y  RG3G2W2  I G32  m2 ( RGx G AGy2  RGy G AGx )
3 2
3 2
2
(3)
575
Therefore, there are three equations and five unknowns for the free-body diagram of link
2. The unknowns are the three reaction forces F32x , F32y , F12y , the distance R12x (that is, the
location of F12y ), and the external force P.
The free-body diagram of link 3 is shown in the next figure.
The sum of the forces acting on link 3 in the x direction can be written as
 F x F23x  F13x  m3 AGx3
(4)
The sum of the forces acting on link 3 in the y direction can be written as
 F y F23y  F13y  FC  W3  m3 AGy3
(5)
Note that the acceleration of the center of gravity of link 3 is the same as the acceleration
of link 2 (that is, AGx3  AG2  10 m/s2 and AGy3  0 ).
The sum of the moments acting on link 3 about the center of mass G3 can be written as
 M R cos F  R sin  F  R cos F  I 
G3
32
y
3 13
32
x
3 13
C
3 C
x
13
G3
3
(6)
y
13
Equations (4), (5) and (6) contain two new unknowns, F and F . Therefore, there are a
total of six equations and seven unknowns.
Since links 1 and 3 have contact at a pin in a slot, the direction of the reaction force must
be perpendicular to the slot. This means only the magnitude of the reaction force F13 is
unknown.
The x and y components of this reaction force can be written as
F13x  F13 cos(3  90) and F13y  F13 sin(3  90)
This provides a seventh equation allowing the seven unknowns to be solved. Also, since
the internal reaction force F13 is perpendicular to the slot then Eq. (6) can be written as
R32 F13  RC cos 3 FC  I G33
(7)
The angular acceleration of link 3 can be written as
3  3 R2  3 R22
where R2  VG2  10 m/s and R2   AG2  10 m/s2 . This agrees with the observation
that the input velocity R2 must be positive; that is, the vector R 2 is increasing in length
for this position. Substituting this information and the known kinematic coefficients for
576
link 3 (that is, 3  0.200 rad/m and 3  0.1386 rad/m2 ) into Eq. (8a), the angular
acceleration of link 3 is
3  (0.200 rad/m)(10 m/s 2 )  (0.1386 rad/m2 )(10 m/s)2  15.86 rad/s2
Note that the angular acceleration of link 3 has a negative value; that is, the angular
acceleration of link 3 is clockwise. Also, note that the angular velocity of link 3 is
counterclockwise for this posture.
Substituting the known values into Eq. (7) gives
(2.5 m) F13  (3.5 m) cos(30)10 N  (7.5 kg  m2 )(15.86 rad/s2 )
Therefore, this reaction force is
Ans.
F13   35.46 N
The negative sign indicates that the internal reaction force F13 is acting downward; that is,
in the opposite direction to the assumed direction shown in the figure. Therefore, the
point of contact between link 3 and link 1 is on the top side of the ground pin O.
Substituting the known values into Eq. (5) gives
F23y   35.46 N  sin(60)  10 N  (5 kg)(9.81 m/s2 )  0
Therefore, this reaction force is
F23y  89.76 N
Substituting the known values into Eq. (4) gives
F23x   35.46 N  cos(60)  (5 kg)(10 m/s2 )
Ans.
Therefore, the reaction force is
F23x  67.73 N
Substituting the known values into Eq. (2) gives
F12y  89.76 N  (3 kg)(9.81 m/s2 )  0
Therefore, this reaction force is
F12y  119.19 N
Substituting the known values into Eq. (1) gives
67.73 N  P  (3 kg)(10 m/s 2 ) )
Therefore, the applied force is
P  97.73 N
Substituting the known values into Eq. (3) gives
R12x (119.19 N)  RG3G2 (3 kg)(9.81 m/s2 )  0
Ans.
Ans.
Ans.
Therefore, the distance is
R12x  0.25RG3G2
The negative sign indicates that the location of the internal reaction force F12y is to the left
of the mass center of link 3. Given that the distance
RG3G2  0.5 m )
Then the distance from the mass center of link 3 to the point of application of the normal
force is
R12x  0.25  0.5 m  0.125 m
Ans.
577
12.34 For the slider-crank linkage in the posture when 2  45 , the angular velocity and
acceleration of the input link 2 are   100kˆ rad/s and   10kˆ rad/s2 , respectively. The
2
2
postures, velocities, and accelerations of links 3 and 4 are provided in the table. Gravity
is in the negative y-direction and the weights and mass moments of inertia of the links
are: w3  3.40 lb, w4  2.86 lb, IG2  0.352 in  lb  s2 , and IG3  0.108 in  lb  s2 . The stiffness
and unstretched length of the spring are K S  20 lb/in and r0  3 in, respectively. The
damping coefficient of the viscous damper is C  7 lb  s/in . An external force FB acts
horizontally at pin B on link 4, and the motor torque is   60kˆ lb.in. Determine: (a) the
2
first- and second-order kinematic coefficients of the linkage, (b) the equivalent mass
moment of inertia, and (c) the external force FB .
RG3 A  4.5 in. .
Table P12.34
3
2
deg
deg
45  10.18
RAO2
RBA
REO2
in
3
in
12
in
5
3
rad/s
 17.96
VB4
3
AB4
in/s
 250.23
rad/s2
 1736.20
in/s 2
 21361.02
Assume: The effects of friction in the mechanism can be neglected.
The vectors for a kinematic analysis of the mechanism are shown in the next figure.
The first-order kinematic coefficient of link 3 can be written as
578
3 17.96 rad/s

 0.1796 rad/rad
2
100 rad/s
The first-order kinematic coefficient of link 4 can be written as
3 
250.23 in/s
(1)
 2.5023 in/rad
2
100 rad/s
The angular acceleration of link 3 can be written as
3  322  3 2
Rearranging this equation, the second-order kinematic coefficient of link 3 can be written
as
   
(1736.20 rad/s 2 )  ( 0.1796 rad/rad)(10 rad/s 2 )
3  3 2 3 2 
 0.1738 rad/rad 2
2
(100 rad/s) 2
Similarly, the linear acceleration of link 4 can be written as
AB4  R422  R42
R4 
VB4

Therefore, the second-order kinematic coefficient of link 4 is
AB  R42 21 361 in/s2  ( 2.5023 in/rad)(10 rad/s 2 )
R4  4 2

 2.1336 in/rad 2
2
2
(100 rad/s)
Since the mass center of link 2, G2, is located at the fixed pivot O2, their first- and
second-order kinematic coefficients are
xG 2  0 , xG2  0
yG 2  0 , and yG2  0 .
To determine the first- and second-order kinematic coefficients for the center of mass of
link 3: The vector loop for point G3, see Fig. 1, can be written as
??
I
C
R G3  R 2  R 33
where the magnitude of the vector R 33 is given as 4.5 in and the angle 33  3 . This
implies that the first-order kinematic coefficient 33  3 and the second-order kinematic
coefficient 33  3 . The x and y components of the above equation are
xG3  R2 cos2  R33 cos3 =6.5504 in
yG3  R2 sin 2  R33 sin 3  1.3258 in
The first-order kinematic coefficients of the center of mass of link 3 are
xG 3   R2 sin 2  R33 sin 33  2.2642 in/rad
yG 3  R2 cos2  R33 cos33  1.3258 in/rad
The second-order kinematic coefficients of the center of mass of link 3 are
xG3   R2 cos2  R33 cos332  R33 sin 33  2.1259 in/rad2
yG3   R2 sin 2  R33 sin 332  R33 cos33  1.3258 in/rad2
The first- and second-order kinematic coefficients of the center of mass of link 4 are
xG4  R4  2.1336 in/rad2
xG 4  R4  2.5023 in/rad
yG 4  0
yG4  0
579
The first-order kinematic coefficient for the damper is
RC  R4  2.5023 in/rad
The first-order kinematic coefficient for the spring can be obtained from the vector loop
for point E, see Fig. 1; that is,
??
C
R E  R 22
where the magnitude of the vector R 22 is given as 5 in and the angle 22  2  180  225 .
This implies that the first-order kinematic coefficient 22  1 rad/rad . The x component
of point E is
xE  R22 cos22 =  3.5355 in
Therefore, the first-order kinematic coefficient of point E is
xE   R22 sin 222  3.5355 in/rad
Note that the first-order kinematic coefficient for the spring is
RS   xE  3.5355 in/rad
4
4
j 2
j 2
To determine  A j : Note that  Aj  I EQ ; (that is, the equivalent mass moment of
inertia) therefore, the units must be lb  in  s2 .
for link 2:
A2  m2 ( xG22  yG22 )  I G2 22
 m2 (0  0)   0.352 lb  in  s 2  (1 rad/rad) 2  0.352 lb  in  s 2
for link 3:
A3  m3 ( xG23  yG23 )  IG332
3.4 lb
[(2.2642 in/rad) 2  (1.3258 in/rad) 2 ]   0.108 lb  in  s2  (0.1796 rad/rad) 2  0.064 lb  in  s 2
2
386.4 in/s
for link 4:
2.86 lb
A4  m4 ( xG24  yG24 )  I G442 
[(2.5023 in/rad) 2  (0) 2 ]  0  0.0463 lb  in  s 2
2
386.4 in/s
Therefore, the sum of the coefficients is

4
 A  A  A  A  0.352 lb  in  s  0.064 lb  in  s  0.0463 lb  in  s  0.462 lb  in  s
2
j 2
j
2
3
2
2
2
4
4
4
j 2
j 2
To determine  B j : Note that  B j 
1 4 dAj
 ; therefore, the units must be lb  in  s2 .
2 j 2 d 2
for link 2:
B2  m2 ( xG 2 xG2  yG 2 yG2 )  IG222  m2 (0  0)   0.352 lb  in  s 2  (1 rad/rad)(0)  0
for link 3:
Ans.
580
B3  m3 ( xG 3 xG3  yG 3 yG3 )  I G333
3.4 lb
[(2.2642 in/rad)(2.1259 in/rad)  (1.3258 in/rad)(1.3258 in/rad)]
386.4 in/s 2
  0.108 lb  in  s 2  (0.1796 rad/rad)(0.1738 rad/rad 2 )

 0.0235 lb  in  s 2
for link 4:
B4  m4 ( xG 4 xG4  yG 4 yG4 )  I G444
2.86 lb
[( 2.5023 in/rad)( 2.1336 in/rad)  0]  0  0.0395 lb  in  s2
2
386.4 in/s
Therefore, the sum of the coefficients is

4
 B  B  B  B  0  0.0235 lb  in  s  0.0395 lb  in  s  0.063 lb  in  s
2
j 2
j
2
3
2
4
The power equation can be written as
dT dU dW f
P


dt
dt
dt
The time rate of change of the kinetic energy can be written as
4
4
dT
  Aj   B j 3
dt j 2
j 2
2
Ans.
(2)
where the generalized inputs for this problem are   2 ,   2  2 , and   2   2 .
Therefore, the time rate of change of the kinetic energy is
4
dT 4
  Aj2 2   B j 23  [0.462 lb  in  s 2 (10 rad/s 2 )  0.063 lb  in  s 2 (100 rad/s)2 ]2
dt j 2
j 2
That is,
dT
  634.62 lb  in  2
dt
The time rate of change of the potential energy can be written as
4
dU
  m j gyG j  K S ( RS  RO ) RS  m3 gyG 2  K S ( RS  RO ) RS 2
3
dt
j 2
The time rate of change of the potential energy due to gravity is
dU g
dt
 m3 gyG 2  w3 yG 2
3
3
 3.40 lb(1.3258 in/rad)2   4.508 lb  in/rad  2
The time rate of change of the potential energy due to the linear spring can be written as
dU S
 K S ( RS  RS 0 ) RS 2
dt
 20 lb/in[ 5 in  cos 45  3 in]( 3.5355 in/rad)2   37.8676 lb  in/rad  2
Therefore, the time rate of change of the total potential energy is
581
dU dU g dU S


dt
dt
dt
  4.508 lb  in/rad  2   37.8676 lb  in/rad  2   33.3596 lb  in/rad  2
The time rate of change of the dissipative effects due to the viscous damper is
dW f
 CRC 2 2  7 lb  s/in(2.5023 in/rad) 2 (100 rad/s)2   4 383.053 lb  in/rad  2
dt
Note that the time rate of change of the dissipative effects due to the viscous damper is a
positive value. This must always be true for this term on the right-hand side of the power
equation.
Therefore, the right-hand side of the power equation, see Eq. (2), can be written as
dT dU dW f


  634.62 lb  in/rad  2   37.8676 lb  in/rad  2   4 383.053 lb  in/rad  2
(3)
dt dt
dt
  4 979.80 lb  in/rad  2
Note that the most influential term is the time rate of change of the dissipative effects due
to the viscous damper. This implies that the damping coefficient C  7 lb  s/in is a large
value.
The left-hand side of the power equation, see Eq. (2), can be written as
(4)
P  T2  ω2  FB  VB   60 in  lb 2  FBx ( xB2 )
Note that the torque T2 is acting in the same direction as the angular velocity of link 2
(that is, counterclockwise) and the external force acting on the piston FB is assumed
positive when acting in the same direction as the velocity of the piston (link 4) (that is, in
the negative x direction). The first-order kinematic coefficient of point B can be obtained
from the point path vector equation, or by noting that xB  R4 . From Eq. (1), the firstorder kinematic coefficient of link 4 is R4  2.5023 in/rad . Therefore, the first-order
kinematic coefficient of point B is
xB  2.5023 in/rad
(5)
Substituting Eq. (5) into Eq. (4) gives
P   60 lb  in/rad  2   2.5023 in/rad  FBx2
(6)
Finally, equating the two equations, Eqs. (3) and (6), the power equation can be written as
(7)
 60 lb  in/rad  2   2.5023 in/rad  FBx2   4 979.80 lb  in/rad  2
The equation of motion is obtained by dividing both sides of the power equation, Eq. (7),
by the input angular velocity 2 . Therefore, the equation of motion can be written as
60 lb  in   2.5023 in  FBx  4979.80 lb  in
Ans.
(8)
Rearranging Eq. (8), the external force acting on the piston is
FBx  1 966.11 lb
Ans.
The negative sign indicates that the external force acting on the piston (link 4) is acting to
the left, that is, in the negative x direction. Therefore, the assumption made in Eq. (4)
was correct; that is, the force is acting in the same direction as the velocity of the piston
(link 4).
582
12.35 For the mechanism in the posture illustrated, the massless link 4 is rolling on the ground
link. The first- and second-order kinematic coefficients of links 3 and 4 are 3  0,
4  1.0 rad/m, 3  1.0 rad/m2 , and 4  3.0 rad/m2 . The velocity and acceleration of
the mass center of the input link 2 are VG2  7ˆi m/s and AG2  2ˆi m/s2 , respectively.
The free length and spring rate of the spring, the damping constant of the viscous damper,
the masses, and mass moments of inertia of links 2 and 3 are provided in the table.
Gravity is in the negative y-direction and a horizontal external force P acts on link 2.
Determine: (a) the first-order kinematic coefficients of the spring and the damper; (b) the
first- and second-order kinematic coefficients of the mass center of link 3; (c) the
equivalent mass, (d) the equation of motion; and (e) the force P acting on link 2.
R4  1 m, RG2 A  6 m, and R1  2 m.
Table P12.35
Ro
K
C
m2
m3
I G2
I G3
m
N/m
N·s/m
kg
kg
3
25
15
1.20
0.80
kg  m2
0.25
kg  m2
0.10
583
The vectors for kinematic analysis of the mechanism are shown in the following figure.
Vectors for a kinematic analysis of the mechanism.
The vector loop equation for the mechanism can be written as
R 2  R3  R 7  0
where link 7 is an arm connecting the ground link to the center of the wheel, link 4. The
x and y components of this equation are
R2 cos 2  R3 cos 3  R7 cos 7  0
R2 sin 2  R3 sin 3  R7 sin 7  0
Differentiating these equations with respect to the input position R2 gives
cos 2  R3 sin 33  R7 sin 77  0
sin 2  R3 cos 33  R7 cos 77  0
Differentiating again with respect to the input position gives
 R3 sin 33  R3 cos3 32 R7 sin 77  R7 cos7 72  0
R3 cos 33  R3 sin 3 32 R7 cos77  R7 sin 7 72  0
Solving for the first-order kinematic coefficients we get
3  0 and 7  0.333 rad/m
Solving for the second-order kinematic coefficients, they are
3  1 rad/m2 and 7  1 rad/m2
The rolling contact equation between the wheel, link 4, and the ground link can be written
as
  7
R1
 4
R4
1  7
The correct sign is negative because there is external contact between link 4 and the
ground link. Differentiating this equation with respect to the input position gives
584
    7
R1
 4
R4
1   7
Substituting the known values into this equation, the first-order kinematic coefficient for
link 4 is
4  1 rad/m
Differentiating the above equation with respect to the input position gives
    
R1
 4 7
R4
1 7
Substituting the known values into this equation, the second-order kinematic coefficient
for link 4 is
4  3 rad/m2
The first-order kinematic coefficient of the spring is
Ans.
RS  R2  1 m/m
Note that the answer is positive because the length of the spring increases for a positive
change in the input position. The first-order kinematic coefficient of the damper can be
written as
Ans. (1)
RC   R2  1 m/m
Note that the answer is negative because the change in the length of the vector RC  RAO1
decreases for positive change in the input position.
Check: Since link 4 is rolling on the ground link at point E then point E is the instant
center I14 . Therefore, the velocity of point A (which is directed in the positive x direction)
can be written as
(2)
VA  4 RI14 A  (4 R2 ) RI14 A  (1 rad/m  7 m/s)(1 m)  7 m/s
The first-order kinematic coefficient of the damper is defined as
V
 7 m/s 
RC   A   
  1 m/m
R2
 7 m/s 
The correct sign is negative because the change in length of the vector RC  RAO1
decreases as the change in length of the input vector R2 increases.
The vector equation for the center of mass of link 3 can be written as
RG3  R 2  R 33
The x and y components of this equation are
xG3  R2 cos  2  R33 cos3 and
yG3  R2 sin 2  R33 sin 3
Differentiating these equations with respect to the input position, the first-order kinematic
coefficients for the center of mass of link 3 are
xG 3  cos 2  R33 sin 33 and yG 3  sin 2  R33 cos 33
(3)
Substituting the known data into this equation, the first-order kinematic coefficients for
the center of mass of link 3 are
xG 3  1 m/m and yG 3  0
Ans.
Differentiating Eq. (3) with respect to the input position, the second-order kinematic
coefficients of the center of mass of link 3 can be written as
585
xG3   R33 sin 33  R33 cos 3 332
and
yG3  R33 cos 33  R33 sin 3 332
Substituting the known data into this equation, the second-order kinematic coefficients
for the center of mass of link 3 are
Ans.
xG3  1.5 m/m2 and
yG3  2.598 m/m2
The x and y components of the acceleration of the mass center of link 3 are
AGx 3  xG3 R22  xG 3 R2  (1.5 m/m2 )(7m/s)2  (1 m/m)( 2 m/s2 )  75.5 m/s2
AGy 3  yG3 R22  yG 3 R2  (2.598 m/m2 )(7 m/s)2  (0)( 2 m/s2 )  127.302 m/s2
Note that the mass center of link 3 has nonzero x and y components; therefore, the path of
the mass center of link 3 is not a horizontal straight line. The magnitude and direction of
the acceleration of the mass center of link 3 are
AG3  148.0 m/s2239.33
The equivalent mass of the mechanism can be written as
n
n
j 2
j 2
mEQ   Aj   m j ( xG j 2  yG j 2 )  I G j  j2
Substituting known data into Eq. (4) gives
A2  1.2 kg[(1 m/m)2  02 ]  0.25 kg  m2 (0)2  1.2 kg
Substituting known data into Eq. (4) gives
A3  0.8 kg[(1 m/m)2  02 ]  0.10 kg  m2 (0)2  0.8 kg
Since link 4 is massless then
A4  0
Therefore, the equivalent mass of the mechanism is
mEQ  1.20 kg  0.80 kg  0  2.00 kg
(4)
Ans.
The power equation for the mechanism can be written as
4
4
 4

P  VG2    Aj R2   B j R22  R2   m j gyG j R2  K ( RS  R0 ) RS R2  CRC2 R22
j 2
j 2
 j=2

Assume that the force P acting on link 2 is positive in the same direction as the positive
input velocity. Canceling the input velocity VG2  R2 , the equation of motion for the
mechanism can be written as
n
4
4
j 2
j 2
j 2
P   Aj R2   B j R22   m j gyG j  K (R S  R0 ) Rs  CRC2 R2
Ans. (5)
The coefficients B j can be written as
B j  m j ( xG j xG j  yG j yG j )  IG  j j
j
Substituting known data into Eq. (6) gives
B2  1.20 kg[(1 m/m)(0)  (0)(0)]  (0.25 kg  m2 )(0)(0)  0
Substituting known data into Eq. (6) gives
B3  0.80 kg[(1 m/m)(1.5 m/m)  (0)(2.598 m/m2 )]  0.10 kg  m2 (0)(1 rad/m2 )  1.20 kg/m
Since link 4 is massless,
B4  0
(6)
586
Therefore, the sum of the coefficients B j is
4
 B 0  1.20 kg/m  0  1.20 kg/m
j
j 2
(7)
The effects of gravity can be written as
4
 m gy  g[m y  m y  m y ]  9.81 m/s [(1.20 kg)(0)  (0)(0)  (4 kg)(0)]  0
2
j 2
j
Gj
2 G2
3 G3
4 G4
The terms for the spring and the damper in Eq. (5) are
K ( RS  R0 )Rs  25 N/m(5.196 m  3 m)(1 m/m)  54.9 N  m/m
CR R2  (15 N  s/m)(1 m/m) (7 m/s)  105 N  m/m
The external force P acting on link 2 can be written from Eq. (5) as
2
C
2
4
4
j 2
j 2
(8a)
(8b)
P  mEQ R2   B j R22   m j gyG j  K (R S  R0 ) Rs  CRc2 R2
Substituting the given data and Eqs. (1), (2), (7), and (8) into this equation, the magnitude
of the external force P acting on link 2 can be written as
Ans.
P  (2.0 kg)(2 m/s2 )  (1.2 kg/m)(7 m/s) 2  0  54.9 N 105 N  97.1 N
The positive sign indicates that the assumption that the external force P is acting to the
right (that is, in the same direction as the input velocity) is correct. Therefore, the
external force P is acting to the right.
587
12.36 For the mechanism in the posture illustrated, the first- and second-order kinematic
coefficients of links 3 and 4 are 3  0.125 rad/rad, R4  1.299 m/rad, 3  0, and
R4  0.094 m/rad 2 . The constant angular velocity of the input link 2, which rolls without
slipping on the inclined plane, is 2  20 rad/s cw. The free length of the spring is 3 m,
the spring rate is k  25 N/m, and the damping constant of the viscous damper is
C  15 N·s/m. The masses and mass moments of inertia of links 2 and 4 are: m2  7 kg,
m4  4 kg, IG  18 kg·m2 , and IG4  22 kg·m2 .
2
The mass of link 3 is negligible
compared to the masses of links 2 and 4, and gravity is in the negative y-direction.
Determine: (a) the first- and second-order kinematic coefficients of the mass centers of
links 2 and 4, (b) the equivalent mass moment of inertia, and (c) the magnitude and
direction of the torque acting on link 2.
2  1.5 m, RBA  RG4G2  6 m, and RAOS  2.5 m.
Vectors for the mass centers of links 2 and 4 are shown in the following figure.
From this figure, the vector equation for the mass center of link 2 can be written as
?


R G2  R 9  R 7
588
The x and y components of this equation are
xG2  R9 cos9  R7 cos 7 and
yG2  R9 sin 9  R7 sin 7
Differentiating with respect to the input position  2 gives
xG 2  R9 cos9 and
yG 2  R9 sin 9
(1)
The rolling contact equation between link 2 and the inclined plane (link 1) can be written
in terms of the first-order kinematic coefficients as
R9   22   2  1.5 m/rad
Therefore, the second-order kinematic coefficient is
R9  0
Substituting 9  150 and R9  1.5 m/rad into Eq. (1) gives
xG 2  1.5cos150o  1.299 m/rad
and
yG 2  1.5sin150o  0.75 m/rad
Ans.
Differentiating Eqs. (1) with respect to the input position  2 gives
xG2  R9 cos 9  0 and
yG2  R9 sin 9  0
Ans. (2)
The vectors for the mass center of link 4 are shown in the above figure. The first-order
kinematic coefficients of the mass center of link 4 can be written as
Ans.
xG 4   R4  xG 2  1.299 m/rad and
yG 4  0
The second-order kinematic coefficients of the mass center of link 4 can be written as
xG4   R4  0.094 m/rad2 and
Ans.
yG4  0
(b) The power equation for the mechanism can be written as
4
4


T2  ω2   I EQ2   B j22  2   m j gyG j 2  K ( rs  r0 )rs2  Crc222
j 2
j 2


The input torque is taken positive in the same direction as the given input angular velocity
(that is, clockwise). Then canceling the input angular velocity, the equation of motion for
the mechanism can be written as
4
4
j 2
j 2
T2  I EQ 2   B j22   m j gyG j  K (rs  r0 )rs  Crc22
(3)
The equivalent mass moment of inertia of the mechanism can be written as
I EQ   A   m j ( xG 2  yG 2 )  I G  2
j
j
j
j
(4)
j
Substituting known data for link 2 into this equation gives
A2  (7 kg)((1.299 m/rad)2  (0.75 m/rad) 2 )  18 kg  m2 (1 rad/rad)2  33.75 kg  m2 /rad 2
Substituting known data for link 3 into Eq. (4) gives
A3  0
Substituting known data for link 4 into Equation (4) gives
A4  (4 kg) ( 1.299 m/rad)2  02   22 kg  m2 (0)2  6.75 kg  m2 /rad2
Therefore, the equivalent mass moment of inertia of the mechanism is
4
I EQ   Aj 33.75 kg  m2  0  6.75 kg  m2  40.5 kg  m2
j 2
Ans.
589
(c) The coefficient B j can be written as
B j  m j ( xG j xG j  yG j yG j )  IG j  j j
(5)
Substituting known data for link 2 into Eq. (5) gives
B2  7 kg ( 1.299 m/rad)(0)  (0.75 m/rad)(0)  (18 kg  m2 )(1 rad/rad)(0)  0
Substituting known values for link 3 into Eq. (5) gives
B3  0
Substituting known data for link 4 into Eq. (5) gives
B4  4 kg ( 1.299 m/rad)( 0.094 m/rad 2 )  (0)(0)   (22 kg  m2 )(0)(0)  0.488 kg  m2
Therefore, the coefficient is
4
 B 0  0  0.488 kg  m  0.488 kg  m
2
j 2
2
j
The effects of gravity can be written as
4
 m gy  [m gy  m gy  m gy ]
j 2
j
Gj
2
G2
3
G3
4
G4
 9.81 m/s 2 [(7 kg)(0.75 m/rad)  (0)( yG 3 )  (4 kg)(0)]  51.5 N  m/rad
The velocity of the mass center of link 2 down the inclined plane is
VG2  R2  1.5 m 20 rad/s  30 m/s
which agrees with the first-order kinematic coefficients in Eq. (2); that is,
 1.299 m/rad    0.75 m/rad  2  1.5 m/rad  2  30 m/s
VG2 
2
2
Therefore, the first-order kinematic coefficient of the spring is
rS  R  1.5 m / rad
and is negative because the length of the spring is decreasing for positive input motion.
Therefore
K (rS  r0 )rS  25 N/m(2.5 m  3 m)(1.5 m/rad)  18.75 N  m/rad
The first-order kinematic coefficient of the damper can be written as
rC  r4  1.299 m/rad
Therefore
CrC22  (15 N  s/m)(1.299 m/rad)2 (20 rad/s)  506.22 N  m/rad
From Eq. (3). the input torque can be written as
4
4
T2  I EQ 2   B j22   m j gyG j  K (rS  r0 )rS  CrC22
j 2
j 2
 (40.5 kg  m )(0)  (0.488 kg  m 2 )( 20 rad/s) 2  51.5 N  m  18.75 N  m  506.22 N  m Ans.
 240.77 N  m
The negative sign indicates that the torque is in the direction opposite to the input angular
velocity (which is specified as clockwise). Therefore, the input torque must be acting
counterclockwise.
2
590
12.37 The two-throw opposed-crank crankshaft is mounted in bearings at A and G. Each crank
has an eccentric weight of 6 lb, which may be considered located at a radius of 2 in from
the axis of rotation, and at the center of each throw (points C and E). It is proposed to
locate weights at B and F to reduce the bearing reactions, caused by the rotating eccentric
cranks, to zero. Determine the magnitudes of these weights, if they are to be mounted 3
in from the axis of rotation.
 M   2 in  m r   8 in  m r    20 in  m r    26 in  m r    28 in  F  0
 M   28 in  F   26 in  m r    20 in  m r   8 in  m r    2 in  m r   0
y
A
2
B B
y
G
2
2
C C
2
B B
A
2
2
E E
2
2
C C
2
2
2
F F
2
E E
2
2
G
2
F F
2
Dividing by  2 in   2 and substituting numeric values gives
9 in  m  144 in  lb   117 in  m  0
 117 in  m  144 in  lb    9 in  m  0
2
2
2
B
F
2
2
B
Solving simultaneously gives
wB  wF  mB  1.333 lb
2
F
Ans.
591
12.38 The two-throw crankshaft, mounted in bearings at A and F, with the cranks spaced 90
apart. Each crank may be considered to have an eccentric weight of 6 lb at the center of
the throw and 2 in from the axis of rotation. It is proposed to eliminate the rotating
bearing reactions, which the crank would cause, by mounting additional correction
weights on 3-in arms at points B and E. Calculate the magnitudes and angular locations
of these weights.
 M   2ˆi in  × sin  ˆj  cos  kˆ  m r   8ˆi in  ×  ˆj m r 
y
A
B
2
B B
B
2
2
C C
2
   
  

 M   26ˆi in  × sin  ˆj  cos  kˆ  m r    20ˆi in  ×  ˆj m r 
  8ˆi in  ×  kˆ  m r    2ˆi in  ×  sin  ˆj  cos  kˆ  m r   0
 20ˆi in × kˆ mD rD2 2  26ˆi in × sin  E ˆj  cos  E kˆ mE rE2 2  0
y
F
B
2
D D
B
2
B B
2
2
C C
2
E
E
2
E E
2
2
Dividing by  2 in   2 , substituting numeric values, and equating vector components
gives

9 in 2  mB cos  B   240 lb  in 2   117 in 2  mE cos  E  0
 9 in  m sin    96 lb  in   117 in  m sin   0
117 in  m cos   96 lb  in    9 in  m cos  0
 117 in  m sin    240 lb  in    9 in  m sin   0
2
2
B
B
B
B
2
2
B
B
2
2
2
2
2
E
E
E
E
E
E
Solving simultaneously gives
mB cos  B  0.667 lb , mB sin  B  2.000 lb , mE cos  E  2.000 lb , mE sin  E  0.667 lb,
mB  2.108 lb ,  B  108.43 , mE  2.108 lb ,  E  161.57
Ans.
592
12.39 Solve Problem 12.38 with the angle between the two throws reduced from 90 to 0.
 M   2ˆi in  × sin  ˆj  cos  kˆ  m r   8ˆi in  × kˆ  m r 
y
A
B
2
B B
B
2
2
C C
2
   
  

 M   26ˆi in  × sin  ˆj  cos  kˆ  m r    20ˆi in  × kˆ  m r 
  8ˆi in  ×  kˆ  m r    2ˆi in  ×  sin  ˆj  cos  kˆ  m r   0
 20ˆi in × kˆ mD rD2 2  26ˆi in × sin  E ˆj  cos  E kˆ mE rE2 2  0
y
F
B
2
D D
B
2
B B
2
2
C C
2
E
2
E E
E
2
2
Dividing by  2 in   2 , substituting numeric values, and equating vector components
gives

9 in 2  mB cos  B   336 lb  in 2   117 in 2  mE cos  E  0
 117 in  m sin   0
 9 in  m sin 
117 in  m cos  336 lb  in    9 in  m cos  0
 117 in  m sin 
  9 in  m sin   0
2
2
B
B
2
2
B
B
2
E
E
E
E
E
E
2
2
B
B
Solving simultaneously gives
mB cos  B  2.667 lb , mB sin  B  0.000 lb , mE cos E  2.667 lb , mE sin  E  0.000 lb
mB  2.667 lb ,  B  180.00 , mE  2.667 lb ,  E  180.00
Ans.
593
12.40 The connecting rod weighs 7.90 lb and is pivoted on a knife edge and caused to oscillate
as a pendulum. The rod is observed to complete 64.5 oscillations in 1 min. Determine
the mass moment of inertia of the rod about the center of mass.
rG  3.125 in, w  7.90 lb ,   60 s 64.5 cycles  0.930 s/cycle
From Eq. (12.101),
2
2
IO  mgrG  2    7.90 lb  3.125 in  0.930 s/cycle 2 rad/cycle   0.541 in  lb  s2
I G  IO  mrG2  0.541 in  lb  s2   7.90 lb 386 in/s2  3.125 in   0.341 in  lb  s2
2
Ans.
594
12.41 The gear is suspended on a knife edge at the rim and caused to oscillate as a pendulum.
The period of oscillation is observed to be 1.08 s. Assume that the center of mass and the
axis of rotation are coincident. If the weight of the gear is 41 lb, find the mass moment of
inertia and the radius of gyration of the gear.
From Eq. (12.101),
2
2
IO  mgrG  2    41.0 lb 8.0 in 1.08 s/cycle 2 rad/cycle   9.691 in  lb  s2
IG  IO  mrG2  9.691 in  lb  s2   41.0 lb 386 in/s 2  8.0 in   2.894 in  lb  s2
Ans.
kG  IG m  2.894 in  lb  s2  41.0 lb 386 in/s 2   5.221 in
Ans.
2
595
12.42 The wheel is mounted on a shaft in bearings with very low frictional resistance to
rotation. At one end of the shaft and on the outboard side of the bearings is connected a
rod with a weight Wb secured to its end. The weight Wb is displaced from equilibrium
and the assembly is permitted to oscillate. If the weight of the pendulum arm is
neglected, show that the mass moment of inertia of the wheel can be written as
 2
l
I  Wbl 
 
 4 2 g 


Using W for the weight of the wheel, the location of the center of mass of the assembly is
rG where WrG  Wb  l  rG  or Wbl  W  Wb  rG and the mass moment of inertia is
IO  I  Wbl 2 g . Now, using Eq. (12.101)
2
W l 2 W  Wb 
   Wbl
I b 
grG 


g
g
4 2
 2 
Rearranging this we get
 2
l
I  Wbl 
 
 4 2 g 


2
Q.E.D.
596
12.43 If the weight of the pendulum arm, Wa , is not neglected in Problem 12.42, but is assumed
to be uniformly distributed over the length l, show that the mass moment of inertia of the
wheel can be written as
 2 
Wa  l 
Wa  
I l
W


W


b
2 b
2  g 
3  
 4 
where Wa is the weight of the arm.
Using W for the weight of the wheel and rG for the location of the center of mass of the
assembly,
WrG  Wb  l  rG   Wa  l 2  rG  or
W  Wb  Wa  rG  Wbl  Wa l 2
The total mass moment of inertia is
2
Wbl 2 Wal 2 Wa  l  
Wbl 2 Wal 2
IO  I 



