solutions to chapter 2 exercise problems problem 2 · solutions to chapter 2 exercise problems...

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Solutions to Chapter 2 Exercise Problems

Problem 2.1

In the mechanism shown below, link 2 is rotating CCW at the rate of 2 rad/s (constant). In theposition shown, link 2 is horizontal and link 4 is vertical. Write the appropriate vector equations,solve them using vector polygons, and

a) Determine vC4, ωωωω3, and ωωωω4.

b) Determine aC4, αααα3, and αααα4.

Link lengths: AB = 75 mm, CD = 100 mm

B

C

2

3

4A

D

50 mm

250 mm

ω2

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

- 49 -

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

ω2 2= rad s CCW/

v r rB A B A B A rad s mm mm s2 2 2 2 75 150/ / /( ) ( / )( ) /= × ⊥ = =ω to

v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C D4 4/ / /( )= × ⊥ω44 to

Solve Eq. (1) graphically with a velocity polygon. From the polygon,

vC B mm s3 3 156/ /=

v vC D C mm s4 4 4 43/ /= =

Now,

ω33 3 156

182 86= = =vrC B

C B

/

/. rad / s

From the directions given in the position and velocity polygons

ω3 86= . rad / s CW

Also,

ω44 4 43

100 43= = =vrC D

C D

/

/. rad / s


ω44 = .43 rad / s CW

Acceleration Analysis:

a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A mm2 2 2 22 2 22 22 75 300/ / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2

- 50 -

in the direction of - rB A2 2/

aB At

2 2 0/ = since link 2 rotates at a constant speed (α2 0= )

a r a rC Br C B C B

r C B mm3 3 3 33 3 32 286 182 134 6/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2

in the direction of - rC B/

a r a r rC Bt C B C B

t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D mm s4 4 4 44 4 42 2 243 100 18 5/ / / / . . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω

in the direction of - rC D/

a r a r rC Dt C D C D

t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,

aC Bt mm s3 3 19 22/ . /= 2

aC Dt mm s4 4 434 70/ . /= 2

Then,

α33 3 67 600

2 42 27 900= = =arC Bt

C B

/

/

,. , rad / s2

α44 4 434 70

100 4 347= = =a

rC Dt

C D

/

/

. . rad / s2

To determine the direction of αααα3, determine the direction that rC B/ must be rotated to be parallel to

aC Bt

3 3/ . This direction is clearly counter-clockwise.

To determine the direction of α4, determine the direction that rC D/ must be rotated to be parallel to

aC Dt


From the acceleration polygon,

aC mm s4 435= / 2

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Problem 2.2

In the mechanism shown below, link 2 is rotating CCW at the rate of 500 rad/s (constant). In theposition shown, link 2 is vertical. Write the appropriate vector equations, solve them using vectorpolygons, and



Link lengths: AB = 1.2 in, BC = 2.42 in, CD = 2 in

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

- 52 -

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

ω2 500= rad s CCW/

v r rB A B A B A rad s in in s2 2 2 500 1 2 600/ / /( ) ( / )( . ) /= × ⊥ = =ω to

v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C D4 4/ / /( )= × ⊥ω44 to


vC B in s3 3 523 5/ . /=

v vC D C in s4 4 4 858/ /= =

Now,

ω33 3 523 5

2 42 216 3= = =vrC B

C B

/

/

.. . rad / s


ω3 216 3= . rad / s CCW

Also,

ω44 4 858

2 429= = =vrC D

C D

/

/rad / s


ω44 =429 rad s CC/ W


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A in s2 2 2 22 2 22 2500 1 2 300000/ / / / . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


aB At


- 53 -

a r a rC Br C B C B

r C B in3 3 3 33 3 32 2216 3 2 42 113 000/ / / / . . ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D in s4 4 4 44 4 42 2 2429 2 368 000/ / / / , /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α


aC Bt in s3 3 67561/ /= 2

aC Dt in s4 4 151437/ /= 2

Then,

α33 3 67561

2 42 27 900= = =arC Bt

C B

/

/ . , rad / s2

α44 4 151437

2 75 700= = =a

rC Dt

C D

/

/, rad / s2


aC Bt

3 3/ . This direction is clearly clockwise.


aC Dt



aC in s4 398 000= , / 2

- 54 -

Problem 2.3

In the mechanism shown below, link 2 is rotating CW at the rate of 10 rad/s (constant). In theposition shown, link 4 is vertical. Write the appropriate vector equations, solve them using vectorpolygons, and



Link lengths: AB = 100 mm, BC = 260 mm, CD = 180 mm

B

C

2

3

4

D

250 mm

ω 2

A

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

- 55 -

Now,

ω2 10= rad s CW/


v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C D4 4/ / /( )= × ⊥ω44 to


vC B mm s3 3 31 3/ . /=

v vC D C mm s4 4 4 990/ /= =

Now,

ω33 3 31 3

260 12= = =vrC B

C B

/

/

. . rad / s


ω3 12=. rad / s CCW

Also,

ω44 4 990

180 5 5= = =vrC D

C D

/

/. rad / s


ω44 =5 5. /rad s CW


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A mm s2 2 2 22 2 22 210 100 10 000/ / / / , /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


aB At


a r a rC Br C B C B

r C B mm3 3 3 33 3 32 212 260 3 744/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2

- 56 -



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D mm s4 4 4 44 4 42 2 25 5 180 5 445/ / / / . , /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α


aC Bt mm s3 3 4784/ /= 2

aC Dt mm s4 4 1778/ /= 2

Then,

α33 3 4785

260 18 4= = =arC Bt

C B

/

/. rad / s2

α44 4 1778

180 9 88= = =a

rC Dt

C D

/

/. rad / s2


aC Bt



aC Dt



aC mm s4

5 700= , / 2

- 57 -

Problem 2.4

In the mechanism shown below, link 2 is rotating CW at the rate of 4 rad/s (constant). In theposition shown, θ is 53˚. Write the appropriate vector equations, solve them using vector polygons,and



Link lengths: AB = 100 mm, BC = 160 mm, CD = 200 mm

B C2

3

4

D

ω 2A

220 mm

160 mm

θ

- 58 -

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

ω2 4= rad s CW/


v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C D4 4/ / /( )= × ⊥ω44 to


vC B mm s3 3 500/ /=

v vC D C mm s4 4 4 300/ /= =

Now,

ω33 3 500

160 3 125= = =vrC B

C B

/

/. rad / s


ω3 3 125= . rad / s CCW

Also,

ω44 4 300

200 1 5= = =vrC D

C D

/

/. rad / s


ω44 =1 5. /rad s CCW


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

- 59 -

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A mm s2 2 2 22 2 22 24 100 1600/ / / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


aB At


a r a rC Br C B C B

r C B mm3 3 3 33 3 32 23 125 160 1560/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D mm s4 4 4 44 4 42 2 21 5 200 450/ / / / . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α


aC Bt mm s3 3 618 5/ . /= 2

aC Dt mm s4 4 3 220/ , /= 2

Then,

α33 3 618 5

160 3 87= = =arC Bt

C B

/

/

. . rad / s2

α44 4 3220

200 16 1= = =a

rC Dt

C D

/

/. rad / s2


aC Bt



aC Dt



aC mm s4

3250= / 2

- 60 -

Problem 2.5

In the mechanism shown below, link 2 is rotating CCW at the rate of 4 rad/s (constant). In theposition shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vectorpolygons, and



