chpt 4.pdf

Solutions to Chapter 4 Exercise Problems Problem 4.1 Locate all of the instant centers in the mechanism shown below. A

4 30˚

AB = 1.5"

2

63˚ 1

B 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-

∞ I12

A

I14

4 2

B

I23

∞

1

3

1

I24 I34

∞

I34

∞

∞

I14

I13

∞

2

4

3

Problem 4.2 Find all of the instant centers of velocity for the mechanism shown below.

B BC = CD BD = 3.06" 3

2

150˚ C

A

50˚

D 4

-2-

I24 I 23

1

4

B

2

3 3

2

C A

D

50˚

4

I12

I34

I13

-3-

I14

Problem 4.3 In the linkage shown below, locate all of the instant centers.

3.3"

AB = 1.35" BD = 3.9" DE = 0.9" BC = 0.9" CF = 2.0"

F B

6

5 C

2 3

55˚

A

E 4 D\

Solution 1 6

2

5

3

I 36

I13

I1 B I2

4

I15

6 F 6

5

I56

3

2

I25 C I 3 5

3

A I26

I12

I 24

I14

E

I46 4 I3

D

4

I4 5

Problem 4.4 Find all of the instant centers of velocity for the mechanism shown below. -4-

I 15 I 45 D I34

I35

E 5

I 25

4

I 24 I 23

C

2

E 5 D

A I 12

4

C

5.0 cm

I14

2 28˚

3

A

1

AB = 8.0 cm AC = 4.5 cm BD = 13.0 cm DE = 2.9 cm

3

5

B I 13 2

B 4

-5-

3 Solution

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" AB = 1.2" BC = 3.0" CD = 2.3" CE = 1.35" EB = 2.05"

C

3 4

B E 2 125˚ A

D

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, 1ω3

= 1ω2

rI23/ I12 = 60 1.2 = 17.7 rad / s 4.07 rI23 /I13

Because the instant center I23 lies between I12 and I13, 1ω3 is in the opposite direction of 1ω2. Therefore, 1ω3 is counterclockwise. Then, 1vC

= 1ω3 × rC/I13 ⇒ 1vC3 = 1ω3 rC/I13 = 17.7⋅2.11 = 37.3 in / s

1vE

= 1ω3 × rE/I13 ⇒ 1v E3 = 1ω3 rE/I13 = 17.7⋅3.25 = 57.5 in / s

3

and 3

The directions for the velocity vectors are shown in the drawing.

-6-

I 13

1

4

I 34

2

3

C

vC 3

4

3 E

vE 3

vB3

I23 B 2

D

I 24

A

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.

3 103˚

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”

ω2 = 900 rpm =

1

900(2π ) = 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 E

4

C

6

100˚ B 3

A 2

AB = 1.1" AC = 0.9" CD = 1.5" 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

1ω 4

= 1ω 2 ⋅

rI24 /I12 = 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 4

I14 C 1v

B2

I 56 I 26 I12

I 23 B I34

2

3

1v 'B

I16

E

6

4

24

I35

I25

1v B

1 v 'I

5

I 36

1 I13

A

2

1v I

6

2

5

3

I 24 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 -

G (2.55", 2.95")

AC = 1.2" AB = 1.35" BC = 0.9" CE = 2.7" BD = 2.6" DE = 2.6" DF = 2.2" EF = 3.1" FG = 2.8" Y

6 D F 5 B 4

2 C A

20˚

X

3 E

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 1ω 6

= 1ω 2 ⋅

rI26 /I12 = 5.24 ⋅ 1.56 = 3.47 rad / s rI26 /I16 2.35

So, 1ω6

= 3.47 rad / s CW

- 12 -

1 6

I 16

2

G

6

I 46 F

D 5

I 45

3 5

I 25

I 24

4

B I 26 2

I 23 C

A I 12

3

- 13 -

E I 34

Problem 4.9

In the operation of this mechanism, link 3 strikes and trips link 5, which is initially at rest. High wear has been observed at the point of contact between links 3 and 5. As an engineer, you are asked to correct this situation. Therefore, you decide to do the following: 1) Determine the direction of the velocity of point C on link 3 at the moment of contact. 2) Relocate the ground pivot of link 4 to make the direction of the velocity of point C perpendicular to link 5 (hence less rubbing at the point of contact) when contact occurs. Y