  I
g
g  2  
g
3g
 12 g
Using Eq. (12.101)
W l2 W l2 
W l   
I  b  a   Wbl  a  

g
3g 
2   2 
which can now be rearranged to read
 2 
W  l
W 
I  l  2  Wb  a    Wb  a  
2  g
3 
 4 
2
Q.E.D.
597
12.44 Wheel 2 is a round disk that rotates about a vertical axis z through its center. The wheel
carries a pin B at a distance R from the axis of rotation of the wheel, about which link 3 is
free to rotate. The center of mass G of link 3 is located at a distance r from the vertical
axis through B, and link 3 has a weight W3 and a mass moment of inertia I G about its
own mass center. The wheel rotates at an angular velocity 2 with link 3 fully extended.
Develop an expression for the angular velocity 3 that link 3 would acquire if the wheel
were suddenly stopped.
Consider link 3 alone. The momentum before and after are
L  m3  R  r  2ˆi
L  m3r3ˆi
The angular momentum before and after about point G are
HG   m3 3 r 23kˆ
HG   m3 3 r 22kˆ
The angular momentum before and after about point B are
H B  H G  rˆj× L
H  H  rˆj× L
B
G
  m3 3 r 22kˆ  m3  Rr  r 2  2kˆ
 m3  Rr  4 r 2 3 2kˆ
  m3 3 r 23kˆ  m3r 23kˆ
  4 m3 3 r 23kˆ
Since there is no angular impulse on link 3 about point B, H B  H B
3  1  3R 4r  2
Ans.
598
12.45 Repeat Problem 12.44, but assume that the wheel rotates with link 3 radially inward.
Under these conditions, is there a value for the distance r for which the resulting angular
velocity 3 is zero?
Consider link 3 alone. The momentum before and after are
L  m3  R  r  2ˆi
L  m3r3ˆi
The angular momentum before and after about point G are
HG   m3 3 r 23kˆ
HG   m3 3 r 22kˆ
The angular momentum before and after about point B are
H B  H G 2  rˆj× L
H B  H G1  rˆj× L1
  m3 3 r 22kˆ  m3  Rr  r 2  2kˆ
  m3 3 r 23kˆ  m3r 23kˆ
 m3   Rr  4 r 2 3 2kˆ
  4 m3 3 r 23kˆ
Since there is no angular impulse on link 3 about point B, H B  H B
3  1  3R 4r  2
3  0 for r  3R 4
Ans.
599
12.46 A planetary gear-reduction unit that utilizes 7-pitch spur gears is cut on the 20 fulldepth system. All parts are steel with density 0.282 lb/in 3 . The arm is rectangular and
is 4 in wide by 14 in long with a 4-in diameter central hub and two 3-in diameter
planetary hubs. The segment separating the planet gears is a 0.5-in by 4-in diameter
cylinder. The inertia of the gears can be obtained by treating them as cylinders equal in
diameter to their respective pitch circles. The input to the reducer is driven with 25 hp at
600 rev/min. The mass moment of inertia of the resisting load is 5.83 in  lb  s2 .
Calculate the bearing reactions on the input, output, and planetary shafts. As a designer,
what forces would you use in designing the mounting bolts? Why?
600
N1
104 teeth

 7.429 in
2 P 2  7 teeth/in 
N
52 teeth
R3  3 
 3.714 in
2 P 2  7 teeth/in 
R1 
R2 
N2
35 teeth

 2.500 in
2 P 2  7 teeth/in 
R4 
N4
17 teeth

 1.214 in
2 P 2  7 teeth/in 
 25 HP  33 000 ft  lb/HP/min 12 in/ft   2 626 in  lb
 600 rev/min  2 rad/rev 
Tout  T1A 
Assume a symmetric arrangement of m (typically m = 3) planets, symmetrically arranged
on an m-pronged planet carrier. Then the tangential component of the force F2 A for each
planet is
T
2 626 in  lb
F2tA  out 
 533 m lb
mRA m  4.929 in 
For equilibrium of each planet
 M 2  R3 F43 cos 20  R2 F12 cos 20  0 R3 F43  R2 F12
 F  F cos 20  F cos 20  F  0
t
2
12
43
t
A2
F12  F43  1  R2 R3  F12  FAt 2 cos 20
F12 
R3
Ft
 R2  R3  cos 20 A2
F12  339 m lb
F43   R2 R3  F12
 F  F sin 20  F sin 20  F  0
r
2
12
43
F43  228 m lb
r
A2
FAr2   F12  F43  sin 20
FAr2  38 m lb
F   F 
FA2  534 m lb
FA2 
2
t
A2
r 2
A2
Ans.
Assuming that the m-pronged planet carrier is arranged symmetrically, there is no net
F14  F1 A  0
force on the input or output shafts;
Ans.
The input torque is
Tin  T14  260 in  lb
Ans.
 M 4  mR4 F34 cos 20  T14  0
Balancing the moments on the casing
 M1  mR2 F21 cos 20  16 in F11  0
F11  50 lb
This force F11 must be absorbed by the mounting bolts.
Ans.
601
12.47 It frequently happens in motor-driven machinery that the greatest torque is exerted when
the motor is first turned on, because of the fact that some motors are capable of delivering
more starting torque than running torque. Analyze the bearing reactions of Prob. 12.46
again, but this time use a starting torque equal to 250% of the full-load torque. Assume a
normal-load torque and a speed of zero. How does this starting condition affect the
forces on the mounting bolts?
Note that data are given for m = 2 planets. The masses of the moving elements are:
mA  0.282 lb/in3  4 14 1.5   220.5  2 1.520.5  2 0.6323.5  29.9 lb
m2  0.282 lb/in 3  2.521.38   3.7121.38   220.5   0.7523.25   24.6 lb
m4  0.282 lb/in 3  1.2121.38  1.8 lb
The centroidal mass moments of inertia are:
I A  0.282 lb/in 3[4 14 1.5 142  42  12   240.5 2  21.540.5 2
 2 1.520.5  4.932  2 0.6343.5 2  2 0.6323.5  4.932 ]  532 lb  in 2
I 2  0.282 lb/in 3  2.541.38 2   3.7141.38 2   240.5 2   0.7543.25 2   142 lb  in 2
I 4  0.282 lb/in 3  1.2141.38 2   1.31 lb  in 2
The angular accelerations are found by the tabular method (see Sec. 7.17):
Step
Frame 1 Arm A Planets 2, 3
Sun 4
Gears fixed to arm




Arm fixed

0
 104 35
104 3552 17 
Total
0

  69 35
 6 003 595
The output torque is Tout  2 626 in  lb .
The input torque is Tin  2.5  260 in  lb   650 in  lb .
Balancing the sun gear and input shaft:
 M 4  2R4 F34t  Tin  I44
2 1.214 in  F34t  650 in  lb  1.31 lb  in 2 386 in/s2   6003 595 
F34t  267.6  0.01410
Balancing the arm and output shaft:
 M A  Tout  2RA F2tA  I A A
2 626 in  lb  2  4.928 in  F2tA = 532 lb  in 2 386 in/s 2  
F2tA  266.4  0.1399
602
Balancing a typical planet:
 M 2  R3 F43t  R2 F12t  I 22
 3.714 in  267.6  0.01410    2.500 in  F12t  142 lb  in2 386 in/s 2   69 35 
F12t  397.6  0.3110
 F F  F  F  m A
t
2
t
12
t
43
t
A2
2
G2
 397.6  0.3110    267.6  0.01410    266.4  0.1399    24.6 lb 386 in/s2   4.928 
  512 rad/s2
Now, reassembling the above results,
F34t  267.6  0.01410  260.4 lb
F34r  F34t tan 20  94.8 lb
F12t  397.6  0.3110  238.4 lb
F12r  F12t tan 20  86.8 lb
F2tA  266.4  0.1399  338.0 lb
F2rA  86.8  94.8  8.0 lb
The input and output bearing reactions are zero.
Ans.
 F    F   338.1 lb
Ans.
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11  2R1F21t 16 in  221.4 lb
Ans.
The load on the planet shaft is
F2 A 
2
t
2A
r 2
2A
603
12.48 The gear-reduction unit of Problem 12.46 is running at 600 rev/min when the motor is
suddenly turned off, without changing the resisting-load torque. Solve Problem 12.46 for
this condition.
Here we can use the free-body diagrams from Prob. 12.46 and the mass data and angular
motion relationships from Prob. 12.47. Then, proceeding as in Prob. 12.47, but with Tin =
0, we balance the sun gear and input shaft:
 M 4  2R4 F34t  Tin  I44
 2.428 in  F34t  1.31 lb  in 2 386 in/s2   6003 595
F34t  0.01410
Balancing the arm and output shaft:
 M A  Tout  2RA F2tA  I A A
2 626 in  lb  2  4.928 in  F2tA = 532 lb  in 2 386 in/s 2  
F2tA  266.4  0.1399
Balancing a typical planet:
 M 2  R3 F43t  R2 F12t  I 22
 3.714 in  0.01410    2.500 in  F12t  142 lb  in 2 386 in/s 2   69 35
F12t  0.3110
 F F  F  F  m A
t
2
t
12
t
43
t
A2
2
t
G2
 0.3110    0.01410    266.4  0.1399    24.6 lb 386 in/s 2   4.928 
  342 rad/s2
A  600 rev/min  62.8 rad/s
AGr 2   RAA2  19 457 in/s2
 F   F  F  F  m A   24.6 lb 386 in/s  19 457 in/s   1 240 lb
r
2
r
12
r
43
r
A2
2
r
G2
2
2
Reassembling the above results,
F34t  0.01410  4.8 lb
F34r  F34t tan 20  1.8 lb
F12t  0.3110  106.4 lb
F12r  F12t tan 20  38.7 lb
F2tA  266.4  0.1399  218.6 lb
F2rA  38.7  1.8  1 277 lb
The input and output bearing reactions are zero.
Ans.
 F    F   1 295 lb
Ans.
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11  2R1F21t 16 in  98.8 lb
Ans.
The load on the planet shaft is
F2 A 
t
2A
2
r
2A
2
604
12.49 The differential gear train has gear 1 fixed and is driven by rotating shaft 5 at 500 rev/min
in the direction shown. Gear 2 has fixed bearings constraining it to rotate about the
positive y axis, which remains vertical; this is the output shaft. Gears 3 and 4 have
bearings connecting them to the ends of the carrier arm, which is integral with shaft 5.
The pitch diameters of gears 1 and 5 are both 8.0 in, while the pitch diameters of gears 3
and 4 are both 6.0 in. All gears have the 20 pressure angles and are each 0.75 in thick,
and all are made of steel with density 0.286 lb/in3. The mass of shaft 5 and all
gravitational loads are negligible. The output shaft torque loading is T  100ˆj ft  lb as
shown. Note that the coordinate axes shown rotate with the input shaft 5. Determine the
driving torque required, and the forces and moments in each of the bearings. (Hint: It is
t
t
reasonable to assume through symmetry that F13
. It is also necessary to recognize
 F14
that only compressive loads, not tension, can be transmitted between gear teeth.)
m2  0.286 lb/in 3  42  0.75  10.78 lb
IGyy2  10.78 lb  42 2  86.26 lb  in 2
m3  0.286 lb/in 3  32  0.75  6.06 lb
IGxx3   6.06 lb  32 2  27.29 lb  in 2
IGyy3   6.06 lb  32 4  13.65 lb  in 2
IGzz3  IGyy3  13.65 lb  in 2
m4  m3  6.06 lb
IGxx4  IGxx3  27.29 lb  in 2
IGyy4  IGyy3  13.65 lb  in 2
IGzz4  IGyy4  13.65 lb  in 2
m5  0
ω5  500ˆj rev/min  52.36ˆj rad/s
ω2  2ω5  104.72ˆj rad/s
Noting that the primary xyz axes rotate at an angular velocity of ω5
ω  69.81ˆi  52.36ˆj rad/s
ω  69.81ˆi  52.36ˆj rad/s
3
4
Recognizing that dˆi dt  ω5 × ˆi  52.36kˆ rad/s , the absolute accelerations are α 2  0 ,
α  69.81 52.36kˆ  3 655kˆ rad/s 2 , α  69.81 52.36kˆ  3 655kˆ rad/s 2 , α  0 .
3


4


5
605
Link 5: Note that we assume the mass of link 5 is negligible. Also, assuming no thrust
bearing at the fixed pivot, F15y  F35y  F45y  0 .
F  F  F  F  0
F  F  F  F  0
M  M  d F  0
 M  M  4F  4F  0
M  M  M  M  d F  0
x
5
x
15
x
35
x
45
z
5
z
15
z
35
z
45
x
5
x
15
z
5 15
y
5
y
15
z
35
z
45
z
5
z
15
z
35
z
45
x
5 15
Link 4: Note that the forces on bevel gear teeth are related by Eq. (11.20) where, in this
case, cos  4  0.8 , sin  4  0.6 , and   20 . Noting that
A  ω × ω × 4ˆi in  4 2ˆi in  10 966ˆi in/s 2
G4

5

5
5
 F  0.218F  0.218F  F  m A   6.06 lb 386 in/s  10 966 in/s   172 lb
F F
 F  0.291F  0.291F  0
F  F  F  F  0
x
4
z
14
z
24
y
4
z
14
z
24
z
4
z
14
z
24
x
54
4
x
G4
2
z
24
2
z
14
z
54
Using Eqs. (12.110) for the moment equations,
 M 4x  3F14z  3F24z  0
M  0
 M  0.218F  0.218F  M  I   ( I  I )   258.4 in/s
y
4
z
4
z
14
z
24
z
54
zz
G4
z
4
xx
G4
yy
G4
x
4
y
4
M 54z  258.4 in/s2
2
606
Link 3: Again cos  3  0.8 , sin  3  0.6 , and   20 and using Eqs. (12.110) we get
similar results
AG3  ω5 × ω5 × 4ˆi in  452ˆi in  10 966ˆi in/s 2


 F  0.218F  0.218F  F  m A   6.06 lb 386 in/s 10 966 in/s   172.1 lb
F F
 F  0.291F  0.291F  0
 F  F  F  F  0
 M  3F  3F  0
M  0
 M  0.218F  0.218F  M  I   ( I  I )   258.4 in/s
x
3
z
13
y
3
z
23
z
13
z
3
z
13
x
3
z
13
x
53
3
x
G3
2
z
23
z
23
z
23
2
z
13
z
53
z
23
y
3
z
3
z
13
z
23
z
53
zz
G3
z
3
xx
G3
yy
G3
x
3
y
3
2
M 53z  258.4 in/s 2
Link 2: This time cos  2  0.6 , sin  2  0.8 , and   20 .
 F  F  0.218F  0.218F  0
 F  F  0.291F  0.291F  0
F  F  F  F  0
F 0
M  d F  0
F  F  1 200 in  lb 4 in  300 lb
 M  T  4F  4F  0
 M  4  0.291F   4  0.291F   d F  0
x
2
x
12
y
2
z
32
y
12
z
2
z
12
x
2
z
2 12
y
2
y
12
z
2
z
42
z
32
z
32
z
42
z
42
z
12
z
32
z
42
z
32
z
32
z
42
Reviewing these again shows
F32z  F42z  150 lb
Finally, collecting all results, we have:
F12  87.35ˆj lb
z
42
x
2 12
F12x  0 , F12y  87.35 lb
Ans.
F15  0
T12  1 200ˆj in  lb
M  2 400ˆj in  lb
F35  237.6ˆi  300kˆ lb
F  237.6ˆi  300kˆ lb
M35  258.4kˆ in  lb
M  258.4kˆ in  lb
Ans.
45
15
45
Ans.
Ans.
607
12.50 Arms 2 and 3 of the flyball governor are pivoted to block 6, which remains at the height
shown but is free to rotate around the y axis. Block 7 also rotates about, and is free to
slide along, the y-axis. Links 4 and 5 are pivoted at both ends between the two arms and
block 7. The two balls at the ends of links 2 and 3 weigh 3.5 lb each, and all other
masses are negligible in comparison; gravity acts in the jˆ direction. The spring
between links 6 and 7 has a stiffness of 1.0 lb/in and would be unloaded if block 7 were
at a height of R D  11ˆj in. All moving links rotate about the y-axis with angular
velocities of  ˆj . Make a graph of the height R versus the rotational speed  in
D
rev/min, assuming that the changes in speed are slow.
The free-body diagram below shows only one of the arms, body 2, containing one of the
two flyballs. Force F42 comes from body 4, which is a two-force member, thus defining
its line of action. The force FA comes from the spring. The position of the bottom of the
spring is
RD  16  2  6cos    16  12cos  in
The total force in the spring is
k  RD  RD 0   1 lb/in 16 in  12cos  in  11 in   5  12cos  lb
Since this total force must balance two flyball arms, force FA of the free-body diagram is
FA  2.5  6cos  lb
608
Taking moments about point B,
 6cos in  m2 12sin    2   6sin  in  m2 g   6sin  in  2.5  6cos  lb   0
Dividing by 6sin  in
12cos  m2 2  m2 g  2.5 lb  6cos lb  0
Substituting values and rearranging we get
 0.1088 lb  s2  cos 2  6  6cos  6 1  cos  lb , cos 2  55.155 1  cos  rad2 /s2
  7.427 rad/s
1  cos 
1  cos 
 70.919 rev/min
,
cos 
cos 
RD  16  12cos  in
Ans.
609
12.51 For the mechanism in the posture illustrated, the pin at the center of the wheel is sliding
in the slot in link 2 and the wheel 3 is rolling without slipping on the ground link. The
constant angular velocity of input link 2 is ω2  1.50kˆ rad/s. The angular acceleration of
the wheel is α   62.354 kˆ rad/s2 , and the acceleration of the center of mass of the
3
wheel 3 is AG3  3.1177 ˆi m/s2 . The masses and mass moments of inertia of link 2 and
the wheel are m2  15 kg, m3  25 kg, IG2  0.105 N  m  s2 , and IG3  0.016 N  m  s 2 ,
respectively. Gravity acts in the negative y direction. The force acting at point C is
FC  75ˆi N and there is a torque T12 acting on link 2 about the crankshaft O2. Determine
the torque T12 and the minimum coefficient of friction between the wheel and the ground.
RG3O2  200 mm, RCO2  350 mm, 3  50 mm.
The free-body diagram of link 2 is shown in the following figure.
Ans.
The sum of forces in the x and y directions can be written as
610
F12x  F32x  FC  0
(1)
F12y  F32y  W2  0
The sum of external moments about the mass center of link 2 can be written as
 M G2   R2x F32y  R2y F32x   RCOy 2 FC  T12  IG22  0
(2)
The x and y components of the contact force F32 are related by the equation
F32y  F32x tan 60
x
12
y
12
(3)
x
32
y
32
Therefore, these 4 equations contain 5 unknown variables, namely, F , F , F , F , and
T12.
The free body diagram of link 3 is shown in the following figure.
Ans.
The sum of external forces in the x and y directions can be written as
F13x  F23x  m3 AGx3   25 kg   3.1177 m/s2   77.94 N
(4)
F13y  F23y  W3  0
The sum of external moments about the center of mass of the wheel can be written as
(5)
 M G3  3F13x  IG33  0.016 N  m  s2  62.354 rad/s2   0.998 N  m
These 3 equations contain 2 new unknown variables, namely: the forces F13x and F13y .
Therefore, we now have a total of 7 equations in 7 unknowns.
From Eq. (5) we obtain
F13x   0.998 N  m 50 mm  19.96 N
Then from Eq. (4a) we find
F23x   77.94 N    19.96 N   97.90 N
Next, from Eq. (3) we find
F32y   97.90 N  tan 60  169.57 N
and Eq. (1b) gives us
F12y  15 kg   9.81 m/s2    169.57 N   316.72 N
611
Equation (4b) gives
F13y   25 kg   9.81 m/s2   169.57 N   75.68N
Once the two components of F13 are known, the minimum coefficient of friction between
the wheel and ground can be found as follows
Fx
19.96 N
  13y 
 0.26
Ans.
F13
75.68 N
The torque T12 can be found from Eq. (2)
y
T12  RCO
F   R2x F32y  R2y F32x 
2 C
  350 mm  sin  30  75 N    200 mm  cos 30  169.57 N    200 mm  sin  30  97.90 N  
 26.04 N  m
Ans.
The positive result indicates that the torque T12 is counterclockwise.
612
12.52 For the mechanism in the posture illustrated, RAO2  68 in , the position of the mass center
of block 3 is 4 in below pin A, and the kinematic coefficients of link 3 are
R3  57.72 in/rad and R3  66.68 in/rad 2 (where R 3 is the vertical vector from point C
fixed in link 3 to the x-axis). The angular velocity and acceleration of the input link 2 are
ω  12 kˆ rad / s and α  15 kˆ rad/s2 , respectively. The masses and mass moments of
2
2
inertia of links 2 and 3 are: m2  19.84 lb, m3  15.43 lb, IG2  7 434 lb  in 2 , and
IG3  6 726 lb  in 2 , respectively. The torque acting on link 2 at the crankshaft O2 is
T   450 kˆ in  lb. Gravity is in the negative y direction. Determine the internal
12
reaction forces and the force P acting at point B on link 3. Is link 3 sliding or tipping on
the ground link in this posture?
Block 3 is 24 in by 8 in and pin A is centrally located.
The free-body diagram of link 2 is shown in the following figure.
613
The sum of the external forces acting on link 2 in the x direction can be written as
F12x  F32n cos  2  90   0
F12y  F32n sin 2  90  W2  0
(1)
since the mass center of link 2 is pinned to ground. Also, since friction in the mechanism
can be neglected then only a normal force acts between links 2 and 3. The sum of the
external moments acting about the center of mass of link 2 can be written as
R2 F32n  T12  I G22
(2)
Equations (1) and (2) contain three unknown variables; namely, the internal reaction
forces F12x , F12y , and F32n . Rearranging Equation (2) and substituting known values gives
 7 434 lb  in / 386 in/s 15 rad/s   450 in  lb  10.866 lb
I  T
F  G 2 2 12 
R2
68 in
2
2
2
n
32
Equation (1a) gives
F12x   F32n cos 2  90   10.866 lbcos  30  90   5.433 lb
and Eq. (1b) gives
F12y   F32n sin 2  90   W2  10.866 lbsin  30  90   19.84 lb  10.430 lb
The free-body diagram for link 3 is shown in the following figure.
For the initial iteration assume that link 3 is sliding and not tipping. Then, the sum of the
external forces acting on link 3 in the x and y directions can be written as
F13x  F23n cos  2  90  0
P  F23n sin  2  90   W3  m3 AGy3
(3)
Note that the unknown force P is assumed to act in the positive y direction (that is,
vertically upward).
The sum of the external moments acting about the center of mass G3 can be written as
614
y
 RAG
F23n cos 2  90  R13y F13x  0
3
(4)
y
13
where R is defined positive when pointing in the positive y direction.
Equations (3a), (3b), and (4) contain three new unknown variables; namely, the internal
reaction force F13x , the external force P, and the distance R13y .
Rearranging Equation (3a) gives
F13x   F23n cos 2  90     10.866 lb  cos  30  90   5.433 lb
Ans.
The negative sign indicates that the normal force F13x is acting to the left. Therefore, the
point of contact between links 1 and 3 is on the right side of link 3.
The acceleration of the center of mass G3 can be written as
2
AG3  R32  R322  57.72 in/rad 15 rad/s2   66.68 in/rad 2 12 rad/s   10 467.72 in/s2
Rearranging Equation (3b) gives
P   F23n sin 2  90   W3  m3 AGy3
   10.866 lb  sin  30  90   15.43 lb  15.43 lb/386 in/s2 10 467.72 in/s2  Ans.
 443.28 lb
The positive sign indicates that the force P is acting in the positive y direction.
Rearranging Equation (4), the vertical location of the normal force F13x can be written as
R13y 
y
RAG
F23n cos 2  90
3
F
x
13

 4 in  10.866 lb cos  30  90  4.0 in
  5.433 lb 
The positive sign indicates that the normal force F13x is acting above the center of gravity
of link 3, as shown in the figure. Note that this force is in line with pin A. From Eq. (4),
the force F13x is applied 4.0 in above G3. Since link 3 extends 16 in above G3 the force
F13x is applied within the length of link 3. Therefore, the assumption that link 3 is
slipping on the ground link (that is, link 3 is not tipping) is correct.
Ans.
615
12.53 For the mechanism in the posture illustrated, the constant angular velocity of the input
link 2 is ω2  26 kˆ rad / s, and the accelerations of the centers of mass of links 3 and 5 are
A   8.45 ˆi  14.64 ˆj m/s 2 , and A  7.00 ˆi  26.12 ˆj m/s2 , respectively. The masses
G3
G5
and mass moments of inertia of the links are: m2  m3  m4  m5  7 kg and
The input torque is
IG  IG  IG  IG  0.013 N  m  s2 , respectively.
5
3
4
2
T  0.50 kˆ N  m and gravity is in the negative z-direction. The link lengths, position
12
variables, and first- and second-order kinematic coefficients are:
O2A
mm
25
O2C
mm
40
CD
mm
45
3
AD
mm
30
AB
mm
50
O4B
mm
60
O2O4
mm
75
 4
5
3
 4
θ3
deg
41.2
θ4
deg
114.5
θ5
deg
4.0
5
2
rad/rad
rad/rad rad/rad rad/rad
rad/rad
rad/rad2
-0.425
0.140
-0.188
0.333
0.563
1.810
Determine the internal reaction force between links 2 and 5 at pin C
Kinematics
The first- and second-order kinematic coefficients are given in the table above. In
addition to these, we will need the angular accelerations of links 3, 4, and 5.
2
3  32  322  0.425 rad/rad  0  0.333 rad/rad 2  26 rad/s   225.11 rad/s2
4  42  422  0.140 rad/rad  0  0.563 rad/rad 2  26 rad/s   380.59 rad/s2
2
5  52  522  0.188 rad/rad  0  1.810 rad/rad 2  26 rad/s   1 223.56 rad/s2
2
616
Dynamics
The free-body diagram of link 2 appears as follows
Ans.
Summing forces in the x- and y- directions gives
F12x  F32x  F52x  0
F12y  F32y  F52y  0
and summing moments about G2 gives
 RAOx 2 F32y  RAOy 2 F32x    RCOx 2 F52y  RCOy 2 F52x   T12  IG22  0
(1)
12.500 mmF32y  21.651 mmF32x    10.353 mmF52y  38.637 mmF52x   500 N  mm  0
(2)
The free-body diagram of link 3 appears as follows
Ans.
Summing forces in the x- and y- directions gives
F23x  F43x  cos 3  90 F53  m3 AGx3  7 kg  8.45 m/s2 
F23x  F43x  0.659 F53  59.15 N
F23y  F43y  sin 3  90  F53  m3 AGx3  7 kg  14.64 m/s2 
F23y  F43y  0.752 F53  102.48 N
and summing moments about G3 gives
(3a)
(3b)
617
R F  R F   R F  I 
37.621 mmF  32.934 mmF   30 mmF  0.013 N  m  s2 225.11 rad/s   2.926 N  m
x
y
BA 43
y
43
y
x
BA 43
DA 53
x
32
G3
3
2
(4)
53
The free-body diagram of link 4 appears as follows
Ans.
Summing forces in the x- and y- directions gives
F14x  F34x  0
(5)
F14y  F34y  0
and summing moments about G4 gives
 RBOx 4 F34y  RBOy 4 F34x   T14  IG44
 24.881 mmF  54.598 mmF   T  0.013 N  m  s2 380.59 rad/s   4.948 N  m
y
34
x
34
2
(6)
14
The free-body diagram of link 4 appears as follows
Ans.
Summing forces in the x- and y- directions gives
F25x  cos 3  90  F35  m5 AGx5  7 kg  7.00 m/s 2 
F25x  0.659 F35  49.00 N
F25y  sin 3  90  F35  m5 AGy5  7 kg  26.12 m/s 2 
(7)
F25y  0.752 F35  182.84 N
and summing moments about G5 gives
RDC F35 sin 3  90  5   I G55
 35.844 mm  F35  0.013 N  m  s2 1 223.56 rad/s 2   15.906 N  m
F35  443.75 N
(8)
618
The positive sign for F35 indicates that link 5 contacts link 3 on the BOTTOM surface of
the slot in link 3 as shown in the free body diagram of link 5, immediately above.
Therefore, the magnitude and direction of the vector F35 are
Ans.
F35  443.75 N131.2
Then, from Eqs. (7),
F25x  49.000  0.659 F35  341.431 N
F25y  182.84  0.752 F35  516.540
Combining these, we find
F25  619.184 N  56.54
Ans.
619
12.54 For the mechanism in the posture illustrated, the angular velocity and acceleration of the
input link 2 are ω2  5kˆ rad/s and α 2  3kˆ rad/s2 , respectively. The kinematic coefficients
are
RAO2   10.40 in/rad,
   14.80 in/rad 2 ,
RCA
  14.80 in/rad ,
RCA
xC   14 in/rad,
yC  14 in/rad,
RAO2   18.00 in/rad 2 ,
xC   14 in/rad 2 ,
and
yC   14 in/rad2 . Gravity is acting in the negative y direction. The distances, the free
length and spring rate of the rectilinear spring, the damping constant of the viscous
damper, and the weights and second moments of mass about the mass centers of links 2,
3, and 4 are:
C
RCO2
RAO2 RCD RS0 K
m2 m3 m4
I G2
I G3
I G4
in
20
in
in
17.93 20
in
24
lb/in
0.28
lb·s/in
0.084
lb
1.1
lb
11
lb lb  in  s2 lb  in  s2 lb  in  s2
8.8
27.4
64
55
Determine: (a) the first-order kinematic coefficients of the spring and the damper; (b) the
kinetic energy of the linkage; and (c) the torque T12
Kinematics:
a) A vector loop-closure equation for the rectilinear spring can be written as follows:
R DO2  RCD  RCO2  0
and the equivalent two scalar equations are
xDO2  RS cosS  RCO2 cos 2  0
and
with the solution
 S  180
and
yDO2  RS sin S  RCO2 sin 2  0
RS  xDO2  RCO2 cos2  20 in
Differentiating Eqs. (1) with respect to the input position variable 2 gives
(1)
620
cos  S  RS sin  S   RS    RCO2 sin 2 


 sin 
RS cos  S   S   RCO2 cos 2 
S

The determinant of the Jacobian is
  RS
and Cramer’s rule shows the solution for the first-order kinematic coefficients to be
RS  RCO2 sin  S   2   14.142 in/rad
 S 
RCO2
cos  S   2   0.707 in/rad
(2)
Ans. (3)
RS
b) A vector equation for the viscous damper can be written as follows:
RC  R AO2
and the equivalent scalar equation is
RC  RAO2  17.93 in
and the derivative gives the first-order kinematic coefficient as
RC  RAO2  10.40 in/rad
Ans. (4)
c) The kinetic energy of the mechanism can be written as
1
T  I EQ  22
2
where the equivalent mass moment of inertia of the mechanism can be written as
I EQ  A2  A3  A4
For Link 2:
(5)
A2  m2  xG 2 2  yG 2 2   I G 222
However, since RG2  0 , this becomes
A2  m2  02  02   27.4 lb  in  s2  27.4 lb  in  s2
For Link 3:


A3  m3 xG 3 2  yG 3 2  I G332
Since RG3  RC and 3  0 , this becomes
11 lb 
2
2
2
14 in/rad   14 in/rad    64 lb  in  s2  0  11.17 lb  in  s2
2 

386 in/s
A4  m4  xG 2  yG 2   I G 42
For Link 4:
A3 
4
4
4
Since RG4  RAO2 and 4  0 , this becomes
8.8 lb 
2
2
2
10.40 in/rad    0   55 lb  in  s2  0  2.47 lb  in  s2
2 

386 in/s
Therefore, from Eq. (5),
I EQ  A2  A3  A4  41.04 lb  in  s2
A4 
and the kinetic energy becomes
1
1
2
T  I EQ 22  41.04 lb  in  s2  5 rad/s   513 in  lb
2
2
d) The power equation can be written as
dT dU dW f
T12  ω2  FB  VB 


dt
dt
dt
Ans.
621
or as
4
4
4
j 2
j 2
j 2
T122  FB RB 2   Aj22   B j23   m j gyG j 2  K S  RS  RS 0  RS2  CRC222
(6)
Note in Eq. (6) that the unknown torque T12 is assumed to be acting in the same direction
as the angular velocity of input link 2, that is, in the clockwise direction.
4
We already have
A  A  A  A  I
j
j 2
2
3
4
EQ
 41.04 lb  in  s2
The B terms can be written as
B j  m j ( xG j xG j  yG j yG j )  I G j j j
B2  m2 ( xG 2 xG2  yG 2 yG2 )  I G222
For Link 2:
But, because G2 and O2 are coincident, and 2 is the input variable,
xG 2  yG 2  0, 2  1,
and 2  0
B2  0
Therefore,
For link 3:
B3  m3 xG 3 xG3  yG 3 yG3  I G333


11 lb
 14 in/rad   14 in/rad 2   14 in/rad   14 in/rad 2   64 lb  in  s2  0  0 

386 in/s2 
B3  0
Therefore,
For link 4:
B4  m4 xG 4 xG4  yG 4 yG4  I G444



8.8 lb
 10.40 in/rad    18.00 in/rad 2    0  0    55 lb  in  s2  0  0 
2 

386 in/s
2
Therefore,
B4  4.268 lb  in  s

4
B j  B2  B3  B4  4.268 lb  in  s2

Finally,
j 2
The changes in gravitational potential energy are:
m2 gyG 2  1.1 lb  0  0
For link 2:
m3 gyG 3  11 lb 14 in/rad   154 in  lb/rad
For link 3:
m4 gyG 4  8.8 lb  0  0
For link 4:
4
The summation gives
 m gy  154 in  lb/rad
j 2
j
Gj
Dividing the power equation (6) by the input velocity 2 we find the equation of motion
4
4
4
j 2
j 2
j 2
T12  FB RB   Aj2   B j22   m j gyG j  K S  RS  RS 0  RS  CRC22
Substituting the given and known values, this becomes
622
T12  56 lb  10.40 in/rad    41.04 lb  in  s 2  3 rad/s 2   4.268 lb  in  s2   5 rad/s 
2
154 in  lb/rad  0.280 lb/in  20 in  24 in 14.142 in/rad 
0.084 lb  s/in  10.40 in/rad   5 rad/s 
Therefore, the torque acting on link 2 is
T12  904.95 in  lb
Recall that the torque acting on link 2 was assumed to be in the same direction as the
angular velocity of input link, that is, in the clockwise direction. Therefore, the positive
sign in the result indicates that the torque can be written as
T12  904.95kˆ in  lb
Ans.
2
623
12.55 For the elliptic trammel linkage in the posture illustrated, the velocity and acceleration of
input link 2 are VA2  3ˆj m/s and A A2  7ˆj m/s2 , respectively. The kinematic
coefficients, masses, and second moments of mass about the mass centers of the links are:
 3
rad/m
0.800
 3
R4
2
rad/m
1.109
R4
m2
2
m/m
m/m
1.732 3.200
kg
3.5
m3
kg
6.0
m4
kg
8.0
I G2
I G3
2
kg·m
2.50
kg·m
7.25
I G4
2
kg·m2
1.75
The stiffness and free length of the spring are K  200 N/m and R0  0.5 m , respectively,
and the coefficient of the viscous damper is C  15 N  s/m . The only friction in the
linkage is between links 1 and 4 where the coefficient of friction is   0.35. The normal
force at the point of contact between links 1 and 4 is F n  500ˆj N and gravity is
41
vertically downward. Determine: (a) the first-order kinematic coefficients of the spring
and the damper; (b) the kinetic energy of the linkage; and (c) the external force P acting at
point H..
AG3  G3 B  1.25 m.
624
Kinematics:
The coordinates of the center of mass of link 3 are
and
xG3  x A  RAG3 cos3  0.625 m
yG3  y A  RAG3 sin 3  1.083 m
Differentiating these with respect to the input position RA, the first-order kinematic
coefficients of the center of mass of link 3 are
xG 3  RAG3 sin 33  0.866 m/m
yG3  1.0  RAG3 cos 33  0.500 m/m
Taking the next derivative, the second-order kinematic coefficient of the center of mass
of link 3 are
xG3  RAG3 sin 33  RAG3 cos 332  1.600 m/m2
yG3   RAG3 cos 33  RAG3 sin 332  0
A vector loop equation for the spring can be written as
??
I
R S  R AG3  R A  0
The rectilinear components of this equation are
RS cos  S  RAG3 cos 3  0
RS sin  S  RAG3 sin 3  RA  0
with the solution RS  1.25 m ,  S  60 and 3   AG3  120 .
The derivative of these equations with respect to input position RA gives
RS cos  S  RS sin  S S  RAG3 sin 33  0
RS sin  S  RS cos  S S  RAG3 cos 33  1  0
Writing these equations in matrix format gives
cos  S  RS sin  s   RS   RAG3 sin 33 


 sin 
RS cos  S   S  1  RAG3 cos 33 
s

The determinant of the Jacobian of this set is   RS  1.25 m and, by Cramer’s rule,
RS  sin  S  RAG3 S sin 3   S   0
 S 
cos  S  RAG33 cos 3   S 
 0.800 rad/m
Ans.