Link lengths: AB = 1.25 in, BC = 2.5 in, CD = 2.5 in

B

C

2

3

4

D

ω 2

A

1.0 in

0.75 in

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

- 61 -

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

ω2 4= rad s CCW/

v r rB A B A B A rad s in in s2 2 2 4 1 25 5/ / /( ) ( / )( . ) /= × ⊥ = =ω to

v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C D4 4/ / /( )= × ⊥ω44 to


vC B in s3 3 6 25/ . /=

v vC D C in s4 4 4 3 75/ . /= =

Now,

ω33 3 6 25

2 5 2 5= = =vrC B

C B

/

/

.. . rad / s


ω3 2 5= . rad / s CCW

Also,

ω44 4 3 75

2 5 1 5= = =vrC D

C D

/

/

.. . rad / s


ω44 =1 5. /rad s CW


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

- 62 -

a r a rB Ar B A B A

r B A in s2 2 2 22 2 22 24 1 25 20/ / / / . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


aB At


a r a rC Br C B C B

r C B in3 3 3 33 3 32 22 5 2 5 15 6/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D in s4 4 4 44 4 42 2 21 5 2 5 5 6/ / / / . . . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α


aC Bt in s3 3 4 69/ . /= 2

aC Dt in s4 4 4 69/ . /= 2

Then,

α33 3 4 69

2 5 1 87= = =arC Bt

C B

/

/

.. . rad / s2

α44 4 4 69

2 5 1 87= = =a

rC Dt

C D

/

/

.. . rad / s2


aC Bt



aC Dt



aC in s4

7 32= . / 2

- 63 -

Problem 2.6

In the mechanism shown below, link 2 is rotating CW at the rate of 100 rad/s (constant). In theposition shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vectorpolygons, and

a) Determine vC4 and ωωωω3

b) Determine aC4 and αααα3

Link lengths: AB = 60 mm, BC = 200 mm

B

C

2

3

4

ω 2

A

120 mm

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

- 64 -

Now,

ω2 100= rad s CW/


v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

vC D4 4/ →parallel to the ground.


vC B mm s3 3 7 500/ , /=

v vC D C mm s4 4 4 4500/ /= =

Now,

ω33 3 7500

200 37 5= = =vrC B

C B

/

/. rad / s


ω3 12=. rad / s CW


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A mm s2 2 2 22 2 22 2100 60 600 000/ / / / , /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


aB At


a r a rC Br C B C B

r C B mm3 3 3 33 3 32 237 5 200 281 000/ / / / . ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a aC D C4 4 4/ = →parallel to ground


- 65 -

aC Bt mm s3 3 211 000 2

/ , /=

a aC D C mm s4 4 4 248 000 2/ , /= =

Then,

α33 3 211 000

200 1060= = =arC Bt

C B

/

/

, rad / s2


aC Bt



aC mm s4

248 000= , / 2

Problem 2.7

In the mechanism shown below, link 4 is moving to the left at the rate of 4 ft/s (constant). Write theappropriate vector equations, solve them using vector polygons, and

a) Determine ωωωω3 and ωωωω4.

b) Determine αααα3 and αααα4.

Link lengths: AB = 10 ft, BC = 20 ft.

B

C

23

4

A

8.5 ft

120˚

vC4

- 66 -

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

vC4 4= ft / s parallel to the ground

v r rB C B C B C3 3 3/ / /( )= × ⊥ω to

v r rB A B A B A2 2 2/ / /( )= × ⊥ω to


vB C ft s3 3 2 3/ . /=

vB A ft s2 2 2 3/ . /=

or

ω33 3 2 3

20 115= = =vrB C

B C

/

/

. . rad / s


- 67 -

ω3 115=. rad / s CW

Also,

ω22 2 2 3

10 23= = =vrB A

B A

/

/

. . rad / s


ω2 23=. rad / s CCW

ω4 0= rad / s since it does not rotate


a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A ft s2 2 2 22 2 22 223 10 529/ / / / . . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2


a r a r rB At B A B A

t B A B A2 2 2 22 2/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Br C B C B

r C B ft3 3 3 33 3 32 2115 20 264/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a C D4 4 0/ = link 4 is moving at a constant velocity


aC Bt ft s3 3 0 045/ . /= 2

a B At ft s2 2 0 017 2

/ . /=

Then,

α33 3 0 45

20 023= = =arC Bt

C B

/

/

. . rad / s2

α22 2 0 017

10 0017= = =arB At

B A

/

/

. . rad / s2

- 68 -


aC Bt


To determine the direction of αααα2, determine the direction that rB A/ must be rotated to be parallel to

aB At


Problem 2.8

In the mechanism shown below, link 4 is moving to the right at the rate of 20 in/s (constant). Writethe appropriate vector equations, solve them using vector polygons, and

a) Determine ωωωω3 and ωωωω4.

b) Determine αααα3 and αααα4.

Link lengths: AB = 5 in, BC = 5 in.

2

A

B

C 3

4

7 in

45˚

vC4

- 69 -

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

vC in s4 20= / parallel to the ground

v r rB C B C B C3 3 3/ / /( )= × ⊥ω to

v r rB A B A B A2 2 2/ / /( )= × ⊥ω to


vB C in s3 3 14 1/ . /=

vB A in s2 2 14 1/ . /=

or

ω33 3 14 1

5 2 82= = =vrB C

B C

/

/

. . rad / s


ω3 2 82= . rad / s CCW

Also,

ω22 2 14 1

5 2 82= = =vrB A

B A

/

/

. . rad / s


ω2 2 82= . rad / s CCW

ω4 0= rad / s since it doesn’t rotate


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

- 70 -

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A in s2 2 2 22 2 22 22 82 5 39 8/ / / / . . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2



t B A B A2 2 2 22 2/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Br C B C B

r C B in3 3 3 33 3 32 22 82 5 39 8/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to



aC Bt in s3 3 38 8/ . /= 2

a B At in s2 2 38 8 2

/ . /=

Then,

α33 3 38 8

5 7 76= = =arC Bt

C B

/

/

. . rad / s2

α22 2 38 8

5 7 76= = =arB At

B A

/

/

. . rad / s2

α4 0 4= ( )link isnotrotating


aC Bt


To determine the direction of α22, determine the direction that rB A/ must be rotated to be parallel toaB A

t2 2/ . This direction is clearly clockwise.

- 71 -

Problem 2.9

In the mechanism shown below, link 4 is moving to the left at the rate of 0.6 ft/s (constant). Writethe appropriate vector equations, solve them using vector polygons, and determine the velocity andacceleration of point A3.

Link lengths: AB = 5 in, BC = 5 in.