Driving Link A (0, 1.37")

2

AE = 0.79" BD = 0.69" ED = 0.74" CD = 0.59"

E -10˚

3 157˚

4

r = 0.125"

D

Compression Spring X

C

B (0.7", 0)

Stop F (0.75", -0.72")

5 6

Solution Before link 3 strikes link 5, link 5 is stationary and can be considered to be fixed to the frame. Hence, link 5 is the same as link 1. For part 1, find the location of the instant center I13. Then find the direction of the velocity of point C3 relative to the frame. This is shown in the drawing. For part 2, we want the velocity of C3 to be perpendicular to the surface of link 5. Knowing the direction of the velocity of C3, we can find the new location of the instant center I13. Knowing the location of the instant center, we can find the new location of the ground pivot for link 4. The solution is shown in the figure.

- 14 -

A

Original I13

E

Desired I13

3 New location for pivot B'

4

Original velocity of point C

Desired velocity of point C

D

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 C

35˚

5 D

E 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

vI 26 I24

vB'6 B

I23 A

2 I26

6

B'

3

I56

A' D

I16 F

∞

∞

5

C I12

E 4 ∞ I35

I34

1 in

1 6

2

5

3 4

I36

- 16 -

I14 ∞

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

0.7" F ω2

7˚

A

1.0" 28˚

4 X

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

D

I56 6 C

I16

I24

E I 13

I 23 1 2

6

2 2 F I12

3 I34 1 in

3

5

I 14

4

I25

- 18 -

A

Problem 4.12

If ω2 = 5 rad/s CCW, find ω5 using instant centers. C

B

4 3

5 ω2

62˚

A

E

2

AE = 4.1" EF = 2.0" AB = 1.5" BC = 1.55" CF = 4.0" DE = 1.0"

D

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 , 1 ω5

= 1ω2 ⋅

rI25/ I12 = 5⋅ 1.83 = 4.03 rad / s rI25 / I15 2.27

So, 1ω5

= 4.03 rad / s CW

- 19 -

C

4

I 34

B I 45

I 35

3

5

A

E I 12

I 25

I15

2 I23

1 D 5

2

I13 F 3

4

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" DF = 1.6"

A

ω2

Y 6

F (3.6", 1.45")

B

2

3

30˚ 5 C

35˚

D

E 4

- 20 -

X

Solution: I 13

I 12 A I46

B 6 I26

F I36

2

1 vA 6 1 v'I

I25 D

26

C

5 1 v'

D2

I 35

E I34

1 vD

I 16 1 vI

3

I 14 4

2

26

1 6

2

5

3 4

I 23

Find necessary instant centers, i.e. I12, I16, and I26 , and the velocity of point D as - 21 -

1vD

2

=1ω 2 × rD2 / F2 ⇒ 1vD2 = 1ω 2 ⋅ rD2 / F2 = 1 ⋅1.6 = 1.6 in / s

Using rotating radius method 1v A

6

= 3.095 in / s in the direction shown.

Problem 4.14

If vA2 = 10 in/s as shown, find vB4 using the instant-center method. AC = 1.95" AD = 2.0" DE = 1.1" BD = 0.9" BE = 1.9"

Y

5

B

vA2 A

3 2

D 4

C (3.35", 0.3") 6 E

X

1.0"

Solution: Find necessary instant centers, i.e. I12, I15 and I25 , then using these instant centers and the rotating radius method, find the velocity of I25 . Because the velocity is the same for all points in link 5, 1vB

5

= 1vB4 = 1vI25 = 27.4 cm / s in the direction shown.

- 22 -

1v2

5 B I15

A

I45

I23

I14 3 I24

I34 D

2

4 E

C 6 I46

I25

I12

I16 1 6

2

5

3 4

Problem 4.15

If vA2 = 10 in/s as shown, find vB4 using the instant-center method. Y

5

CB = 1.1" BE = 2.0" BD = 1.1" DE = 1.2" AD = 1.6" AF = 1.6"

X

15˚

C

vA2

A

B 3 D

2

4 1.5"

F (4.2, -1.3) E

6

Solution: Find necessary instant centers, i.e. I12, I14 and I24 , then using these instant centers and the rotating radius method, find the velocity of I25 . Because velocity of link 5 makes linear movement, 1vB4

=11.8 cm / s in the direction shown.