Note that the first-order kinematic coefficient of the spring RS is zero because the
velocity of G3 is perpendicular to the line of action of the spring. Also, note that the sign
of  S is positive because the spring rotates counterclockwise for a positive change in the
input position.
The length of the viscous damper can be written as
RC  REO  RDO  1.25 m
Differentiating this with respect to input position RA gives
RC   RD   R4  1.732 m/m
Ans.
Note that the sign is positive because the length of the damper is increasing for a positive
change in the input position.
625
Dynamics:
The equivalent mass of the mechanism is

4

mEQ   m j xG2j  yG2j  I G j  j 2
j 2


 m2 yG22  m3 xG23  yG2  I G332  m4 xG2
3
4
2
2
2
 3.5 kg 1.0 m/m   6.0 kg  0.866 m/m    0.500 m/m  


7.25kg  m 2  0.800 rad/m   8.0 kg  1.732 m/m 
2
2
 38.14 kg
Therefore, the kinetic energy of the mechanism is
1
38.14 kg
2
Ans.
T  mEQ RA2 
 3 m/s  171.63 N  m
2
2
The power equation for the mechanism can be written as
4
4


P  VA   mEQ RA   B j RA2  RA   m j gyG j RA  K  RS  R0  RS RA  CRC2 RA2   F41n Rf RA
j 2
j 2


Consider the left-hand side of this equation. If we write this as P  VA  PVA then we are
implicitly defining that positive P acts in the same direction as RA , that is, in the negative
y direction. Therefore, dividing the power equation by RA gives us the equation of
motion:
4
4
j 2
j 2
P  mEQ RA   B j RA2   m j gyG j  K  RS  R0  RS  CRC2 RA   F41n Rf
The individual terms of the equation of motion are
mEQ RA  38.14 kg  7 m/s2   266.98 N
The Bj coefficient for link j can be written as
B j  m j xG j xG j  yG j yG j  I G j j j

For link 2:
For link 3:
For link 4:

B2  m2 1 m/m 0   0 0  I G2  0 0  0
B3  6.0 kg  0.866 m/m   1.600 m/m 2    0.500 m/m  0  
7.25 kg  m2  0.800 rad/m   1.109 rad/m2   14.746 kg/m
B4  8.0 kg  1.732 m/m   3.200 m/m2    0  0    I G4  0  0 
 44.339 kg/m
4
 B  B  B  B  59.085 kg/m
j 2
4
j
2
3
4
 B R  59.085 kg/m  3 m/s  531.77 N
j 2
j
2
A
2
(1)
626
The rate of change of gravitational potential energy can be written as
dU G
 m2 gyG 2  m3 gyG 3  m4 gyG 4
dRA
 3.5 kg  9.81 m/s2 1.0 m/m+6.0 kg  9.81 m/s2  0.5 m/m+8.0 kg  9.81 m/s2  0.0 m/m
 63.77 N
The rate of change of energy stored in the spring can be written as
dU S
 K  Rs  R0  Rs  200 N/m 1.25 m  0.50 m  0 m/m  0
dRA
The rate of energy dissipated by the viscous damper can be written as
dWC
2
 CRC2 RA  15 N  s/m  1.732 m/m   3 m/s   134.99 N
dRA
The rate of energy dissipated by Coulomb friction can be written as
dW f
  F41n Rf  0.35  500 N  1.732 m/m   303.10 N
dRA
Summing all terms of Eq. (1) gives the equation of motion as
P  266.98 N  531.81 N  63.77 N  134.99 N  303.10 N  424.47 N
Therefore the external force at point H on link 2 is
Ans.
P  424.47 N
Since this answer is positive, the force P is vertically downward per the above
convention.
627
12.56 For the linkage in the posture illustrated, the angular velocity and acceleration of the input
link 2 are ω2   5 kˆ rad/s and α 2  3 kˆ rad/s2 , respectively, and the angular acceleration and
the acceleration of the mass center of link 3 are α  15.75 kˆ rad/s2 and
3
AG
3
 9.60 ˆi  15.20 ˆj in/s2 , respectively. The masses and second moments of mass of
IG2  IG3  45.69 lb  in  s2 , and
IG4  4.56 lb  in  s2 . The torque acting on link 4 is T14   17.7 kˆ in  lb and there is an
the
links
are:
m2  m3  13.2 lb,
m4  2.2 lb,
unknown torque T12 acting on link 2. Gravity is in the negative y-direction. Determine:
(a) the magnitude, direction, and location of the reaction force between links 3 and 4; and
(b) the magnitude and direction of the torque T12 .
O2 A  2.0in, AO4  4.0in, and AG3  2.40in.
Kinematic Analysis
A set of vectors for kinematic analysis of the mechanism are shown in the following
figure:
From this figure, the x and y components of the loop-closure equation are
R2 cos 2  R3 cos 3  R1  0
R2 sin 2  R3 sin 3  0
628
At the posture illustrated 2  90, the given dimensions are R1  3.464 in, R2  2.0 in,
and the solution gives R3  4.0 in, and 3  30.
Derivatives with respect to the input variable 2 give
 R2 sin 2  R3 cos 3  R3 sin 33  0
R2 cos 2  R3 sin 3  R3 cos 33  0
In matrix format these equations appear as
cos 3  R3 sin 3   R3   R2 sin 2  2.0 in 


 sin 
R3 cos 3  3    R2 cos 2   0 
3

(1)
The determinant of the Jacobian is   R3  4.0 in, and Cramer’s rule gives the solution
for the first-order kinematic coefficients as
R3  R2 sin 2  3   1.732 in/rad
3   R2 cos 2  3  R3  0.250 rad/rad
Taking the next derivative of Eqs. (1) with respect to input variable 2 gives
cos 3  R3 sin 3   R3  R2 cos 2  R3 cos 332  2 R3 sin 33   0.216 51 in/rad 2 


 sin 
R3 cos 3  3  R2 sin 2  R3 sin 332  2 R3 cos33   1.125 00 in/rad 2 
3

and Cramer’s rule gives the solution for the second-order kinematic coefficients as
and
3  0.216 51 rad/rad2
R3  0.750 in/rad 2
The angular acceleration of link 3 is
3  3  3 2  322
 0.250 rad/rad  3 rad/s2   0.216 51 rad/rad 2  5 rad/s 
2
 6.163 rad/s2
A set of vectors for the position of the center of mass of link 3 is shown in the figure
The corresponding scalar equations for the center of mass of link 3 are
xG3  R2 cos 2  R33 cos 3  2.078 in
yG3  R2 sin 2  R33 sin 3  0.800 in
629
Differentiating these equations with respect to the input position  2 , the first-order
kinematic coefficients of the mass center of link 3 are
xG 3   R2 sin 2  R33 sin 33  1.700 in/rad
yG3  R2 cos 2  R33 cos 33  0.520 in/rad
Differentiating again with respect to input position  2 , the second-order kinematic
coefficients of the mass center of link 3 are
xG3   R2 cos 2  R33 sin 33  R33 cos 332  0.130 in/rad 2
yG3   R2 sin 2  R33 cos 33  R33 sin 332  1.475 in/rad 2
Therefore, the acceleration of the mass center of link 3 is
A G3  xG 3 2  xG322 ˆi  yG 3 2  yG322 ˆj

 

  1.700 in/rad  3 rad/s   0.130 in/rad  5 rad/s   ˆi


 0.520 in/rad  3 rad/s   1.475 in/rad  5 rad/s   ˆj


2
2
2
2
2
2
 1.850 in/s2ˆi  35.315 in/s2ˆj
Dynamics:
The free-body diagram of link 2 is shown in the following figure:
The governing dynamic equations for link 2 are
 F x  F12x  F32x  0
F  F  F W  0
y
M
G2
y
12
y
32
2
x
y
 RAG
F32y  RAG
F32x  T12  I G22  45.69 lb  in  s2  3 rad/s2   137.07 in lb
2
2
The free-body diagram of link 3 is shown in the following figure:
(2)
630
The governing dynamic equations for link 3 are
 F x  F23x  F43n sin 3  m3 AGx3  13.2 lb 386 in/s2  9.60 in/s2
(3)
F23x  0.500F43n  0.328 lb
 F  F  F cos  W  m A  13.2 lb 386 in/s  15.20 in/s 
y
y
23
n
43
3
3
3
y
G3
2
2
F23y  0.866 F43n  13.20 lb  0.520 lb
(4)
F  0.866 F  12.680 lb
y
23
n
43
 M  R F  R F  R F  I   45.69 lb  in  s 15.75 rad/s   719.62 in lb
G3
x
AG3
y
23
y
AG3
x
23
n
34 43
2
G3
2
3
 2.078F23y  1.200 F23x  R34 F43n  719.62 in lb
Solving Eqs. (3) and (4) for components of F23 yields
and F23y  12.680 lb  0.866F43n
F23x  0.328 lb  0.500F43n
and substituting these into Eq. (5) gives
 R34  2.400 in  F43n  746.36 in  lb
The free-body diagram of link 4 is shown in the following figure:
(5)
(6)
(7)
631
The governing dynamic equations for link 4 are
 F x F14x  F34n sin 30  m4 AGx4  0
F14x  0.500 F34n  0
 F  F  F cos30  W  m A  0
y
M
y
14
n
34
4
4
y
G4
F14y  0.866F34n  2.2 lb  0
G4
R43F43n  17.7 in  lb  I G4 4  4.56lb in s2 15.75 rad/s2   71.82 in lb
R43 F43n  89.52 in lb
with the geometric constraint, that
R34  R43  4.0 in  2.4 in  1.6 in
(8)
Recognizing that F34n  F43n , adding Eqs. (7) and (8) gives
 R34  R43  2.4 in  F43n   4.0 in  F43n  746.36 in lb  89.52 in lb  835.88 in lb
Ans.
F43n  208.97 lb
From this, Eq. (7) shows that R43  0.428 in . Since this is positive and less that 0.600 in,
it confirms that link 3 is slipping within block 4 and not tipping and that the forces F43n
and F34n actually follow as they appear in the last two figures.
Now, Eq. (6a) gives
F23x  0.328 lb  0.500  208.97 lb   104.16 lb
and Eq. (2) gives
T12  137.07 in  lb  (0) F32y   2 in 104.16 lb  345.39 in  lb ccw
In vector format this shows
T12  345.39kˆ in  lb
Ans.
632
12.57 For the mechanism in the posture illustrated, a force F2 = 15 N is applied at point A on
the input link 2 which causes the link to move up the inclined plane with a velocity
VG2  0.5 m/s. The first and second-order kinematic coefficients (where R3  RG2O3 ) are:
3 rad/m
R3 m/m
3 rad/m2
12.4
0.500
180
R3 m/m2
10.7
A linear spring with a free length RS0 = 10 mm, a spring constant K = 2500 N/m, is
attached to link 2 at point B, and has a current length of 30 mm. A viscous damper with a
damping coefficient C = 80 N·s/m is attached to link 2 at point D. The masses and mass
moments of inertia of links 2 and 3 are m2 = 3 kg, m3 = 5 kg, IG2  0.000 035kg  m2 , and
IG3  0.000 070 kg  m2 . Gravity is in the negative y-direction. Determine the first- and
second-order kinematic coefficients of the mass center of link 2, and the first-order
kinematic coefficients of the spring and damper. Write the equation of motion for the
mechanism in symbolic form and then determine the acceleration of link 2.
G3G2 = 70 mm and BG2 = 10 mm
Kinematic Analysis
A set of vectors for analysis of the mechanism are shown in the figure below:
The two scalar equations for loop closure are
 R1  R2 cos30  R3 cos 3  0
R11  R2 sin 30  R3 sin 3  0
633
At the posture shown, the lengths of the vectors are R1 = 95.26 mm, R11 = 15 mm, R2 = 40
mm, and R3 = 70 mm. The variable angle is 3 = 150°.
The vector equation for the center of mass of link 2 can be written as
RG2  R1  R11  R 2
with x and y components of
xG2   R1  R2 cos30  60.62 mm
yG2  R11  R2 sin 30  35.00 mm
The derivative with respect to input variable R2 gives the first-order kinematic
coefficients
xG 2  cos30  0.866 m/m
Ans.
yG2  sin 30  0.500 m/m
Another derivative with respect to input variable R2 gives the second-order kinematic
coefficients
xG2  yG2  0
Ans.
The length of the spring is given by
RS  R2  10 mm
Therefore the first- and second-order kinematic coefficients are
RS  1.000 m/m and RS  0
The length of the damper is given by
RC  constant  R2
Therefore the first- and second-order kinematic coefficients are
RC  1.000 m/m and RC  0
Dynamic Analysis
The power equation for the mechanism can be written as
dT dU dW f
P


dt
dt
dt
The left-hand side of this equation can be written as
P  F2  V2  F2 R2  15 N  R2
Ans.
Ans.
(1)
(2)
where the sign is positive because positive force and positive velocity are in the same
direction.
The equivalent mass for link 2 is
A2  m2 ( xG22  yG22 )  I G222  3 kg(0.8662  0.5002 )  0.000 035 kg  mm2 (0) 2  3 kg
and for link 3
A3  m3 ( xG23  yG23 )  I G332
 5 kg(02  02 )  0.000 070 kg  m2 ( 12.4 rad/m)2  0.0 108 kg
Therefore, the total equivalent mass of the mechanism is
3
A  m
j 2
j
EQ
Th Bj coefficient for link 2 is
 A2  A3  3 kg  0.0108 kg  3.011 kg
634
B2  m2 ( xG 2 xG2  yG 2 yG2 )  I G222
 3 kg 0.866(0)  0.500(0)  0.000 035 kg  m2 (0)(0)  0
and for link 3
B3  m3 ( xG 3 xG3  yG 3 yG3 )  I G333
 5 kg (0)(0)  (0)(0)   0.000 070 kg  m 2 ( 12.4 rad/m)( 180 rad/m2 )
 0.156 kg/m
Therefore, the sum of the Bj coefficients is
3
 B  B  B  0  0.156 kg/m  0.156 kg/m
j 2
j
2
3
The time rate of change of the kinetic energy can be written as
3
3
dT
  Aj R2 R2   B j R23  3.011 kgR2 R2  0.156 kg/mR23
dt j 2
j 2
The time rate of change of the potential energy due to gravity is
3
dU
  m j gyG j R2  m2 gyG 2 R2  m3 gyG 3 R2
dt
j 2
 3 kg  9.81 m/s   0.500 m/m  R2  5 kg  9.81 m/s   0  R2  14.715 N  R2
2
(3)
(4)
2
The time rate of change of the potential energy in the spring is
dU S
(5)
 K ( RS  RS 0 ) RS  R2  2500 N/m  30 mm  10 mm 1.0 m/mR2  50 N  R2
dt
The time rate of change of the energy dissipated by the damper is
dW f
2
(6)
 CRC 2 R22  80 N  s/m  1.0 m/m  R22  80 N  s/m R22
dt
Substituting Equations (2)-(6) into Equation (1), the power equation can be written as
15 N  R2  3.011 kgR2 R2  0.156 kg/mR23  14.715 N  R2  50 N  R2  80 N  s/m R22
Dividing by the input velocity, the equation of motion can be written as
15 N  3.011 kgR2  0.156 kg/mR22  14.715 N  50 N  80 N  s/m R2
Substituting the known input velocity, this becomes
15 N  3.011 kgR2  0.156 kg/m  0.5 m/s   14.715 N  50 N  80 N  s/m  0.5 m/s 
2
15 N  3.011 kgR2  0.039 N  14.715 N  50 N  40 N
3.011 kgR2  89.754 N  0
Therefore, the input acceleration is
R2  29.809 m/s2
Ans.
635
12.58 For the mechanism in the posture illustrated, the constant angular velocity of input link 2
is ω2  3kˆ rad/s . The free length of the linear spring attached between ground pin O1 and
pin A is 20 in and a viscous damper is attached between pin O and link 3. The torque
acting on link 2 is T12  13.3kˆ in  lb , there is a horizontal force P acting at pin A, and
gravity is in the negative y-direction. The known data (where R2 is the vector from
bearing O2 to the mass center G3 and R3 is the vector from ground pin O to the mass
center G3) are:
C
I G2
I G3
K
R2
m3
R2
R3
m2
R3
lb/in lb·s/in
in/rad in/rad in/rad2 in/rad2
lb
lb
lb·in2
lb·in2
0.0827
0.413
2.080 4.160 6.000 4.800 15.44 26.46 2 646
7 938
Determine: (a) the first-order kinematic coefficients of the spring and the damper; (b) the
potential energy of the spring; (c) the equivalent mass moment of inertia; and (d) the
force P .
RAO2  4.8 in, RG3O2  3.6 in, and RG2O2  1.8 in.
Kinematic Analysis:
The coordinates of the center of mass of link 2 can be written as
xG2 
RG2O2 cos 2 
1.8 in  cos  60  0.900 in
yG2  yO2O  RG2O2 sin 2  3.118 in  1.8 in  sin  60  1.559 in
Differentiating these equations with respect to the input position θ2 gives
636
xG 2   RG2O2 sin 2   1.8 in  sin  60  1.559 in/rad
yG 2  RG2O2 cos 2  1.8 in  cos 60  0.900 in/rad
Then differentiating these equations again with respect to the input θ2 gives
xG2   RG2O2 cos 2   1.8 in  cos 60  0.900 in/rad 2
yG2   RG2O2 sin 2   1.8 in  sin  60  1.559 in/rad 2
The coordinates of the center of mass of link 3, G 3 , can be written as
xG3  x3  1.800 in and
yG3  y3  0
Differentiating these equations with respect to the input position θ2 gives
xG 3  x3  4.160 in/rad and
yG 3  y3  0
Then differentiating these equations again with respect to the input θ2 gives
xG3  x3  4.800 in/rad2 and yG3  y3  0
A vector loop equation for the spring can be written as
??
I
R S  R AO2  RO2O  RO1O  0
The x and y component equations can be written as
RS cos  S  RAO2 cos  2  0
RS sin  S  RAO2 sin 2  yO2O  yO1O  0
which, for the current posture, has the solution RS = 2.400 in and S = 0°.
Differentiating the above equations with respect to the input θ2 gives
RS cos  S  RS sin  S S  RAO2 sin 2  0
RS sin  S  RS cos  S S  RAO2 cos 2  0
In matrix format, these become
cos  S  RS sin  S   RS    RAO2 sin 2 


 sin 
RS cos  S   S   RAO2 cos 2 
S

The determinant of the Jacobian matrix is   RS  2.400 in. Using Cramer’s rule, the
solution for the first-order kinematic coefficients of the spring are
RS  RAO2 sin  S   2    4.800 in  sin  0  60   4.157 in/rad
 RAO2 
 4.800 in 
 cos  S   2   
 cos  0  60   1.000 rad/rad
R
2.400
in


 S 
A vector loop equation for the viscous damper can be written as
 S  
?
I
R C  R 2  RO2O  0
The x and y component equations can be written as
RC cos C  R2 cos 2  0
RC sin C  R2 sin 2  yO2O  0
which, for the current posture, has the solution RC = 1.800 in and R2 = 3.600 in.
Differentiating the above equations with respect to the input θ2 gives
Ans.
637
RC cos C  R2 cos 2  R2 sin 2  0
RC sin C  R2 sin  2  R2 cos 2  0
In matrix format, these become
cos C  cos 2   RC    R2 sin 2 

 sin 
 sin 2   R2   R2 cos 2 
C

The determinant of the Jacobian matrix is   sin C  2   0.866.
Using Cramer’s
rule, the solution for the first-order kinematic coefficients of the spring are
RC  R2 sin C  2    3.600 in  sin  0  60   4.157 in/rad
R2  R2 cos C  2  sin C   2 
Ans.
  3.600 in  cos  0  60  sin  0  60   2.078 in/rad
Dynamic Analysis:
The potential energy stored in the spring can be written as
2
2
U S  ½ K  RS  RS 0   ½  0.0827 lb/in  2.400 in  20.000 in   12.809 in  lb
Differentiating with respect to the input position θ2 gives the time rate of change of the
potential energy in the spring, that is
dU S
 K  RS  RS 0  RS2
dt
(1)
 0.0827 lb/in  2.400 in  20.000 in  4.157 in/rad  2  6.051 in  lb/rad2
The equivalent mass moment inertia of the mechanism can be written as

3



I EQ   Aj  m2 xG22  yG22  I G2 22  m3 xG23  yG23  I G332
j 2
2
2
2
 15.44 lb 1.559 in/rad    0.900 in/rad    2 646 lb  in 2 1.000 rad/rad 


26.46 lb  4.160 in/rad    0    7 938 lb  in 2  0 


2
 3 154 lb  in
The Bj coefficients for the mechanism can be written as
2
2
2
 B  m  x x  y y   I     m  x x  y y   I   
3
j 2
j
2
G2 G2
G2
G2
G2 2 2
3
G3 G3
G3
G3
G3 3 3
 15.44 lb 1.559 in/rad   0.900 in/rad 2    0.900 in/rad  1.559 in/rad 2 
2 646 lb  in 2 1.000 rad/rad  0 
26.46 lb  4.160 in/rad   4.800 in/rad 2    0  0    7 938 lb  in 2  0  0 
 528.35 lb  in 2 / rad 3
The time rate of change of kinetic energy is
Ans.
638
3

dT 
  I EQ 2   B j22  2
dt 
j 2

 3 154 lb  in 2
528.35 lb  in 2
2

 0 
 3 rad/s   2  12.33 in  lb2
2
2
386 in/s
 386 in/s

The time rate of change of gravitational potential energy is
dUG
 m2 gyG 22  m3 gyG 32  15.44 lb  0.900 in/rad   26.46 lb 0  2  13.90 in  lb2
dt
The energy dissipated by the viscous damper can be written as
dW f
2
 CRC222  0.413 lb  s/in  4.157 in/rad   3 rad/s  2  21.41 in  lb  s/rad2
dt
The power input to the mechanism by the external force P can be written as
P  VA  PxA2   PRAO2 sin 22   4.157 in/rad  P2
(2)
(3)
(4)
(5)
Similarly, the power dissipated by the load torque T12 can be written as
T12  ω2  13.3 in  lb2
(6)
The power equation for the mechanism can be written as
3
3


P  VA  T12  ω2   I EQ2   B j22  2   m j gyG j  K  RS  RS 0  RS2  CRC222
j 2
j 2


Substituting the terms from Eqs. (1)-(6) and dividing by the input velocity, the equation
of motion for the mechanism can be written as
 4.157 in  P  13.3 in  lb  12.33 in  lb  13.90 in  lb  6.051 in  lb  21.41 in  lb
P  13.204 lb
Therefore, the applied external force vector is
P  13.204ˆi lb
Ans.
639
12.59 For the linkage in the posture illustrated, the kinematic coefficients of link 3 are
3  0.25 rad/rad and 3  0.60 rad/rad 2 . The angular velocity and acceleration of the
input link 2 are ω2   5 kˆ rad/s and α 2  3 kˆ rad/s2 , respectively. The free length and
stiffness of the horizontal spring are RS 0  30 mm and K  2 000 N/m, respectively.
The damping coefficient of the viscous damper is C  150 N  s/m. The masses and mass
moments of inertia of the links are m2  m3  6 kg, m4  1 kg, IG2  IG3  5 kg  m2 , and
I  4 kg  m2 . The input torque T   10 kˆ N  m and there is a horizontal force F
G4
12
acting at the mass center of link 3. Gravity is in the negative y-direction. Determine: (a)
the first-order kinematic coefficients of the horizontal spring and the viscous damper; (b)
the kinetic energy of this linkage; and (c) the force F
O2 A  50mm, AOS  52 mm, AO4  80 mm, AD  100mm, and AG3  60 mm.
Kinematic Analysis:
A set of vectors for the kinematic analysis of the spring are shown in the following figure:
The vector loop-closure equation and the corresponding scalar component equations are:
??
I
R1S  R S  R 2  0
640
R1S cos 1S  RS cos  S  R2 cos 2  0
R1S sin 1S  RS sin  S  R2 sin 2  0
with the corresonding solution RS  52 mm and  S  180.
Taking derivatives with respect to input angle 2 gives the pair of equations
cos  S  RS sin  S   RS    R2 sin  2 

 sin 
RS cos  S   S   R2 cos 2 
S

The determinant of the Jacobian is   RS  52 mm and Cramer’s rule gives the firstorder kinematic coefficients for the spring as RS  R2  50 mm/rad and  S  0.
Ans.
A set of vectors for the kinematic analysis of the spring are shown in the following figure:
The vector loop-closure equation and the corresponding scalar component equations are:
??
R 2  R3  RC  0
R2 cos 2  R3 cos 3  RC cos C  0
R2 sin 2  R3 sin 3  RC sin C  0
with the corresonding solution RC  86.60 mm and C  0.
Taking derivatives with respect to input angle 2 gives the pair of equations
  cos C RC sin C   RC   R2 sin 2  R3 sin 33   37.5 mm/rad 
  sin 
      R cos  R cos     21.65 mm/rad 

R
cos



C
C
C C 
2
3
3 3

 2
The determinant of the Jacobian is   RC  86.60 mm and Cramer’s rule gives the firstorder kinematic coefficients for the spring as RC  37.5 mm/rad and C  0.25 rad/rad. Ans.
Using the same vector symbols of the previous figure, the position of the mass center of
link 3 is given by the following scalar equations:
xG3  R2 cos 2  RG3 A cos 3  51.96 mm
yG3  R2 sin 2  RG3 A sin 3  20.00 mm
The derivative with respect to input angle 2 gives the first-order kinematic coefficients:
xG 3   R2 sin 2  RG3 A sin 33  42.50 mm/rad
yG 3  R2 cos 2  RG3 A cos 33  12.99 mm/rad
641
Another derivative with respect to input angle 2 gives the second-order kinematic
coefficients:
xG3   R2 cos 2  RG3 A cos 332  RG3 A sin 33  14.752 mm/rad 2
yG3   R2 sin 2  RG3 A sin 332  RG3 A cos 33  16.948 mm/rad 2
Dynamic Analysis:
The equivalent mass coefficients of the links are
2
A2  m2 xG22  yG22  I G222  6 kg  0  0  5 kg  m2 1 rad/rad   5 kg  m2



A3  m3 x  y 
2
G3
2
G3
  I 
2
G3 3
 6 kg  0.042 50 m/rad    0.012 99 m/rad    5 kg  m 2 0.25 rad/rad 


2
 0.324 kg  m
2
2
2


B  m  x x  y y   I     6 kg  0  0  0  0  5 kg  m 1 rad/rad  0  0
B  m  x x  y  y    I   
 6 kg  0.042 50 m/rad   0.014 752 m/rad    0.012 99 m/rad   0.016 948 m/rad 
5 kg  m  0.25 rad/rad   0.60 rad/rad 
A4  m2 xG24  yG24  I G442  1 kg  0  0  4 kg  m2  0.25 rad/rad   0.25 kg  m2
2
2
2
2
G2 G2
3
3
G3 G3
G2
G3
G2
G3
G2 2 2
G3 3 3
2
2
2
2
 0.744 92 kg  m 2


B4  m4 xG 4 xG4  yG 4 yG4  I G444
 1 kg  0  0  0  0   4 kg  m2  0.25 rad/rad   0.60 rad/rad 2   0.60 kg  m2
 B  0  0.744 92 kg  m   0.60 kg  m  1.344 92 kg  m
4
2
j 2
2
2
j
The equivalent mass moment of inertia of the mechanism is
4
I EQ   Aj  5 kg  m2 0.324 kg  m2  0.25 kg  m2  5.574 kg  m2
j 2
The kinetic energy of the mechanism is
2
T  ½ I EQ22  ½  5.574 kg  m2   5 rad/s   69.675 N  m
Ans.
The power equaion for the mechanism can be written as
dT dU dW f
T12  ω2  F  VG3 


dt
dt
dt
4
4
4
j 2
j 2
j 2
T122  FxG 32   Aj22   B j23   m j gyG j 2  K  RS  RS 0  RS2  CRC222
Dividing by the input angular velocity w2 gives the equation of motion of the mechanism,
that is
4
4
4
j 2
j 2
j 2
T12  FxG 3   Aj2   B j22   m j gyG j  K  RS  RS 0  RS  CRC22
642
Substituting the known data, this becomes
2
10 N  m  F  0.042 50 m/rad   5.574 kg  m 2  3 rad/s 2   1.344 92 kg  m 2  5 rad/s 
 6 kg  0   6 kg  0.012 99 m/rad   1 kg  0    9.81 m/s 2 
2 000 N/m  0.052 m  0.030 m  0.050 m/rad 
150 N  s/m  0.0375 m/rad   5 rad/s 
2
Finally, this reduces to
F
62.254 90 N  m
 1 465 N
0.042 50 m/rad
or
F  1 465ˆi N
Ans.
643
12.60 For the mechanism in the posture illustrated, the velocity and acceleration of the input
link 2 down the slope EA are VA  8 in/s and AA  24 in/s2 , respectively. The wheel,
link 4, is rolling without slipping on the ground link. The free length of the linear spring
is RS0 = 8 in and the damping coefficient of the viscous damper is C = 1.4 lb·s/in. Gravity
acts vertically downward. The dimensions, first- and second-order kinematic coefficients
(R44 is the vector from point O to pin B), masses, and second moments of mass about the
mass centers of links 2, 3, and 4 are:
m2 m3 m4 I G2


I G3
I G4
 3
 3
R44
R44
EA AB BD 4
in in in in rad/in rad/in2 in/in in/in2 lb lb lb lb·in2 lb·in2 lb·in2
211 158
10 24 14 6 0.0723 0.0090 2.0 0.25 1.1 4.4 8.8 12
Determine: (a) the first-order kinematic coefficients of the linear spring and the damper;
(b) the kinetic energy of the mechanism; and (c) the stiffness of the spring.
Note that R44 is the vector from point O to pin B. Gravity acts vertically downward.
There are no external forces or torques, and the effects of friction can be neglected.
Kinematic Analysis:
The vectors for a kinematic analysis are shown in the following figure.
644
The loop-closure equation is
I
?
?
R1  R 2  R 3  R 44  0
with the posture solution of 3 = 150° and R44 = 17 in.
In addition, there is the rolling contact constraint equation
R44  4 4
Two derivatives of this constraint give the first- and second-order kinematic coefficients
R
2 in/in
 4  44 
 0.333 rad/in
4
6 in

R44
0.25 in/in 2
 4 

 0.0417 rad/in 2
4
6 in
The coordinates of the center of mass of link 2 are
xG2  R1  R2 cos150  20.785 in
yG2 
R2 sin150  5.000 in
Two derivatives with respect to R2 give the first- and second-order kinematic coefficients
xG 2  cos150  0.866 in/in
yG 2  sin150  0.500 in/in
xG2  yG2  0
The coordinates of the centers of mass of links 3 and 4 are
xG3  xG4  0
yG3  yG4  R44  17.000 in
Two derivatives with respect to R2 give the first- and second-order kinematic coefficients
xG 3  xG 4  0
  2.0 in/in
yG 3  yG 4  R44
xG3  xG4  0
  0.25 in/in
yG3  yG4  R44
The length of the spring can be written as RS  R2  10 in. Two derivatives with respect
to input R2 give the first- and second-order kinematic coefficients
RS  1 in/in and RS  0.
Ans.
The length of the viscous damper can be written as RC  OD  R44  14 in. Two
derivatives with respect to input R2 give the first- and second-order kinematic coefficients
  2.0 in/in and RC   R44
  0.25 in/in 2 .
RC   R44
Ans.
Dynamic Analysis:
The equivalent mass values are
A2  m2 xG22  yG22  I G2 22


1.1 lb 
12 lb  in 2
2
2
2


0.866 in/in    0.500 in/in  
0  0.002 85 lb  s 2 /in
2 
2  

386 in/s
386 in/s
645


A3  m3 xG23  yG23  I G332
4.4 lb  2
211 lb  in 2
2
2


0   2.000 in/in  
0.0723 rad/in   0.048 45 lb  s 2 /in
2  
2 

386 in/s
386 in/s
2
2
2
A4  m4 xG 4  yG 4  I G4 4


8.8 lb  2
158 lb  in 2
2
2


0   2.000 in/in  
0.333 rad/in   0.136 58 lb  s2 /in
2  
2 

386 in/s
386 in/s
4
mEQ   Aj  0.187 88 lb  s2 /in

j 2

B2  m2 xG 2 xG2  yG 2 yG2  I G222

1.1 lb
12 lb  in 2

0.866
in/in
0

0.500
in/in
0




  
  
 0  0   0
386 in/s2 
386 in/s2
B3  m3 xG 3 xG3  yG 3 yG3  I G333


4.4 lb
 0  0    2.000 in/in   0.250 in/in 2 

386 in/s2 
2
211 lb  in

0.0723 rad/in   0.0090 rad/in 2 
2 
386 in/s
 0.006 06 lb  s2 /in
B4  m4 xG 4 xG4  yG 4 yG4  I G4 4 4



8.8 lb
 0 0   2.000 in/in   0.250 in/in 2 
2    

386 in/s
2
158 lb  in

0.333 rad/in   0.0417 rad/in 2 
2 
386 in/s
 0.017 08 lb  s2 /in

4
 B  0.023 14 lb  s /in
2
j 2
2
j
The kinetic energy of the mechanism can be written as
2
T  ½mEQ R22  ½  0.187 88 lb  s2 /in   8 in/s   6.012 in  lb
The rate of change of gravitational potential energy is
dU G
 m2 gyG 2 R2  m3 gyG 3 R2  m4 gyG 4 R2
dt
 1.1 lb  0.500 in/in  R2  4.4 lb  2.000 in/in  R2  4.4 lb  2.000 in/in  R2
  26.95 in  lb/in  R2
The rate of change of potential energy stored in the spring is
dUS
 K  RS  RS0  RS R2  K 10 in  8 in 1 in/in  R2  2 inKR2
dt
Ans.
646
The rate of energy dissipated by the viscous damper is
dW f
2
 CRC2 R22  1.4 lb  s/in  2.000 in/in  R22   5.600 lb  s/in  R22
dt
The power equation for the mechanism can be written as
4


F  V  T  ω   mEQ R2   B j R22  R2  26.950 in  lb/inR2  2.000 inKR2  5.600 lb  s/inR22
j 2