2

A

B

C

3

4vC4

135˚

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

v v vB A B A3 3 3 3= + /

Therefore,

v v v vC B C A B A3 3 3 3 3 3+ = +/ / (1)

- 72 -

Now,

vC4 6= . ft / sparallel to the ground

v r rB C B C B C3 3 3/ / /( )= × ⊥ω to

v r rB A B A B A3 3 3/ / /( )= × ⊥ω to


vB C ft s3 3 85/ . /=

or

ω33 3 85

5 12 2 04= = =vrB C

B C

/

/

.( / ) . rad / s


ω3 2 04= . rad / s CW

Now,

v r rB A B A B A ft s3 3 3 2 04 5 12 85/ / /( ) ( . )( / ) . /= × ⊥ = =ω to

Using velocity image,

vA ft s3 1 34= . /


a aC C4 3 0= =

a a a a aB B B C r B C t B C3 2 3 3 3 3 3 3= = = +/ / / (2)

Now,

a r a rB Cr B C B C

r B C ft s3 3 3 33 3 32 22 04 5 12 1 73/ / / / . ( / ) . /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2

in the direction of - rB C3 3/

a r a r rB Ct B C B C

t B C B C3 3 3 3 3/ / / / /( )= × ⇒ = ⋅ ⊥α α33 to



aB Ct ft s3 3 1 73/ . /= 2

Then,

- 73 -

α33 3 1 73

5 12 4 15= = =arB Ct

B C

/

/

.( / ) . rad / s2

To determine the direction of αααα3, determine the direction that rB C/ must be rotated to be parallel to

aB Ct


Using acceleration image,

aA ft s3

4 93= . / 2

Problem 2.10

In the mechanism shown below, link 4 moves to the right with a constant velocity of 75 ft/s. Writethe appropriate vector equations, solve them using vector polygons, and

a) Determine vB2, vG3, ωωωω2, and ωωωω3.

b) Determine aB2, aG3, αααα2, and αααα3.

Link lengths: AB in= 4 8. , BC in= 16 0. , BG in= 6 0.

A

B

C

G

2 3

442˚

Position Analysis: Draw the linkage to scale.

- 74 -

B

G

2 3

42˚

A C

AB = 4.8"BC = 16.0"BG = 6.0"AC = 19.33"

g3

a1 a 2,

ov c3 c4,

b2 b3,

25 ft/sec

Velocity Polygon

Velocity Analysis:

v v vB C B C3 3 3 3= + /

v vB B3 2=

- 75 -

v v vB A B A2 2 2 2= + /

vA2 0=

Therefore,

v v v vC B C A B A3 3 3 2 2 2+ = +/ / (1)

Now,

vC3 75= ft / sin the direction of rC A/

v r rB C B C B C3 3 3/ / /( )= × ⊥ω to

v r rB A B A B A2 2 2/ / /( )= × ⊥ω to


vB C ft s3 3 69 4/ . /=

or

ω33 3 69 4

16 1 12 52= = =vrB C

B C

/

/

.( / ) rad / s


ω3 52= rad / s CCW

Also,

ω22 2 91 5

4 8 1 12 228= = =vrB A

B A

/

/

.. ( / ) rad / s


ω2 228= rad / s CW

To compute the velocity of G3,

v v v v rG B G B B G B3 3 3 3 3 3 33= + = + ×/ /ω

Using the values computed previously

ω3 3 3 52 6 0 312× = =rG B/ ( . ) in / s

and from the directions given in the velocity and position diagrams

ω3 3 3 3 3312× = ⊥r rG B G B/ /in / s

Now draw vG3 on the velocity diagram

vG3 79 0= . ft / s in the direction shown.

- 76 -


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A ft s2 2 2 22 2 22 2228 4 8 12 20 900/ / / / ( . / ) , /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω 2



t B A B A2 2 2 22 2/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Br C B C B

r C B ft3 3 3 33 3 32 252 16 12 3605/ / / / ( / )= × ×( )⇒ = ⋅ = ⋅ =ω ω ω / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to



aC Bt ft s3 3 28 700/ , /= 2

a B At ft s2 2 20 000 2

/ , /=

Then,

α33 3 28 700

16 12 21 500= = =arC Bt

C B

/

/

,( / ) , rad / s2

α22 2 20 000

4 8 12 50 000= = =arB At

B A

/

/

,( . / ) , rad / s2

To determine the direction of α3 , determine the direction that rC B/ must be rotated to be parallel to

aC Bt


To determine the direction of α22, determine the direction that rB A/ must be rotated to be parallel toaB C

t2 2/ . This direction is clearly counter-clockwise.


aB ft s2

28 900= , / 2

- 77 -

To compute the acceleration of G3, use acceleration image. From the acceleration polygon,

aG ft s3

18 000= , / 2

Problem 2.11

For the four-bar linkage, assume that ωωωω2 = 50 rad/s CW and αααα2 = 1600 rad/s2 CW. Write theappropriate vector equations, solve them using vector polygons, and

a) Determine vB2, vC3, vE3, ωωωω3, and ωωωω4.

b) Determine aB2, aC3, aE3,αααα3, and αααα4.

B

E

A D

C

2

3

4

120˚

AB = 1.75"AD = 3.55"CD = 2.75"BC = 5.15"BE = 2.5"EC = 4.0"

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B andthen C. Then locate E.

Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,

v r v r rB A B A B A B A B A2 2 2 2 50 1 75 87 5/ / / / /. . ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to

v r v r rC D C D C D C D C D4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

v r v r rC B C B C B C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

- 78 -

3 3aC / B

t

aC3 / B3r

aC4 /D4t

aC 4 /D4

r

c'3

b'3

o' d'4

C

B

D

2

3

4

A

E

50 in/s

Velocity Scale

b3

o

c3

e3

2000 in/s

Acceleration Scale

2

aB 2/A2

t

aB2/ A2r

e'3

Solve Eq. (1) graphically with a velocity polygon and locate the velocity of point E3 by image.From the polygon,

vC B3 3 65 2/ .= in / s

vC D4 4 92 6/ .= in / s

and

vE3 107 8= . in / s

in the direction shown.

Now

ω33 3 65 2

5 15 12 7= = =vrC B

C B

/

/

.. . rad / s

- 79 -

and

ω44 4 92 6

2 75 33 7= = =vrC D

C D

/

/

.. . rad / s

To determine the direction of ω3, determine the direction that rC B/ must be rotated to be parallel tovC B3 3/ . This direction is clearly clockwise.

To determine the direction of ω4 , determine the direction that rC D/ must be rotated to be parallel tovC D4 4/ . This direction is clearly clockwise.


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A2 2 2 22 2 22 250 1 75 4375/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



tB A B A2 2 2 22 2 1600 1 75 2800/ / / / /. ( )= × ⇒ = = ⋅ = ⊥α α in / s to2

a r a rC Br C B C B

r C B3 3 3 33 3 32 212 7 5 15 830 6/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D in4 4 4 44 4 42 2 233 7 2 75 3123/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 byimage. From the acceleration polygon,

aC Bt

3 3 1563/ = in / s2

aC Dt

4 4 4881/ = in / s2

Then,

- 80 -

α33 3 1563

5 15 303= = =arC Bt

C B

/

/ . rad / s2

α44 4 4881

2 75 1775= = =a

rC Dt

C D

/

/ . rad / s2


aC Bt



aC Dt


Also

aE3 5958= in / s2

Problem 2.12

Resolve Problem 2.11 if ωωωω2 = 50 rad/s CCW and αααα2 = 0 .

Position Analysis


Velocity Analysis:

The velocity analysis is similar to that in Problem 2.18.

