Solutions to Chapter 4 Exercise Problems
Problem 4.1
Locate all of the instant centers in the mechanism shown below.
A 4
30˚ 63˚
2 B 1
AB = 1.5"
3
Position Analysis
Draw the linkage to scale. Start by locating point B. Then draw the line on which B must slide
and draw a horizontal line on which link 4 must slide. This will locate all of the ponts and
directions required for the analysis.
Instant Center Locations
Locate the obvious instant centers ( I12, I23, I34, I14 ). Then I24 is found in a straight forward
manner using the procedures given in Section 2.15. To locate I13, note that it must lie on the line
AB. It also lies on the line through I14 and I43. However, both points are at infinity and, the line
through the two ponits lies at infinity. Therefore, the line through AB intersects this line at
infinity meaning that I13 must be at infinity in the direction indicated.
-1-
∞
A I12
4 I14
∞
2 I23 1
B
3 1
I24
I34
I34 ∞ 4 2
∞ ∞ ∞ I13
I14
Problem 4.2
Find all of the instant centers of velocity for the mechanism shown below.
BC = CD
BD = 3.06"
3
2
150˚
C
D
A 50˚
4
-2-
I24
I 23 1
B 4 2
3
3
2
C D
A 50˚
4 I14
I12
I34
I13
-3-
Problem 4.3
In the linkage shown below, locate all of the instant centers.
3.3"
6
AB = 1.35" F
BD = 3.9"
DE = 0.9" B 5
BC = 0.9"
CF = 2.0" C
2
55˚ 3
A E
4
D\
Solution
1
I13 I 36
6 2
I1
5 3 6
F 6
I15 I56
B I2 5
4 3
I25 C I
2 3
5
A 3 I 24 I14 E I46
I26
I12 4
D
I3
4
I4
5
Problem 4.4
Find all of the instant centers of velocity for the mechanism shown below.
-4-
I 15
E I35
I 45
5
D I 25
4
I34 I 24
I 23
C
2
E
A
5
D I 12
4
5.0 cm I14
C
2
3
28˚ A
1
B
AB = 8.0 cm
AC = 4.5 cm I 13
BD = 13.0 cm 3 5
DE = 2.9 cm 2
B
4 3
Solution
-5-
Problem 4.5
Locate all of the instant centers in the mechanism shown below. If link 2 is turning CW at the
rate of 60 rad/s, determine the linear velocity of points C and E using instant centers.
AD = 3.8"
C AB = 1.2"
BC = 3.0"
CD = 2.3"
CE = 1.35"
EB = 2.05"
3
B
4 E
2
125˚
D A
Velocity Analysis
The two points of interest are on link 3. To find the angular velocity of link 3, use I13 and I23.
Then
1vI
23 = 1ω2 × rI23 /I12 = 1ω3 × rI23/I13
Therefore,
rI23/ I12
1ω3 = 1ω2 = 60 1.2 = 17.7 rad / s
rI23 /I13 4.07
Because the instant center I23 lies between I12 and I13, 1ω3 is in the opposite direction of 1ω2.
Therefore, 1ω3 is counterclockwise.
Then,
1vC
3 = 1ω3 × rC/I13 ⇒ 1vC3 = 1ω3 rC/I13 = 17.7⋅2.11 = 37.3 in / s
and
1vE
3 = 1ω3 × rE/I13 ⇒ 1v E3 = 1ω3 rE/I13 = 17.7⋅3.25 = 57.5 in / s
The directions for the velocity vectors are shown in the drawing.
-6-
I 13 1
4 2
C 3
I 34 vC 3
4 3
vE 3 I23 vB3
E
B
2
D A I 24
I 14 I12
-7-
Problem 4.6
Locate all of the instant centers in the mechanism shown below. If the cam (link 2) is turning
CW at the rate of 900 rpm, determine the linear velocity of the follower using instant centers.
103˚
R
B
AB = 1.5"
R = 0.75" 2
70˚
A
Instant Centers
-8-
Velocity of the Follower
Convert the angular velocity from “rpm” to “rad/s”
900(2π )
ω2 = 900 rpm =
1
= 94.25 rad / s CW
60sec
At the point I 23 the linear velocity of follower and cam is same.
1
v I 23 = 1 v A2 + 1 v I23 / A2 = 0 + 1ω2 × rI 23 / A = (94.25 rad / s )(0.82 in) = 77.285 in / s Down
Problem 4.7
Locate all of the instant centers in the mechanism shown below. If link 2 is turning CW at the
rate of 36 rad/s, determine the linear velocity of point B4 by use of instant centers. Determine the
angular velocity of link 4 in rad/s and indicate the direction. Points C and E have the same
vertical coordinate, and points A and C have the same horizontal coordinate.
D
5
C 4 E
6
100˚
AB = 1.1"
B AC = 0.9"
A CD = 1.5"
3 2 DE = 3.25"
Solution:
Find all instant centers and linear velocity of point B2.