- 23 -

Y I14 I 45 I 15

1vA

B

5

D

4 1.5"

1vB

3

X I 34

15˚

C

I 23

A

2

1 6

2

5

3

I 24 2

4

4

I 46 E

I 12

F (4.2, -1.3)

I 16

6

Problem 4.16

If vA6 = 10 in/s as shown, determine the velocity vector (direction and magnitude) for point B on link 3 using the instant-center method. G 3

B

H

4 2 F

C

I 57˚

6 A

vA 6

D

CD = 0.8" CA = 0.6" ED = 1.85" EF = FG = 1.35" GH = 1.5" HI = 0.95" CI = 2.1" CF = 0.65"

5 E

Solution: Find necessary instant centers, i.e. I13, I16, and I36 . Using rotating radius method as shown in the figure above, 1v B

3

= 17.5 in / s in the direction shown.

- 24 -

I 13

G I 34 4

B

1

3

I 36

I 23 H

6

2

5

3 4

2 F

C

I 16 I 14

I 1v A6

6

I 12

I 46

A I 56

I45

D

5 E

Problem 4.17

In the mechanism below, ω2 is 20 rad/s CCW. Find I26 and use it to find the angular velocity of link 6. Y A

X 2 AB = 1.5" BC = 4.9" CE = 4.3" EF = 1.2" (XD , Y D) = (0.95, -4.45) (X F, Y F) = (2.5, -4.85)

B

3

E 5 4

6 D

F

C

- 25 -

65˚

Solution: A I12

Y X 2

B I23

1 6

2

5

3

3

I24

4 I 26 I 46

I45 C I 34

4

5 I15

E

6

D

F I56 I16 65.0°

I14

Find necessary instant centers as shown in the sketch above, i.e. I12, I16, and I26 . Then using the relationship 1ω2

× rI26/ I12 = 1ω6 × rI26 / I16

we can find the angular velocity of link 6 as 1ω6

= 1ω2 ⋅

rI26 / I12 = 20 ⋅ 3.40 = 33.2 rad / s 2.05 rI26 / I16

- 26 -

Because I26 is between I12 and I16 , 1ω6

= 33.1 rad / s CW

Problem 4.18

If vB2 = 10 in/s as shown, determine the velocity vector (direction and magnitude) of point C4 using the instant center method. D 5

Y B 2

A

3 4

C

E

50˚ X vB 2

AB = 0.75" BE = 3.4" EF = 1.6" FD = 2.85" CD = 1.35"

6

F (4.7, 0.75)

Solution: Find the necessary instant centers, i.e. I12, I14 and I24 . Using rotating radius method as shown in the figure below, 1vC

4

= 26 in / s

- 27 -

I14

1 6

2

5

3 4

I46 I24

I 56

D

5 I 26 B 2

I23

3 I34

I45 4 C

E I36

A 6 I12 F I16

- 28 -

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" CE = 1.45"

5 6

B 98˚

3 C

E

4

A

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 6 6

= 1ω 6 × rD6 / E6 ⇒ 1ω6 =

1vD / E 6 6

rD6 / E6

= 13.2 = 5.87rad / s 2.25

Therefore, 1ω6

= 5.87 rad / s CW

- 29 -

1 I56 D

6

2

5

3

I26 vA' 2 5

6

4 I34 , I45

I24

B C

E I16

3

4

A I 46

I14

I23

vA2

2

I12

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 O4 B = 30.1 cm 4

3 D

C O2

A B

O4

= 29.5 cm

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 -

A = 14.75 cm D D = 7.5 cm C 7.5 cm O 2 O4 =

1vC

3

= 0.464 m / s

Now, using the relations between instant centers distances and angular velocities, we can find the angular velocities as, 1ω2

/ I12 × rI23/ I12 = 1ω3 × rI23/ I13 ⇒ 1ω3 = 1ω 2 ⋅ rII23 = 60 ⋅ 3.03 = 43.9 rad / s 23 /I13 4.14