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,

v r v r rB A B A B A B A B A2 2 2 2 50 1 75 87 5/ / / / /. . ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to



Solve Eq. (1) graphically with a velocity polygon and locate the velocity of point E3 by image.From the polygon,

vC D4 4 103 1/ .= in / s

and

vE3 116= in / s

- 81 -


Now

ω33 3 88 8

5 15 17 2= = =vrC B

C B

/

/

.. . rad / s

and

ω44 4 103 1

2 75 37 5= = =vrC D

C D

/

/

.. . rad / s

To determine the direction of ω3, determine the direction that rC B/ must be rotated to be parallel tovC B3 3/ . This direction is clearly counterclockwise.

To determine the direction of ω4 , determine the direction that rC D/ must be rotated to be parallel tovC D4 4/ . This direction is clearly counterclockwise.

- 82 -

3 3aC /B

t

aC3 / B3

r

aC4 /D4t

aC4 / D4

r

c'3

b'3

o' d'4

C

B

D

2

3

4

A

E

50 in/s

Velocity Scale

c3

b3

o

g3

1000 in/s

Acceleration Scale

2

aB 2/ A2r

e'3

- 83 -


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A2 2 2 22 2 22 250 1 75 4375/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2


a r a rB At B A B A

tB A2 2 2 22 2 0 1 75 0/ / / / .= × ⇒ = = ⋅ =α α

a r a rC Br C B C B

r C B3 3 3 33 3 32 217 24 5 15 1530/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D4 4 4 44 4 42 237 49 2 75 3865/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 byimage. From the acceleration polygon,

a

aC Bt

C Dt

3 3

4 4

2751

1405/

/

=

=

in / s

in / s

2

2

Then,

α

α

3 2

4

3 3

4 4

27515 15 534

14052 75 511

= = =

= = =

ar

ar

C Bt

C B

C Dt

C D

rad s/

/

/

/

. /

. rad / s2

To determine the direction of α3 , determine the direction that rC B/ must be rotated to be parallel toaC B


To determine the direction of α4, determine the direction that rC D/ must be rotated to be parallel toaC D


- 84 -

Also

aE3 2784= in / s2

Problem 2.13

In the mechanism shown below, link 2 is rotating CW at the rate of 180 rad/s. Write theappropriate vector equations, solve them using vector polygons, and

a) Determine vB2, vC3, vE3, ωωωω3, and ωωωω4.

b) Determine aB2, aC3, aE3, αααα3, and αααα4.

Link lengths: AB = 4.6 in, BC = 12.0 in, AD = 15.2 in, CD = 9.2 in, EB = 8.0 in, CE = 5.48 in.

A

B

C

D

E

2

4

120˚

3

X

Y

Position Analysis: Draw the linkage to scale.

Velocity Analysis:

v v r v rB B A B A B B A2 2 2 2 2 2 2 22 2 180 4 6 828= = × ⇒ = = =/ / / ( . )ω ω in / s

v vB B3 2=

v v vC B C B3 3 3 3= + /

v v v vC C D C D3 4 4 4 4= = + /

and

vD4 0=

- 85 -

A

B

C

D

E

2

4

120˚

3

AD = 15.2"DC = 9.2"BC = 12.0"AB = 4.6"EC = 5.48"EB = 8.0"

a1 a2,

o v

c3 c4,

b2 b3,

400 in/sec

Velocity Polygon

e3

Therefore,

v v vC D B C B4 4 3 3 3/ /= + (1)

Now,

vB3 828= ft / s ( )/⊥ to rB A

v r rC B C B C B3 3 3/ / /( )= × ⊥ω to

v r rC D C D C Dto4 4 4/ / /( )= × ⊥ω

Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directlyfrom the polygon. The magnitudes are given by:

vvrC BC B

C B3 3

3 3583 58312 48 63/

/

/.= ⇒ = = =in / s rad / sω


ω3 48 6= . /rad s CCW

Also,

- 86 -

v vrC DC D

C D4 4

4 4475 4759 2 51 64/

/

/ . .= ⇒ = = =rad / s rad / sω


ω4 51 6= . rad / s CW

To compute the velocity of E3,

v v v v vE B E B C E C3 3 3 3 3 3 3= + = +/ / (1)

Because two points in the same link are involved in the relative velocity terms

v r rE B E B E Bto3 3 3/ / /( )= × ⊥ω

and

v r rE C E C E Cto3 3 3/ / /( )= × ⊥ω

Equation (2) can now be solved to give

vE3 695= in / s


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A2 2 2 22 2 22 2180 4 6 149 000/ / / / . ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2


a r a rB At B A B A

tB A2 2 2 22 2 0 4 6 0/ / / / .= × ⇒ = = ⋅ =α α

a r a rC Br C B C B

r C B3 3 3 33 3 32 248 99 12 28 800/ / / / . ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D4 4 4 44 4 42 250 4 9 2 23 370/ / / / . . ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

- 87 -

Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 byimage.


aC Bt

3 3 96 880/ ,= in / s2

aC Dt

4 4 9785/ = in / s2

Then,

α3 23 3 9687612 8073= = =

arC Bt

C Brad s/

//

α44 4 9785 5

9 2 1063 6= = =a

rC Dt

C D

/

/

.. . rad / s2

To determine the direction of α3 , determine the direction that rC B/ must be rotated to be parallel toaC B


- 88 -

To determine the direction of α4, determine the direction that rC D/ must be rotated to be parallel toaC D


Also

aE3 123 700= , in / s2

and

aC3 149 780= , in / s2

Problem 2.14

The accelerations of points A and B in the coupler below are as given. Determine the acceleration ofthe center of mass G and the angular acceleration of the body. Draw the vector representing aGfrom G.

G

B

A

a B

22˚

50˚

- 63˚

Aa

Aa = 7000 in/s 2

aB = 7000 in/s 2

AG = 1.5"BG = 1.5"AB = 2.8"


Draw the accelerations of points A and B on an acceleration polygon. Then locate the accleration ofpoint G by image.

For the angular acceleration of the body, resolve the acceleration aB At

/ in terms of componentsalong and perpendicular to rB A/ . The tangential component is perpendicular to rB A/ .

a r a rB At B A B A

tB A/ / / /= × ⇒ =α α

and

α = = =ar

B At

B A

/

/ .31412 8 1122 rad / s2

- 89 -

o'

2000 in/s

Acceleration Scale2

A

G

B a'

b'

g'

aB/ At

aG

To determine the direction of α , determine the direction that rB A/ must be rotated to be parallel toaB A

t/ . This direction is clearly clockwise.

Also

aG = 6980 in / s2


- 90 -

Problem 2.15

Crank 2 of the push-link mechanism shown in the figure is driven at a constant angular velocity ωωωω2= 60 rad/s (CW). Find the velocity and acceleration of point F and the angular velocity andacceleration of links 3 and 4.

Y

XA

B2

3 4

30˚

C

D

EF

AB = 15 cmBC = 29.5 cmCD = 30.1 cmAD = 7.5 cmBE = 14.75 cmEF = 7.5 cm

Position Analysis:

Draw the linkage to scale. First located the pivots A and D. Next locate B, then C, then E, then F.

Velocity Analysis:

v v v r v rA B B A B A B B A2 3 2 2 2 2 2 2 22 2 60 0 15 9= = = · � = = � =/ / / ( . )ω ω m / s

v v vC B C B3 3 3 3= + / (1)

v v v rC C C D C D3 4 4 4 4= = = ·/ /ω

Now,

v rB B Am s3 9= ^/ ( )/to

v r rC B C B C B3 3 3/ / /( )= · ^ω to

v r rC C D C D4 4= · ^ωω / /( )to


v

vrC BC B

C Bm s3 3

3 312 82 12 820 295

43 453//

/. / .