Recognizing that there are no applied external forces or torques, and dividing by the input
velocity R2 gives the equation of motion for the mechanism.
4


0   mEQ R2   B j R22   26.950 in  lb/in  2.000 inK  5.600 lb  s/inR2
j 2


0  0.187 88 lb  s2 /in  24.0 in/s2   0.023 14 lb  s2 /in 2  8.0 in/s 
2
26.950 in  lb/in  2.000 inK  5.600 lb  s/in  8.0 in/s 
0  4.509 lb  1.481 lb  26.950 lb  2.000 inK  44.800 lb
0  2.000 inK  23.840 lb
K  11.92 lb/in
Ans.
647
Chapter 13
Vibration Analysis
13.1
Derive the differential equation of motion for each system and write the formula for the
natural frequency n for each system.
(a)
 F  k  x  y   mx
(b)
 F  F  cx  kx  mx
(c)
 F  cx  k1x  k2 x  mx
mx  kx  ky
n  k m
Ans.
mx  cx  kx  F
n  k m
mx  cx   k1  k2  x  0
n 
 k1  k2  m
Ans.
Ans.
648
(d)
 F  k x  k  x  y   mx
(e)
 F  c  x  y   kx  mx
(f)
1
2
mx   k1  k2  x  k2 y
n 
 k1  k2  m
Ans.
mx  cx  kx  cy
Ans.
n  k m
Both springs 3 and 4 experience the same spring force F34, and each is deflected
by an amount consistent with its own rate, F34/k3 or F34/k4, respectively. The total
deflection is x   F34 k3    F34 k4  or F34   k3k4 x   k3  k4 
k3k4
 F  k x  k x  k  k x  mx
1

kk 
mx   k1  k2  3 4  x  0
k3  k 4 

2
3
4
n 
k1  k2 
m
k3 k 4
k3  k 4
Ans.
649
13.2
Evaluate the constants of integration of the solution to the differential equation for an
undamped free system, using the following sets of starting conditions:
(a)
x  x0 , x  0
(b)
x  0, x  v0
(c)
x  x0 , x  a0
(d)
x  x0 , x  b0
For each case, transform the solution to a form containing a single trigonometric term.
For each case we use a trial solution of:
x  A sin nt  B cos nt
with initial value of:
x(0)  B
(1)
x  n A cos nt  n B sin nt
x  0   n A
(2)
x  n2 A sin nt  n2 B cos nt
x  0   n2 B
(3)
x   A cos nt   B sin nt
x  0    A
(4)
3
n
(a)
3
n
3
n
x  x0 , x  0 . Use Eqs. (1) and (2); A  0, B  x0
x  x0 cos nt
(b)
Ans.
x  0, x  v0 . Use Eqs. (1) and (2); A  v0 n , B  0
x   v0 n  sin nt
(c)
Ans.
x  x0 , x  a0 . Use Eqs. (1) and (3); B  x0 , B   a0 n2
These are inconsistent unless a0  n2 x0 . Second given condition is not useful.
One more initial condition required, such as x  0   v0 from which A  v0 n .
x   v0 n  sin nt  x0 cos nt
x  x02   v0 n  sin nt   where   tan 1  x0n v0 
2
(d)
Ans.
x  x0 , x  b0 . Use Eqs. (1) and (4); B  x0 , A   b0 n3
x   b0 n3  sin nt  x0 cos nt
x  x02   b0 n3  sin nt   where   tan 1  x0n3 b0 
2
Ans.
650
13.3
A system like Figure 13.5 has m = 1 kg and an equation of motion
x  20cos 8 t   / 4  mm. Determine: (a) spring constant k; (b) static deflection  st ;
(c) period; (d) frequency in hertz; and (e) velocity, acceleration, and spring force at t =
0.20 s. Plot a phase diagram to scale showing the displacement, velocity, acceleration,
and spring-force phasors at this instant.
(a)
n  k m  8 rad/s
k  n2 m  8 rad/s  1 kg   631.65 N/m
2
Ans.
(c)
2
F mg 1 kg   9.81 m/s 
 st  

 0.015 53 m  15.53 mm
k
k
631.65 N/m
  2 n   2 rad/rev  8 rad/s   0.250 s/rev
(d)
f  1   4 rev/s  4 Hz
(e)
  8 t   4  8t  0.25   8  0.20   0.25   1.35 rad  243
(b)
Ans.
Ans.
Ans.
x   8 rad/s  0.020 m  sin 8 t   4   0.503sin  243 m/s  0.448 m/s Ans.
x   8 rad/s   0.020 m  cos 8 t   4   12.633cos  243 m/s 2  5.735 m/s2 Ans.
2
(f)
F  kx   631.65 N/m  0.020 m  cos 243  12.633cos 243 N  5.735 N
x  0.020cos 243 m , x  0.503sin 243 m/s , x  12.633cos 243 m/s
F  12.633cos 243 N
2
Ans.
651
13.4
The weight W1 drops through the distance h and collides with W2 with plastic impact (a
coefficient of restitution of zero). Derive the differential equation of motion of the
system, and determine the amplitude of the resulting motion of W2 .
Define t   0 at the instant of impact. At the beginning of impact we have v1  2 gh .
By conservation of momentum, m1v1   m1  m2  v2 . Thus W1 g  2 gh  W1  W2  v2 g
or v2  2 ghW1 W1  W2  . Therefore, at t   0 , x  0 , x  v2 .
Note that, at x  0 , the spring force includes a reaction to W2 .
And so,
 F  kx  W  W g  x where, for convenience, we have defined W  W  W . From
1
1
2
the force balance we get the differential equation of motion
W g  x  kx  W1
Ans.
with natural frequency of n  kg W . Then
x  A cos nt   B sin nt   W1 k
x   An sin nt   Bn cos nt 
and with the initial conditions stated above A  W1 k and B  v2 n . Therefore
x   W1 k  cos nt    v2 n  sin nt   W1 k 
Transforming to a single transient term we get
x  X sin nt     W1 k where
X
W1 k    v2 n 
2
W k 
and   tan 1  1 
 v2  n 
2
Ans.
652
13.5
The vibrating system has k1  k3  875 N/m , k2  1 750 N/m , and W  40 N . What is
the natural frequency in hertz?
Springs 1 and 2 both experience the same spring force F12, and each is deflected by an
amount consistent with its own rate, F12/k1 or F12/k2, respectively. The total deflection is
x   F12 k1    F12 k2  or F12   k1k2 x   k1  k2 
 F  F  k x  mx
12
3
 kk

mx   1 2  k3  x  0
 k1  k2

The natural frequency is
 k1k2

 k3 

k k

n   1 2
W g
875 N/m 1750 N/m   875 N/m
875 N/m  1750 N/m
40 N 9.81 m/s 2
 18.91 rad/s  3.010 Hz
Ans.
653
13.6
Weight W = 15 lb is connected to a pivoted rod which is assumed to be weightless but
rigid. A spring having a rate of k = 60 lb/in is connected to the center of the rod and
holds the system in static equilibrium at the position shown. Assuming that the rod can
vibrate with a small amplitude, determine the period of the motion.
 12 in
 M  a   ka   W  W g  
2
n 

2
n
ka 2
a kg 6 in


2
W g
W 12 in

W
2
g    ka 2   W
 60 lb/in   386 in/s2 
2 rad/rev
 0.320 s/rev
19.65 rad/s
15 lb
 19.65 rad/s
Ans.
654
13.7
The upside-down pendulum of length l is retained by two springs connected a distance a
from the pivot. The springs have been positioned such that the pendulum is in static
equilibrium when it is in the vertical posture. (a) For small amplitudes, find the natural
frequency of this system. (b) Find the ratio l/a at which the system becomes unstable.
(a)  M  W    a  k1a  a  k2a  W g  2
W
2
g    k1  k2  a 2  W   0
2

g   k1  k2  a
 
 1   0
W


n 
2

g   k1  k2  a
 1

W


Ans.
(b) The system becomes unstable whenever the natural frequency becomes the square
root of a negative number (imaginary). At such values the system is not oscillatory.
This happens whenever
a   k1  k2  a W
Ans.
655
13.8
Write the differential equation for the system and find the natural frequency. Find the
response x if y is a step input of height y0 . Find the relative response z  x  y to this
step input.
 F  k x  k  x  y   mx
1
2
mx   k1  k2  x  k2 y
n 
 k1  k2  m
Ans.
The complementary solution is
x  A cos nt  B sin nt
For a particular solution, arrange the equation to the form
x  n2 x   k2 m  y
Since, for a step input the right-hand side is constant, try x  C , x  0 .
n2C   k2 m  y
C  k2 y  k1  k2 
The complete solution is x  x  x
x  A cos nt  B sin nt  k2 y  k1  k2 
At t  0 , x  0 and y  y0 .
0  A 1  B  0   k2 y0  k1  k2 
x  n A sin nt  n B cos nt
At t  0 , x  0
0  n A  0   n B 1
A   k2 y0  k1  k2 
B0
Therefore, the complete response is
x   k2 y0  k1  k2  cos nt  k2 y0  k1  k2 
x
k2 y0
1  cos nt 
k1  k2
k  k  y
k2 y0
1  cos nt   1 2 0
k1  k2
k1  k2
k y  k y cos nt
z 1 0 2 0
k1  k2
Ans.
z  x y 
Ans.
656
13.9
An undamped vibrating system consists of a spring whose scale is 35 kN/m and a mass of
1.2 kg. A step force F = 50 N is exerted on the mass for 0.040 s. (a) Write the equations
of motion of the system for the era in which the force acts and for the era that follows.
(b) What are the amplitudes in each era? (c) Sketch a time plot of the displacement.
(a)
n 
 35 000 N/m  1.2 kg   171 rad/s
1.2 kg  x  35 000 N/m  x  50 N
First era: 0  t  0.040 s :
From Eq. (13.21)
x   F k 1  cos nt    50 N 35 000 N/m 1  cos171t 
x   0.001 429 m 1  cos171t 
x  1.429 mm 1  cos171t 
Ans.
Also x   0.244 m/s  sin171t   244 mm/s  sin171t
At the end of the first era t  0.040 s , nt  6.831 rad  391.4
x  0.000 209 m  0.209 mm ,
x  0.127 m/s  127 mm/s
Second era: t  0.040 s :
From Eqs. (13.16) and (13.17)
X 0  x02   v0 n  
2
1.2 x  35 000x  0
 0.209 mm 2  127 mm/s 171 rad/s 2  0.773 mm
  tan 1  v0 n x0   tan 1 127 mm/s  171 rad/s  0.209 mm   74.3
(b)
(c)
x  X 0 cos nt      0.773 mm  cos 171t  74.3
Ans.
First era; 0  t  0.040 s :
Second era; t  0.040 s :
Ans.
Ans.
X  1.429 mm
X  0.773 mm
657
13.10
A round shaft whose torsional spring constant is kt in  lb/rad connecting two wheels
having mass moments of inertia I1 and I 2 . Show that the system is likely to vibrate
torsionally with a frequency of
n 
kt  I1  I 2 
I1 I 2
Designating the angular positions of the two wheels by 1 and  2 , respectively, and
summing moments on each, we get
I11  kt1  kt 2  0
 M1  kt 1 2   I11
 M  k     I 
2
t
1
2
2 2
I 2 2  kt1  kt 2  0
Next, we assume a solution of the form
 j  C j cos nt   
for each inertia with j = 1, 2.
Substituting these gives
n2 I1C1 cos nt     kt C1 cos nt     kt C2 cos nt     0
n2 I 2C2 cos nt     kt C1 cos nt     kt C2 cos nt     0
Dividing each by cos nt    and writing these in matrix form they become
  n2 I1  kt
  C1 
kt

  0
 n2 I 2  kt  C2 
 kt
For this set of equations to have a non-trivial solution for C1 and C2, the determinant of
the coefficient matrix must vanish. Therefore,
 n2 I1  kt  n2 I2  kt    kt  kt   0
This expands to a quadratic equation in  n2
I1I 2 n2   kt  I1  I 2  n2  0
2
Solving this, we get four roots:
n  0
n  
kt  I1  I 2 
I1 I 2
Q.E.D.
Note that the other frequency of n  0 shows the capability for rigid body motion
since the entire shaft with wheels is free to rotate.
658
13.11
A motor is connected to a flywheel by a 5/8-in diameter steel shaft 36 in long. Using the
methods of this chapter, it can be demonstrated that the torsional spring rate of the shaft is
4 700 in  lb/rad. The mass moments of inertia of the motor and flywheel are 24.0 and
56.0 in  lb  s2 , respectively. The motor is turned on for 2 s, and during this period it
exerts a constant torque of 200 in  lb on the shaft. (a) What speed in revolutions per
minute does the shaft attain? (b) What is the natural circular frequency of vibration of the
system? (c) Assuming no damping, what is the amplitude of the vibration of the system
in degrees during the first era? During the second era?
This is a difficult problem, but too interesting and challenging not to include.
(a)
The angular impulse equation is
t
H  H 0   Tdt
0
Since the motor starts from rest, its initial angular momentum is H 0  0 . We also
see that H   I1  I 2   , T  200 in  lb , and t  2 s . Substituting,
 I1  I 2    H 0  0  200 in  lb  dt
80.0 in  lb  s2     200 in  lb  t
t
   2.5 rad/s2  t for 0  t  2 s
(b)
Ans.
From Prob. 13.10
n 
(c)
At t = 2 s   5.0 rad/s  47.75 rev/min
k t  I1  I 2 
I1 I 2

 4 700 lb·in/rad   24 in  lb·s 2  56 in  lb·s 2 
 24 in  lb·s  56 in  lb·s 
2
2
 16.726 rad/s
First era; 0  t  2 s :
The differential equations are:
I11  kt1  kt 2  T
I 2 2  kt1  kt 2  0
After using the conditions that, at t = 0, 1   2  0 the solutions become
I
I1
1  A  2 B cos nt 
T  I2 t 2 
  
I1  I 2  kt 2 
Ans.
659
 2  A  B cos nt 
Tt 2
2  I1  I 2 
Then using the conditions that, at t = 0, 1   2  0 , we find
I1I 2
T
B  A 
kt  I1  I 2 2
and the solutions become
I2
T
T  I2 t 2 
1  
 I1  I 2 cos nt  
  
kt  I1  I 2 2
I1  I 2  kt 2 
2  
I1 I 2
T
Tt 2
1

cos

t



n
kt  I1  I 2 2
2  I1  I 2 
But we are interested in the relative motion, the twist in the shaft, which is
T I 2 1  cos  nt 
1   2 
kt
 I1  I 2 
So the amplitude during the first era is
2
200 in  lb   56.0 in  lb·s 

I2
T


 0.029 rad  1.707
kt  I1  I 2   4 700 in  lb/rad   80.0 in  lb·s 2 
Ans.
For the completion of the first era we can compute that, at t = 2.0 s,
nt  16.7 rad/s  2.0 s   33.45 rad  1 916.7 and cos nt  0.449 .
Thus,
1  5.0302 rad , and 1  5.0302 rad . These are the initial displacements for the
second era.
Second era; t  2 s :
The differential equations now become:
I11  kt1  kt 2  0
I 2 2  kt1  kt 2  0
Following a similar procedure to that above, we eventually obtain
1   2  0.0555cos nt   19.6
Therefore the amplitude of the second era is
  0.0555 rad  3.180
Ans.
660
13.12
The weight of the mass of a vibrating system is 10 lb, and it has a natural frequency of 1
Hz. Using the phase-plane method, plot the response of the system to the given force
function. What is the final amplitude of the motion?
k  mn2  10 lb 386 in/s2   2 rad/s   1.023 lb/in
2
F1 k  12 lb 1.023 lb/in  11.736 in
F2 k  6 lb 1.023 lb/in  5.868 in
n t1   2 rad/s  0.25 s   1.571 rad  90.0 , n t2   2 rad/s  0.25 s   1.571 rad  90.0
The final amplitude is X = 18.56 in.
Ans.
661
13.13
An undamped vibrating system has a spring rate of 200 lb/in and a weight of 50 lb. Find
the response and the final amplitude of vibration of the system if it is acted upon by the
given forcing function. Use the phase-plane method.
n  k m  200 lb/in  50 lb 386 in/s2   39.298 rad/s
n t   39.298 rad/s  0.040 s   1.572 rad  90 , F k  12.5 lb 200 lb/in  0.0625 in
The final amplitude is 0.1768 in.
Ans.
662
13.14
A vibrating system has a spring rate of k = 400 lb/in and a weight of W = 80 lb. Plot the
response of this system to the given forcing function using: (a) three steps, and (b) six
steps
Fmax = 200 lb.
n  k m  400 lb/in 80 lb 386 in/s2   43.937 rad/s
(a)
Three-step solution: n t   43.937 rad/s  0.1 s 3  1.465 rad  83.91
F1 k  0.083 in , F2 k  0.250 in , F3 k  0.417 in , F k  0.500 in
(b)
Six-step solution: n t   43.937 rad/s  0.1 s 6   0.732 rad  41.96
F1 k  0.042 in , F2 k  0.125 in , F3 k  0.208 in , F4 k  0.292 in ,
F5 k  0.375 in , F6 k  0.458 in , F k  0.500 in
663
13.15 What is the value of the coefficient of critical damping for a spring-mass-damper system
in which k = 56 kN/m and m = 40 kg? If the actual damping is 20% of critical, what is
the natural frequency of the system? What is the period of the damped system? What is
the value of the logarithmic decrement?
n  k m  56 000 N/m 40 kg  37.417 rad/s
cc  2mn  2  40 kg  37.417 rad/s   2993.3 N  s/m
Ans.
d  n 1   2  37.417 rad/s 1  0.202  36.661 rad/s
  2 d  2 36.661 rad/s  0.171 s/cycle
Ans.
Ans.
  2
13.16
1   2  2  0.2 
1  0.202  1.283
Ans.
A vibrating system has a spring rate of k = 3.5 kN/m and a mass m = 15 kg. When
disturbed, it is observed that the amplitude decayed to one-fourth of its original value in
4.80 s. Find the damping coefficient and the damping factor.
n  k m  3 500 N/m 15 kg  15.275 rad/s
Using Eq. (13.32) with  N  ln 1.0 0.25  1.386 and N  4.80 s
13.17
   N  Nn   1.386  4.80 s 15.275 rad/s   0.0189
Ans.
c   2mn  0.0189  2 15 kg 15.275 rad/s  8.664 N  s/m
Ans.
A vibrating system has k = 300 lb/in, W = 90 lb, and damping equal to 20% of critical.
(a) What is the damped natural frequency d of the system? (b) What are the period and
the logarithmic decrement?
n  k m  300 lb/in  90 lb 386 in/s2   35.87 rad/s
(a) d  n 1   2  35.87 rad/s 1  0.202  35.15 rad/s
(b)   2 d  2 35.15 rad/s  0.179 s/cycle
Ans.
Ans.
  2
Ans.
1   2  2  0.20
1  0.202  1.283
664
13.18 Solve Problem 13.14 using damping equal to 15% of critical.
n  k m  400 lb/in 80 lb 386 in/s2   43.937 rad/s
d  n 1   2  43.937 rad/s 1  0.152  43.440 rad/s
Six-step solution: d t   43.440 rad/s  0.10 s 6   0.724 rad  41.48
F1 k  0.042 in , F2 k  0.125 in , F3 k  0.208 in , F4 k  0.292 in ,
F5 k  0.375 in , F6 k  0.458 in , F k  0.500 in
In each step of d t  0.724 rad the reduction in amplitude is
X n 41.48 X n  e  0.724 rad  1  0.896 . Therefore,
2
x0  0.0417 in , 0  90
x1  0.896  0.0417 in   0.0376 in
x1  0.1138 in , 1  77.34
x2  0.896  0.1138 in   0.1020 in
x2  0.1653 in , 2  59.98
x3  0.896  0.1653 in   0.1481 in
x3  0.1916 in , 3  42.86
x4  0.896  0.1916 in   0.1716 in
x4  0.1926 in , 4  27.01
x5  0.896  0.1926 in   0.1726 in
x5  0.1719 in , 5  13.53
x6  0.896  0.1719 in   0.1539 in
x6  0.1394 in , 6  12.64
665
13.19
A damped vibrating system has an undamped natural frequency of 10 Hz and a weight of
800 lb. The damping ratio is 0.15. Using the phase-plane method, determine the
response of the system to the given forcing function.
n  10 revs/s  2 rad/rev   62.832 rad/s
d  n 1   2  62.832 rad/s 1  0.152  62.121 rad/s
d t   62.121 rad/s  0.01 s   0.621 rad  35.59
k  mn2  800 lb 386 in/s2   62.832 rad/s   1 905 lb/in
2
F1 k  1.050 in , F2 k  1.575 in , F3 k  0.525 in , F4 k  0.525 in ,
In each step of d t  0.621 rad the reduction in amplitude is
X n35.59 X n  e  0.621 rad  1  0.910 . Therefore,
x0  1.050 in , 0  90
x1  0.955 in
x1  1.188 in , 1  43.53
x2  1.081 in
x2  1.740 in , 2  59.95
x3  1.584 in
x3  2.336 in , 3  113.94
x4  2.126 in
x4  2.050 in , 4  199.90
2
x1  0.910 in
x2  0.984 in
x3  1.441 in
x4  1.935 in
666
13.20
A vibrating system has a spring rate of 3 000 lb/in, a damping factor of 55 lb  s / in, and a
weight of 800 lb. It is excited by a harmonically varying force F0  100 lb at a frequency
of 435 cycles per minute. (a) Calculate the amplitude of the forced vibration and the
phase angle between the vibration and the force. (b) Plot several cycles of the
displacement-time and force-time diagrams.
(a)
m  800 lb 386 in/s2  2.072 lb  s2 /in
n  k m 
 3 000 lb/in   2.072 lb  s2 /in   38.050 rad/s
  c  2mn   55 lb  s/in  2  2.072 lb  s2 /in  38.050 rad/s   0.349
 n   435 cycles/min  2 rad/cycle 60 s/min  38.050 rad/s  1.197
F0 k
X
1       2   
2
2
n
100 lb 3 000 lb/in
X
1  1.197    2  0.349 1.197 
2 2
  tan 1
13.21
2 2
n
 0.035 in
Ans.
2
2  n
2  0.349 1.197
 tan 1
 117.42
2
2
1   n
1  1.1972
Ans.
A spring-mounted mass has k = 525 kN/m, c = 9 640 N  s / m, and m = 360 kg. This
system is excited by a force having an amplitude of 450 N at a frequency of 4.80 Hz.
Find the amplitude and phase angle of the resulting vibration and plot several cycles of
the force-time and displacement-time diagrams.
 525 000 N/m  360 kg  38.188 rad/s
  c  2mn    9 640 N  s/m   2  360 kg  38.188 rad/s   0.351
 n   4.80 Hz  2 rad/cycle  38.188 rad/s  0.790
n  k m 
X
F0 k
1       2   
2
X
2 2
n
2
n
450 N 525 000 N/m
1  0.790    2  0.351 0.790
  tan 1
2 2
 0.001 280 m  1.280 mm
Ans.
2
2  n
2  0.351 0.790
 tan 1
 55.80
2
2
1   n
1  0.7902
Ans.
667
13.22 When a 6 000-lb press is mounted upon structural-steel floor beams, it causes them to
deflect 0.75 in. If the press has a reciprocating unbalance of 420 lb and it operates at a
speed of 80 rev/min, how much of the force will be transmitted from the floor beams to
other parts of the building? Assume no damping. Can this mounting be improved?
k  6 000 lb 0.75 in  8 000 lb/in
m  6 000 lb 386 in/s2  15.54 lb  s2 /in
n  k m  8 000 lb/in 15.54 lb  s 2 /in  22.689 rad/s
 n  80 rev/min  2 rad/rev 60 s/min  22.689 rad/s  0.369
Assuming no damping, Eq. (13.62) gives
1
1
1
T


 1.158
2
2
2
1



1

0.369
2
2 2
n
1 

n

Ftr  TF0  1.158  420 lb  486 lb
Figure 13.37 shows that, with  n  0.369 , small changes in either damping or  n will
do little to reduce transmissibility. Therefore the mounting cannot be improved. The
primary opportunity for improvement would be to reduce the unbalance.
668
13.23 Four vibration mounts are used to support a 450-kg machine that has a rotating unbalance
of 0.35 kg  m and runs at 300 rev/min. The vibration mounts have damping equal to
30% of critical. What must the spring constant of the mounting be if 20% of the exciting
force is transmitted to the foundation? What is the resulting amplitude of motion of the
machine?
From Eq. (13.63)
 /   1   2  0.30 /     /   1  0.36  /  
T  0.20 
1   /      2  0.30 /  
1   /     0.36  /  




2
2
2
n
2
2
n
2
2
n
2
2
2
0.04 1   / n    0.36  / n 


2
2
   


/
2
n
2
2
n
2
n
0.20 1   / n    0.36  / n    / n 


2
2
n
n
1  0.36  / n 
4
2
n
2
1  0.36  / n 2 


0.36  / n   0.96  / n   0.0656  / n   0.04  0
6
4
2
Numerically searching for the root we find
2
 / n   0.168 423
  300 rev/min  31.415 93 rad/s
 / n  0.410 39
n  76.550 70 rad/s
k  mn2  450 kg  76.550 70 rad/s   2 637 000 N/m
2
Ans.
Now, from Eq. (13.57)
 / n 
 0.168 423
mX


 0.194 20
2
mu e
2 2
2
1   / n    0.36  / n 
1   0.168 423  0.36  0.168 423


X  0.194 20  0.35 kg  m  450 kg  0.151 mm
Ans.
2
669
13.24 A 600-mm long steel shaft is simply supported by two bearings at A and C. Flywheels 1
and 2 are attached to the shaft at locations B and D, respectively. Flywheel 1 at location
B weighs 50 N, flywheel 2 at location D weighs 20 N, and the weight of the shaft can be
neglected. The known stiffness coefficients are k11  25 000 N/m, k12  50 000 N/m,
and k22  40 000 N/m. Determine: (a) the first and second critical speeds of the shaft
using the exact solution and (b) the first critical speed using the Dunkerley and (c)
Rayleigh-Ritz approximations. (d) If flywheel 2 is then placed at location B and flywheel
1 is placed at location D, determine the first critical speed of the new system using the
Dunkerley approximation.
(a)
The exact solutions for the first and second critical speeds of the shaft are
(a11m1  a22 m2 )  (a11m1  a22 m2 ) 2  4m1m2 (a11a22  a12 a21 )
,

12 22
2
The influence coefficients are the reciprocals of the stiffness coefficients; that is,
and a jk  akj  1 k jk
aii  1 kii
1
1
Therefore, the influence coefficients are
a11  1  2.5 104 N/m   4 105 m/N ,
and
a22  1  4 104 N/m   2.5 105 m/N
a12  1  5 104 N/m   2 105 m/N
The masses of the two flywheels are
50 N
20 N
m1 
 5.10 kg and m2 
 2.04 kg
2
9.81 m/s
9.81 m/s 2
Substituting Eqs. (3) and (4) into Eq. (1) gives
(1)
(2)
(3a)
(3b)
(4)
25.5 105 s 2  (25.52 s 4  249.696 s 4 ) 10 10
 (12.75  10.01) 105 s 2
1 2
2
Using the positive sign for the first critical speed and the negative sign for the second
critical speed gives
1  66.28 rad/s and 2  191.04 rad/s
Ans.
1
,
2
1

2
670
(b)
Using the Dunkerley approximation, the first critical speed of the shaft with the
two flywheels can be written as
1
(5)
 a11m1  a 22 m2
12
Substituting Eqs. (3) and the masses into Eq. (5) gives
1
 4 105 m/N  5.10 kg   2.5 105 m/N  2.04 kg   2.25 104 s 2
2
1
Therefore, the first critical speed of the shaft is
1  62.62 rad/s
Ans.
(c)
Using the Rayleigh-Ritz approximation, the first critical speed of the shaft with
the two flywheels can be written as
g (W1 x1  W2 x2 )
(6)
12 
(W1 x12  W2 x22 )
The total deflections of the shaft at the mass particles can be written as
(7)
x1  a11W1  a12W2 and x2  a12W1  a22W2
Substituting the known values and Eq. (2) into Eqs. (7) the total deflections are
(8)
x1  2.4 103 m and x2  1.5 103 m
Substituting Eqs. (8) and the known values into Eq. (6) gives
9.81 m/s 2 [(50 N)(2.4 103 m)  (20 N)(1.5 103 m)]
12 
[(50 N)(2.4 103 m)2  (20 N)(1.5 103 m)2 ]
1.47 m/s2
 4 414 rad 2 /s 2
333 106 m
Therefore, the first critical speed of the shaft is
1  66.44 rad/s
or
12 
Ans.
(d)
When the two flywheels are interchanged then the Dunkerley approximation, see
Eq. (5), can be written as
1
(9)
 a11m1new  a 22 m2new
2
1
Note that the influence coefficients of the shaft do not change (even though the two
flywheels were interchanged). Therefore, substituting these values into Eq. (9) gives
1
 4 105 m/N  2.04 kg   2.5 105 m/N  5.10 kg   2.09 104 s2
2
1
Therefore, the first critical speed of the shaft is
1  69.15 rad/s
Ans.
671
13.25 The first critical speeds of a rotating shaft with two mass disks, obtained from three
different mathematical techniques, are 110 rad/s, 112 rad/s, and 100 rad/s, respectively.
(a) Which values correspond to the first critical speed of the shaft from the exact solution,
the Dunkerley approximation, and the Rayleigh-Ritz approximation? (b) If the influence
coefficients are a11  a22  104 m/N and the masses of the two disks are the same (that is,
m1 = m2 = m) then use the Dunkerley approximation to calculate the mass m. (c) If the
influence coefficients are a11  a22  104 m/N and the masses of the two disks are
specified as m1 = m2 = m = 0.5 kg, use the Rayleigh-Ritz approximation to calculate the
influence coefficient a12.
(a) The first critical speed from the Rayleigh-Ritz approximation is an upper bound;
therefore, the value 1  112 rad/s corresponds to the answer from the Rayleigh-Ritz
approximation. The Dunkerley approximation gives a lower limit to the first critical
speed; therefore, the value 1  100 rad/s corresponds to the answer from the Dunkerley
approximation. The value of the first critical speed from the exact method is
Ans.
1  110 rad/s .
(b) Since the first critical speed and the influence coefficients are given then the
Dunkerley approximation can be used to calculate the two masses; that is,
(1)
1 12  a11m1  a22 m2
Substituting the given information and the first critical speed from the table into Eq. (1)
gives
2
1 100 rad/s   104 m/N  m  104 m/N  m
Therefore, the mass is
1
Ans.
m
 0.5 kg
 2 10
4
m/N  100 rad/s 
2
(c) The Rayleigh-Ritz equation can be written as
g (W1 x1  W2 x2 )
12 
(W1 x12  W2 x22 )
Since the two masses are the same then this equation can be written as
12  g ( x1  x2 )  x12  x22 
(2)
The total deflections of the shaft at locations 1 and 2 can be written as
x1  a11W1  a12W2 and x2  a12W1  a22W2
(3)
Since the influence coefficients a11  a22 and a12  a21 and the masses m1 = m2 = m then
the deflections x1  x2  x . Therefore, Eq. (2) can be written as
12  g x
Substituting the known data and Eqs. (3) into Eq. (4) gives
9.81 m/s 2
2
112 rad/s  
(0.5 kg)(9.81 m/s 2 )(104 m/N  a12 )
Solving for the influence coefficient gives a12  5.94 105 m/N
(4)
Ans.
672
13.26 A steel shaft is simply supported by two rolling element bearings at A and B. The length
of the shaft is 1.45 m and two flywheels with weight 300 N are attached to the shaft at the
locations shown. One flywheel is 0.35 m to the right of the left bearing at A and the other
flywheel is 0.35 m to the left of the right bearing at B. The weight of the shaft can be
neglected. The influence coefficients are specified as a11  126 105 mm/N and
a21  92.5 105 mm/N. (a) Determine the first and second critical speeds of the shaft
using the exact solution. Determine the first critical speed of the shaft using: (b) the
Dunkerley approximation and (c) the Rayleigh-Ritz equation.
(a) The exact solutions for the first and second critical speeds of the shaft can be written
(a11m1  a22 m2 )  (a11m1  a22 m2 )2  4(a11a22  a12 a21 )m1m2
1 2
2
From the symmetry of the loading, we find the influence coefficients
a11  a22  1.26 106 m/N
From Maxwell's reciprocity theorem, we get the influence coefficients
a21  a12  0.925 106 m/N
The mass of the flywheels are
300 N
m  m1  m2 
 30.581 N  s 2 /m
2
9.81 m/s
Substituting Eqs. (2), (3), and (4) into Eq. (1), the exact solutions can be written as
1 1
,
  a11  a21  m
12 22
Equation (5) can be written as
1
12 , 22 
 a11  a21  m
Substituting the numerical values into Eq. (6), the exact solutions can be written as
106
12 , 22 
(1.26 m/N  0.925 m/N)  30.581 N  s 2 /m 
1
,
2
1

2
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Using the positive sign in the denominator of Eq. (7), the first critical speed of the shaft is
obtained from the relation
106
106
12 

 1.4966 104 rad 2 / s2
2
2
 2.185 m/N  30.581 N  s /m 66.819 s


673
Therefore, the first critical speed of the shaft is
ω1  122.3 rad / s
Ans.
(8)
Similarly, using the negative sign in the denominator of Eq. (7), the second critical speed
of the shaft can be obtained from the relation
106
106
2
1 

 9.761104 rad 2 / s 2
2
2
 0.335 m/N  30.581 N  s /m 10.245 s


Therefore, the second critical speed of the shaft is
Ans.
2  312.4 rad/s
Note that the second critical speed is about three times the first critical speed.
(b) The Dunkerley approximation to the first critical speed of the shaft can be written as
1
  a11  a22  m  2a11m
(9)
2
1
Substituting the numerical values into Eq. (9), the Dunkerley approximation to the first
critical speed of the shaft is
1
 2 1.26 106 m/N  30.581 N  s 2 /m   77.064 106 s 2
12
Therefore, the Dunkerley approximation to the first critical speed of the shaft is
Ans.
1  113.9 rad/s
Note that the the Dunkerley approximation to the first critical speed of the shaft is less
than the exact answer, see Eq. (8); that is, the Dunkerley approximation always gives a
lower bound.
(c) The Rayleigh-Ritz equation can be written as
 W x  W2 x2 
12  g  1 12
(10)
2
W1 x1  W2 x2 
where the deflections are
x1  a11W1  a12W2  300 N(1.260  0.925) 106 m/N  655.5 106 m
and
x2  a21W1  a22W2  300 N(0.925  1.260) 106 m/N  655.5 106 m
Substituting these values into Eq. (10), the Rayleigh-Ritz equation can be written as
9.81 m/s 2
 2Wx 
1
ω12  g 

g

 1.4966 104 rad 2 /s 2
2
6


 2Wx 
 x  655.5 10 m
Therefore, the Rayleigh-Ritz equation to the first critical speed of the shaft is
1  122.3 rad/s
Ans.
Note that the Rayleigh-Ritz approximation to the first critical speed of the shaft gives the
same as the exact answer, see Eq. (8). In general, the Rayleigh-Ritz equation will give a
slightly greater value than the exact answer; that is, the Rayleigh-Ritz equation will give
an upper bound.
674
13.27 A steel shaft is simply supported by two rolling element bearings at A and C. The length
of the shaft is 0.6 m and two flywheels are attached to the shaft at the locations B and D
as shown. The flywheel at location B weighs 200 N and the flywheel at location D
weighs 90 N. The weight of the shaft can be neglected. It was determined that with
flywheel 1 alone, the first critical speed of the shaft is 800 rad/s and with flywheel 2
alone, the first critical speed of the shaft is 1 200 rad/s. (a) Determine the first critical
speed for the two mass system. (b) If the two flywheels are interchanged (that is,
flywheel 2 is placed at location B and flywheel 1 is placed at location D), determine the
first critical speed of the new system using the Dunkerley approximation.
(a) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
1 ω12  a11m1  a22 m2
(1)
or as
1
1
1
 2  2
(2)
2
1
11 22
From the given data, the critical speeds are 11  800 rad/s and 22  1200 rad/s.
Therefore, Eq. (2) can be written as
1
1
1
1 
 1
4 2