1v B
2 = 1ω2 × rB2 /A2 ⇒ 1v B2 = 1ω 2 ⋅ rB2 /A2 = 36⋅1.1 = 39.6 in / s
Using rotating radius method,
-9-
1v B
4 = 32.5 in / s
To calculate the angular velocity of link 4, we can use the relations between related instant
centers.
1ω 2 × rI 24 /I12 =1ω 4 × rI24 /I14
rI24 /I12
1ω 4 = 1ω 2 ⋅ = 36 ⋅ 1.283 = 21.1 rad / s
rI24 /I14 2.186
Therefore,
1ω 4 = 21.1 rad / s CW
- 10 -
I15
I 46
I45
D
5
I 36
4 I35
I14 I16
I25 E
C 6
1v
B2 I 56
1v B I 26
4 1
I 23 I12
I13
B 2 A 6 2
3
1 v 'I I34
24
1v 'B
2 5 3
I 24
1v I
24 4
Problem 4.8
Using the instant-center method, find angular velocity of link 6 if link 2 is rotating at 50 rpm
CCW.
- 11 -
AC = 1.2" G (2.55", 2.95")
AB = 1.35"
BC = 0.9" 6
CE = 2.7"
BD = 2.6"
DE = 2.6" D
DF = 2.2" F
EF = 3.1"
FG = 2.8" 5
Y B
4
2
C
A 20˚
X
3
Position Analysis:
Draw linkage to scale. This is a trial and error process because the linkage is a Stephenson II
linkage. First draw link 2 to locate points B and C. Draw a circle centered at B of radius 2.6 ".
Draw a second circle centered at C and of radius 2.7". Draw a third circle centered at G and of
radius 2.8". Next construct the triangle CDF to scale and manipulate the triangle until points D,
E, and F intersect their respective circles. Alternatively, the procedure given in Section 2.10 can
be used.
Velocity Analysis:
The angular velocity of link 2 is
ω2 = 2 ⋅ π ⋅50 = 5.24 rad / s
60
Using the instant centers I12, I16, and I26 ., we can write the relationship between ω2 and ω 6 as
1ω2 × rI26/ I12 = 1ω6 × rI26 /I16 (1)
Solve Eq. (1) for 1ω6
rI26 /I12
1ω 6 = 1ω 2 ⋅ = 5.24 ⋅ 1.56 = 3.47 rad / s
rI26 /I16 2.35
So,
1ω6 = 3.47 rad / s CW
- 12 -
1
G
6 2 I 16 6 I 46
F
D
5 3 I 45
5 4
B
I 25
I 26
2 I 23
I 24
C
3
A I
12 E I 34
- 13 -
Original
A E Desired
I13
I13 Original velocity
3 of point C
New location
for pivot B' Desired velocity
D
4 of point C
C
B
Problem 4.10
For the linkage given, ω2 = 1 rad/s CCW. Find I26 using the circle-diagram method. Using vA2
and I26, determine the magnitude and direction of vB6 using the rotating radius method.
AC = 1.4"
AE = 3.15"
DF = 1.6"
BF = 1.25"
BD = 0.8" B
Y F (3.6", 1.45")
A 6
3
2
5
D E
C 35˚
4 X
Solution:
Draw the linkage to scale. Start by locating the pivots C and F and line of motion of E. Next
locate link 2 and point A. Then locate point E and draw the line AE. Next locate point D and
finally E.
Find the necessary instant centers, and locate I26. Find the velocity of A2 which is given by
1v A2 = 1ω2 × rA2 /I12 ⇒ 1vA2 = 1ω2 ⋅ rA2 /I12 = 1(1.4) = 1.4 in / s
Rotate point A onto the line defined by I12 and I16 to get A'2 and draw the velocity of A'2. From
the proportionality relationship, find the velocity of I26.
- 15 -
I13
vA2 vB 6
vA'2 vB'6
I16 F
vI 26
I23 B
I24 6
A B'
3 ∞
I56 ∞
2 A'
I26 5 E
D I14
4
C
I12 ∞ I34
I35 ∞
1 in
1
6 2
5 3
4
I36
- 16 -
Rotate point B onto the line defined by I12 and I16 to get B'6. From the proportionality
relationship, using I26 and I16, find the velocity of B'6. This point will have the same velocity
magnitude as will B6. Show the velocity vector at B6 perpendicular to BI16 . The magnitude of
the velocity of B6 is given by
1vB
6 = 0.563 in / s
Problem 4.11
Find the velocity of point C given that the angular velocity of gear 2 is 10 rad/s CW. B is a pin
joint connecting links 4 and 5 . Point A is a pin in link 3 that engages a slot in link 4.
B
AE = 0.85"
BD = 1.65"
BC = 3.0"
5
D (2.05", 1.3")
C Y
6 0.95"
3 E
7˚
1.0" A
0.7" 28˚ 4
F X
ω2 2
Solution:
To find the velocity of point C, considered as a point in link 5, from the angular velocity of link
2 relative to link 1, the instant centers I12, I15, and I25 are needed. These may be located as
shown in the figure
Then,
1ω5 = 1ω2 × (I25I12) / (I25I15) = 10(1.28) / (9.57) = 1.34 rad / s CW
vC5 = 1ω5 × (I15C) = 1.34(7.21) = 9.66 in / s to the left.