..= � = = =ω rad / s


vF m s3 4 94= . /

- 91 -


o

2 m/s

Velocity Polygon

b2 b3, c3 c4,o'

100 m/s

Acceleration Polygon

2b2' b3',

f 3'

A

B

C

2

3 4

30°

5 cm

arB2 /A2

D

E F

a2

arC3 /B3

atC3 /B3

c 3' c 4',

atC4 /D4

arC4 /D4


ωω .. //3 43 45 rad s CW=

Also,

v v

rC DC D

C D4 4

4 411 39 11 390 301

37 844//

/. .

..= � = = =m / s rad / sωω


ωω4 37 84= . rad / s CW


a a a rB B rB A B A2 3 2 2 2 2= = = · ·( )/ /ωω ωω

- 92 -

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a atC D rC D rB A tB A rC B tC B4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a rrC D C D m s4 4 42 2 237 84 0 301 430 99/ / . . . /= = � =ωω in the direction opposite to rC D/ )

a rrC B C B m s3 3 32 2 243 45 0 295 556 93/ / . . . /= = � =ω in the direction opposite to rC B/ )

a rrB A B A m s2 2 22 2 260 0 15 540/ / . /= = � =ωω in the direction opposite to rB A/ )

a r

ar

rtC B C BtC B

C BC B3 3

3 33 3/ /

/

//( )= · � = ^α α to

a r

ar

rtC D C DtC D

C DC D4 4

4 44 4/ /

/

//( )= · � = ^α α to

Solve Eq. (2) graphically with an acceleration polygon. The acceleration directions can be gottendirectly from the polygon. The magnitudes are given by:

α3

3 3 142 790 295

484= = =a

r

tC B

C B

/

/

..

rad / s CW2

Also,

α4

4 4 41 010 301

136= = =a

rtC D

C D

/

/

..

rad / s CCW2

Using acceleration polygon,

aF m s3 256 2= /


Problem 2.16

For the straight-line mechanism shown in the figure, ωωωω2 = 20 rad/s (CW) and αααα2 = 140 rad/s2

(CW). Determine the velocity and acceleration of point B and the angular acceleration of link 3.

A C

B

2

4

3

15oD

DA = 2.0"AC = 2.0"AB = 2.0"

- 93 -

Velocity Analysis:

v v v rA A D A A D2 2 2 3 2 22= = = ·/ /ωω

v v v vC C A C A3 4 3 3 3= = + / (1)

Now,

v r rA A D A D3 2 2 2 22 20 2 40= = ⋅ = ⊥ω / /( )in / s to

vC3 in horizontal direction

v r v r rC A C AC A C A C A3 3 3 3 3 3 3 3 3 33 3/ // / /( )= × ⇒ = ⊥ω ω to


vB3 77 3= . in / s

Also,

vC A3 3 40/ = in / s

or

ω33 3

3 3

402 20= = =

vrC A

C A

/

/rad / s CCW

Also,

vC3 20 7= . in / s


a a a aC C A C A3 4 3 3 3= = + /

a a a a aC A Dr

A Dt

C Ar

C At

3 2 2 2 2 3 3 3 3= + + +/ / / / (2)

Now,

aC3 in horizontal direction

a r a rA Dr A D A D

r A D2 2 2 2 2 22 2 22 220 2 800/ / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction opposite to rA D/

a r a r rA Dt A D A D

t A D A D2 2 2 2 2 22 2 140 2 280/ / / / /( )= × ⇒ = ⋅ = ⋅ = ⊥α α in / s to2

a r a rC Ar C A C A

r C A3 3 3 3 3 3 3 33 3 32 220 2 800/ / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction opposite to rC A3 3/

- 94 -

a r a r rC At C A C A

t C A C A3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

Solve Eq. (2) graphically with a acceleration polygon. From the polygon,

aB3 955= in / s2

aC A3 3 280/ = in / s2

Also,

α33 3

3 3

2802 140= = =

ar

C At

C A

/

/rad / s CCW2

A C

B

2

4

3

D

o

20 in/s

Velocity Polygon

a2 a3,

c3 , c4

b 3

vC3

v A3 vC3 / A3

v B3

v B3 / A3

o'

400 in/sAcceleration Polygon

2

a2' a3',

aA2 / D 2r

aA2/D 2t

aC3c3'

aC3 / A3r

aC3 / A3t

b3'

aB3

- 95 -

Problem 2.17

For the data given in the figure below, find the velocity and acceleration of points B and C. AssumevA = 20 ft/s, aA = 400 ft/s2, ωωωω2 = 24 rad/s (CW), and αααα2 = 160 rad/s2 (CCW).

Bω215o

vA

aA

α 2

A

C

90˚

AB = 4.05"AC = 2.5"BC = 2.0"

Position Analysis

Draw the link to scale

o

a2

b2

c2

5 ft/sec

Velocity Polygon

o'

a2'

b2'

2'

80 ft/sec

Acceleration Polygon2

aB2/A2r

a B2/A2t

c

Velocity Analysis:

v v vB A B A2 2 2 2= + / (1)

Now,

- 96 -

vA ft2 20= / sec in the positive vertical direction

v r v rB A B A B A B A in2 2 2 22 2 24 4 05 97 2/ / / / . . / sec= × ⇒ = ⋅ = ⋅ =ω ω

v rB A B Aft to2 2 8 1/ /. / sec( )= ⊥ in the positive vertical direction


vB ft2 11 9= . / sec

Also, from the velocity polygon,

vC ft2 15 55= . / sec in the direction shown


a a a a a aB A B A A B At

B Ar

2 2 2 2 2 2 2 2 2= + = + +/ / / (2)

Now,

aA ft2 400 2= / sec in the given direction

aB At B Ar in ft2 2 2 2 2160 4 05 648 54/ / . / sec / sec= ⋅ = ⋅ = =α

aB Ar B Ar in ft2 2 2

2 2 2 224 4 05 2332 194/ / . / sec / sec= ⋅ = ⋅ = =ω

Solve Eq. (2) graphically with an acceleration polygon. From the polygon,

aB ft2 198 64 2= . / sec

in the direction shown. Determine the acceleration of point C by image. From the accelerationimage,

aC ft2 289 4 2= . / sec


- 97 -

Problem 2.18

In the mechanism shown below, link 2 is turning CCW at the rate of 10 rad/s (constant). Draw thevelocity and acceleration polygons for the mechanism, and determine aG3 and αααα4.

2

3

4

A

C

B

D

G

αω2 2,

90˚

AB = 1.0"BC = 2.0"BG = 1.0"CD = 3.0"AD = 3.0"

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B andthen C. Then locate G.

Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,

v r v r rB A B A B A B A B A2 2 2 2 10 2 20/ / / / /( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to



- 98 -

Solve Eq. (1) graphically with a velocity polygon.