 
(3a)
 10 s
2
2
2
1  800 rad/s  1 200 rad/s   64 144 
or as
12  44.308 104 rad2 /s2
Therefore, the first critical speed of the shaft is
1  665.6 rad/s
(3b)
Ans.
(4)
(b) When the two flywheels are interchanged then Eq. (1) can be written as
1 12  a11m1new  a22 m2new
(5)
Note that the influence coefficients of the shaft (by definition) do not change (even
though the two flywheels were interchanged). Therefore, the influence coefficient
a11  1 112 m1 
(6a)
675
which can be written as
1
a11 
 7.6641108 m/N
2
2
800 rad/s  (200 N / 9.81 m/s )
Similarly, the influence coefficient
2
a22  1 22
m2 
(6b)
(7a)
which can be written as
1
a22 
 7.5694 108 m/N
2
2
1 200 rad/s  (90 N / 9.81 m/s )
(7b)
Substituting Eqs. (6b) and (7b) into Eq. (5) gives
1
 90 N 
 200 N 
 (7.6641108 m/N) 
 (7.5694 108 m/N) 
2
2 
2 
ω1
 9.81 m/s 
 9.81 m/s 
From this equation, the first critical speed of the shaft is 1  667.2 rad/s. Ans.
(8)
676
13.28 A steel shaft, which is 50 inches in length, is simply supported by two bearings at B and
D. Flywheels 1 and 2 are attached to the shaft at A and C, respectively. The flywheel at
location A weighs 15 lbs, the flywheel at location C weighs 30 lbs, and the weight of the
shaft can be neglected. The stiffness coefficients are specified as k11  2.5 104 lb/in
and k22  4.0 104 lb/in . (a) Determine the first critical speed for the two-mass system
using the Dunkerley approximation. (b) If the two flywheels are interchanged (that is,
flywheel 2 is placed at location A and flywheel 1 is placed at location C), determine the
first critical speed of the new system.
(a) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
(1)
1 12  a11m1  a22 m2
The influence coefficients are the inverse of the spring stiffness coefficients.
aii  1 kii
Therefore, the influence coefficients are
a11  4 105 in/lb and a22  2.5 105 in/lb
Substituting these values and the masses into Eq. (1) gives
1
 15 lb 
 30 lb 
  4 105 in/lb  
 2.5 105 in/lb  
 3.5 106 s 2
2
2  
2 
1
 386 in/s 
 386 in/s 
which gives
1  534.8 rad/s
Ans.
(b)
When the two flywheels are interchanged then Eq. (1) becomes
2
(2)
1 1  a11m1new  a22 m2new
Note that the influence coefficients of the shaft do not change (even though the two
flywheels are interchanged). Therefore, substituting values into Eq. (2) gives
1
 30 lb 
 15 lb 
  4 105 in/lb  
 2.5 105 in/lb  
 4.08 106 s 2
2
2  
2 
1
386.1
in/s
386
in/s




which gives
1  495.1 rad/s
Ans.
677
13.29 The weights of two gears rigidly attached to a shaft at two different locations, denoted as
1 and 2, are W1  200 N and W2  350 N, respectively. The shaft is rotating
counterclockwise with a constant operating speed   100 rad/s . From a deflection
analysis of the shaft, the known influence coefficients are a11  35  106 mm/N,
a12  50  106 mm/N, and a22  90  106 mm/N. Neglecting gravity and the mass of the
shaft determine the first and second critical speeds of the shaft using the exact equation.
Is the operating speed of the shaft acceptable? Determine the first critical speed of the
shaft using: (a) the Rayleigh-Ritz method; and (b) the Dunkerley approximation.
The exact equation for the first and second critical speeds can be written as
( a11 W1  a22 W2 )  ( a11 W1  a22 W2 )2  4 ( a11 a22  a12 a21 ) W1 W2
,

12 22
2g
Substituting the masses and influence coefficients gives
1
1
9
9
9
9
9
9
8
8
1 1 (35 10 ) (200)  ( 90 10 )(350)  [(35 10 ) (200)  ( 90 10 )(350)]  4 [ (35 10 )( 90 10 )  (5 10 )( 5 10 )] (200)(350)
, 
2 x 9.81
ω12 ω22
2
Simplifying, this equation gives
1 1
, 2  3.800 152  106 rad 2 / s2 ,0.124 415  106 rad 2 / s2
2
1 2
Therefore, the first critical speed is
1  512.98 rad/s
and the second critical speed is
Ans.
2  2 835.07 rad/s
Ans.
The operating speed of the shaft is much less than the first critical speed. Therefore, the
operating speed of the shaft is acceptable.
Ans.
(a) The Rayleigh-Ritz equation can be written as
 W x  W2 x2 
12  g  1 12
2
W1 x1  W2 x2 
where the deflections are
x1  a11W1  a12W2   35  109 m/N  200 N  50  109 m/N  350 N  24.5  10 6 m
x2  a21W1  a22W2   50  109 m/N  200 N   90  109 m/N  350 N  41.5  106 m
Substituting values, the Rayleigh-Ritz equation gives
 200 N  24.5  106 m   350 N  41.5  106 m  
2
2
  263 626.68 rad 2 /s2
1  9.81 m/s 
2
2

6

6
 200 N  24.5  10 m   350 N  41.5  10 m  


Therefore, the first critical speed of the shaft is
1  513.45 rad/s
Ans.
Note that the Rayleigh-Ritz approximation for the first critical speed of the shaft is
greater than the exact value. This is consistent with the fact that the Rayleigh-Ritz
approximation is an upper bound to the first critical speed.
678
(b) The Dunkerley approximation to the first critical speed of the shaft can be written as
1
200 N
350 N
 a11m1  a22m2   35  109 m/N 
  90  109 m/N 
2
2
1
9.81 m/s
9.81 m/s2
Therefore, the Dunkerley approximation to the first critical speed of the shaft is
Ans.
1  504.78 rad/s
Note that the Dunkerley approximation for the first critical speed of the shaft is less than
the exact value. This is consistent with the fact that the Dunkerley approximation is a
lower bound on the first critical speed.
679
13.30 The weights of the two masses m1 and m2 which are rigidly attached to the rotating shaft
are 31.5 lb and 13.5 lb, respectively. The shaft is rotating counterclockwise with a
constant angular velocity   100 rad/s . From a deflection analysis, the influence
coefficients for the shaft are a11  3.56  106 in/lb, a22  21.36  106 in/lb, and
a12  a21  7.12  106 in/lb. Neglecting the mass of the shaft determine the first and
second critical speeds of the shaft using the exact equation. Is the operating speed of the
shaft acceptable? Determine the first critical speed of the shaft using the Rayleigh-Ritz
equation and the Dunkerley approximation. Determine the first critical speed of the shaft
if the mass m1 is moved to location 2 and the mass m2 is moved to location 1.
The exact equation for the first and second critical speeds can be written as
(a11m1  a22m2 )  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2
,

12 22
2
Substituting the known values gives
1
1
1 1 (112.14 106 )  (288.36 106 )  [(112.14 106 )  (288.36 106 )]2  4[(76.0416  1012 )  (50.6944  1012 )](425.25)
, 
12 22
2  386
or as
1
,
1
12 22

(400.5  106 )  160 400.25 10 12  101.39 10 12  (425.25)
2  386
Further simplifying this gives
1
,
2
1

2
1 2
(400.5  106 in)  117 284.15  1012 in 2
2  386 in/s2
680
1
,
1
 
2
1
2
2
 0.962  106 s2 , 0.075  106 s2
Therefore, the first two critical speeds can be written as
Ans.
1  1 019.56 rad/s , 2  3 651.48 rad/s
The operating speed of the shaft is much less than the first critical speed. Therefore, the
operating speed of the shaft is acceptable.
The Rayleigh-Ritz equation can be written as
 W x  W2 x2 
12  g  1 12
2
W1 x1  W2 x2 
where the deflections are
x1  a11W1  a12W2   3.56  106  31.5   7.12  106 13.5  208.26  10 6 in
x2  a21W1  a22W2   7.12  106  31.5   21.36  106 13.5  512.64  10 6 in
Substituting values into the Rayleigh-Ritz equation gives
  31.5 lb   208.26  106 in   13.5 lb   512.64  10 6 in  

12  386 in/s2 
2
2
  31.5 lb   208.26  106 in   13.5 lb  512.64  10 6 in  


  6 560.19  106 lb  in    6 920.64  10 6 lb  in  
 386 in/s 

6
2
6
2
 1.366  10 lb  in    3.548  10 lb  in  
 1 058 933.74 rad 2 /s 2
Therefore, the Rayleigh-Ritz approximation for the first critical speed of the shaft is
1  1 029.05 rad/s
Ans.
Note that the Rayleigh-Ritz equation for the first critical speed of the shaft gives a greater
value than the exact equation. The Rayleigh-Ritz equation is an upper bound.
The Dunkerley approximation to the first critical speed of the shaft can be written as
1
 a11m1  a22m2
2
2
1
Substituting the numerical values, the Dunkerley approximation gives
1
31.5 lb
13.5 lb
  3.56  106 in/lb 
  21.36  106 in/lb 
2
2
1
386 in/s
386 in/s2
 0.290 518 13  106 s2  0.747 046 63  106 s2
 1.037 564 76  106 s2
Therefore, the Dunkerley approximation to the first critical speed of the shaft is
1  981.73 rad/s.
Note that the the Dunkerley approximation to the first critical speed of the shaft is less
than the exact value. This is consistent with the fact that the Dunkerley approximation is
a lower bound to the first critical speed.
When the two masses are interchanged, then the exact equation can be written as
681
(a11m2  a22m1 )  (a11m2  a22m1 )2  4(a11a22  a12a21 )m1m2
,

12 22
2
Substituting the known values gives
1
1
1 1 (48.06 106 )  (672.84 106 )  [(48.06 106 )  (672.84 106 )]2  4[(76.0416 1012 )  (50.6944 1012 )](425.25)
, 
or
12 22
2  386 in/s2 
as
1
,
1
12 22

(1 152.90 106 )  1 329 178.4110 12  101.39 10 12  (425.25)
2  386 in/s 2 
Further simplifying this gives
1
,
2
1

2
1 2
(1 152.90 106 in)  1 286 062.311012 in 2
2  386 in/s 2 
1
,
1
 
2
1
2
2
 2.962  106 s2 , 0.024  106 s2
Therefore, the first critical speed is
1  581.01 rad/s .
Note that this answer is less than the answer before the masses were interchanged.
Ans.
682
13.31 The first and second critical speeds of a rotating shaft with two flywheels rigidly attached
are 1  375 rad/s and 2  615 rad/s. The weights of the flywheels are W1  65 N and
W2  80 N and the known influence coefficients of the shaft are a11  5.90  104 mm/N
and a21  2.74  104 mm/N. Determine the influence coefficient a22 .
From the exact equation, Eq. (13.75), the sum of the first two reciprocal critical speeds
squared of a rotating shaft is
1
1
 2  a11m1  a22 m2
2
1
2
Substituting the given data, this gives
1
1
65 N
80 N

  5.90  107 m/N 
 a22
2
2
2
9.81 m/s
9.81 m/s2
 375 rad/s  615 rad/s 
80 N
7.11111  106 s2  2.64393  106 s2  3.90928  106 s2  a22
9.81 m/s2
This yields the value
9.81 m/s2
a22 
5.84576  106 s2  0.71684  106 m/N
80 N
Therefore, the influence coefficient a22 is
a22  7.17  104 mm/N
683
13.32 A shaft rotating with a constant angular velocity is simply supported at A and B. Gears C
and D are rigidly attached to the shaft at locations 1 and 2, respectively. The weight of
gear C at location 1 is 13 N and the weight of gear D at location 2 is 23 N. The known
influence coefficients of the shaft are a11  6.8 106 m/N, a12  5.3 106 m/N, and
a22  7.9 106 m/N. Determine the first critical speed of the shaft using: (a) the
Dunkerley approximation; and (b) the Rayleigh-Ritz method. If gear C is moved to
location 2 and gear D is moved to location 1 then calculate the new first critical speed
using: (c) the Dunkerley approximation; and (d) the Rayleigh-Ritz method.
(a) The first critical speed of the shaft, using the Dunkerley approximation, can be written
as
1
1
1
 2  2  m1a11  m2a22
2
1
11
22
Substituting the given masses and influence coefficients gives
1
13 N
23 N

6.8  106 m/N 
7.9  106 m/N  27.533  106 s 2
2
2
2
1 9.81 m/s
9.81 m/s
Therefore, the Dunkerley approximation for the first critical speed of the shaft is
1  190.58 rad/s
(b) The deflections of the shaft at locations 1 and 2 can be written as
y1  a11W1  a12W2   6.8  106 m/N 13 N  5.3  106 m/N  23 N  210.3  106 m
Ans.
y2  a21W1  a22W2   5.3  106 m/N 13 N   7.9  106 m/N  23 N  250.6  106 m
The first critical speed of the shaft using the Rayleigh-Ritz method can be written as
(W y  W2 y2 )
12  g 1 12
(W1 y1  W2 y22 )
 9.81 m/s
2
13 N  210.3  106 m   23 N  250.6  106 m 
13 N  210.3  106 m   23 N  250.6  106 m 
2
2
 41 281.869 05 rad 2 /s2
Therefore, the Rayleigh-Ritz approximation for the first critical speed is
1  203.18 rad/s
Ans.
684
(c) With the two gears interchanged the Dunkerley approximation for the first critical
speed can be written as
1
1
1
 2  2  m2a11  m1a22
2
1
11
22
23 N
13 N
6.8  106 m/N 
7.9  106 m/N  26.4118  10 6 s 2
2
9.81 m/s
9.81 m/s 2
Therefore, the first critical speed of the shaft is
1  194.58 rad/s
Note that this answer is less than the previous answer.
(d) The deflections with the gears interchanged can be written as
y1  a11W2  a12W1   6.8  106 m/N  23 N  5.3  106 m/N 13 N  225.3  106 m

Ans.
y2  a21W2  a22W1   5.3  106 m/N  23 N   7.9  106 m/N 13 N  224.6  106 m
The first critical speed of the shaft using the Rayleigh-Ritz method can be written as
(W y  W1 y2 )
12  g 2 12
(W2 y1  W1 y22 )
 9.81 m/s
2
23 N  225.3  106 m   13 N  224.6  106 m 
23 N  225.3  106 m   13 N  224.6  106 m 
2
2
 43 590.754 rad 2 /s2
Therefore, with the gears interchanged, the Rayleigh-Ritz approximation for the first
critical speed of the shaft is
Ans.
1  208.78 rad/s
685
13.33 A shaft is simply supported at A and B and is rotating with constant angular velocity  .
Two identical flywheels C and D of unknown mass are rigidly attached to the shaft (at
locations 1 and 2). The known influence coefficients are a11  41.1  106 in/lb,
a12  29.4  106 in/lb, and a22  53.3  106 in/lb. The first critical speed of the shaft,
obtained from the exact equation, is 135 rad/s. Determine: (a) the masses of the two
flywheels, and (b) the second critical speed of the shaft.
The exact solution for the first and second critical speeds can be written as
(a11m1  a22m2 )  (a11m1  a22m2 )2  4m1m2 (a11a22  a12a21 )
1 2
2
Substituting the known influence coefficients, and noting that m1  m2  m, gives
1
,
2
1

2
(41.1  53.3)  (41.1  53.3) 2  4  41.153.3   29.4  

 6
, 2 
10 m
2
1 2
2
2
1
1
  47.2  30.0   106 in/lb  m
  77.2  106 in/lb  m, 17.2  10 6 in/lb  m
Substituting the first critical speed of the shaft and using the first of these two roots, we
can solve for the value of the mass of each flywheel
386 in/s2 

Ans.
m
 274.35 lb
2
77.2  106 in/lb 135 rad/s 
Next, substituting this value of mass into the second of the two roots, we find the second
critical speed of the shaft
1
274.35 lb
 17.2  106 in/lb 
 12.224 922 s2
2
2
2
386 in/s
2  286.0 rad/s
Ans.
686
13.34 The 600 mm long shaft is simply supported by the bearings at A and D. Flywheel 1 at
location B weighs 30 N and flywheel 2 at location C weighs 15 N. The weight of the
shaft can be neglected. The stiffness coefficients of the shaft are k11  1.0  106 N/m and
k22  5.0  106 N/m. (a) Using the Dunkerley approximation, determine the first critical
speed of the shaft. (b) If the flywheels are interchanged (that is, if flywheel 2 is at
location B and flywheel 1 is at location C), then use the Dunkerley approximation to
determine the first critical speed of the new system.
Since the influence coefficients are the reciprocal of the spring stiffness coefficients then
a11 
1
1

 1.0  106 m/N
k11 1.0  106 N/m
1
1

 0.2  106 m/N
k22 5.0  106 N/m
(a) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
1
 a11m1  a22m2
2
a22 
1
 1.0  106 m/N 
30 N
15 N
  0.2  106 m/N 
2
9.81 m/s
9.81 m/s2
 3.363 914 373 rad 2 /s2
1  545.23 rad/s
Ans.
(b) Note that, when the two flywheels are interchanged, the two influence coefficients do
not change. Therefore, the Dunkerley approximation for the first critical speed of the
687
shaft with the two flywheels becomes
1
 a11m1new  a22m2new
2
1
 1.0  106 m/N 
15 N
30 N
  0.2  106 m/N 
2
9.81 m/s
9.81 m/s2
 2.140 672 783 rad 2 /s2
1  683.48 rad/s
Ans.
688
13.35 Three identical flywheels, each weighing 19 lb, are rigidly attached to a rotating shaft.
The known influence coefficients of the shaft are: a11  2.25  104 in/lb,
a22  10.75  104 in/lb, a33  6.25  104 in/lb, a12  0.80  104 in/lb, a13  0.15  104 in/lb,
and a23  1.0  104 in/lb. Use the Rayleigh-Ritz method to determine the first critical
speed of the shaft.
The deflection of the shaft at flywheel i can be written as
3
xi   aijW j
j 1
Note from Maxwell’s reciprocity theorem that aij  a ji . Substituting the known influence
coefficients and the weights of the flywheels, the deflections at each flywheel are
x1  2.25  104 19 lb   0.80  104 19 lb   0.15  10 4 19 lb   0.608  10 2 in
x2  0.80  104 19 lb   10.75  104 19 lb   1.0  10 4 19 lb   2.3845  102 in
x3  0.15  104 19 lb   1.0  104 19 lb   6.25  10 4 19 lb   1.406  10 2 in
Since the three flywheel weights are equal, the Rayleigh-Ritz equation can be written as
 W x  W2 x2  W3 x3 
 x x x 
12  g  1 21
 g  21 22 32 
2
2
W1 x1  W2 x2  W3 x3 
 x1  x2  x3 
Substituting the deflection values, this becomes
  0.608  2.3845  1.406  102 in 
2
2
1  386 in/s 
  21 137 rad 2 /s2
2
2
2
4
2
  0.608  2.3845  1.406   10 in 
which gives the first critical speed of the shaft as
1  145.4 rad/s
Ans.
689
13.36 The first and second critical speeds of a rotating shaft supporting two identical gears,
each with a mass of 6 kg, are 1  500 rad/s and 2  1 120 rad/s, respectively. If the
influence coefficient a11  2a22 then determine the numerical values of the influence
coefficients of the shaft.
The exact solution for the first critical speed of the shaft can be written as
(a11m1  a22m2 )  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2
1
2
and the exact solution for the second critical speed of the shaft can be written as
1

2
(a11m1  a22m2 )  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2
2
2
Adding Eqs. (1a) and (1b), the sum of the roots can be written as
1
1
 2  a11m1  a22m2
2
(1a)
1

2
1
2
(1b)
(2)
Since we are given that a11  2a22 , this can be written as
1
1
 2  2a22m1  a22m2  (2m1  m2 )a22
2
1
2
or as
1
1
1

2
 
(500 rad/s) (1 120 rad/s)2
a22 

 2.665  107 m/N
2m1  m2
3(6 kg)
and we can now write that
a11  2a22  2(2.665  107 m/N)  5.330  107 m/N
Subtracting Eq. (1b) from Eq. (1a)we can write that
1
1
 2  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2
2
2
1
1

1
2
2
2
Squaring both sides and substituting Eq. (2) from above gives
2
2
 1
1   1
1 
 2  2    2  2   4(a11a22  a12a21 )m1m2
 1 2   1 2 
Recognizing, from Maxwell’s reciprocity theorem, that a12  a21 , this becomes
2
2
 1
1   1
1 
4
2
 2  2    2  2   2 2  4(a11a22  a12 )m1m2
 1 2   1 2  1 2
Rearranging this we get
1
a122  a11a22 
m1m21222
Substituting known values gives
Ans.
Ans.
690
a122  a11a22 
1
(6 kg)(6 kg)1222
 (5.330  107 m/N)(2.665  107 m/N) 
1
(6 kg)(6 kg)(500 rad/s) 2 (1120 rad/s) 2
1
 5.347  1014 m 2 /N 2
13
2 2
1.129  10 kg /s
Finally, the influence coefficients for the shaft are
a21  a12  2.312  107 m/N
 14.204  1014 m 2 /N 2 
Ans.
691
13.37 The shaft is simply supported by the bearings at A and D. Flywheel 1 at location B
weighs 5.6 lb and flywheel 2 at location C weighs 17.9 lb and the weight of the shaft can
be neglected. The influence coefficients of the shaft are a11  1.8  104 in/lb,
a12  0.7  104 in/lb, and a22  9.8  104 in/lb. Determine the first critical speed of the
shaft using: (a) the Dunkerley approximation; and (b) the Rayleigh-Ritz approximation.
(a) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
1
 a11m1  a22m2
2
1
 1.8  104 in/lb 
5.6 lb
17.9 lb
  9.8  104 in/lb 
2
386 in/s
386 in/s2
 0.480 569 948  104 s2
Therefore, the first critical speed of the shaft by the Dunkerley approximation is
1  144.25 rad/s
Ans.
(b) The deflection of the shaft at location I can be written as
2
xi   aijW j
j 1
From Maxwell’s reciprocity theorem, aij  a ji . Therefore, substituting the known data
values, the deflections of the shaft at the two flywheels are
x1  1.8  104 in/lb  5.6 lb   0.7  104 in/lb 17.9 lb  22.61  104 in
x2   0.7  104 in/lb  5.6 lb   9.8  104 in/lb 17.9 lb  179.34  104 in
The Rayleigh-Ritz equation can be written as
692
 W1 x1  W2 x2 
2
2
W1 x1  W2 x2 
12  g 
Substituting the data values gives
 5.6 lb  22.61  104 in   17.9 lb 179.34  104 in  

12  386 in/s2 
2
2
 5.6 lb  22.61  104 in   17.9 lb 179.34  104 in  


 2.226 158 936  104 rad 2 /s2
Therefore, by the Rayleigh-Ritz equation, the first critical speed of the shaft is
Ans.
1  149.20 rad/s
Note that the first critical speed of the shaft from the Rayleigh-Ritz approximation is
greater than the first critical speed of the shaft from the Dunkerley approximation. The
first critical speed of the shaft from the exact solution lies somewhere between these two
bounds and closer to the Rayleigh-Ritz approximation.
693
13.38 A shaft with negligible mass is simply supported by two bearings. When a gear with a
mass of 7 kg is attached to the shaft at location 1, the first critical speed is measured as 1
200 rad/s. After a second identical gear is attached to the shaft at location 2, the first and
second critical speeds of the shaft are measured as 500 rad/s and 1 800 rad/s, respectively.
Using the exact solution to the first and second critical speeds of a rotating shaft,
determine the four influence coefficients of this shaft.
The critical speed for a shaft when only the first gear is attached can be written as
1
 a11m1
2

Rearranging and substituting known data, the influence coefficient is found as
1
1
a11 

 9.921  108 m/N
2
2
m1
7 kg 1 200 rad/s 
Ans.
The exact solutions for the first two critical speeds of a shaft with two masses can be
written as
(a11m1  a22m2 )  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2

12
2
The sum of the two roots in Eq. (1) is
1
1
 2  a11m1  a22m2
2
1
1
2
(1)
(2)
Rearranging and substituting known data, the influence coefficient a22 is found as
1
1
 2  a11m1
2
 2
a22  1
m2
1

 500 rad/s 
2

1
1 800 rad/s 
2
  9.921  108 m/N   7 kg 
 5.163  107 m/N
7 kg
Subtracting the second root of Eq. (1) from the first gives
1
1
 2  (a11m1  a22 m2 )2  4 (a11a22  a12 a21 )m1m2
2
1
2
Squaring both sides of this equation and substituting Eq. (2) into the result gives
Ans.
694
2
2
 1
1   1
1 
 2  2    2  2    4 (a11a22  a12 a21 )m1m2
 1 2   1 2 
Then rearranging this equation and substituting a12  a21 (from Maxwell’s reciprocity
theorem), into the result gives
2
2
 1
1   1
1 
 2  2    2  2 
1
2 
2 
 1
a122  a11a22   1
 a11a22 
4m1m2
m1m21222
Substituting known values gives
1
a122   9.921  108 m/N 5.163  107 m/N  
2
2
 7 kg  7 kg 500 rad/s  1 800 rad/s 
which becomes
a122  2.603  1014 m2 /N 2
Therefore, the two influence coefficients of the shaft are
a12  a21  1.613 107 m/N
Ans.
695
13.39 The 72 in long shaft is simply supported by the bearings at B and D. Flywheel 1 at
location A weighs 9 lb and flywheel 2 at location C weighs 16 lb. The weight of the shaft
can be neglected. The stiffness coefficients of the shaft are k11  5.6  103 lb/in,
k12  33.6  103 lb/in, and k22  22.4  103 lb/in. Determine the first and second critical
speeds of the shaft using the exact equation. Then determine the first critical speed of the
shaft using: (a) the Rayleigh-Ritz method; and (b) the Dunkerley approximation.
The masses of the flywheels are
m1  W1 g  9 lb 386 in/s2  0.023 32 lb  s2 /in
m2  W2 g  16 lb 386 in/s2  0.041 45 lb  s2 /in
The influence coefficients are the reciprocals of the stiffness coefficients; that is
a11  1 k11  1  5.6  103 lb/in   1.7857  104 in/lb
a22  1 k22  1  22.4  103 lb/in   0.4464  104 in/lb
a12  a21  1 k12  1  33.6  103 lb/in   0.2976  104 in/lb
The exact equation for the first and second critical speeds of the shaft can be written as
(a11m1  a22m2 )  (a11m1  a22m2 )2  4(a11a22  a12a21 )m1m2
1 2
2
Substituting the known data, this becomes
 1.7857  104 in/lb  0.023 32 lb  s2 /in    0.4464  104 in/lb  0.041 45 lb  s2 /in 




2


4
2
4
2
 1.7857  10 in/lb  0.023 32 lb  s /in    0.4464  10 in/lb  0.041 45 lb  s /in 



2
 4 1.7857  104 in/lb  0.4464  104 in/lb    0.2976  104 in/lb    0.023 32 lb  s2 /in  0.041 45 lb  s2 /in  
1 1 



, 2
2
1 2
2
1
,
2
1

2
696
2

4 2
4 2
12 4 
  0.06015  10 s    0.06015  10 s   4 6.8497  10 s  
1 1 

,

12 22
2
1 1
, 2   0.03008  104 s2    0.01482  104 s2 
2
1 2
Therefore, the first two critical speeds of the shaft are
Ans.
1  471.95 rad/s , and 2  809.42 rad/s .
(a) The deflection of the shaft at locations 1 and 2 can be written as
x1  a11W1  a12W2  1.7857  104 in/lb  9 lb   0.2976  104 in/lb 16 lb  20.8329  104 in
x2  a21W1  a22W   0.2976  104 in/lb  9 lb   0.4464  104 in/lb 16 lb  9.8208  104 in
Using the Rayleigh-Ritz approximation, the first critical speed of the shaft is
W x  W2 x2
1  g 1 12
W1 x1  W2 x22
 386 in/s
2
9 lb  20.8329  104 in   16 lb  9.8208  104 in 
9 lb  20.8329  104 in   16 lb  9.8208  104 in 
2
2
Ans.
 494.08 rad/s
We note that the Rayleigh-Ritz equation predits a greater first critical speed than the exact
equation. This is consistent with the fact that the Rayleigh-Ritz equation is an upper
bound solution.
(b) The Dunkerley approximation for the first critical speed can be written as
1
1 
a11m1  a22m2

1.7857  10
4
1
in/lb  0.023 32 lb  s /in    0.4464  10 4 in/lb  0.041 45 lb  s2 /in 
2
 407.75 rad/s
Ans.
Note that the Dunkerley approximation underestimates the first critical speed.
697
13.40 The steel shaft is simply supported by two bearings at A and D. Two gears are rigidly
attached to the shaft at locations B and C. Gear 1 at location B weighs 150 N, gear 2 at
location C weighs 90 N, and the weight of the shaft can be neglected. The stiffness
coefficients are k11  2.5  104 N/m, k12  5.0  104 N/m, and k22  4.0  104 N/m.
Determine the first critical speed of the shaft using: (a) the Dunkerley approximation; and
(b) the Rayleigh-Ritz method. Determine the first and second critical speeds of the shaft
using the exact solution. If gear 2 is now placed at location B and gear 1 is placed at
location C, then determine the first critical speed of the new system using the Dunkerley
approximation.
.
The masses of the two gears are
W
150 N
W
90 N
and m1  1 
m1  1 
 15.29 kg
 9.17 kg
2
g 9.81 m/s
g 9.81 m/s2
The influence coefficients are the reciprocals of the stiffnesses; that is
a11  1 k11  1  2.5  104 N/m   4  105 m/N
a12  a21  1 k12  1  5.0  104 N/m   2  105 m/N
a22  1 k22  1  4.0  104 N/m   2.5  105 m/N
(a) Using the Dunkerley approximation, the first critical speed of the shaft can be written
as
1
 a11m1  a22m2
2
1
  4  105 m/N  15.29 kg    2.5  105 m/N   9.17 kg 
 84.085  105 s2
1  34.49 rad/s
Ans.
Note that the Dunkerley approximation is a lower bound to the first critical speed.
(b) The deflections of the shat at the locations of the two gears are
x1  a11W1  a12W2   4  105 m/N 150 N   2  105 m/N  90 N  7.8  10 3 m
x2  a21W1  a22W2   2  105 m/N 150 N   2.5  105 m/N  90 N  5.25  10 3 m
Using the Rayleigh-Ritz approximation, the first critical speed of the shaft with the two
698
gears can be written as
 W x  W2 x2 
12  g  1 12
2
W1 x1  W2 x2 
 150 N  7.8  103 m   90 N 5.25  10 3 m  
  1 388.25 s 2
 9.81 m/s 
2
2

3

3
150 N  7.8  10 m   90 N 5.25  10 m  


2
Therefore, the first critical speed of the shaft is
Ans.
1  37.26 rad/s
Note that the Rayleigh-Ritz approximation is an upper bound to the first critical speed.
(c)The exact solution for the first and second critical speeds of the shaft can be written as
2
1 1 (a11m1  a22 m2 )  (a11m1  a22 m2 )  4(a11a22  a12 a21 ) m1m2
,

12 22
2
  4 105 m/N 15.29 kg   2.5 10 5 m/N  9.17 kg 




2


 4 105 m/N 15.29 kg   2.5 10 5 m/N  9.17 kg 




 

2

4  4 105 m/N  2.5 105 m/N    2 105 m/N   15.29 kg  9.17 kg  





2
5 2
5 2
 42.0425 10 s  30.4354 10 s
 7.2478 104 s 2 , 1.1607 10 4 s 2
Therefore, the first and second critical speeds of the shaft are
Ans.
37.14 rad/s and 92.82 rad/s
(d) When the two gears are interchangedthen the Dunkerley approximation can be written
as
1
 a11m1new  a22m2new
2
1
  4  105 m/N  9.17 kg   2.5  105 m/N 15.29 kg
 74.905  105 s2
Therefore, with these values, the first critical speed of the shaft is
36.54 rad/s
Ans.
699
13.41 The first critical speeds of a rotating shaft with two mass disks, obtained from three
different mathematical techniques, are 150 rad/s, 152 rad/s, and 153 rad/s, respectively.
The influence coefficients a11  a22  3  104 m/N. (a) Specify which value correspond
to the first critical speed from the exact solution, the Rayleigh-Ritz method, and the
Dunkerley approximation. (b) If the masses of the two disks are identical then use the
Dunkerley approximation to calculate the mass of each disk. (c) If the masses of the two
disks are specified as m1  m2  0.10 kg , then use the Rayleigh-Ritz method to calculate
the influence coefficient a12 .
(a) The Dunkerley approximation gives a lower bound to the first critical speed; therefore,
1 = 150 rad/s corresponds to the answer from the Dunkerley approximation. The
Rayleigh-Ritz approximation is an upper bound to the first critical speed; therefore, 1 =
153 rad/s corresponds to the answer from the Rayleigh-Ritz approximation. The value of
the first critical speed from the exact equation is 1 = 152 rad/s.
Ans.
(b) The Dunkerley approximation can be written as
1
 a11m1  a22 m2
2
1
Substituting the given information and the first critical speed from above gives
1
 3  104 m/N m  3  104 m/N m
2
150 rad/s
Therefore, the mass is
1
m
 0.074 kg
2
4
2   3  10 m/N  150 rad/s 