1ω2

/ I12 × rI24 / I12 = 1ω4 × rI24 / I14 ⇒ 1ω4 = 1ω2 ⋅ rII24 = 60⋅ 2.62 = 38.4 rad / s 24 / I14 4.09

r

r

Therefore, 1ω3

= 44.0 rad / s CW

1ω4

= 38.5 rad / s CW

1 vB 3

I34

1

B 4

1v ' A2

3

2

4 3

C

D

1v C3

2 A

O4

O2

I 24 1 vA 2

I13

I 12

I 23

- 31 -

I 14

Problem 4.21

The circular cam shown is driven at an angular velocity ω2 = 15 rad/s (CW). There is rolling contact between the cam and roller, link 3. Using the instant center method, find the angular velocity of the oscillating follower, link 4. 3 4

D

E (3.0", 3.0")

C AB = 1.22" DE = 3.50" BC = 2.00" CD = 0.50"

Y 2

B

135˚ X

A

Solution: Find the necessary instant centers, i.e. I12, I14, and I24 . Using the relation among these instant centers, we can write 1ω 2 × rI

24 /I12

= 1ω 4 × rI 24 /I14

Now, 1ω4

= 1ω2 ⋅

rI24/ I12 = 15⋅ 2.08 = 4.18 rad / s 7.46 rI24/ I14

Therefore, 1ω 4

= 4.18 rad / s CW

- 32 -

3

I34

D

4

E

I23

C

I 14 1 B

2 4

A

2

I12 3

I 24

Problem 4.22

If ω3 = 1 rad/s CCW, find the velocity of points E and F using the instant center method. Show the velocity vectors vF3 and vE4 on the figure.

168˚

F

ω3

C 3

B

0.48"

1.01"

E 4

2 A

28˚

1

D

Solution Find the instant centers I13, I14, and I34. These are shown on the figure below. Then, 1vF

3

= 1ω3 × rF / I13 ⇒ 1vF3 = 1ω3 rF/ I13 = 1(2.84) = 2.84 in / s

The vector is shown on the figure. To find 1ω 4 , note that 1ω3 × rI

34 / I13

= 1ω4 × rI34 / I14

Therefore, 1ω4

= 1ω3

rI34 /I13 = 1 0.292 = 0.343 rad / s CW 0.85 rI34/I14

- 33 -

AB = 1.65" BC = 0.88" CD = 0.85" AD = 2.46" CE = 1.26" DE = 1.56" BF = 1.94"

Then, 1vE

4

= 1ω4 × rE / I14 ⇒ 1vE4 = 1ω4 rE/ I14 = 0.343(1.56) = 0.536 in / s

The direction is shown on the figure I13 F

C

I23 1v F3

B

E

I34

3

1v E4

4

2 I14

I12 A

1

D

Problem 4.23

In the eight-link mechanism, most of the linkage is contained in the black box and some of the instant centers are located as shown. The velocity of point B is 100 in/s in the direction shown. Compute the velocity of point D8 and determine the angular velocity of link 2. Y vB2

I58 (2.46, 1.74)

(0.40, 1.48)

I47

B

C

3

7

Black Box

2

ω2

AB AE EC ED

= 1.30" = 4.25" = 1.30" = 0.86"

D 8

65˚

65˚ E

A I27

(1.07, -0.45) I52

X

I 67 (3.58, -0.50)

(1.85, -0.78)

Solution To find the velocity of D8, we need the instant centers I12, I18, and I28 . I12 and I18 are determined by inspection. I28 can be found using I12 and I18 and I52 and I58 . The velocity of D8 can then be found by the rotating radius method. The result is 1vD

8

= 59.7 in / s

in the direction shown.

- 34 -

1v I 28 1 vB 2

I37 I 23

I58 I 78

1v B'2

B 3

7

Black Box

2

C

I 12

B'

8

8

1v D'8

A

1 vD

D

I 18

E

D'

I28

I27 I52

Problem 4.24

If the velocity of point A on link 2 is 10 in/s as shown, use the instant center method to find the velocity of point C on link 5.