From the polygon,

vC B3 3 14 4/ .= in / s

vC D4 4 13 7/ .= in / s


Now

ω33 3 11 4

2 5 7= = =vrC B

C B

/

/

. . rad / s

and

ω44 4 13 7

3 4 57= = =vrC D

C D

/

/

. . rad / s



The velocity of point G3

v v v r rG B G B G B G B3 3 3 3 3 3 5 7 1 5 7= + = × = ⋅ = ⋅ =/ / / . .ω ω in / s


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

- 99 -

a r a rB Ar B A B A

r B A2 2 2 22 2 22 210 1 100/ / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2


a r a rB At B A B A

tB A2 2 2 22 2 0/ / / /= × ⇒ = =α α

a r a rC Br C B C B

r C B3 3 3 33 3 32 25 7 2 64 98/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D in4 4 4 44 4 42 2 24 57 3 62 66/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically for the accelerations.

- 100 -


aC Bt

3 3 38/ = in / s2

aC Dt

4 4 128/ = in / s2

Then,

α33 3 38

2 19= = =arC Bt

C B

/

/rad / s2

α44 4 128

3 42 67= = =a

rC Dt

C D

/

/. rad / s2


aC Bt

3 3/ . This direction is clearly counterclockwise.


aC Dt


Determine the acceleration of point G3

a a a a a aG B G B B G Br

G Bt

3 3 3 3 2 3 3 3 3= + = + +/ / /

a r a rG Bt G B G B

t G B in3 3 3 33 3 219 1 19/ / / / / sec= × ⇒ = ⋅ = ⋅ =α α

a r a rG Br G B G B

r G B in3 3 3 33 3 32 2 25 7 1 32 49/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω


aG3 116= in / s2

Problem 2.19

If ωωωω2 = 100 rad/s CCW (constant) find the velocity and acceleration of point E.

A

C

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

ω2

B

D

E

70˚2

3

4

- 101 -

Position Analysis


Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,





From the polygon,

vC B3 3 77 5/ .= in / s

vC D4 4 71/ = in / s


- 102 -

Now

ω33 3 77 5

1 75 44 29= = =vrC B

C B

/

/

.. . rad / s

and

ω44 4 71

2 35 5= = =vrC D

C D

/

/. rad / s



The velocity of point E3

v v v r rE D E D D E D E4 4 4 4 4 4 35 5 0 8 28 4= + = × = ⋅ = ⋅ =/ / / . . .ω ω in / s


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A2 2 2 22 2 22 2100 1 10 000/ / / / ,= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2


a r a rB At B A B A

tB A2 2 2 22 2 0/ / / /= × ⇒ = =α α

a r a rC Br C B C B

r C B3 3 3 33 3 32 244 29 1 75 3432 8/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D in4 4 4 44 4 42 2 235 5 2 2520 5/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically with acceleration.

- 103 -


aC Bt

3 3 3500/ = in / s2

aC Dt

4 4 10 900/ ,= in / s2

Then,

α33 3 3500

1 75 2000= = =arC Bt

C B

/

/ . rad / s2

α44 4 10 900

2 5450= = =a

rC Dt

C D

/

/

, rad / s2


aC Bt



aC Dt


Determine the acceleration of point E4

a a a a aE D E D E Dr

E Dt

4 4 4 4 4 4 4 4= + = +/ / /

- 104 -

a r a rE Dt E D E D

t E D in4 4 4 44 4 25450 0 8 4360/ / / / . / sec= × ⇒ = ⋅ = ⋅ =α α

a r a rE Dr E D E D

r E D in4 4 4 44 4 42 2 235 5 0 8 1008 2/ / / / . . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω


aE4 4600= in / s2

Problem 2.20

Draw the velocity polygon to determine the velocity of link 6. Points A, C, and E have the samevertical coordinate.

2

4

5

A

B

C

D

E

3

6

= 6ω2rads

1

AB = 1.80"BC = 1.95"CD = 0.75"DE = 2.10"

50˚

Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v vC C B C B4 3 3 3 3= = + / (1)

v vD D5 3=

v v v vE E D E D5 6 5 5 5= = + / (2)

Now,

v r v r rB A B A B A B A B A2 2 2 2 2 2 2 2 2 22 2 6 1 8 10 8/ / / / /. . ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to

vC3 is in the vertical direction. Then,

v r v r rC B C B C B C B C B3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image,

v in sD3 18 7= . /

- 105 -

2

45

A

B

C

D

E

3

6

b3

c3

d3

o

10 in/sec

Velocity Polygon

e5

d5

Now,

vE5 in horizontal direction

v r v r rE D E D E D E D E D5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to


v v in sE E5 6 8 0= = . /

- 106 -

Problem 2.21

Link 2 of the linkage shown in the figure has an angular velocity of 10 rad/s CCW. Find theangular velocity of link 6 and the velocities of points B, C, and D.

A

D

C

B

2

4

6

5

ω2

EF

3

X

Y

0.3"

AE = 0.7"AB = 2.5"AC = 1.0"BC = 2.0"EF = 2.0"CD = 1.0"DF = 1.5"

θ 2

= 135˚θ 2

Position Analysis

Locate points E and F and the slider line for B. Draw link 2 and locate A. Then locate B. Nextlocate C and then D.

Velocity Analysis:

v v vA A A E3 2 2 2= = /

v v v vB B A B A4 3 3 3 3= = + / (1)

Find vC3 by image.

v vC C5 3=

v v v v vD D D F C D C5 6 6 6 5 5 5= = = +/ / (2)

Now,

vB3 in horizontal direction

v r v r rA E A E A E A E A E2 2 2 22 2 10 0 7 7/ / / / /. ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to

v r v r rB A B A B A B A B A3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to


- 107 -

E

A

B

CD

F

2

34

6

Slider Line

Velocity Polygon2.5 in/s

a3

o

f 6b3

c3c5

d5

vB3 3 29= . in / s


v vC C5 3 6 78= = . in/ s

Now,

v r v r rD F D F D F D F D Fto6 6 6 66 6/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

v r v r rD C D C D C D C D Cto5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


v v vD D D F5 6 6 6 6 78= = =/ . in / s

or

ω66 6

6 6

6 781 5 4 52= = =

vr

D F

D F

/

/

.. . rad / s CCW

- 108 -

Problem 2.22

The linkage shown is used to raise the fabric roof on convertible automobiles. The dimensions atgiven as shown. Link 2 is driven by a DC motor through a gear reduction. If the angular velocity,ωωωω2 = 2 rad/s, CCW, determine the linear velocity of point J, which is the point where the linkageconnects to the automobile near the windshield.

AB = 3.5"AC = 15.37"BD = 16"CD = 3"CE = 3.62"EG = 13.94"GF = 3.62"HF = 3"FC = 13.62"HI = 3.12"GI = 3.62"HL = 0.75"KC = 0.19"JH = 17"

BA

E

C

D

F

G

H

I

JH

FC D

KL

2

3

4

6

7

8 5

3

12ω

Detail of Link 3

110˚

Position Analysis:

Draw linkage to scale. Start with link 2 and locate points C and E. Then locate point D. Thenlocate points F and H. Next locate point G. Then locate point I and finally locate J.