Ans.
(c) The Rayleigh-Ritz equation can be written as
 W x  W2 x2 
12  g  1 12
2
W1 x1  W2 x2 
Since the two mass disks have the same weights, then this equation can be written as
x x
12  g [ 12 22 ]
x1  x2
The total deflections of the shaft at locations 1 and 2 can be written as
x1  a11W1  a12W2
x2  a12W1  a22W2
and
Since the influence coefficients a11  a22 and a12  a21 and the mass disks m1 = m2 then
the deflections x1  x2  x . Therefore, this equation can be written as
12 
g
x
Substituting the known data gives
153 rad/s 
2
9.81 m/s2
(0.10 kg)(9.81 m/s2 )(3  104 m/N  a12 )
700
Solving for the influence coefficient gives
a12  1.27 x 104 m / N
Ans.
701
Chapter 14
Dynamics of Reciprocating Engines
14.1
A one-cylinder, four-stroke engine has a compression ratio of 7.6 and develops brake
power of 2.25 kW at 3 000 rev/min. The crank length is 22 mm with a 60-mm bore.
Develop and plot a rounded indicator diagram using a card factor of 0.90, a mechanical
efficiency of 72%, a suction pressure of 100 kPa and a polytropic exponent of 1.30.
A   D2 4    0.060 m  4  0.002 827 m2
2
v  2rA  2  0.022 m  0.002 827 m2 1 000 L/m3  0.124 4 L  124.4 mL
v1  v R  R  1  124.4 mL   7.6  7.6  1  143.2 mL
v2  v1  v  143.2 mL  124.4 mL  18.8 mL
C  v2 v  18.8 mL 124.4 mL  0.1511  15.11%
pb 
 2.25 kW  60 s/min  1 000 N  m/  kW  s   0.001 kPa  m 2 / N 
 724 kPa
 0.044 m   0.002 827 m 2  3 000 rev/min 2 rev/work stroke 
pi  pb em  724 kPa 0.72  1 005 kPa
p1  100 kPa
p4   k  1
R  1 pi
7.6  1 1005 kPa
 p1  1.30  1 1.3
 100 kPa  447 kPa
k
R  R fc
7.6  7.6 0.90
702
As in Example 14.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
18.8
1401
6266
5
25.0
966
4321
10
31.2
724
3238
15
37.5
572
2557
20
43.7
468
2094
25
49.9
394
1761
30
56.1
338
1512
35
62.3
295
1319
40
68.6
261
1165
45
74.8
233
1041
50
81.0
210
938
55
87.2
191
852
60
93.4
174
779
65
99.7
160
717
70 105.9
148
662
75 112.1
138
615
80 118.3
128
573
85 124.5
120
536
90 130.8
113
503
95 137.0
106
474
100 143.2
100
447
Then we sketch and round the following diagram:
4500
4000
3500
Pressure p, kPa
3000
2500
2000
1500
1000
500
0
0
20
40
60
Displacement X, %
80
100
120
703
14.2
Construct a rounded indicator diagram for a four-cylinder, four-stroke gasoline engine
having a 85-mm bore, a 90-mm stroke, and a compression ratio of 6.25. The operating
conditions to be used are 22.4 kW at 1 900 rev/min. Use a mechanical efficiency of 72%,
a card factor of 0.90, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.085 m  4  0.001 806 m2
2
v  A  0.090 m  0.001 806 m2 1 000 L/m3  0.162 54 L  162.54 mL
v1  v R  R  1  162.54 mL  6.25  6.25  1  193.5 mL
v2  v1  v  193.5 mL  162.54 mL  30.96 mL
C  v2 v  30.96 mL 162.54 mL  0.1905  19.05%
 22.4 kW  60 s/min  1 000 N  m/  kW  s   0.001 kPa  m 2 / N 
pb 
 8 704 kPa
 0.090 m   0.001 806 m 2  1 900 rev/min 2 rev/work stroke 
pi  pb em  8 704 kPa 0.72  12 090 kPa
p1  100 kPa
R  1 pi
6.25  1 12 090 kPa
 p1  1.30  1
 100 kPa  4 719 kPa
k
R  R fc
6.251.3  6.25
0.90
As in Example 14.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
31.0
1 083 51 102
5
39.1
800 37 744
10
47.2
626 29 526
15
55.3
509 24 019
20
63.5
426 20 100
25
71.6
364 17 186
30
79.7
317 14 944
35
87.8
279 13 172
40
96.0
249 11 741
45 104.1
224 10 564
50 112.2
203
9 580
55 120.4
185
8 748
60 128.5
170
8 036
65 136.6
157
7 420
70 144.7
146
6 883
75 152.9
136
6 411
80 161.0
127
5 994
85 169.1
119
5 622
90 177.2
112
5 289
95 185.4
106
4 990
100 193.5
100
4 719
p4   k  1
704
Then we sketch and round the following diagram:
45000
40000
35000
Pressure p, kPa
30000
25000
20000
15000
10000
5000
0
0
10
20
30
40
50
60
Displacement X, %
70
80
90
100
705
14.3
Construct an indicator diagram for a V6 four-stroke gasoline engine having a 100-mm
bore, a 90-mm stroke, and a compression ratio of 8.40. The engine develops 150 kW at
4 400 rev/min. Use a mechanical efficiency of 72%, a card factor of 0.88, a suction
pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.100 m  4  0.007 854 m2
2
v  A  0.090 m  0.007 854 m2 1 000 L/m3  0.707 L  707 mL
v1  v R  R  1  707 mL  8.40 8.40  1  803 mL
v2  v1  v  803 mL  707 mL  96 mL
C  v2 v  96 mL 707 mL  0.1358  13.58%
150 000 W 6 cyl  60 s/min  0.001 kPa  m 2 / N 
pb 
 965 kPa
 0.090 m   0.007 854 m 2   4 400 rev/min 2 rev/work stroke 
pi  pb em  965 kPa 0.72  1 340 kPa
p1  100 kPa
R  1 pi
8.40  1 1 340 kPa
 p1  1.30  1
 100 kPa  550 kPa
k
R  R fc
8.401.3  8.40 0.88
As in Example 14.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
96
1 590
8 751
5
131
1 056
5 812
10
167
774
4 259
15
202
603
3 315
20
237
488
2 687
25
272
408
2 243
30
308
348
1 914
35
343
302
1 661
40
378
266
1 462
45
414
237
1 302
50
449
213
1 170
55
484
193
1 061
60
520
176
968
65
555
161
888
70
590
149
820
75
626
138
760
80
661
129
708
85
696
120
661
90
732
113
620
95
767
106
583
100
802
100
550
p4   k  1
706
Then we sketch and round the following diagram:
707
14.4
A single-cylinder, two-stroke gasoline engine develops 30 kW at 4 500 rev/min. The
engine has an 80-mm bore, a stroke of 70 mm, and a compression ratio of 7.0. Develop a
rounded indicator diagram for this engine using a card factor of 0.990, a mechanical
efficiency of 65%, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.080 m  4  0.005 027 m2
2
v  A  0.070 m  0.005 027 m2 1 000 L/m3  0.352 L  352 mL
v1  v R  R  1  352 mL  7.0  7.0  1  411 mL
v2  v1  v  411 mL  352 mL  59 mL
C  v2 v  59 mL 352 mL  0.1662  16.62%
pb 
30 000 W  60 s/min   0.001 kPa  m 2 / N 
 0.070 m   0.005 027 m 2   4 500 rev/min 1 rev/work stroke 
 1 137 kPa
pi  pb em  1 137 kPa 0.65  1 749 kPa
p1  100 kPa
R  1 pi
7.0  1 1 749 kPa
 p1  1.30  1 1.3
 100 kPa  673 kPa
k
R  R fc
7.0  7.0 0.990
As in Example 14.1, we calculate the values:
x(%) v(mL) pc(kPa) pe(kPa)
0
59
1 259
8 473
5
76
894
6019
10
94
682
4 592
15
111
546
3 672
20
129
451
3 034
25
147
382
2 569
30
164
329
2 217
35
182
288
1 942
40
199
256
1 722
45
217
229
1 542
50
235
207
1 394
55
252
188
1 268
60
270
173
1 162
65
287
159
1 070
70
305
147
991
75
323
137
921
80
340
128
860
85
358
120
805
90
375
112
756
95
393
106
712
100
411
100
673
p4   k  1
708
Then we sketch and round the following diagram:
709
14.5
The engine of Prob. 14.1 has a connecting rod 80 mm long and a mass of 0.100 kg, with
the mass center 10 mm from the crankpin end. Piston mass is 0.180 kg. Find the bearing
reactions and the crankshaft torque during the expansion stroke corresponding to a piston
displacement of X = 30% ( t  60 ). To find pe , see the answer to Problem 14.1 in
Appendix B.
 0.080 m , m3  0.100 kg , m4  0.180 kg ,
A
 0.010 m ,
m3 A  m3 B
  0.100 kg  0.070 m  0.080 m  0.087 5 kg ,
m3B  m3 A
  0.100 kg  0.010 m  0.080 m  0.012 5 kg ,
B
  A  0.070 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s2 ,
2
  t  60 ,
 sin     0.022 m  cos60   0.080 m  1  0.275sin 60  0.088 7 m ,
2
X  30% , pe  1 512 kPa (from Prob. 14.1), A    0.060 m  4  0.002 827 m2 ,
P  pe A  1 512 kPa   0.002 827 m2   4 274 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    2 171 m/s 2   cos 60  0.275cos120   787 m/s 2


 0.2752

r2
2
tan   sin t 1  2 sin t   0.275sin 60 1 
sin 2 60   0.244 9
2
 2



ˆ
F41    m3B  m4  x  P  tan  j
r


   0.012 5 kg  0.180 kg  787 m/s2  4 274 N  0.244 9ˆj


 1 010ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj

Ans.

  0.180 kg  787 m/s2  4 274 N  ˆi  1 010ˆj N


 4 132ˆi  1 010ˆj N  4 254 N 13.7


Ans.
F32  [m3 Ar 2 cos t   m3B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3B  m4  x  P  tan  ˆj


  0.087 5 kg  2 171 m/s 2 cos 60  4 274 N  ˆi  1 010ˆj N


 4 179ˆi  1 010ˆj N  4 299 N166.4
T21  x  m3B  m4  x  P  tan kˆ   0.088 7 m 1 010 N  kˆ  89.59kˆ N  m
Ans.
Ans.
710
14.6
Repeat Problem 14.5, but do the computations for the compression cycle ( ωt  660 ).
 0.080 m , m3  0.100 kg , m4  0.180 kg ,
A
 0.010 m ,
m3 A  m3 B
  0.100 kg  0.070 m  0.080 m  0.087 5 kg ,
m3B  m3 A
  0.100 kg  0.010 m  0.080 m  0.012 5 kg ,
B
  A  0.070 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s2 ,
2
  t  660 ,
 sin     0.022 m  cos660   0.080 m  1  0.275sin 660  0.088 7 m ,
2
X  30% , pc  338 kPa (from Prob. 14.1), A    0.060 m  4  0.002 827 m2 ,
P  pc A   338 kPa   0.002 827 m2   956 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    2 171 m/s 2   cos 660  0.275cos1320   787 m/s2


 0.2752

r
r2
tan   sin t 1  2 sin 2 t   0.275sin 660 1 
sin 2 660   0.244 9
2
 2



ˆ
F41    m3B  m4  x  P  tan  j


   0.012 5 kg  0.180 kg  787 m/s2  956 N   0.244 9  ˆj


 197ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj

Ans.

  0.180 kg  787 m/s2  956 N  ˆi  197ˆj N


 814ˆi  197ˆj N  837 N13.6


Ans.
F32  [m3 Ar 2 cos t   m3B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3B  m4  x  P  tan  ˆj


  0.087 5 kg  2 171 m/s 2 cos 660  804 N  ˆi  157 ˆj N


 709ˆi  157ˆj N  726 N  167.5
T21  x  m3B  m4  x  P  tan kˆ   0.088 7 m  197 N  kˆ  17.47kˆ N  m
Ans.
Ans.
711
14.7
Make a complete force analysis of the engine of Problem 14.5. Plot a graph of the
crankshaft torque versus crank angle for 720 of crank rotation.
 0.080 m , m3  0.100 kg , m4  0.180 kg ,
m3 A  m3
B
 0.010 m ,
  0.100 kg  0.070 m  0.080 m  0.087 5 kg ,
m3 B  m3
A
  0.100 kg  0.010 m  0.080 m  0.012 5 kg ,
A
B
  A  0.070 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s2 ,
2
A    0.060 m  4  0.002 827 m2 , p  pe and/or pc as taken from Prob. 14.1,
2
P  Ap   0.002 827 m2  p ,
x  r cos t 
2
2
r

1   sin t    0.022 m  cos    0.080 m  1   0.275sin   ,


r
x  r 2 (cos t  cos 2t )    2 171 m/s 2   cos   0.275cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.275sin  1  0.037 8sin 2  
 2

F14   m3 B  m4  x  P  tan 
  0.192 5 kg  x  P  tan 
T21   m3B  m4  x  P  x tan   F14 x
t, 
x, m
X,%
P, N
x, m/s 2
tan 
F14 , N
T21 , N·m
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
0.102
0.101
0.098
0.094
0.089
0.083
0.077
0.071
0.067
0.063
0.060
0.059
0.058
0.059
0.060
0.063
0.067
0
2.27
8.43
18.12
30.23
43.59
57.01
69.47
80.23
88.83
95.03
98.76
100.00
98.76
95.03
88.83
80.23
14 276
15 218
10 115
6 412
4 249
3 042
2 326
1 888
1 615
1 444
1 340
1 283
1 264
1 264
1 264
1 264
1 264
-2 768
-2 614
-2 179
-1 535
-787
-45
597
1 078
1 384
1 535
1 582
1 580
1 574
1 580
1 582
1 535
1 384
0
0.071 36
0.138 80
0.198 13
0.244 91
0.275 00
0.285 40
0.275 00
0.244 91
0.198 13
0.138 80
0.071 36
0
-0.071 36
-0.138 80
-0.198 13
-0.244 91
0
1 050
1 346
1 212
1 004
834
697
576
461
345
228
113
0
-112
-218
-309
-375
0
106
132
114
89
69
54
41
31
22
14
7
0
-7
-13
-19
-25
712
255
270
285
300
315
330
345
360
375
390
405
420
435
450
465
480
495
510
525
540
555
570
585
600
615
630
645
660
675
690
705
720
0.071
0.077
0.083
0.089
0.094
0.098
0.101
0.102
0.101
0.098
0.094
0.089
0.083
0.077
0.071
0.067
0.063
0.060
0.059
0.058
0.059
0.060
0.063
0.067
0.071
0.077
0.083
0.089
0.094
0.098
0.101
0.102
69.47
57.01
43.59
30.23
18.12
8.43
2.27
0
2.27
8.43
18.12
30.23
43.59
57.01
69.47
80.23
88.83
95.03
98.76
100.00
98.76
95.03
88.83
80.23
69.47
57.01
43.59
30.23
18.12
8.43
2.27
0
1 264
1 264
1 264
1 264
1 264
1 264
1 264
774
283
283
283
283
283
283
283
283
283
283
283
283
287
300
324
361
422
521
681
950
1 434
2 262
3 402
3 961
1 078
597
-45
-787
-1 535
-2 179
-2 614
-2 768
-2 614
-2 179
-1 535
-787
-45
597
1 078
1 384
1 535
1 582
1 580
1 574
1 580
1 582
1 535
1 384
1 078
597
-45
-787
-1 535
-2 179
-2 614
-2 768
-0.275 00
-0.285 40
-0.275 00
-0.244 91
-0.198 13
-0.138 80
-0.071 36
0
0.071 36
0.138 80
0.198 13
0.244 91
0.275 00
0.285 40
0.275 00
0.244 91
0.198 13
0.138 80
0.071 36
0
-0.071 36
-0.138 80
-0.198 13
-0.244 91
-0.275 00
-0.285 40
-0.275 00
-0.244 91
-0.198 13
-0.138 80
-0.071 36
0
T21 (N∙m) vs. ωt (deg.)
-405
-394
-345
-272
-192
-117
-54
0
-16
-19
-2
32
75
114
135
135
115
82
42
0
-42
-84
-123
-154
-173
-181
-185
-196
-226
-256
-207
0
-29
-30
-29
-24
-18
-11
-5
0
-2
-2
0
3
6
9
10
9
7
5
2
0
-2
-5
-8
-10
-12
-14
-15
-17
-21
-25
-21
0
713
14.8 The engine of Problem 14.3 uses a connecting rod 300 mm long. The masses are
m3 A  0.80 kg, m3B  0.38 kg, and m4  1.64 kg. Find all the bearing reactions and the
crankshaft torque for one cylinder of the engine during the expansion stroke at a piston
displacement of X = 30% ( t  63.2 ). The pressure should be obtained from the
indicator diagram, Figure AP14.3 in Appendix B.
 0.300 m , m3 A  0.80 kg , m3B  0.38 kg , m4  1.64 kg ,
   4 400 rev/min  2 rad/rev   60s/min   460.8 rad/s , r  0.045 m ,
r
  0.045 m   0.300 m   0.150 , r 2   0.045 m  460.8 rad/s   9 554 m/s 2 ,
2
  t  63.2 ,
 sin     0.045 m  cos63.2   0.300 m  1  0.15sin 63.2  0.317 6 m ,
2
X  30.0% , pe  1 914 kPa (from Prob. 16.3), A    0.100 m  4  0.007 854 m2 ,
P  pe A  1 914 kPa   0.007 854 m2   15 033 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    9 554 m/s 2   cos 63.2  0.150cos126.4   3 457 m/s 2


 0.152

r2
2
tan   sin t 1  2 sin t   0.15sin 63.2 1 
sin 2 63.2   0.135
2
 2



ˆ
F41    m3B  m4  x  P  tan  j
r


   0.38 kg  1.64 kg  3 457 m/s 2  15 033 N  0.135ˆj


 1 087ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
   1.64 kg  3 457 m/s2  15 033 N  ˆi  1 087ˆj N


ˆ
ˆ
 9 364i 1 087 j N  9 426 N  6.6

Ans.

Ans.
F32  [m3 Ar 2 cos  t   m3B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3B  m4  x  P  tan  ˆj






  0.80 kg  9 554 m/s 2 cos 63.2  8 050 N  ˆi   0.80 kg  9 554 m/s 2 sin 63.2  1 087 N ˆj


ˆ
ˆ
 4 604i  7 909 j N  9 152 N120.2
T21  x  m3B  m4  x  P  tan  kˆ   0.317 6 m 1 087 N  kˆ  345kˆ N  m
Ans.
Ans.
714
14.9
Repeat Problem 14.8, but do the computations for the same position in the compression
cycle ( t  656.8 ).
 0.300 m , m3 A  0.80 kg , m3B  0.38 kg , m4  1.64 kg ,
   4 400 rev/min  2 rad/rev   60s/min   460.8 rad/s , r  0.045 m ,
r
  0.045 m   0.300 m   0.150 , r 2   0.045 m  460.8 rad/s   9 554 m/s 2 ,
2
  t  656.8 ,
x  r cos 
1 r
  sin     0.045 m  cos 656.8   0.300 m  1   0.15sin 656.8  0.317 6 m ,
2
2
2
X  30.0% , pc  348 kPa (from Prob. 16.3), A    0.100 m  4  0.007 854 m2 ,
2
P  pc A   348 kPa   0.007 854 m2   2 733 N
x  r 2 (cos t  cos 2t )    9 554 m/s 2   cos 656.8  0.150 cos1313.6   3 457 m/s 2
r


 0.152

r
r2
tan   sin t 1  2 sin 2 t   0.15sin 656.8 1 
sin 2 656.8   0.135
2
 2



F41    m3B  m4  x  P  tan  ˆj
     0.38 kg  1.64 kg  3 457 m/s 2  2 733 N   0.135  ˆj


ˆ
 574 j N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
   1.64 kg  3 457 m/s 2  2 733 N  ˆi  574ˆj N


ˆ
ˆ
 2 936i  574 j N  2 992 N191.0


Ans.
Ans.
F32  [m3 Ar 2 cos  t   m3B  m4  x  P ]ˆi  m3 Ar 2 sin  t   m3B  m4  x  P  tan  ˆj






  0.80 kg  9 554 m/s 2 cos 656.8  4 250 N  ˆi   0.80 kg  9 554 m/s 2 sin 656.8  574 N ˆj


ˆ
ˆ
 7 696i  5 534 j N  9 479 N  35.7
T21  x  m3B  m4  x  P  tan  kˆ   0.317 6 m  574 N  kˆ  182kˆ N  m
Ans.
Ans.
715
14.10 Additional data for the engine of Problem 14.4 are l3  110 mm, RG 3 A  15 mm,
m4  0.24 kg, and m3  0.13 kg. Make a complete force analysis of the engine and plot
a graph of the crankshaft torque versus crank angle for 360 of crank rotation.
 0.110 m , m3  0.13 kg , m4  0.24 kg ,
A
 0.015 m ,
m3 A  m3 B
  0.13 kg  0.095 m  0.110 m  0.112 kg ,
m3B  m3 A
  0.13 kg  0.015 m  0.110 m  0.018 kg ,
B
  A  0.095 m ,
   4 500 rev/min  2 rad/rev   60 s/min   471.24 rad/s , r  0.035 m ,
r
  0.035 m   0.110 m   0.318 , r 2   0.035 m  471.24 rad/s   7 772 m/s 2 ,
2
A   D2 4    0.080 m  4  0.005 027 m2 , p  pe or pc as taken from Prob. 14.4,
2
P  Ap   0.005 027 m2  p ,
x  r cos t 
1 r
 sin 2 t   0.035 m  cos    0.110 m  1   0.318sin   ,
2
r
x  r 2 (cos t  cos 2t )    7 772 m/s 2   cos   0.318cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.318sin  1  0.050 62sin 2  
 2

F14   m3B  m4  x  P  tan 
T21   m3B  m4  x  P  x tan 


  0.005 027 m2 p   2005 N  cos   0.318cos 2   x tan 


2
716
t, 
P, N
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
31 945
35 849
24 949
16 102
10 845
7 812
6 022
4 943
4 263
3 831
3 567
3 429
1 943
510
531
568
635
733
897
1 160
1 609
2 394
3 706
5 508
31 945
x, m/s 2
-10 245
-9 649
-7 968
-5 496
-2 650
130
2 473
4 153
5 123
5 496
5 495
5 366
5 299
5 366
5 495
5 496
5 123
4 153
2 473
130
-2 650
-5 496
-7 968
-9 649
-10 245
x, m
tan 
F14 , N
T21 , N·m
0.145 00
0.143 43
0.138 91
0.131 93
0.123 24
0.113 74
0.104 28
0.095 62
0.088 24
0.082 43
0.078 29
0.075 82
0.075 00
0.075 82
0.078 29
0.082 43
0.088 24
0.095 62
0.104 28
0.113 74
0.123 24
0.131 93
0.138 91
0.143 43
0.145 00
0
0.082 63
0.161 10
0.230 68
0.286 02
0.321 86
0.334 29
0.321 86
0.286 02
0.230 68
0.161 10
0.082 63
0
-0.082 63
-0.161 10
-0.230 68
-0.286 02
-0.321 86
-0.334 29
-0.321 86
-0.286 02
-0.230 68
-0.161 10
-0.082 63
0
0
2 757
3 688
3 387
2 906
2 525
2 226
1 936
1 597
1 211
803
398
0
-157
-314
-458
-560
-581
-513
-384
-265
-225
-266
-249
0
0
395.4
512.3
446.9
358.2
287.2
232.2
185.1
140.9
99.8
62.9
30.2
0
-11.9
-24.6
-37.8
-49.4
-55.5
-53.5
-43.7
-32.6
-29.7
-36.9
-33.8
0
600
500
T21, N.m
400
300
200
100
0
0
90
180
270
-100
wt, deg
T21 (N∙m) vs. ωt (deg)
360
717
14.11 The four-stroke engine of Problem 14.1 has a stroke of 66 mm and a connecting rod
length of 183 mm. The mass of the rod is 0.386 kg, and the center of mass is 42 mm
from the crankpin. The piston assembly has mass of 0.576 kg. Make a complete force
analysis for one cylinder of this engine for 720 of crank rotation. Use 110 kPa for the
exhaust pressure and 70 kPa for the suction pressure. Plot a graph to show the variation
of the crankshaft torque with the crank angle. Use Figure 14.23 for the pressures.
 0.183 m , m3  0.386 kg , m4  0.576 kg ,
A
 0.042 m ,
m3 A  m3 B
  0.386 kg  0.141 m  0.183 m  0.297 4 kg ,
m3B  m3 A
  0.386 kg  0.042 m  0.183 m  0.088 6 kg ,
B
  A  0.141 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.033 m ,
r
  0.033 m   0.183 m   0.180 , r 2   0.033 m  314.16 rad/s   3 257 m/s 2 ,
2
A    0.060 m  4  0.002 83 m2 , p  pe or pc as taken from Example 14.1,
2
P  Ap   0.002 83 m2  p ,
2
2
r

x  r cos t  1   sin t    0.033 m  cos    0.183 m  1   0.180sin   ,


r
x  r 2 (cos t  cos 2t )    3 257 m/s 2   cos   0.180cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.180sin  1  0.016 2sin 2  
 2

F14   m3B  m4  x  P  tan 
T21   m3B  m4  x  P  x tan 
t, 
P, N
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
14 279
16 066
10 688
6 792
4 445
3 230
2 432
1 976
1 649
1 461
1 357
1 288
1 263
1 263
1 263
x, m/s 2
-3 843
-3 654
-3 114
-2 303
-1 335
-335
586
1 351
1 922
2 303
2 528
2 638
2 671
2 638
2 528
x, m
tan 
F14 , N
T21 ,N·m
0.216
0.215
0.211
0.205
0.197
0.189
0.180
0.172
0.164
0.158
0.154
0.151
0.150
0.151
0.154
0
0.046 64
0.090 36
0.128 31
0.157 78
0.176 49
0.182 92
0.176 49
0.157 78
0.128 31
0.090 36
0.046 64
0
-0.046 64
-0.090 36
0
636
778
675
561
531
516
507
462
384
274
142
0
-137
-266
0
136.8
164.3
138.4
110.6
100.3
92.9
87.2
75.7
60.6
42.3
21.4
0
-20.7
-40.9
718
225
240
255
270
285
300
315
330
345
360
375
390
405
420
435
450
465
480
495
510
525
540
555
570
585
600
615
630
645
660
675
690
705
720
200
1 263
1 263
1 263
1 263
1 263
1 263
1 263
1 263
1 263
773
283
283
283
283
283
283
283
283
283
283
283
283
289
305
327
371
440
543
723
993
1 521
2 378
3 588
3 962
2 303
1 922
1 351
586
-335
-1 335
-2 303
-3 114
-3 654
-3 843
-3 654
-3 114
-2 303
-1 335
-335
586
1 351
1 922
2 303
2 528
2 638
2 671
2 638
2 528
2 303
1 922
1 351
586
-335
-1 335
-2 303
-3 114
-3 654
-3 843
0.158
0.164
0.172
0.180
0.189
0.197
0.205
0.211
0.215
0.216
0.215
0.211
0.205
0.197
0.189
0.180
0.172
0.164
0.158
0.154
0.151
0.150
0.151
0.154
0.158
0.164
0.172
0.180
0.189
0.197
0.205
0.211
0.215
0.216
-0.128 31
-0.157 78
-0.176 49
-0.182 92
-0.176 49
-0.157 78
-0.128 31
-0.090 36
-0.046 64
0
0.046 64
0.090 36
0.128 31
0.157 78
0.176 49
0.182 92
0.176 49
0.157 78
0.128 31
0.090 36
0.046 64
0
-0.046 64
-0.090 36
-0.128 31
-0.157 78
-0.176 49
-0.182 92
-0.176 49
-0.157 78
-0.128 31
-0.090 36
-0.046 64
0
-358
-401
-381
-302
-184
-59
34
73
54
0
-100
-161
-160
-95
11
123
208
246
233
177
95
0
-95
-179
-238
-260
-236
-171
-88
-17
1
-28
-54
0
-56.6
-65.7
-65.6
-54.4
-34. 7
-11.7
7.0
15.4
11.7
0
-21.5
-34.1
-32.8
-18.8
2.0
22.1
35.8
40.4
36.8
27.3
14.3
0
-14.4
-27.6
-37.6
-42.6
-40.6
-30.7
-16.7
-3.3
0.3
-5.9
-11.6
0
150
T21, N.m
100
50
0
0
90
180
270
360
450
-50
-100
T21 (N∙m) vs. ωt (deg)
wt, deg
540
630
720
810
719
Chapter 15
Balancing
15.1
Determine the bearing reactions at A and B if the speed of the shaft is 300 rev/min. Also
determine the magnitude and orientation of the balancing mass if it is located at a radius
of 50 mm.
a  800 mm, b  200 mm, R1  25 mm, R2  35 mm, R3  40 mm, m1  2 kg,
m2  1.5 kg, and m3  3 kg.
   300 rev/min  2 rad/rev   60 s/min   31.416 rad/s
F1  m1R1 2   2 kg  0.025 m  31.416 rad/s   49.348 N
2
F2  m2 R2 2  1.5 kg  0.035 m  31.416 rad/s   51.815 N
2
F3  m3 R3 2   3 kg  0.040 m  31.416 rad/s   118.435 N
F  49.348 N90  49.349ˆj N
2
1
F2  51.815 N  165  50.050ˆi  13.411ˆj N
F  118.435 N  75  30.653ˆi  114.400ˆj N
3
 F  F  F  F  19.397ˆi  78.462ˆj N  80.824 N  103.9
1
2
3
720
Since all rotating masses are in a single plane, the correction mass must be in that plane.
 M A  0.200kˆ m × 80.824 N 103.9  1.000kˆ m ×FB  0
FB  16.276.1 N
Ans.
 M  0.800kˆ m × 80.824 N 103.9 1.000kˆ m ×F  0
A
A
FA  64.776.1 N
Ans.
FC   F  80.824 N76.1
FC  mC RC 2  mC  0.050 m  31.416 rad/s   80.824 N
2
mC  FC  RC 2   80.824 N   0.050 m  31.416 rad/s    1.638 kg


C  76.1
2
Ans.
Ans.
721
15.2
Three weights are connected to a shaft that rotates in bearings at A and B. Determine the
magnitudes and orientations of the bearing reactions if the speed of the shaft is 300
rev/min. Also, determine the magnitude and orientation of a counterweight that is to be
located at a radius of 10 in.
a  6 in, b  12 in, R1  8 in, R2  12 in, R3  6 in, w1  2 oz, w2  1.5 oz, and w3  3 oz.
   300 rev/min  2 rad/rev   60 s/min   31.416 rad/s
 2 oz 8 in  31.416 rad/s   2.556 lb
F1  m1 R1 
16 oz/lb   386 in/s2 
2
1.5 oz 12 in  31.416 rad/s 

2
F2  m2 R2 
 2.876 lb
16 oz/lb   386 in/s2 
2
3 oz  6 in  31.416 rad/s 

2
F3  m3 R3 
 2.876 lb
16 oz/lb   386 in/s2 
2
2
F1  2.556 lb90  2.556ˆj lb
F2  2.876 lb  135  2.034ˆi  2.034ˆj lb
F  2.876 lb  30  2.491ˆi  1.438ˆj lb
3
 F  F  F  F  0.457ˆi  0.915ˆj lb  1.023 lb  63.5
1
2
3
Since all rotating masses are in a single plane, the correction mass must be in that plane.
FC   F  1.023 lb116.5
FC  mC RC 2  mC 10.0 in  31.416 rad/s   1.023 lb
2
1.023 lb  386 in/s 2 
FC
mC 

 0.040 lb
RC 2 10.0 in  31.416 rad/s 2
Ans.
722
C  116.5
Ans.
Without the correction mass
 M A 18.0 in kˆ ×FB  12.0 in kˆ ×  F  0
FB  0.682 lb116.5
 M  18.0 in kˆ  F  6.0 in kˆ   F  0
Ans.
FA  0.341 lb116.5
Ans.
B
A
723
15.3
Two weights are connected to a rotating shaft and mounted outboard of bearings A and B.
If the speed of the shaft is 120 rev/min, what are the magnitudes and orientations of the
bearing reactions at A and B? Suppose the system is to be balanced by removing weight
at a radius of 5 in. Determine the magnitude and orientation of the weight to be removed.
l  4 in, a  2 in, R1  4 in, R2  6 in, w1  4 lb, and w2  3 lb.
  120 rev/min  2 rad/rev   60 s/min   12.566 rad/s
 4 lb  4 in 12.566 rad/s   6.544 lb
F  m R 
2
2
1
1
1
386 in/s 2
 3 lb  6 in 12.566 rad/s   7.362 lb
F m R 
2
2
2
2
2
386 in/s 2
F1  6.544 lb90  6.544ˆj lb
F2  7.362 lb  135  5.206ˆi  5.206ˆj lb
F  F  F  5.206ˆi  1.338ˆj lb  5.375 lb165.6

 M   6 in  kˆ ×  F   4 in  kˆ × F
1
2
B
A


  8.030 in  lb  ˆi   31.235 in  lb  ˆj   4 in  kˆ × FAx ˆi  FAy ˆj  0
FA  7.809ˆi  2.007ˆj lb  8.063 lb 14.4
F    F  F   2.603ˆi  0.669ˆj lb  2.688 lb165.6
B
A

Ans.
Ans.
The correction force should be
FC   F  5.375 lb165.6
This can be done by mass removal at a radius of 5.0 in of
5.376 lb  386 in/s 2 
FC
mC 

 2.628 lb
RC 2 5.0 in 12.566 rad/s 2
C  14.4
Ans.
Ans.
724
15.4
If the speed of the shaft is 220 rev/min, calculate the magnitudes and orientations of the
bearing reactions at A and B for the two-mass system.
a  b  250 mm, c  50 mm, R1  60 mm, R2  40 mm, m1  2 kg, and m2  1.5 kg.
   220 rev/min  2 rad/rev   60 s/min   23.038 rad/s
F1  m1R1 2   2 kg  0.060 m  23.038 rad/s   63.692 N
2
F2  m2 R2 2  1.5 kg  0.040 m  23.038 rad/s   31.846 N
F  63.692 N90  63.692ˆj N
2
1
F2  31.846 N  90  31.846ˆj N
 M  0.250kˆ m × 63.692ˆj N  0.550kˆ m × 31.846ˆj N  0.500kˆ m ×F  0
B
 

 
 
FA  3.185ˆj N  3.185 N90.0
F    F  F  F   35.031ˆj N  35.031 N  90.0
B
1
2
A
 

A
Ans.
Ans.
725
15.5
Determine the bearing reactions at A and B and their orientations if the shaft speed is 100
rev/min.
a  c  300 mm, b  600 mm, R1  R2  60 mm, m1  2 kg, and m2  1.5 kg.
  100 rev/min  2 rad/rev   60 s/min   10.472 rad/s
F1  m1R1 2  1 kg  0.060 m 10.472 rad/s   6.580 N
2
F2  m2 R2 2   3 kg  0.060 m  23.038 rad/s   19.739 N
F  6.580 N90  6.580ˆj N
2
1
F2  19.739 N  90  19.739ˆj N
 M  0.300kˆ m × 6.580ˆj N  0.900kˆ m × 19.739ˆj N  1.200kˆ m ×F  0
B

 
FA  13.159ˆj N  13.159 N90.0
FB    F1  F2  FA   0
 
 
 

A
Ans.
Ans.
726
15.6
The rotating shaft illustrated in Figure P15.5 supports two masses m1 and m2 whose
weights are 4 lb and 5 lb, respectively. The dimensions are a  2 in, b  8 in,
c  3 in, R1  4 in, R2  3 in, Find the magnitudes of the rotating-bearing reactions at A and
B and their orientations if the shaft speed is 360 rev/min.
   360 rev/min  2 rad/rev   60 s/min   37.699 rad/s
 4 lb  4 in  37.699 rad/s   58.897 lb
F  m R 
2
2
1
1 1
386 in/s 2
 5 lb  3 in  37.699 rad/s   55.216 lb
F m R 
2
2
2
2
2
386 in/s 2
F1  58.897 lb90  58.897ˆj lb
F  55.216 lb  90  55.216ˆj lb
2
 M   2 in  kˆ × F   10 in  kˆ × F   13 in  kˆ × F
B
1
2
A


 117.947 in  lb  ˆi   552.163 in  lb  ˆi   13 in  kˆ × FAx ˆi  FAy ˆj  0
FA  33.413ˆj lb  33.413 lb90
F    F  F  F   37.094ˆj lb  37.094 lb  90
B
1
2
A
Ans.
Ans.
727
15.7
The shaft is to be balanced by placing masses in the correction planes L and R. Calculate
the magnitudes and orientations of the correction masses.
a  1 in, b  e  8 in, c  10 in, d  9 in, R1  R3  5 in, R2  4 in, w1  w3  4 oz, and
w2  3 oz.