F (3.15", 1.9") B

Y

6

0.25"

vA DE = 2.5" AD = 0.75" AB = 1.75" BE = 1.5" GF = 1.5"

A

3 0.45"

2 D

5 G

C5

5 4

75˚

E

Solution: Find necessary instant centers, i.e. I12, I15 and I25 . Using rotating radius method, we find the velocity of point C5 as shown below. 1vC

5

= 11.15 in / sec

- 35 -

X

I 16 1 1 v 'I

2

6

1v C5 1 vA 2

3

5

A I 25

4

F

25

I34 6

B

3 I 56

I23

5 G 5

2

I24

C5 I45

4

D

E

I12 1v

I25

I14

1 v' A2

I 15

Problem 4.25

Assume that link 7 rolls on link 3 without slipping, and find the following instant centers: I13, I15, and I27. For the given value for ω2, find ω7 using instant centers. AB = 1.8" BG = 0.85" GF = 1.7" BD = 3.9" DE = 3.25" AE = 2.0"

Y

7

0.75"

5

D

1.0" G

C B

4

3

6

F (-2.0", 0.95")

2

90˚

E

A

X

ω 2 = 2 rad/s

Solution: Draw linkage to scale (shown below) and find necessary instant centers, i.e. I13, I15, I27, I12 and I17. - 36 -

Having the relationship between 1ω2 and 1ω7 using instant centers, we can write, 1ω 2 × rI

27 /I12 =

1ω7 × rI

(1)

27 /I17

Solve Eq. (1) for 1ω7,

rI27 / I12 = 2⋅ 3.199 = 6.385 rad / s rI27 / I17 1.002

1ω7

= 1ω 2 = ⋅

1ω7

= 6.385 rad / s CW

So

I13 D I 56

5 7 G I 34 4

I 57

I 37 I 23 I 25

I 27 C I 17

B 3

6

F I14 2 E

I12

I 16

A

1 7

2

6 3 5

I 15

- 37 -

4

Problem 4.26 If vA2 = 10 in/s as shown, find vC5 using the instant-center method. BE = 0.85" EG = 2.2" EC = 1.2" CG = 1.25"

DA = 0.95" DF = 2.45" AB = 1.45" BF = 1.8"

G C

6

B

vA 2 A

5

3

1.9"

E 2 4 D

70˚

F

Solution: Find necessary instant centers, i.e. I12, I15, and I25 . Using rotating radius method, get the velocity of point C5 as 1vC

5

= 3.81cm / sec

- 38 -

I 13 I 16 I 56 G

C

6

B 1v A2

I 34 5

3

A I 23

I 45

E

2 I25 I 24

4 F

D I14

I12 1 6

2

5

3 4 I 15

- 39 -

Problem 4.27 If ω2 = 10 rad/s CCW, find the velocity of point B using the instant-center method. Show the velocity vector vB3 on the figure.

110˚

E CA = 1.5" DE = 2.5" CD = 4.0" AB = 1.6"

3

4 Β

Α 18˚ 1ω 2

C

2 45˚

D

Solution: Find necessary instant centers, i.e. I12, I13, and I23 .and velocity of point A as, 2=

1vA

1 ω2 × rA / C 2 2

⇒ 1 vA2 = 1ω 2 ⋅ rA 2 /C2 = 10⋅1.5 = 15in / sec

Then, using rotating radius method, 1vB

3

= 9.5 in / sec in the direction shown.

- 40 -

I13

I34 E

1 vA

2

3

4

1 Β

Α 4

2

1 vB3

I23 2

3

1 v' A2

C

D

I12

I 14

Problem 4.28 If ω2 = 100 rad/s CCW, find the velocity of point E using the instant center method. Show the velocity vector vE4 on the figure.

ω2

D

A 70˚

2

E 4

B 3

C

AB = 1.0" BC = 1.75" CD = 2.0" DE = 0.8" AD = 3.0"

Solution: I , I , and I24 and velocity of point B as Find necessary instant centers, i.e. 12 14 1vB

2

= 1ω2 × rB2 /A 2 ⇒ 1vB2 = 1ω2 ⋅ rB2 /A 2 = 100 ⋅1 = 100 in / sec

Therefore, using rotating radius method, 1vE

4

= 27.78 in / s in the direction shown.