- 17 -
I 15
I25 I15 = 9.57
I25 I12 = 1.29
I15C = 7.21
B
I45
4
5
I56 D I 14
I16 6
C
I24 E I 13
A
I 23
1
2 3
2 2 F I12
6 I34
1 in
5 3
4 I25
- 18 -
Problem 4.12
If ω2 = 5 rad/s CCW, find ω5 using instant centers.
C
B
4
3
5
62˚ ω2 E
A
AE = 4.1" 2
EF = 2.0"
AB = 1.5" D
BC = 1.55"
CF = 4.0"
DE = 1.0"
F
Solution:
Draw linkage to scale and find necessary instant centers ( I12, I15, and I25 ).
The relationshp between 1ω 2 and 1ω 5 is
1ω2 × rI25/I12 = 1ω5 × rI25 /I15 (1)
Solve Eq. (1) for 1ω5 ,
rI25/ I12
1 ω5 = 1ω2 ⋅ = 5⋅ 1.83 = 4.03 rad / s
rI25 / I15 2.27
So,
1ω5 = 4.03 rad / s CW
- 19 -
C
4
B I 34
I 35 I 45 3
A E
I 12
I15 I 25
2
1 I23
D
5 2
I13
F
4 3
Problem 4.13
If ω 2 = 1 rad/s CCW, find the velocity of point A on link 6 using the instant center method.
Show vA6 on the drawing.
AC = BC = 1.4"
BE = 3.15"
A DF = 1.6" ω2
Y
F (3.6", 1.45")
6 B 2
3
30˚
5
D E
C 35˚
4 X
- 20 -
Solution:
I 13
I 12
A
F
I46 B 2
I36 1 vA 6
6 I25
D I 35
I26 1 v'I
26
5 3 E
I 14
C I34 4
1 v'
D2 1 vD
2
I 16
1 vI
26
6 2
5 3
I 23
Find necessary instant centers, i.e. I12, I16, and I26 , and the velocity of point D as
- 21 -
Problem 4.19
If the velocity of A2 is 10 in/s to the right, find ω6 using instant centers.
D
AB = 1.75"
BC = 1"
BD = 3"
ED = 2.25"
5 CE = 1.45"
6
B
98˚ 3
4 A
E C vA2
2
Solution:
Find necessary instant centers as shown in the sketch above, i.e. I12, I16, and I26 . All points in
link 2 have the same velocity; therefore,
1v A
2 = 1v' A2 = 1v I26
Using the rotating radius method,
1vD
6 = 1v D6 / E6 = 13.2 in / s
Now,
1vD / E
1vD / E
6 6 = 1ω 6 × rD6 / E6 ⇒ 1ω6 = 6 6
= 13.2 = 5.87rad / s
rD6 / E6 2.25
Therefore,
1ω6 = 5.87 rad / s CW
- 29 -
1
I56
6 2
D
I26
vA' 2
5 3
6 5
4
I24 I34 , I45
B 3
4 A
E C I23 vA2 I12
I 46 2
I16 I14
Problem 4.20
Crank 2 of the push-link mechanism shown in the figure is driven at ω2 =60 rad/s (CW). Find
the velocity of points B and C and the angular velocity of links 3 and 4 using the instant center
method.
B Y
O2 A = 15 cm A = 14.75 cm
D
O4 B = 30.1 cm D = 7.5 cm
4 C
3 A = 29.5 cm O2 O4 =
7.5 cm
C B
D
O2 O4 X
30˚
2
A
Solution:
Find all instant centers and velocity of point A
1v A
2 = 1ω2 × rA2 / O2 ⇒ 1vA2 = 1ω2 × rA2 /O2 = 60⋅0.015 = 0.9 m / s
Using rotating radius method,
1v B
3 = 1.15 m / s
- 30 -
1vC
3 = 0.464 m / s
Now, using the relations between instant centers distances and angular velocities, we can find the
angular velocities as,
= 60 ⋅ 3.03 = 43.9 rad / s
r
1ω2 × rI23/ I12 = 1ω3 × rI23/ I13 ⇒ 1ω3 = 1ω 2 ⋅ rII23/ I12
23 /I13 4.14
= 60⋅ 2.62 = 38.4 rad / s
r
1ω2 × rI24 / I12 = 1ω4 × rI24 / I14 ⇒ 1ω4 = 1ω2 ⋅ rII24 / I12
24 / I14 4.09
Therefore,
1ω3 = 44.0 rad / s CW
1ω4 = 38.5 rad / s CW
1 vB
3 1
I34
B
1v '
A2 4 2
3
4
3
C
D 1v
C3
I13
I 24 O2 O4
1 vA
2
I 12 I 14
2
A
I 23
- 31 -