m1R1   4 oz  5 inˆj  20.000ˆj oz  in
m2 R 2   3 oz  4 in  150  12 oz  in  150  10.392ˆi  6.000ˆj oz  in
m R   4 oz  5 in  60   20 oz  in  60  10.000ˆi  17.321ˆj oz  in
3
3
Using Eqs. (15.6) and (1.7),
m1R1 18 in 35 in  10.286ˆj oz  in
m R 27 in 35 in  8.017ˆi  4.629ˆj oz  in
2
2
m3R3 8 in 35 in  2.286ˆi  3.959ˆj oz  in
m R  5.731ˆi  1.698ˆj oz  in  5.978 oz  in 16.5
Ans.
mR R R  5.339ˆi  5.019ˆj oz  in  7.328 oz  in136.8
Ans.
L
L
728
15.8
The shaft of Prob. 15.7 is to be balanced by removing weight from the two correction
planes. Determine the correction masses and their orientations.
mL R L  5.978 oz  in163.5
mR R R  7.328 oz  in  43.2
Ans.
Ans.
729
15.9
The shaft illustrated in Figure P15.7 is to be balanced by removing masses in the two
correction planes L and R. The three masses are m1  6 g, m2  7 g, and m3  5 g . The
dimensions are a  25 mm, b  300 mm, c  600 mm, d  150 mm, e  75 mm,
R3  100 mm. Calculate the magnitudes and
R1  125 mm, R2  150 mm, and
orientations of the correction masses.


m1R1   6 g  125 mmˆj  750.000ˆj g  mm
m2 R 2   7 g 150 mm 150  1 050 g  mm 150  909.327ˆi  525.000ˆj g  mm
m R   5 g 100 mm  60  500 g  mm  60  250.000ˆi  433.013ˆj g  mm
3
3
Using Eqs. (17.6) and (17.7),
m1R1 900 mm 1 125 mm  600.000ˆj g  mm
m R 1 050 mm 1 125 mm  848.705ˆi  490.000ˆj g  mm
2
2
m3R3 300 mm 1 125 mm  66.667ˆi  115.470ˆj g  mm
The masses to be removed are:
mL R L  782.038ˆi  5.470ˆj g  mm  782.057 g  mm  179.6
m R  122.711ˆi  202.543ˆj g  mm  236.816 g  mm  58.8
R
R
Ans.
Ans.
730
15.10 Repeat Problem 15.9 if masses are to be added in the two correction planes.
mL R L  782.038ˆi  5.470ˆj g  mm  782.057 g  mm0.4
m R  122.711ˆi  202.543ˆj g  mm  236.816 g  mm121.2
R
R
Ans.
Ans.
731
15.11 Solve the two-plane balancing problem as stated in Sec. 15.8.
This is an experimental procedure and is explained in Sec. 15.8; no further solution
process is shown here.
15.12 A rotor to be balanced in the field yielded an amplitude of 5 at an angle of 142 at the
left-hand bearing and an amplitude of 3 at an angle of 22 at the right-hand bearing
because of unbalance. To correct this, a trial mass of 12 was added to the left-hand
correction plane at an angle of 210 from the rotating reference. A second run then gave
left-hand and right-hand responses of 8160 and 4260 , respectively. The first trial
mass was then removed and a second mass of 6 was added to the right-hand correction
plane at an angle of 70. The responses to this were 274 and 4.5  80 for the leftand right-hand bearings, respectively. Determine the original unbalances.
X A  5142 , XB  3  22 ,
m L  12210 , X AL  8160 , XBL  4260 ,
m R  6  70 , X AR  274 , XBR  4.5  80 .
These gave the following results from a programmable calculator run:
M L  6.05234 , M R  5.9865.2
Ans.
732
15.13 A rotor is rotating with a constant angular velocity   50 rad/s, and is dynamically
balanced by the two correcting masses. A decision has been made to use planes 1 and 3
instead of planes 1 and 2. Determine the magnitudes and orientations of the new masses
(m1 )new and (m3 )new at the same radii R1  50 mm and R3  150 mm
a  d  60 mm, b  100 mm, c  80 mm, R1  50 mm, R2  150 mm, m1  3 kg, and m2  2 kg.
The inertial forces due to the original correcting masses are
F1  m1R1 2  (3 kg)(0.05 m) 2  0.15 2 N
F2  m2 R2 2  (2 kg)(0.15 m) 2  0.3 2 N
These can be written as
F1  0.15 2 N90  0ˆi  0.15 2ˆj N
F  0.30 2 N300  0.15 2ˆi  0.26 2ˆj N
2
The sum of the inertial forces due to the two original correcting masses can be written as
 F  F1  F2  0.152ˆi  0.112ˆj N  0.1862 N323.75
Therefore, the sum of the reaction forces at the two bearings can be written as
FA  FB    F1  F2   0.15 2ˆi  0.11 2ˆj N
Two Procedures: (i) For dynamic balance, the sum of the moments about the correcting
planes (1) and (3) due to the new correcting masses must be the same as the sum of the
moments about the old correcting planes (1) and (2) due to the original correcting masses.
(ii) Also, for dynamic balance, the sum of the moments about any point in the shaft must
be zero. Therefore, we can use the sum of the moments to determine the reaction forces
at A and B for the unbalanced system. Then we can use these answers to find the
correcting masses in planes (1) and (3).
For example, use procedure (i). Consider the sum of the moments about the correcting
plane (3).
0.18kˆ  (F1 )new  0.18kˆ  F1  0.08kˆ  F2
Substituting Eq. (1) and performing the cross-products gives
  F1y  ˆi   F1x  ˆj  0.0344 2ˆi  0.0666 2ˆj N
new
new
Rearranging this, the inertial force in correcting plane (1) can be written as
733
 F1 new  0.06662ˆi  0.03442ˆj  0.0752 N27.3
The correcting mass in plane (1) can be written as
F1
0.075 2 N
 1.5 kg
 m1 new  2  2
 R1   0.050 m 
Ans.
The orientation of the correcting mass in plane (1) is 1  27.3.
Ans.
The sum of the moments about the correcting plane (1) due to the original correcting
masses is equal to the sum of the moments about the same correcting plane due to the
new correcting masses, that is
0.18kˆ  (F3 ) new  0.10kˆ  F2
( F y ) ˆi  ( F x ) ˆj  0.1444 2ˆi  0.0833 2ˆj N
3
new
3
new
(F3 ) new  0.0833 2ˆi  0.1444 2ˆj  0.167 2 N300
The correcting mass in correction plane (3) can be written as
( F3 ) new 0.167 2 N
(m3 ) new  2

 1.113 kg
Ans.
 R3  2 (0.15 m)
The orientation of the correcting mass in plane (3) is 3  300.
Ans.
Check: Use Procedure (ii). For dynamic balance, the sum of the two new inertial forces
must be the same as the sum of the two original inertial forces, that is
 F1 new   F3 new  F1  F2
 F1 new   F3 new  0.0666 2ˆi  0.0334 2ˆj  0.0833 2ˆi  0.1444 2ˆj N
 F1 new   F3 new  0.15 2ˆi  0.11 2ˆj N
Note that this agrees with the answer given above.
734
15.14 The shaft is rotating with a constant angular velocity   80 rad/s. For dynamic balance,
determine the magnitudes and angular locations of the correcting masses to be removed in
planes (1) and (2) at radii RC1  RC 2  2.6 in.
a  10 in, b  16 in, c  24 in, R1  1.2 in, R2  2 in, m1  8.8 lb, and m2  4.4 lb.
The inertial forces of the rotating particles are
 8.8 lb 
F1  m1R1 2  
(1.2 in)(80 rad/s) 2  175.1 lb180  175.1ˆi lb
2 
 386 in/s 
 4.4 lb 
F2  m2 R2 2  
(2 in)(80 rad/s)2  145.9 lb120  72.95ˆi  126.35ˆj lb
2 
 386 in/s 
The sum of the moments about bearing A can be written as
10kˆ in  FBx ˆi  FBy ˆj  16kˆ in  ( 175.1 ˆi lb)  24kˆ in  ( 72.95 ˆi  126.35 ˆj lb)  0


10 FBx ˆj  10 FBy ˆi  2801.60ˆj  1750.80ˆj  3032.40ˆi  0
from which we find
FBx  455.24 lb
and
FBy  303.24 lb
F  455.24ˆi  303.24ˆj lb  547.0 lb33.67
B
The sum of the inertial forces of the two rotating masses can be written as
F1  F2  248.05ˆi  126.35ˆj lb  278.38 lb153.0
Therefore, the force at bearing A can be written as
FA  FB   F1  F2   0
FA    F1  F2   FB  207.19ˆi  176.89ˆj lb  272.43 lb139.51
The sum of the two correcting forces can be written as
FC1  FC 2    F1  F2 
The sum of the moments about correction plane (2) can be written as
735
8kˆ in  F1  8kˆ in  FC1  0
8 in 175.1 lbˆj  8 in F x ˆj  8 in F y ˆi  0
C1
C1
F  175.1 lb
and
F 0
F  175.1ˆi lb  175.1 lb0  175.1 lb180
x
C1
y
C1
C1
The correcting mass in plane (1) can be written as
F
175.1 lb
mC1  2 C1 
 0.01052 lb  s2 /in  4.062 lb
2
 RC1 80 rad/s  2.6 in
The orientation of the correcting mass to be removed from correction plane (1) is
C1  180
The sum of the correction forces can be written as
FC1  FC 2    F1  F2   248.05ˆi  126.35ˆj lb
Ans.
Ans.
From this, the correction force in plane (2) can be written as
FC 2    F1  F2   FC1  248.05ˆi  126.35ˆj lb  175.1ˆi lb
=72.95ˆi  126.35ˆj lb  145.90 lb  60  145.90 lb120
The correcting mass in plane (2) can be written as
F
145.9 lb
mC 2  2 C 2 
 0.00877 lb  s2 /in  3.384 lb
2
 RC 2 80 rad/s  2.6 in
The orientation of the correcting mass to be removed from correction plane (2) is
C 2  120
These answers are shown in the following figure.
The locations of the correction masses.
Ans.
Ans.
736
15.15 The shaft is simply supported by the bearings at A and B and is rotating with a constant
angular velocity   50 rad/s. Determine the magnitudes and angular locations of the
correcting masses that must be removed in the correcting planes (1) and (2) to
dynamically balance the system. The radial distances from the shaft axis are specified as
RC1  RC2  70 mm. Show the orientations of the correcting masses on the right-hand
figure.
a  400 mm, b  150 mm, c  300 mm, R1  60 mm, R2  35 mm, m1  12 kg, and m2  15 kg.
The inertial forces of the two rotating mass particles are
F1  m1R1 2  (12 kg)(0.060 m)(50 rad/s)2  1800 N
F2  m2 R2 2  (15 kg)(0.035 m)(50 rad/s)2  1312.5 N
Therefore, the inertial forces of the two rotating mass particles can be written as
F1  1800 N75  465.87ˆi  1738.67ˆj N
F  1312.5 N240  656.25ˆi  1136.66ˆj N
2
The sum of the inertial forces of the two rotating mass particles can be written as
 F  F1  F2  190.38 ˆi  602.01ˆj N  631.39 N107.55
Therefore, the reaction forces at bearings A and B can be written as
FA  FB    F1  F2   190.38 ˆi  602.01ˆj N
(1)
The sum of the moments about bearing A can be written as
0.4kˆ m  F  0.55kˆ m  (465.87ˆi  1738.67ˆj N)  0.7kˆ m  ( 656.25ˆi  1136.66ˆj N)  0
B
The x and y components of this equation can be written as
0.4 mFBy  ( 0.55 m)(1738.67 N)  ( 0.7 m)(1136.66 N)  0
0.4 mFBx  ( 0.55 m)(465.87 N)  ( 0.7 m)( 656.25 N)  0
Therefore, the x and y components of the force at bearing B are
FBy  4379.83 N
and
FBx  507.87 N
Therefore, the force at bearing B can be written as
FB  507.87ˆi  4379.83ˆj N  4409.18 N  83.39
Substituting this into Eq. (1) gives
737
FA  507.87ˆi  4379.83ˆj N  190.38 ˆi  602.01ˆj N
Therefore, the force at bearing A can be written as
FA  317.49ˆi  3777.82ˆj N  3791.14 N94.80
The sum of the two correcting forces can be written from Eq. (1) as
FC1  FC 2    F1  F2   190.38 ˆi  602.01ˆj N
(2)
The sum of the moments about correcting plane (2) can be written as
0.15 mkˆ  FC1  0.15 mkˆ  F1  0
0.15 mF x ˆj  0.15 mF y ˆi  0.15 m(465.87 N)ˆj  0.15 m(1738.67 N)ˆi  0
C1
C1
From this, the inertial force in correcting plane (1) can be written as
FC1  465.87ˆi  1738.67ˆj N  1800 N  105  1800 N75
The correcting mass removal in correcting plane (1) can be written as
F
1800 N
mC1  2 C1 
 10.286 kg
 RC1  50 rad/s 2  0.07 m 
The orientation of the correcting mass removal in correcting plane (1) is
C1  75
Substituting Eq. (3) into Eq. (2), the force in the second correcting plane is
FC 2  190.38 Nˆi  602.01 Nˆj+465.87 Nˆi  1738.67 Nˆj
 656.25ˆi  1136.66ˆj N  1312.50 N60  1312.50 N  120
(3)
Ans.
Ans.
The correcting mass removal in correcting plane (2) can be written as
F
1312.50 N
mC 2  2 C 2 
 7.500 kg
 RC 2  50 rad/s 2  0.07 m 
Ans.
The orientation of the correcting mass removal in correcting plane (2) is
C 2  120
These answers are shown in the figure below.
Ans.
The locations of the correction masses.
738
15.16 The shaft simply supported by bearings at A and B is rotating clockwise with a constant
angular velocity   75 rad/s . Tto dynamically balance the system, determine the
magnitudes and orientations of the correcting masses that must be removed in the
correcting planes (1) and (2), at the radial distances RC1  RC 2  55 mm.
a  250 mm, b  400 mm, c  600 mm, R1  25 mm, R2  40 mm, m1  5 kg, and m2  7 kg.
The inertial forces of the rotating particles are
F1  m1R1 2  5 kg(0.025 m)(75 rad/s)2  703.13 N180  703.13ˆi N
F  m R  2  7 kg(0.040 m)(75 rad/s)2  1575 N120  787.50ˆi  1364.00ˆj N
2
2
2
The sum of the moments about bearing A can be written as
0.25kˆ m  FBx ˆi  FBy ˆj  0.4kˆ m  ( 703.13ˆi N)  0.6kˆ m  ( 787.50ˆi  1364ˆj N)  0


0.25FBx ˆj  0.25FBy ˆi  281.25ˆj  472.50ˆj  818.40ˆi  0
from which we find
FBx  3015 N
and
FBy  3273.60 N
F  3015ˆi  3273.60ˆj N  4450.47 N  47.35
B
The sum of the inertial forces of the two rotating masses can be written as
F1  F2  1490.63ˆi  1364.00ˆj N  2020.51 N137.53
Therefore, the force at bearing A can be written as
FA  FB   F1  F2   0
FA    F1  F2   FB  1697.82ˆi  1540.89ˆj N  2292.80 N137.77
The sum of the two correcting forces can be written as
FC1  FC 2    F1  F2 
The sum of the moments about correction plane (2) can be written as
739
0.2kˆ m  F1  0.2kˆ m  FC1  0
0.2 m 703.13 Nˆj  0.2 m F x ˆj  0.2 m F y ˆi  0
C1
C1
F  703.13 N
and
F 0
F  703.13ˆi N  703.13 N0  703.13 N180
x
C1
y
C1
C1
The correcting mass in plane (1) can be written as
F
703.13 N
mC1  2 C1 
 2.273 kg
 RC1  75 rad/s 2 0.055 m
The orientation of the correcting mass to be removed from correction plane (1) is
C1  180
The sum of the correction forces can be written as
FC1  FC 2    F1  F2   1490.63ˆi  1364.00ˆj N
Ans.
Ans.
From this, the correction force in plane (2) can be written as
FC 2    F1  F2   FC1  1490.63ˆi  1364ˆj N  703.13ˆi N
=787.50ˆi  1364.00ˆj N  1575.00 N  60  1575.00 N120
The correcting mass in plane (2) can be written as
F
1575.00 N
mC 2  2 C 2 
 5.091 kg
 RC 2  75 rad/s 2 0.055 m
The orientation of the correcting mass to be removed from correction plane (2) is
C 2  120
These answers are shown in the following figure.
The locations of the correction masses.
Ans.
Ans.
740
15.17 Three mass particles are rigidly attached to a simply supported shaft which is rotating
with a constant angular velocity   15 rad/s . Determine the magnitudes and orientations
of the bearing reaction forces at A and B. Then determine the magnitudes and
orientations of the correcting masses that must be added in the two correcting planes (1)
and (2) to dynamically balance the system. The correcting masses are to be placed at
radial distances RC1 = RC2 = 25 mm from the shaft axis.
a = 50 mm, b = 215 mm, c = 275 mm, d = 100 mm, e = 150 mm, f = 125 mm R1 = 15
mm, R2 = 35 mm, R3 = 20 mm, m1  3 kg, m2  1kg, and m3  2 kg.
The magnitudes of the inertial forces caused by the rotating mass particles are
F1  m1 R1 2   3 kg  (0.015 m)(15 rad/s) 2  10.125 N
F2  m2 R2 2  1 kg  (0.035 m)(15 rad/s) 2  7.875 N
F3  m3 R3 2   2 kg  (0.020 m)(15 rad/s) 2  9.000 N
In vector form, these inertial forces can be written as
F1  10.125 N60  5.063ˆi  8.769ˆj N
F  7.875 lb  150  6.820ˆi  3.938ˆj N
2
F3  9.000 lb  30  7.794ˆi  4.500ˆj N
The sum of the moments about bearing A can be written as
 M  0  0.215 mkˆ  5.063ˆi  8.769ˆj N  0.490 mkˆ  6.820ˆi  3.938ˆj N
A






0.590 mkˆ  7.794ˆi  4.500ˆj N  0.740 mkˆ  FBx ˆi  FBy ˆj

 

  4.598ˆj  2.655ˆi N  m    0.740 F ˆj  0.740 F ˆi m 


0  1.089ˆj  1.885ˆi N  m  3.342ˆj  1.930ˆi N  m
x
B
y
B
Separating components of this equation and rearranging gives
FBx  3.169 N
FBy  3.649 N
and
Therefore, the bearing reaction at B is
FB  3.169ˆi  3.649ˆj N  4.833 N130.98
The sum of the moments about bearing B can be written as
Ans.
741
 M  0  0.525 mkˆ  5.063ˆi  8.769ˆj N   0.250 mkˆ   6.820ˆi  3.938ˆj N 
B



0.150 mkˆ  7.794ˆi  4.500ˆj N  0.740 mkˆ  FAx ˆi  FAy ˆj

 

 1.169ˆj  0.675ˆi N  m    0.740 F ˆj  0.740 F ˆi m 

0  2.658ˆj  4.604ˆi N  m  1.705ˆj  0.985ˆi N  m
x
A
y
A
Separating components of this equation and rearranging gives
and
FAx  2.868 N
FAy  3.978 N
Therefore, the bearing reaction at B is
FA  2.868ˆi  3.978ˆj N  4.904 N  125.79
The sum of the moments about correction plane (2) can be written as
M  0  0.650 mkˆ  5.063ˆi  8.769ˆj N  0.375 mkˆ  6.820ˆi  3.938ˆj N

C2






Ans.

0.275 mkˆ  7.794ˆi  4.500ˆj lb  0.915 inkˆ  FCx1 cos C1ˆi  FCy1 sin C1ˆj

 

  2.143ˆj  1.238ˆi in  lb    0.915F cos  ˆj  0.915 F sin  ˆi in 

0  3.291ˆj  5.700ˆi in  lb  26.655ˆj  2.558ˆi in  lb
x
C1
y
C1
C1
C1
Separating components of this equation and rearranging gives
and
FCx1 cos C1  35.070 N
FCy1 sin C1  7.650 N
Therefore, the the correction force required in plane (1) is
FC1  35.070ˆi  7.650ˆj N  35.895 N  167.69
The correction mass in correction plane (1) is
F
35.895 N
mC1  C1 2 
 6.381 kg
2
RC1
0.025 m 15 rad/s 
Ans.
The sum of the moments about correction plane (1) can be written as
M  0  0.265 mkˆ  5.063ˆi  8.769ˆj N  0.540 mkˆ  6.820ˆi  3.938ˆj N



0.640 mkˆ   7.794ˆi  4.500ˆj N   0.915 mkˆ   F cos  ˆi  F sin  ˆj
0   1.342ˆj  2.324ˆi N  m    3.683ˆj  2.127ˆi N  m 
  4.988ˆj  2.880ˆi N  m    0.915 F cos  ˆj  0.915 F sin  ˆi m 
C1


x
C2
x
C1
C1
C2
y
C2
y
C1
C1
C2
Separating components of this equation and rearranging gives
FCx2 cos C 2  2.893 N
FCy1 sin C1  2.932 N
and
Therefore, the the correction force required in plane (2) is
FC 2  2.893ˆi  2.932ˆj N  4.119 lb134.61
The correction mass in correction plane (2) is
mC 2 
FC 2
4.119 N

 0.732 kg
2
2
RC 2
0.025 m 15 rad/s 
Ans.
742
15.18 The three mass particles are rigidly attached to the shaft which is rotating with a constant
angular velocity ω  350kˆ rev/min. Determine the magnitudes and directions of the
reaction forces at the bearings A and B. Then determine the correcting masses that must
be removed in the correction planes (1) and (2) to dynamically balance the system if the
radial distances of the correcting masses are specified as RC1  RC 2  3 in.
a  10.6 in, b=9.4 in, c  8 in, d  5.4 in, e  3 in, R1  95 mm, R2  2.2 in,
R3  3.6 in, m1  8.8 lb, m2  7 lb, and m3  1.1 lb.
The constant angular velocity of the rotating shaft is
(2 rad/rev)(350 rev/min)

 36.65 rad/s
(60 s/min)
The inertial forces of the three rotating mass particles are
 8.8 lb 
F1  m1 R1 2  
(3.8 in)(36.65 rad/s) 2  116.37 lb
2 
 386 in/s 
 7 lb 
F2  m2 R2 2  
(2.2 in)(36.65 rad/s) 2  53.59 lb
2 
386
in/s


 1.1 lb 
F3  m3 R3 2  
(3.6 in)(36.65 rad/s) 2  13.78 lb
2 
386
in/s


Therefore, the inertial forces of the three rotating mass particles can be written as
F1  116.37lb60  58.19ˆi  100.78ˆj lb
F  53.59N90  53.59ˆj lb
2
F3  13.78N  135  9.74ˆi  9.74ˆj lb
The sum of the inertial forces of the three rotating mass particles can be written as
743
 F  F  F  F  48.45ˆi  144.63ˆj lb  152.53 lb71.48
1
2
(1)
3
The sum of the moments about bearing A can be written as
 M  2.6 inkˆ  58.19ˆi  100.78ˆj lb  5.2 inkˆ  53.59ˆj N

5.2 inkˆ   9.74ˆi  9.74ˆj lb   10.6 inkˆ   F ˆi  F ˆj  0
A


x
B

y
B
Therefore the reaction force at bearing B is
FB  9.49ˆi  46.23ˆj lb  47.20 lb 101.60
For static balance, the sum of the forces must be zero; that is,
FA  FB   F1  F2  F3   0
Therefrore, using Eq. (1), the reaction force at being A is
FA  38.96ˆi  98.40ˆj lb  105.83 lb  111.60
The sum of the moments about correction plane (2) can be written as
 M(2)  5.0 inkˆ  58.19ˆi  100.78ˆj lb  2.4 inkˆ  53.59ˆj lb



2.4 inkˆ   9.74ˆi  9.74ˆj lb   6.4 inkˆ   F ˆi  F ˆj  0
x
C1
Ans.
Ans.

y
C1
Therefore, the required inertial force in the correction plane (1) is
FC1  41.81ˆi  95.18ˆj N  103.96 lb  113.71  103.96 lb66.29
The correction mass to be removed in correction plane (1) can be written as
106.96 lb  386 in/s 2 
FC1
mC1 

 10.25 lb66.29
RC1 2
(3 in)(36.65 rad/s) 2
For static balance, the sum of the forces must be zero; that is,
FA  FB  FC1  FC 2  0
Therefrore, the required inertial force in the correction plane (2) is
FC 2  FC1   FA  FB   90.26ˆi  239.81ˆj lb N
 256.23 lb69.37  256.23 lb  110.63
The correction mass to be removed in correction plane (2) can be written as
256.23 lb  386 in/s2 
FC1
mC1 

 24.54 lb  110.63
RC1 2
(3 in)(36.65 rad/s)2
Ans.
Ans.
Ans.
Ans.
744
15.19 The shaft of the distributed mass system is simply supported by the bearings at A and B,
and has a constant speed of 550 rev/min. The reaction forces acting on the bearings at A
and B are FA  17.5ˆi  30.3ˆj N and FB  25.0ˆi  43.3ˆj N, respectively. Using the
graphical approach determine the magnitudes and angular orientations of the reaction
forces at bearings A and B.
a = 1.2 in, b = c = 1.0 in, d = 0.8 in, R1  0.80 in, R2  0.80 in, R3  0.60 in,
m1  15.4 lb, m2  26.5 lb, and m3  17.6 lb.
The angular velocity of the shaft is 550 rev/min = 57.60 rad/s.
The centrifugal forces of the three rotating mass particles are
2
 15.4 lb 
F1  m1R1 2  
0.80 in  57.60 rad/s   105.89 lb
2 
 386 in/s 
2
 26.5 lb 
F2  m2 R2 2  
0.80 in  57.60 rad/s   182.22 lb
2 
 386 in/s 
2
 17.6 lb 
F3  m3 R3 2  
0.60 in  57.60 rad/s   90.77 lb
2 
 386 in/s 
The moments of these about bearing B can be wtitten as
M 1B  z1B m1R1 2  z1B F1  0.80 in 105.89 lb   84.71 in  lb
M 2 B  z2 B m2 R2 2  z2 B F2  2.80 in 182.22 lb   510.22 in  lb
M 3B  z3B m3 R3 2  z3B F3  1.80 in  90.77 lb   163.39 in  lb
The moment polygon for the moments about bearing B appears in the following figure
745
The polygon for moments about bearing B.
From the polygon shown, the moment due to the reaction force from the ground link
acting on the shaft, at bearing A, is M AB  420.78 in  lb  30.
Therefore, the reaction force from the ground link acting on the shaft, at bearing A, is
M
420.78 in  lb
Ans.
FA  AB 
 105.20 lb  30
z AB
4.00 in
The moments of the three centrifugal forces about bearing A can be wtitten as
M 1 A  z1 Am1R1 2  z1 A F1  3.20 in 105.89 lb   338.85 in  lb
M 2 A  z2 Am2 R2 2  z2 A F2  1.20 in 182.22 lb   218.66 in  lb
M 3 A  z3 Am3R3 2  z3 A F3  2.20 in  90.77 lb   199.69 in  lb
The moment polygon for the moments about bearing A appears in the following figure
The polygon for moments about bearing A.
From the polygon shown, the moment due to the reaction force from the ground link
acting on the shaft, at bearing B, is M BA  107.60 in  lb39.
Therefore, the reaction force from the ground link acting on the shaft, at bearing A, is
M
107.60 in  lb
FB  BA 
 26.90 lb  141
Ans.
zBA
4.00 in
746
15.20 The distributed mass system, denoted as body 2 is simply supported in the bearings at A
and B, and has a constant speed of 240 rev/min. The forces from the system acting on the
ground at bearings A and B are  F21  A  250ˆi  75ˆj N and  F21 B  80ˆi  125ˆj N,
respectively. Determine the magnitudes and orientations of the correcting masses that
must be removed in the correction planes (1) and (2) to ensure moment (dynamic) balance
of the system.
a  250 mm, b  200 mm, c  75 mm, and RC1  RC 2  30 mm.
The sum of the reaction forces at the bearings must be equal to the sum of the inertial
forces (that is, the negative of the correcting forces). Therefore, Newton’s seconds law
can be written as
FA  FB  FC1  FC 2  0
(1)
The sum of the moments about correction plane (2) is
0.075 mkˆ  FA  0.125 mk  FC1  0.175 mk  FB  0
0.075 mkˆ  (250ˆi  75ˆj N)  0.125 mkˆ  F  0.175 mkˆ  (80ˆi  125ˆj N)  0
C1
FC 1 ˆi  F ˆj  220ˆi  38ˆj N
y
x
C1
Therefore, the force required in the first correction plane (1) for dynamic balance is
FC1  38ˆi  220ˆj N  223.26 N80.20  223.26 N  99.80
and the mass to be removed from the first correction plane is
F
223.26 N  99.80
Ans.
mC1  C1 2 
 11.78 kg  99.80
RC1
(0.03 m)(25.13 rad/s)2
Substituting known values into Eq. (1) gives
(250ˆi  75ˆj N)  (80ˆi  125ˆj N)  (38ˆi  220ˆj N)  FC 2  0
Therefore, the force required in the second correction plane (2) for dynamic balance is
FC 2  368ˆi  170ˆj N  405.37 N204.79  405.37 N24.79
and the mass to be removed from the second correction plane is
F
405.37 N24.79
mC 2  C 2 2 
 21.40 kg24.79
Ans.
RC 2
(0.03 m)(25.13 rad/s)2
747
15.21 The shaft, simply supported by bearings at A and B, has a constant speed of 300 rev/min.
Using the graphic approach determine the magnitudes and orientations of the reaction
forces at bearings A and B.
a  40 mm, b  c  15 mm, d =25 mm, R1  10 mm, R2  35 mm, R3  15 mm, m1  17 kg, m2  4 kg, and m3  8 kg.
The constant angular velocity of the rotating shaft is
(2 rad/rev)(300 rev/min)

 31.42 rad/s
(60 s/min)
The magnitudes of the inertial forces of the three rotating particles are
F1  m1 R1 2  17 kg  (10 mm)(31.42 rad/s) 2  167.8 N
F2  m2 R2 2   4 kg  (35 mm)(31.42 rad/s)2  138.2 N
F3  m3 R3 2   8 kg  (15 mm)(31.42 rad/s)2  118.5 N
The three inertial forces can be written as
F1  167.8 N30  145.3ˆi  83.9ˆj N
F  138.2 N170  136.1ˆi  24.0ˆj N
2
F3  118.5 N  120  59.3ˆi  102.6ˆj N
The moments of the three inertial forces about bearing B can be written as
M 1B  z1B F1   25 mm 167.8 N   4.195 N  m30
M 2 B  z2 B F2   40 mm 138.2 N   5.528 N  m170
M 3 B  z3 B F3   55 mm 118.5 N   6.518 N  m  120
The moment polygon based on these moments about bearing B is shown in the following
figure
748
The moment polygon for moments about bearing B
The moment due to the force from the ground link acting on the shaft at bearing A is
measured from the moment polygon as
M AB  z AB FA  5.748 N  m27
Therefore, the force from the ground link acting on the shaft at bearing A is
M
5.748 N  m
FA  B 
 60.51 N
z AB
0.095 m
that is,
FA  60.51 N27
The moments of the three inertial forces about bearing A can be written as
M 1 A  z1 A F1   70 mm 167.8 N   11.75 N  m  150
M 2 A  z A F2   55 mm 138.2 N   7.60 N  m  10
M 3 A  z3 A F3   40 mm 118.5 N   4.74 N  m60
The moment polygon based on these moments about bearing A is shown in the following
figure
The moment polygon for moments about bearing A
749
The moment due to the force from the ground link acting on the shaft at bearing B is
measured from the moment polygon as
M BA  zBA FB  3.165 N  m85
Therefore, the force from the ground link acting on the shaft at bearing B is
M
3.165 N  m
FB  BA 
 33.32 N
zBA
0.095 m
that is,
FB  33.32 N  95
The shaking forces at A and B are the forces acting from the shaft onto the ground link.
Therefore,
FSA  F21 A  FA  60.51 N  153
Ans.
FSB  F21B  FB  33.32 N85
750
15.22 The angular speed of the continuous mass system, denoted as 2, in the simply supported
bearings at A and B is a constant 360 rev/min. The forces from the system acting on the
ground at bearings A and B are specified as (F21 ) A  73ˆi  79ˆj lb and
(F )  63ˆi  118ˆj lb, respectively. Determine the magnitudes and orientations of the
21 B
masses that must be removed in the correcting planes (1) and (2) to ensure dynamic
balance of the system.
a  30 in, b  27 in, c  7 in, and RC1  RC 2  2 in. .
The angular speed of the system is
  360 rev/min  2 rad/rev  /  60 s/min   37.70 rad/s
Taking moments applied on shaft 2 about correction plane (2) gives
27 inkˆ  73ˆi  79ˆj lb  20 inkˆ  F x ˆi  F y ˆj  3 inkˆ  63ˆi  118ˆj lb  0






1 971ˆj  2 133ˆi in  lb    20 inF ˆj  20 inF ˆi    189ˆj  354ˆi in  lb   0
 2 487ˆi  1 782ˆj in  lb    20 inF ˆi  20 inF ˆj  0
C1
C1
x
C1
y
C1
y
C1
x
C1
From this, the correction force required in plane (1) is
FC1  89.10ˆi  124.35ˆj lb  152.98 lb  54.38  152.98 lb125.62
Therefore, the correction mass for plane (1) is
152.98 lb  386 in/s 2 
FC1
mC1 

 20.77 lb125.62
2
RC1 2
2 in  37.70 rad/s 
Ans.
Taking moments applied on shaft 2 about correction plane (1) gives
7 inkˆ  73ˆi  79ˆj lb  20 inkˆ  F x ˆi  F y ˆj  23 inkˆ  63ˆi  118ˆj lb  0






511ˆj  553ˆi in  lb    20 inF ˆj  20 inF ˆi    1 449ˆj  2 714ˆi in  lb   0
3 267ˆi  938ˆj in  lb    20 inF ˆi  20 inF ˆj  0
C1
x
C2
C1
y
C2
y
C2
x
C2
From this, the correction force required in plane (2) is
FC 2  46.9ˆi  163.35ˆj lb  169.95 lb73.98  169.95 lb  106.02
Therefore, the correction mass for plane (2) is
169.95 lb  386 in/s 2 
FC 2
mC 2 

 23.08 lb  106.02
2
RC 2 2
2 in  37.70 rad/s 
Ans.
751
15.23 The constant angular velocity of the two-cylinder engine crankshaft is   200 rad/s
counterclockwise. Determine the x and y components of the primary shaking force acting
on the crankshaft bearing in terms of the crank angle θ. Then determine the magnitudes
and orientations of the correcting masses which must be added at the radial distance
RC  40 mm from the crankshaft axis. Determine the answers when the reference line
(attached to crank 1) is specified at the crank angle θ = 30°, as shown on the figure to the
right.
R1  R2  R  80 mm, L1  L2  L  160 mm, and m1  m2  m  15 kg.
Following Sec. 15.11 we find
P1  P2  P  mR 2  15 kg(0.08 m)(200 rad/s)2  48 000 N
 1  45,
 2  300,
1  0,
and
2  240
2
A   Pi cos( i  i )cos i  P cos45 cos45  P cos60 cos300  0.5P  0.25P  36 000 N
i 1
2
B   Pi sin( i  i )cos i  P sin 45 cos45  P sin 60 cos300  0.5P  0.433P  44 785 N
i 1
2
C   Pi cos( i  i )sin i  P cos45 sin 45  P cos60 sin 300  0.5P  0.433P  3 215.4 N
i 1
752
2
D   Pi sin( i  i )sin i  P sin 45 sin 45  P sin 60 sin 300  0.5P  0.75P  12 000 N
i 1
Substituting these values into Eqs. (15.20) gives
FPx  36 000 cos  44 785 sin  N
FPy  3 215.4 cos  12 000 sin  N Ans.(1)
and
(b) The magnitudes of the correcting forces can be written from Eqs (15.25) as
1
1
2
2
F1 
 A  D    B  C   (0.75P  0.25P)2  (0.933P  0.067P)2  0.5P  24 000 N
2
2
1
1
2
2
F2 
 A  D    B  C   (0.75P  0.25P)2  (0.933P  0.067P)2  0.707 P  33 941 N
2
2
The correcting masses are
F
24 000 N
mC1  1 2 
 15.0 kg
RC
(0.04 m)(200 rad / s) 2
Ans.
F2
33 941 N
mC 2 