- 41 -

I24

I 14

A I12

B'

D

E 1 vB

2

' 1 vB 2

1 vI

1 vE

I 23

4

B I 34

C

24

Problem 4.29 If ω2 = 5 rad/s CCW, find ω6 using instant centers. D

E 5

4

6

A

F AC = 1.0" AB = 2.0" BD = 4.0" DE = 1.5" EF = 1.5" AF = 4.1"

50˚ 2 3 B

ω2

C

Solution: Find necessary instant centers as shown in the sketch above, i.e. I12, I16, and I26 . Then using the relationship 1ω 2 × rI

26 /I12 =

1ω6 × rI

26 / I16

we can find the angular velocity of link 6 as 1ω6

= 1ω 2 ⋅

rI26 /I12 = 5⋅ 1.9695 = 5.054 rad / sec rI26 /I16 1.9486

Because I26 is between I12 and I16 , - 42 -

1ω6

= 5.054 rad / sec CW

I 45

D

E

I 56

5

I 46

4

6

A

F

I12

I 26 I 16

1 6

I 34

2 3

2

I 24

I 23 C

5

3

4 B

Problem 4.30 If ω2 = 100 rad/s CCW, find vB4 using instant centers and the rotating radius method. AD = 1.8" CD = 0.75" AE = 0.7" CF = 0.45" FG = 1.75" CB = 1.0" DB = 1.65"

G 125˚ B4 3 C

2

F

E

4 D

A

Solution: - 43 -

95˚

I 14

Find necessary instant centers, i.e. I12, I14, and I24 and the velocity of I24 as 1vI

24

= 1ω2 × rI 24 /I12 ⇒ 1vI24 = 1ω 2 ⋅ rI24 / I12 = 100 ⋅ 2.885 = 288.5 in / s

Using rotating radius method, 1vB

4

= 460 in / s

I 23

1

4

G 1v B4

2

B4 3

3

1 v I'

24

I 34

2 E

1 vI

C 4

F

24

D

A

I 14

I12

I 24

Problem 4.31 If vA2 = 10 in/s as shown, find the angular velocity (ω6) of link 6 using the instant-center method. vA

AB = 1.0" AD = 2.0" AC = 0.95" CE = 2.0" EF = 1.25" BF = 3.85"

2

E 5

6

A C

2 B

27˚

3

D 4

F

Solution: Find necessary instant centers, i.e. I12, I16, and I26 and using rotating radius method, 1vE

6

= 2.65 in / sec

- 44 -

or 1vE

6

1vE 6

= 1 ω6 × rE/ F ⇒ 1ω 6 =

rE /F

= 2.65 = 1.325 rad / s CCW 2 1 v 'I26

1 6

2

5

3 4

I13

I56

1v A2

1vE

6

E

I36 I23

A

I35

2

3

B I26

1 vI

26

6 5 4

C

I12

' 1 vA 2

- 45 -

D I34

I 14

F I 16

Problem 4.32 If ω2 = 50 rad/s CCW, find the velocity of point G using the instant center method. Show the velocity vector vG5 on the figure. 3

C

E

4

B

5

32˚ 2

AB = 1.16" BC = 0.70" CD = 1.45" DE = 1.16" AD = 1.30" DF = 1.30 EG = 2.20"

75˚

A

D

ω2

F

6

G

Solution: Find the necessary instant centers, i.e. I12, I15, and I25 . Using rotating radius method, 1vG

5

= 33in / sec

I 34 1 vB

I 23

2

3

4

B

1v ' B2

I 45

C

E

I 25

A

D I 12

1

I56

2

1 vI 25

I 24

5

6

F I16

I 14 1v ' I 25

6

2

5

3

I 15

4

- 46 -

1v G5

G

Problem 4.33 If ω2 = 100 rad/s CCW, find ω6. E (2.0", 6.0")

6

AB = 1.2" BC = 6.0" CD = 3.0" AD = 4.0" BF = 3.0"

C

Y 3 ω2

B

5

F

A

D

2

Solution: Find necessary instant centers, i.e. I12, I16, and I26 . 1ω 2 × rI

1ω6

26 /I12

= 1ω 2 ⋅

= 1ω 6 × rI26 / I16

rI26 / I12 = 100⋅ 5.988 = 1777 rad / sec rI26 / I16 0.337

Because I26 is between I12 and I16 , 1ω 6 = 1777

4

rad / s CW

- 47 -

X

I 16 E

1 6

2

5

3

6 I34 C

F 5 I 35 I 26 B 2 I 23

3 4 I 36

I 56

D

A I 12

I 14 I 13

- 48 -

4