Velocity Analysis:

The equations required for the analysis are:

v v r v rC C A C A C C A2 2 2 2 2 2 2 22 2 2 15 37 30 74= = × ⇒ = = ⋅ =/ / / ( . ) .ω ω in / s

v vC C3 2=

v v v v vD D D B C D C3 4 4 4 3 3 3= = = +/ / (1)

v v v v v v vG G G F G F E G E5 6 7 5 5 5 6 6 6= = = + = +/ /

v vF F5 3=

v vE E6 2=

So,

- 109 -

v v v vF G F E G E5 5 5 6 6 6+ = +/ / (2)

v v v v v vI I G I G H I H7 8 7 7 7 8 8 8= = + = +/ /

v vH H8 3=

So,

v v v vG I G H I H7 7 7 8 8 8+ = +/ / (3)

c3

d3

f3

h3

e2

g6

i8

j8

AB

CD

E

G

FH

I

J

2 4

3

6

57

8

10 in/s

Velocity Polygon

o

- 110 -

Now,

v rC C A3 30 74= ⊥. ( )/in / s to

v r rD C D C D C3 3 3/ / /( )= × ⊥ω to

v r rD B D B D B4 4 4/ / /( )= × ⊥ω to


vD4 30 9= . in / s

Using velocity image of link 3, find the velocity of points F and H and of link 2, find the velocity ofpoint E.

vF5 30 5= . in / s

vH3 30 3= . in / s

and

vE6 3 80= . in / s

Now,

v r rG F G F G F5 5 5/ / /( )= × ⊥ω to

v r rG E G E G E6 6 6/ / /( )= × ⊥ω to


vG6 37 8= . in / s

Now,

v r rI G I G I G7 7 7/ / /( )= × ⊥ω to

v r rI H I H I H8 8 8/ / /( )= × ⊥ω to

Solve Eq. (3) graphically with a velocity polygon. Using velocity polygon of link 8

vJ8 73 6= . in / s

- 111 -

Problem 2.23

In the mechanism shown, determine the sliding velocity of link 6 and the angular velocities of links3 and 5.

2

4

56

B

C

D

E

F

= 3ω2rads

50˚

10.4"

2.0"

29.5"

A

AB = 12.5"BC = 22.4"DC = 27.9"CE = 28.0"DF = 21.5"

3

34˚

Position Analysis

First locate Points A and E. Next draw link 2 and locate B. Then locate point C by drawing a circlecentered at B and 22.4 inches in radius, and finding the intersection with a circle centered at E andof 28 inches in radius. Find D by drawing a line 27.9 inches long at an angle of 34˚ relative to lineBC. Locate the slider line 2 inches above point E. Draw a circle centered at D and 21.5 inches inradius and find the intersections of the circle with the slider line. Choose the proper intersectioncorresponding to the position in the sketch.

Velocity Analysis

Compute the velocity of the points in the same order that they were drawn. The equations for thefour bar linkage are:

v v rB B A B A2 2 2 2= = ×/ /ω

v vB B3 2=

v v vC B C B3 3 3 3= + /

- 112 -

A

B

C

D

10 in/sec

Velocity Polygon

o

E

c 3

F

b 3

d 3

f 3

Also,

v v v v vC C E C E C E3 4 4 4 4 4 4= = + =/ /

where,

v rB B A in2 2 3 12 5 37 5= = ⋅ =ω / . . / sec

v rC B C B to CB3 3 3/ / ( )= × ⊥ω

v rC E C E to CE4 4 4/ / ( )= × ⊥ω

The velocity of C3 (and C4) can then be found using the velocity polygon. After the velocity of C3is found, find the velocity of D3 by image. Then,

- 113 -

v vD D5 3=

v v vF D F D5 5 5 5= + /

and

v vF F5 6=

where

v r v rF D F D C F D to FD5 5 55 5/ / / ( )= × ⇒ = ⊥ω ω

and vF6 is along the slide direction.. Then the velocity of F5 (and F6) can be found using thevelocity polygon. From the polygon,

vF6 = 43.33 in/sec

vC B in3 3 26 6/ . / sec=

vF D in5 5 18 54/ . / sec=

ω33 3 26 6

22 4 1 187= = =vrC B

C Brad/

/

.

. . / sec

ω55 5 18 54

21 5 0 862= = =vrF D

F Drad/

/

.. . / sec

To determine the direction for ω3, determine the direction that rC B/ must be rotated to be in thedirection of vC B3 3/ . From the polygon, this direction is CCW.

To determine the direction for ω5, determine the direction that rF D/ must be rotated to be in thedirection of vF D5 5/ . From the polygon, this direction is CW.

- 114 -

Problem 2.24

In the mechanism shown, vA2 = 15 m/s. Draw the velocity polygon, and determine the velocity ofpoint D on link 6 and the angular velocity of link 5.

1vA2

= 15 m/s

2

3

4

5

A

C

B

D

45˚

2.05"

AC = 2.4"BD = 3.7"BC = 1.2"

X

Y

2.4"

6

Velocity Analysis:

v vA A3 2=

v v v vC C A C A4 3 3 3 3= = + / (1)

v vB B3 5=

v v v vD D B D B5 6 5 5 5= = + / (2)

Now,

vA m3 15= / sec in vertical direction

vC3 in horizontal direction

v r v r rC A C A C A C A C Ato3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


v vB B m3 5 14 44= = . / sec

Now,

- 115 -

3

4

5

C

B

D

2 A6

a3

c3o

10 m/sec

Velocity Polygon

b3b5b

d5

vD5 along the inclined path

v r v r rD B D B D B D B D Bto5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


vD m6 12 31= . / sec

Also,

vD B m5 5 16 61/ . / sec=

or

ω55 5 16 605

3 7 4 488= = =vrD B

D Brad CCW/

/

.. . / sec

- 116 -

Problem 2.25

In the mechanism shown below, points E and B have the same vertical coordinate. Find thevelocities of points B, C, and D of the double-slider mechanism shown in the figure if Crank 2rotates at 42 rad/s CCW.

A

D

C

2

6

5

B

4

33

E

ω2

0.75"

60˚

EA = 0.55"AB = 2.5"AC = 1.0"CB = 1.75"CD = 2.05"

Position Analysis

Locate point E and draw the slider line for B. Also draw the slider line for D relative to E. Drawlink 2 and locate A. Then locate B. Next locate C and then D.

Velocity Analysis:

v v vA A A E3 2 2 2= = /

v v v vB B A B A4 3 3 3 3= = + / (1)

v vC C5 3=

v v v vD D C D C5 6 5 5 5= = + / (2)

Now,

vB3 in horizontal direction

- 117 -

A

D

C

2

6

5

B

4

3

3

E

10 in/sec

Velocity Polygon

b3

a3

c3 d5

o

v r v r rA E A E A E A E A Ein to2 2 2 2 2 22 2 42 0 55 23 1/ / / / /. . / sec ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω

v r v r rB A B A B A B A B Ato3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


vB in4 17 76= . / sec


vC in3 18 615= . / sec

Now,

vD5 in vertical direction

v r v r rD C D C D C D C D Cto5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

- 118 -


vD in5 5 63= . / sec

Problem 2.26

Given vA4 = 1.0 ft/s to the left, find vB6.

A

B

C

D

2

4

5E

6

3X

Y

157.5˚

DE = 1.9"CD = 1.45"BC = 1.1"AD = 3.5"AC = 2.3"

1.0"

0.5"

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of A, B and E. Next locate C andD.