 21.21 kg
RC 2 (0.04 m)(200 rad/s) 2
The corresponding angles are given by Eqs. (15.26)
BC
0.933P  0.067P
0.866
 1  120
or
tan  1 


( D  A) ( 0.25P  0.75P)
0.5
Ans.
BC
0.933P  0.067P 1.0
 2  135
or
tan  2 


( D  A) ( 0.25P  0.75P) 1.0
(c) Substituting   30 into Eq. (1), the primary shaking force components are
and
Ans.
FPx  53 569 N
FPy  3 215.4 N
The location of the correction masses for   30 .
753
15.24 The two-cylinder engine crankshaft is rotating counterclockwise with a constant angular
velocity   45 rad/s. Determine the magnitude and direction of the primary shaking
force in terms of crank angle  . If correcting masses are required to balance the primary
shaking force then determine: (a) the magnitudes and orientations of the inertial forces
created by these correcting masses, and (b) the magnitudes and orientations of the
correcting masses if RC1  RC 2  6 in. Determine the answers for Parts (a) and (b) when
the reference line attached to crank 1 is at the crank angle   210o.
R1  R2  R  3 in, L1  L2  L  24 in, and m1  m2  m  13.25 lb.
(a) Recognizing that P1  P2  P  mR 2 , The x and y components of the primary
shaking force for the two cylinders can be written as
S x  P cos(  1 ) cos 1  P cos(  2  2 ) cos 2
S y  P cos(  1 )sin 1  P cos(  2  2 )sin 2
Given that the angles are
and
 1   225,
1  0 ,
 2  300,
2  150,
the components of the primary shaking force can be written as
S x  P cos(  225) cos 225  P cos(  150) cos300  0.067 P cos   0.75P sin 
S y  P cos(  225)sin 225  P cos(  150)sin 300  1.25P cos   0.067 P sin 
The magnitude of the primary shaking force can be written as
S
 S    S   P 1.567cos   0.567sin   0.268sin  cos
x 2
y 2
2
2
The direction of the primary shaking force can be written as
y
1.25cos   0.067sin 
1 S
  tan
 tan 1
x
S
0.067cos   0.75sin 
The force magnitude is
(1)
754
2
 13.25 lb 
P  mR 2  
3 in  55 rad/s   311.51 lb
2 
 386 in/s 
Substituting this and the crank position   210o we find the primary shaking force as
SP  311.51 lb 1.567cos2 (210)  0.567sin 2 (210)  0.268sin(210)cos(210)  372.91 lb Ans.
and
1.25cos 210  0.067sin 210
Ans.
  tan 1
 68.79
0.067cos 210  0.75sin 210
(b) Comparing Eqs (1) above with Eqs. (15.20) of the text, the coefficients are
A  0.067P,
B  0.75P,
C  1.25P,
and
D  0.067P
Substituting these into Eqs. (15.25), the two correcting forces are
1
2
2
F1 
 0.067 P  0.067 P    0.75P  1.25P   0.259P  80.68 lb
2
Ans.
1
2
2
F2 
 0.067 P  0.067 P   1.25P  0.75P   P  311.51 lb
2
To achieve these two correcting forces, the correction masses are
80.68 lb  386 in/s2 
F1
mC1 

 1.716 lb
2
RC1 2
6 in  55 rad/s 
Ans.
2
311.51
lb
386
in/s

  6.625 lb
F2
mC 2 

2
2
RC 2
6 in  55 rad/s 
Substituting values into Eqs. (15.26), the orientation angles are
0.75P  1.25P
0.5
 1  255o
tan  1 

 3.73
(0.067 P  0.067 P) 0.134
0.75P  1.25P
2
 2  90o
tan  2 
 
(0.067 P 0.067 P) 0
Ans.
Ans.
755
15.25 The crankshaft of the two-cylinder engine is rotating counterclockwise with a constant
angular velocity     45 rad/s. Determine the magnitude and orientation of the
primary shaking force in terms of crank angle  . If correcting masses are required to
balance the primary shaking force then determine (a) the magnitudes and orientations of
the inertial forces created by these correcting masses, (b) the magnitudes and orientations
of the correcting masses if RC1  RC2  200 mm, and (c) determine the answers when the
reference line (attached to crank 1) is specified at the crank angle   0o.
R1  R2  R  100 mm, L1  L2  L  550 mm, and . m1  m2  m  5 kg.
(a) Recognizing that P1  P2  P  mR 2 , The x and y components of the primary
shaking force for the two cylinders can be written as
S x  P cos(  1 ) cos 1  P cos(  2  2 ) cos 2
S y  P cos(  1 )sin 1  P cos(  2  2 )sin 2
Given that the angles are
and
 1  45,
1  0 ,
 2  135,
2  180,
the components of the primary shaking force can be written as
S x  P cos(  45) cos 45  P cos(  45) cos135  0 P cos   1P sin 
S y  P cos(  45)sin 45  P cos(  45)sin135  1P cos   0 P sin 
The magnitude of the primary shaking force can be written as
S
S   S   P
x 2
y 2
The direction of the primary shaking force can be written as
(1)
756
  tan 1
Sy
cos 
 tan 1
 90  
x
S
sin 
The force magnitude is
2
P  mR 2   5 kg  0.100 m 45 rad/s   1 012.5 N
Substituting this and the crank position   0o we find the primary shaking force as
Ans.
SP  P  1 012.5 N
and
Ans.
  90
Recall that the x and y components of the resultant of the primary shaking force for
any multicylinder reciprocating engine [see Eq. (15.20) in the Uicker, et al., text] can be
written in the form
S x  A cos  B sin 
and
S y  C cos  D sin 
(b) Comparing Eqs (1) above with Eqs. (15.20) of the text, the coefficients are
A  0,
B  P,
C  P,
and
D0
Substituting these into Eqs. (15.25), the two correcting forces are
1
2
2
F1 
 0  0   0P  P   0
2
Ans.
1
2
2
F2 
 0  0   P  P   P  1 012.5 N
2
To achieve these two correcting forces, the correction masses are
F1
0
mC1 

0
2
2
RC1
0.200 m  45 rad/s 
Ans.
F2
1 012.5 N
mC 2 

 2.5 kg
RC 2 2 0.200 m  45 rad/s  2
Substituting values into Eqs. (15.26), the orientation angles are
PP
0
 1  undefined
Ans.
tan  1 

(0  0) 0
This is consistent with the first mass being zero; that is, the first correcting mass is not
needed.
tan  2 
P  P 2P


(0  0)
0
 2  90o
Ans.
757
15.26 The crankshaft of the three-cylinder engine is rotating with a constant angular velocity
ω  50kˆ rad/s. Determine the x and y components of the primary shaking force, and the
magnitude(s) of the correcting force (or forces) created by the correcting mass (or
masses), and the orientation(s) of the correcting force (or forces). Determine the answers
when the reference line (attached to crank 1) is specified at the crank angle   0o.
R1  R2  R3  R  6 in, L1  L2  L3  L  30 in, m1  2m  22 lb, and
m2  m3  m  11lb. .
(a) Recognizing that P1  2 P  2mR 2 and P2  P3  P  mR 2 , and that the angles are
and
 1  90,
 2  225,
2  3  180,
 3  315,
the x component of the primary shaking force for the three cylinders can be written as
S x  2 P cos(  90) cos90  P cos(  225  180) cos 225  P cos(  315  180) cos315
 2 P cos  cos90 cos90  2 P sin  sin 90 cos90  P cos  cos 45 cos 225
 P sin  sin 45 cos 225  P cos  cos135 cos315  P sin  sin135 cos315
 2 P cos  (0)(0)  2 P sin  (1)(0)  P cos  ( 2 / 2)(  2 / 2)
 P sin  ( 2 / 2)(  2 / 2)  P cos  (  2 / 2)( 2 / 2)  P sin  ( 2 / 2)( 2 / 2)
S x  ( 1P ) cos   (0 P)sin 
Similarly, the y component of the primary shaking force for the three cylinders can be
written as
758
S y  2 P cos(  90)sin 90  P cos(  225  180)sin(225)  P cos(  315  180)sin(315)
S y  (0P)cos   (1P)sin 
From these, the magnitude of the resultant of the primary shaking force is
SP 
 S    S   P cos2   sin2   P
2
2
and the direction of the primary shaking force is
Sy
1P sin 
  tan 1 x  tan 1
 180o  
S
1P cos 
However, we have
2
 11 lb 
P  mR 2  
6 in  50 rad/s   427.5 lb
2 
 386 in/s 
For the crank position   0o. The magnitude of the primary shaking force is
SP  P  427.5 lb
Ans.
and the direction of the primary shaking force is
Ans.
  180
(b) Comparing these results with Eq. (15.20) in the text, we see that
A  1P,
B  0,
C  0,
and
D  1P
Therefore, the two correcting forces are given by Eqs. (15.25) and (15.26) as
1
1
2
2
2
2
F1 
 A  D    B  C    1P  1P    0  0  0
2
2
Ans.
1
1
2
2
2
2
F2 
 A  D    B  C    1P  1P    0  0  1P  427.5 lb
2
2
B C
[0  0]
0
tan  1 

  undefined
(D A) [(1P)  ( 1P)] 0
Ans.
BC
[0  0]
0
o
tan  2 


0
2  0
(D A) [(1P )  ( 1P)] 2 P
Note that only one correcting force is required; that is, only one correcting mass, which is
rotating with the same angular speed as the crankshaft but in the opposite direction to the
crankshaft.
759
15.27 The crankshaft of the two-cylinder engine is rotating counterclockwise with a constant
angular speed   330 rev/min. Determine the x and y components of the primary
shaking force on the crankshaft bearing, and the magnitudes and orientations of the
inertial forces created by the correcting masses which balance the primary shaking force.
Determine the magnitude and orientation of the primary shaking force when the reference
line (attached to crank 1) is specified at the crank angle crank angle   0o.
R1  R2  R  120 mm, L1  L2  L  450 mm, m1  3m  60 kg, and m2  m  20 kg.
(a) Recognizing that P1  3P  3mR 2 and P2  P  mR 2 , and that the angles are
and
 1  120,
2  210,
 2  180,
the x component of the primary shaking force for the two cylinders can be written as
S Px  3P cos(  120) cos120  P cos(  180  210) cos(180)
 0.116 P cos   0.790 P sin 
Ans.
S Py  3P cos(  120)sin120  P cos(  180  210)sin(180)
 1.299 P cos   2.250 P sin 
The magnitude of the primary shaking force is
SP 
S   S 
x 2
P
y 2
P
 ( 0.116 P cos   0.790 P sin  ) 2  ( 1.299 P cos   2.250 P sin  ) 2
 P 1.701cos2   5.662 cos  sin   5.687sin 2 
The direction of the primary shaking force is
Ans.
760
SPy
 1.299 P cos   2.250P sin  
 tan 1 
x
SP
 0.116P cos   0.790P sin  
(b) From Eqs. (15.21) we find
  tan 1
2
A   Pi cos( i  i ) cos i  0.116 P
i 1
2
B   Pi sin( i  i ) cos i  0.790 P
i 1
2
C   Pi cos ( i  i )sin i  1.299 P
i 1
2
D   Pi sin ( i  i )sin i  2.250 P
i 1
From these and Eqs. (15.25), the two correcting forces can be written as
1
1
2
2
F1 
 A  D    B  C   P 4.813  1.097 P
2
2
1
1
2
2
F2 
 A  D    B  C   P 9.962  1.578P
2
2
Also, from Eqs. (15.26) the direction of the correcting forces are
B C
0.790P  1.299 P
 1  166.58
tan  1 

 0.239
( D  A) (2.250P  0.116P)
B  C 0.790P  1.299 P
 2  318.56
tan  2 

 0.883
D  A 2.25P  ( 0.116) P
The magnitude of the force P is
2
 (330 rev/min)(2 rad/rev) 
2
P  mR   20 kg  0.120 m  
  2 866 N
60 s/min


Therefore, the magnitudes of the two correcting forces are
F1  1.097  2 866 N   3 144 N
and F2  1.578  2 866 N   4 522 N
Ans.
Ans.
Ans.
Ans.
Ans.
For the arbitrary crank position  , the primary shaking force (or first harmonic force) and
the location of the correcting masses are shown in the figure below.
761
For the crank position   0o , the magnitude of the resultant of the first harmonic force is
SP  P 1.701  1.304 P  3 738 N
At the crank position   0o the direction of the resultant of the first harmonic forces can
be written as
  1.299 P 
1
o
  tan 1 
  tan ( 11.20)   95.10
  0.116 P 
The primary shaking force and the location of the correcting masses are shown in the
figure.
Magnitude and orientation of the correcting mass for the crank position   0o.
762
15.28 The two cranks of the two-cylinder engine are oriented at 240o to each other and the
crankshaft is rotating counterclockwise with a constant angular speed   690 rev/min.
Both pistons are in the same x-y plane. Determine the x and y components of the primary
shaking force acting on the ground bearing, and the magnitudes and orientations of the
correcting masses which balance the primary shaking force. The radial distances of the
correcting masses are RC1  RC2  16 in. Determine the answers when the reference line
(attached to crank 1) is specified at the crank angle   60o.
R1  12 in, R2  6 in, L1  L2  L  26 in, m1  22 lb, and m2  110 lb.
(a) Recognizing that P1  2 P  m  2R   2 and P2  5P  5m  R 2 , where m  22 lb and
R  6 in, and that the angles are
and
 1  135,
2  240,
 2  180,
the x component of the primary shaking force for the two cylinders can be written as
S Px  2 P cos(  135) cos135  5P cos(  180  240) cos(180)
 1.5P cos   3.330 P sin 
Ans.
S Py  2 P cos(  135)sin120  5P cos(  180  240)sin(180)
  P cos   P sin 
The magnitude of the primary shaking force is
Ans.
763
SP 
S   S 
x 2
P
y 2
P
 ( 1.5P cos   3.330 P sin  ) 2  (  P cos   P sin  ) 2
 P 3.250cos2   11.990cos  sin   12.089sin 2 
The direction of the primary shaking force is
Sy
 P cos   P sin 


  tan 1 Px  tan 1 
SP
 1.5P cos   3.330 P sin  
(b) From Eqs. (15.21) we find
2
A   Pi cos( i  i ) cos i  1.5P
i 1
2
B   Pi sin( i  i ) cos i  3.330 P
i 1
2
C   Pi cos ( i  i )sin i   P
i 1
2
D   Pi sin ( i  i )sin i  P
i 1
From these and Eqs. (15.25), the two correcting forces can be written as
1
1
2
2
F1 
 A  D    B  C   P 18.999  2.179 P
2
2
1
1
2
2
F2 
 A  D    B  C   P 11.679  1.709 P
2
2
Also, from Eqs. (15.26) the direction of the correcting forces are
3.330P    P  4.330
B C
 1  83.41
tan  1 


 8.660
( D  A)
( P  1.5P)
0.5
B  C 3.330P  P 2.330
 2  42.98
tan  2 


 0.932
D  A P  ( 1.5) P
2.5
The magnitude of the force P is
2
 22 lb 
 (690 rev/min)(2 rad/rev) 
P  mR 2  
6
in



  1 785 lb
2 
60 s/min
 386 in/s 


Therefore, the magnitudes of the two correcting forces are
F1  2.179 1 785 lb   3 890 lb
and F2  1.709 1 785 lb   3 051 lb
The two correcting masses are
mC1 
mC 2 
F1
RC1 2
F2

16 in  72.257 rad/s 
2
3 051 lb  386 in/s2 
 17.82 lb
Ans.
Ans.
 14.10 lb
Ans.
2
16 in  72.257 rad/s 
For the arbitrary crank position  , the primary shaking force (or first harmonic force)
and the location of the correcting masses are shown in the figure below.
RC 2 2

3 890 lb  386 in/s2 
Ans.
764
Magnitude and location of the correcting mass for an arbitrary crank position.
For the crank position   60o , the magnitude of the resultant of the first harmonic
forces is
SP  P 4.687  3 864 lb
Substituting the crank position   60o , the direction of the resultant of the first
harmonic forces can be written as
 0.366 P 
 0.366 
 tan 1 
 9.73

 2.134 P 
 2.134 
The primary shaking force and the locations of the correcting masses are shown in the
figure below.
  tan 1 
Locations of the correcting masses for the crank position   60o.
765
15.29 The constant angular velocity of the crankshaft of the two-cylinder engine is
ω  250kˆ rad/s. Determine the x and y components of the primary shaking force in terms
of the crank angle θ; and the magnitudes and orientations of the correction masses. The
correction masses are to be added at a radial distance RC  50 mm from the crankshaft
axis. Determine the answers when the reference line (attached to crank 1) is specified at
the crank angle θ = 30°.
R1  R2  R  150 mm, L1  L2  L  300 mm, and m1  m2  m  20 kg.
(a) Recognizing that P  mR 2 and R  6 in, and that the angles are
and
 2  315,
 1  240,
2  75,
From Eqs. (15.21) the coefficients of the primary shaking force can be written as
A  P cos 315 cos 315  P cos165 cos 240  0.5P  0.483P  0.983P
B  P sin 315 cos 315  P sin165 cos 240  0.5P  0.129 P  0.629P
C  P cos 315 sin 315  P cos165 sin 240  0.5P  0.837 P  0.337 P
D  P sin 315 sin 315  P sin165 sin 240  0.5P  0.224 P  0.276 P
P  mR 2  (20 kg)(0.15 m)(250 rad/s)2  187 500 N .
where
Substituting these into Eqs. (1), the coefficients are
A  184 312.5 N,
B  117 937.5 N,
C  63 187.5 N,
and
D  51 750 N.
Substituting these into Eqs. (15.20), the x and y components of the primary shaking force
can be written as
766
SPx  0.983P cos   0.629 P sin   184 312.5 cos   117 937.5sin  N
S  0.337 P cos   0.276 P sin   63 187.5 cos   51 750sin  N
The magnitude of the primary shaking force is
y
P
SP 
Ans.
Ans.
S   S 
x 2
P
y 2
P
 ( 1.5P cos   3.330 P sin  ) 2  (  P cos   P sin  ) 2
 P 3.250cos2   11.990cos  sin   12.089sin 2 
The direction of the primary shaking force is
Sy
 P cos   P sin 


  tan 1 Px  tan 1 
SP
 1.5P cos   3.330 P sin  
(b) The magnitudes of the correcting forces can be written as
1
1
2
2
 A  D    B  C   (0.983P  0.276 P)2  (0.629 P  0.337 P)2
2
2
 0.7935P  148 781.25 N
1
1
2
2
F2 
 A  D    B  C   (0.983P  0.276P)2  ( 0.629P  0.337 P)2
2
2
 0.3825P  71 718.75 N
F1 
The two correction masses are
148 781.25 N
 47.61 kg
RC1
(0.05 m)(250 rad/s)2
F2
71 718.75 N
mC 2 

 22.95 kg
2
RC 2
(0.05 m)(250 rad/s)2
mC1 
F1
2

The angles of the correcting forces can be written as
B C
0.629P  0.337 P 0.966
tan  1 


( A  D) (0.983P  0.276P) 1.259
BC
0.629 P  0.337 P 0.292
tan  2 


( A  D) (0.983P  0.276P) 0.707
Ans.
Ans.
 1   217.5
Ans.
 2   202.4
Ans.
The locations of the two correction masses for an arbitrary crank angle are shown in the
figure below.
767
The locations of the correction masses for an arbitrary crank angle.
(c) Substituting   30 , the x and y components of the resultant primary shaking force
are
SPx  100.65 kN
SPy  80.50 kN
and
Ans.
Therefore, the magnitude of primary shaking force on the crankshaft bearing is
SP 
 S    S   100.65 kN   80.50 kN   128.88 kN
x 2
P
y 2
P
2
2
Ans.
The direction of the primary shaking force is
 SPy 
 80.50 
 tan 1 
  38.65
x 
 100.65 
 SP 
  tan 1 
Ans.
The primary shaking force on the crankshaft bearing is shown in the figure below. The
locations of the two correction masses are also shown on the figure.
The primary shaking force on the crankshaft bearing for   30 .
768
Page intentionally blank.
769
Chapter 16
Flywheels, Governors, and Gyroscopes
16.1
Table P16.1 lists the output torque for a one cylinder engine running at 4 600 rev/min.
(a) Find the mean output torque.
(b) Determine the mass moment of inertia of an appropriate flywheel using Cs  0.025 .
Table P19.1 Torque data for Problem 16.1
i
deg
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
(a)
Ti
Nm
0
17
812
963
1 016
937
774
641
697
849
1 031
1 027
902
712
607
594
544
345
i
deg
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
Ti
Nm
0
-344
-540
-576
-570
-638
-785
-879
-814
-571
-324
-190
-203
-235
-164
-7
150
145
i
deg
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
Ti
Nm
0
-145
-150
7
164
235
203
490
424
571
814
879
785
638
570
576
540
344
i
deg
540
550
560
570
580
590
600
610
620
630
640
650
660
670
680
690
700
710
Ti
Nm
0
-344
-540
-577
-572
-643
-793
-893
-836
-605
-379
-264
-300
-368
-334
-198
-56
-2
Using n = 72 and h = 4π/72, we enter the data from Table P16.1 into Simpson’s
rule to find U 2  U1  890.7 N  m .
Tm  U 2  U1   4   890.7 N  m   4 rad   70.88 N  m
(b)
Ans.
  4 600 rev/min  481.7 rad/s
2
I  U 2  U1   Cs 2    890.7 N  m   0.025  481.7 rad/s    0.154 kg  m2
Ans.
770
16.2
Using the data of Table 16.2, determine the moment of inertia for a flywheel for a fourcylinder 90 V engine having a single crank. Use Cs = 0.012 5 and a nominal speed of
4 600 rev/min. If a cylindrical or disk-type flywheel is to be used, what should be the
thickness if it is made of steel and has an outside diameter of 400 mm? Use  = 7.8
Mg/m3 as the density of steel.
Table 19.2 Torque data for a four-cylinder, four-stroke internal combustion engine
i
T
T 180
T +360
T 540
Ttotal
deg
in  lb
in  lb
in  lb
in  lb
in  lb
0
0
0
0
0
15
2 800
-107
-85
-107
2 501
30
2 090
-206
-125
-206
1 553
45
2 430
-280
-89
-292
1 769
60
2 160
-323
8
-355
1 490
75
1 840
-310
126
-371
1 285
90
1 590
-242
242
-362
1 228
105
1 210
-126
310
-312
1 082
120
1 066
-8
323
-272
1 109
135
803
89
280
-274
898
150
532
125
206
-548
315
165
184
85
107
-760
-384
Using n = 48 and h = 4π/48, we integrate the data from columns 2-5 of Table 16.2 by
Simpson’s rule to find U2  U1  3 490.1 in  lb  394.38N  m .
  4 600 rev/min  481.7 rad/s
2
I  U 2  U1   Cs 2    394.38 N  m   0.0125  481.7 rad/s    0.135 97 kg  m2
m  2I R2  2  0.135 97 kg  m2   0.200 m   6.7985 kg
V  m /   6.7985 kg 7 800 kg/m3  0.000 872 m3
2
t  V A  0.000 872 m3    0.200 m    6.94 mm


2
Ans.
771
16.3
Using the data of Table 16.1, find the mean output torque and the flywheel inertia
required for a three-cylinder in-line engine corresponding to a nominal speed of 2 400
rev/min. Use Cs = 0.03.
. Table 16.1 Example 16.1: Torque data for Figure 16.3
i
i
i
i
i
Ti
Ti
Ti
Ti
Ti
deg
deg
deg
deg
in  lb deg
in  lb
in  lb
in  lb
in  lb
0
0
150
532
300
-8
450
242
600
-355
15
2 800 165
184
315
89
465
310
615
-371
30
2 090 180
0
330
125
480
323
630
-362
45
2 430 195
-107
345
85
495
280
645
-312
60
2 160 210
-206
360
0
510
206
660
-272
75
1 840 225
-280
375
-85
525
107
675
- 274
90
1 590 240
-323
390
-125
540
0
690
-548
105
1 210 255
-310
405
-89
555
-107
705
-760
120
1 066 270
-242
420
8
570
-206
135
803
285
-126
435
126
585
-292
Using n = 48 and h = 4π/48, we integrate the data from Table 16.1 by Simpson’s rule to
find U 2  U1  3 490.1 in  lb .
Tm  U 2  U1   4    3 490.1 in  lb   4 rad   277.7 in  lb
Ans.
  2 400 rev/min  251.3 rad/s
I  U 2  U1   Cs 2    3490.1 in  lb   0.03 251.3 rad/s    1.842 in  lb  s 2
2
Ans.
772
16.4
The load torque required by a 200-ton punch press is displayed in Table P16.4 for one
revolution of the flywheel. The flywheel is to have a nominal angular velocity of 2 400
rev/min and to be designed for a coefficient of speed fluctuation of 0.075.
(a) Determine the mean motor torque required at the flywheel shaft and the motor
horsepower needed, assuming a constant torque-speed characteristic for the motor.
(b) Find the moment of inertia needed for the flywheel.
i
deg
0
10
20
30
40
50
60
70
80
(a)
Table P16.4 Torque data for Problem 16.4
i
i
Ti
Ti
Ti
deg
deg
in  lb
in  lb
in  lb
857
90
7 888
180
1 801
857
100
8 317
190
1 629
857
110
8 488
200
1 458
857
120
8 574
210
1 372
857
130
8 403
220
1 115
1 287
140
7 717
230
1 029
2 572
150
3 515
240
943
5 144
160
2 144
250
857
6 859
170
1 972
260
857
i
deg
270
280
290
300
310
320
330
340
350
Ti
in  lb
857
857
857
857
857
857
857
857
857
Using n = 36 and h = 2π/36, we integrate the data from Table P16.4 by Simpson’s
rule to find U  16 700 in  lb .
Tm  U  2   16 700 in  lb   2 rad   2 658 in  lb  221.5 ft  lb
Ans.
  2 400 rev/min  251.3 rad/s
P  T 
(b)
 221.5 ft  lb  2 400 rev/min  2 rad/rev   6 073 HP
Ans.
550 ft  lb/min/HP
The torque data show a constant requirement of 857 in·lb, probably friction, in
addition to the torque for the punching operation. If this constant torque is
subtracted from the data in the table (for speed fluctuation), and the integration
repeated, then we get U  14 905 in  lb
2
I  U  Cs 2   14 905 in  lb   0.075 251.3 rad/s    3.146 in  lb  s 2
Ans.


773
16.5
Find Tm for the four-cylinder engine whose torque displacement is that of Figure 16.4.
Table 16.2 Torque data for a four-cylinder, four-stroke internal combustion engine
i
T
T 180
T +360
T 540
Ttotal
deg
in  lb
in  lb
in  lb
in  lb
in  lb
0
0
0
0
0
15
2 800
-107
-85
-107
2 501
30
2 090
-206
-125
-206
1 553
45
2 430
-280
-89
-292
1 769
60
2 160
-323
8
-355
1 490
75
1 840
-310
126
-371
1 285
90
1 590
-242
242
-362
1 228
105
1 210
-126
310
-312
1 082
120
1 066
-8
323
-272
1 109
135
803
89
280
-274
898
150
532
125
206
-548
315
165
184
85
107
-760
-384
Using n = 12 and h = π/12, we integrate the data from column 6 of Table 16.2 by
Simpson’s rule to find U 2  U1  3 490.1 in  lb .
Tm  U 2  U1      3 490.1 in  lb   rad   1 111 in  lb
Ans.
774
16.6
In a pendulum mill, illustrated schematically in Figure P16.6, the grinding is by a conical
muller that is free to spin about a pendulous axle that, in turn, is connected to a powered
vertical shaft by a Hooke universal joint. The muller presses against the inner wall of a
heavy steel pan, and it rolls around the inside of the pan without slipping. The weight of
the muller is W = 980 lb; its principal mass moments of inertia are I s  121 in  lb  s2 and
I  88 in  lb  s2 . The length of the muller axle is l  RGA  40 in and the radius of the
muller at its center of mass is RGB  10 in . Assuming that the vertical shaft is to be
inclined at   30 and will be driven at a constant angular velocity of
 p  240 rev/min , find the crushing force between the muller and the pan. Also
determine the minimum angular velocity p required to ensure contact between the muller
and the pan.


 p  240 rev/min  25.133 rad/s
ωs  ω4/ 3  s sin  ˆi  cos ˆj
ω3   p ˆj  25.133 rad/sˆj
ω4  ω3  ω4/ 3  s sin  ˆi   p  s cos  ˆj
775
VG  ω3  R GA

  p ˆj  40sin  ˆi  40 cos  ˆj
VG  ω 4  R GB

  s sin  ˆi   p   s cos   ˆj   10 cos  ˆi  10sin  ˆj
 10 s  10cos p  kˆ
 40sin  p kˆ
Equating these with   30 we find s    4sin   cos   p  2.866 p . Then
ω ×ω    ˆj×   sin  ˆi  cos  ˆj    sin  kˆ   4 sin   sin  cos   kˆ  1.433 kˆ
2
I S  mr 2 2   980 lb 386 in/s2  10 in  2  126.9 in  lb  s 2
2
p
2
p
s
p
s
p
s
2
p
2
2
I  mr 2 4  ml 2   980 lb 386 in/s 2  10 in  4   40 in    3 997.8 in  lb  s 2


Now Eq. (16.36) shows
p
 s

s
cos   p   s
 M  I  I  I
s








 p cos
 4sin 2   sin  cos  in  lb  s 2  2 kˆ
 M  126.9  3 870.9
p

4sin


cos



 p 




2 2
 M  126.9sin   4sin   cos    3 870.9sin  cos  in  lb  s  pkˆ
 M  1 557.75 in  lb  s  pkˆ  98 396 000 in  lb kˆ
2 2
Now formulating the externally applied moments,
 M  Wˆj× RGA  Fcˆi × R BA
 M  980 lbˆj× sin  ˆi  cos ˆj 40 in  F ˆi × sin  ˆi  cos ˆj 40 in  cos  ˆi  sin  ˆj10 in 
 M   39 200sin in  lb kˆ  F  4cos  sin  10 in kˆ 
 M  19 600 in  lb kˆ  29.641F in kˆ
c
c
c
and setting the two expressions equal
 M  19 600 in  lb kˆ  29.641Fc in kˆ  98 396 000 in  lb kˆ
we can now solve for the crushing force
Fc  3 319 000 lb
If we start before setting the angular velocity, then we have
 M  1 557.75 in  lb  s2 2pkˆ  19 600 in  lb kˆ  29.641Fc in kˆ
Ans.
Now by setting Fc to zero we can determine the minimum angular velocity p required to
ensure contact between the muller and the pan:
 p  19 600 in  lb 1 557.75 in  lb  s2  3.547 rad/s  33.87 rev/min
Ans.
776
16.7
Using the gyroscopic formulae, Eqs. (16.33)-(16.35), solve the problem presented in
Example 12.9 of Chapter 12.
From the given data we can identify
ω p  ω2  5kˆ rad/s
ωs  ω3  350ˆi  5kˆ rad/s
I s  mk 2   4.5 kg  0.050 m   0.011 3 kg  m2
2
From Eq. (16.37)
 M  I ω ×ω   0.011 3 kg  m 5kˆ rad/s  350ˆi  5kˆ rad/s   19.8ˆj N  m
s
2
p
s
This brings us precisely to the formulation of Eq. (2) of Example 12.9. From there on the
solution procedure, and the results, are identical. Notice how much more simply this
approach can be accomplished.
Q.E.D.
777
16.8
The oscillating fan precesses sinusoidally according to the equation  p   sin1.5t ,
where   30 ; the fan blade spins at  s  1800ˆi rev/min . The weight of the fan and
motor armature is 5.25 lb, and other masses can be assumed negligible; gravity acts in the
jˆ direction. The principal mass moments of inertia are I s  0.065 in  lb  s2 and
I  0.025 in  lb  s2 ; the center of mass is located at RGC  4 in to the front of the
precession axis. Determine the maximum moment M z that must be accounted for in the
clamped tilting pivot at C.


ω p  1.5 sin ˆi  cos  ˆj rad/s
 s  1800ˆi rev/min  188.5ˆi rad/s
ω  ω   1.5 sin ˆi  cos  ˆj rad/s  188.5ˆi rad/s   282.7 cos  kˆ rad/s  244.9kˆ rad/s
2
p
s
p
 s

s
cos   p   s
 M  I  I  I
s










1.5 rad/s


2
2
cos   244.9kˆ rad/s 2  15.98kˆ in  lb
 M  0.065 in  lb  s  0.040 in  lb  s
188.5 rad/s


On the other side of the equation, the external moments are
z
z
z
 M  M kˆ  RGC W  sin  ˆi  cos  ˆj  M kˆ  4 in W cos  kˆ  M kˆ 18.19 in  lb kˆ


Now, equating the two we can solve for the moment M z .
M z kˆ  18.19 in  lb kˆ  15.98kˆ in  lb
M z  2.20kˆ in  lb
Ans.
Here we see that the gyroscopic moment is almost large enough to support the weight of
the fan motor.
2
778
16.9
The propeller of an outboard motorboat is spinning at high speed and is caused to precess
by steering to the right or left. Do the gyroscopic effects tend to raise or lower the rear of
the boat? What is the effect and is it of noticeable size?
In this case ω s refers to the angular velocity of the propeller and is directed fore or aft
depending on the direction of rotation. ω p is vertical and refers to the angular velocity of
the turn. The moment required to maintain the turn is proportional to  ω p  ω s  as
shown in Eq. (16.36) or (16.37) and this axis is lateral on the boat. Therefore the moment
(or its reaction) can tend to raise or lower the rear of the boat. The direction depends on
both the direction of rotation of the engine and the direction of the turn. Eq. (16.37)
shows that the effect is likely to be very small since, for any reasonable rate of turn,
ω p  ωs  will be at least an order of magnitude smaller than  s2 , which is the order of
the usual accelerations of the engine. In a very extreme case, a knowledgeable person
might be able to detect this moment, but most would not. It would never be a danger.
779
16.10 A large and very high-speed turbine is to operate at an angular velocity of
  18 000 rev/min and will have a rotor with a principal mass moment of inertia of
I s  225 in  lb  s2 . It has been suggested that because this turbine will be installed at the
North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic
loads on its bearings. Estimate the size of these additional loads.
ωs  18 000ˆi rev/min  1885ˆi rad/s
ω  1.0ˆj rev/day  0.0000115ˆj rad/s
p
ω  ω    0.0000115ˆj rad/s   1885ˆi rad/s  0.022kˆ rad/s
I  ω  ω    225 in  lb  s   0.022kˆ rad/s   4.91kˆ in  lb
p
2
s
s
2
p
s
2
Ans.
Thus, if the bearings were separated by only 5.0 inches, they would experience less than
one additional pound of loading. This is totally negligible.
780
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