Velocity Analysis:

v vA A4 3=

v v v vC C A C A5 3 3 3 3= = + / (1)

v vD D3 2=

v v vD E D E2 2 2 2= + / (2)

- 119 -

v v vB C B C5 5 5 5= + /

Now,

vA ft4 1 0= . / sec in horizontal direction

v r v r rC A C A C A C A C Ato3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

v r v r rD E D E D E D E D Eto2 2 2 22 2/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


From the polygon, using velocity image,

vC A ft3 3 1 28/ . / sec=

and,

ω33 3 1 28

2 3 0 56= = =vrC A

C A

/

/

.. . rad / s

To determine the direction of ω3, determine the direction that rC A/ must be rotated to be parallel tovC A3 3/ . This direction is clearly clockwise.

Now,

vB6 is horizontal direction

v r v r rB C B C B C B C B Cto5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

- 120 -


vB ft6 1 23= . / sec

Problem 2.27

If vA2 = 10 cm/s as shown, find vC5.

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of D, F and G. Next locate A andB. Then locate E and C.

Velocity Analysis:

v vA A3 2=

v v vB A B A3 3 3 3= + / (1)

v vB B4 3=

v v v vB F B F B F4 4 4 4 4 40= + = +/ / (2)

v vE E5 4=

v v vG E G E5 5 5 5= + /

- 121 -

v vG G6 5=

Now,

vA cm2 10= / sec ( )/⊥ to A Cr

v r v r rB A B A B A B A B Ato3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

v r v r rB F B F B F B F B Fto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

From the polygon,

vB cm4 6 6= . / sec


vE cm4 3 12= . / sec

Now,

vG6 is horizontal direction

v r v r rG E G E G E G E G Eto5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

For the velocity image

draw a line ⊥ to C Er / at e

draw a line ⊥ to C Er / at g

and find the point “c”

From the velocity polygon

vB cm6 3 65= . / sec

- 122 -

Problem 2.28

If vA2 = 10 in/s as shown, find the angular velocity of link 6.

23

4

56

A

v2A

27˚

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

BC

D

E

F

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of B, D and F. Next locate A andC. Then locate E.

Velocity Analysis:

v vA A3 2=

v v vD A D A3 3 3 3= + / (1)

v vD D4 3=

v vC C5 3=

v v vE C E C5 5 5 5= + / (2)

v vE E6 5=

v v v vE F E F E F6 6 6 6 6 60= + = +/ /

Now,

vA in2 10= / sec

- 123 -

v r v r rD A D A D A D A D Ato3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

From the polygon,

vD A in3 3 9 1/ . / sec=


r r v vD A C A D A C A/ / / /: := 3 3 3 3

vC A in3 3 4 32/ . / sec=

Now,

v r v r rE C E C E C E C E Cto5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ωv r v r rE F E F E F E F E Fto6 6 6 66 6/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

from the velocity polygon

vE in6 2 75= . / sec

and

ω66 6 2 75

2 1 375= = =vrE F

E F

/

/

. . rad / s

To determine the direction of ω6 , determine the direction that rE F/ must be rotated to be parallel tovE F6 6/ . This direction is clearly counterclockwise.

- 124 -

Problem 2.29

The angular velocity of link 2 of the mechanism shown is 20 rad/s, and the angular acceleration is100 rad/s2 at the instant being considered. Determine the linear velocity and acceleration of pointF6.

D2

3

4

5

6

C

B

EF

ω 2 α 2,

A

115˚

2"

0.1"

2.44"

EF = 2.5"CD = 0.95"AB = 0.5"BC = 2.0"CE = 2.4"BE = 1.8"

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B andthen C. Then locate E and finally F.

Velocity Analysis:

The required equations for the velocity analysis are:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

v vE E5 3=

v v v vF F E F E5 6 5 5 5= = + / (2)

Now,

v r v r rB A B A B A B A B A2 2 2 2 2 2 2 2 2 22 2 20 0 5 10/ / / / /. ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to

v r v r rC D C D C D C D C D4 4 4 4 4 4 4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

v r v r rC B C B C B C B C B3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

Solve Eq. (1) graphically with a velocity polygon. From the polygon and using velocity image,

- 125 -

vC B3 3 6 59/ .= in / s

or

ω33 3

3 3

6 592 3 29= = =

vrC B

C B

/

/

. . rad / s CCW

Also,

vC D4 4 8 19/ .= in / s

or

ω44 4

4 4

8 190 95 8 62= = =

vrC D

C D

/

/

.

. . rad / s CW

And,

vE5 5 09= . in / s

Now,

vF5 in horizontal direction

v r v r rF E F E F E F E F E5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to


vF E5 5 3 97/ .= in / s

or

ω55 5

5 5

3 972 5 1 59= = =

vr

F E

F E

/

/

.. . rad/ s CCW

Also,

vF6 3 79= . in / s


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (3)

- 126 -

b3e5

o

5 in/sec

Velocity Polygon

f 5

b'3

c'3

e'5

o'

50 in/sAcceleration Polygon

2

f'5

1 aF5 /E5t

1 aF5

1a E5

D

2

3

4

5

C

B

EF

A

1a F5 /E5r

1aC3 /B3t

1aC3 /B3r

1 aC4/D4t

1aC4 /D4r

1aB2 /A2r

1aB2 /A2t

c3

a aE E5 3=

a a a aF F E F E5 6 5 5 5= = + /

a a a aF E F Er

F Et

5 5 5 5 5 5= + +/ / (4)

Now,

a r a rB Ar B A B A

r B A2 2 2 2 2 2 2 22 2 22 220 0 5 200/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

- 127 -



t B A B A2 2 2 2 2 2 2 2 2 22 2 100 0 5 50/ / / / /. ( )= × ⇒ = ⋅ = ⋅ = ⊥α α in / s to2

a r a rC Br C B C B

r C B3 3 3 3 3 3 3 33 3 32 23 29 2 21 6/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rC B3 3/


t C B C B3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rC Dr C D C D

r C D4 4 4 4 4 4 4 44 4 42 28 62 0 95 70 6/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rC D4 4/


t C D C D4 4 4 4 4 4 4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α to

Solve Eq. (3) graphically with an acceleration polygon. From the polygon, using accelerationimage,

aE5 252 0= . in / s2

Now,

aF5 in horizontal direction

a r a rF Er F E F E

r F E5 5 5 5 5 5 5 55 5 52 21 59 2 5 6 32/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rF E5 5/

a r a r rF Et F E F E

t F E F E5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥α α to


aF6 152 7= . in / s2

- 128 -

Problem 2.30

In the drag-link mechanism shown, link 2 is turning CW at the rate of 130 rpm. Construct thevelocity and acceleration polygons and compute the following: aE5, aF6, and the angular accelerationof link 5.

AB = 1.8'BC = 3.75'CD = 3.75'AD = 4.5'AE = 4.35'DE = 6.0'EF = 11.1'

60˚

A

C

BD

E

F3

2

45

6

Velocity Analysis:

ω π2 130 1302

60 13 614= = =rpm rad / s.

v v vC C C B3 2 2 2= = /

v v v v vD D D A C D C3 4 4 4 3 3 3= = = +/ / (1)

v vE E5 4=

v v v vF F E F E5 6 5 5 5= = + / (2)

Now,

v r v r rC B C B C B C B C B2 2 2 22 2 13 614 3 75 51 053/ / / / /. . . ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω ft / s to

v r v r rD C D C D C D C D C3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to

v r v r rD A D A D A D A D A4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to


vD C3 3 41 3/ .= ft / s

or

- 129 -

20 ft/s

Velocity Scale

c3

d3

o

e4

f 5

1 aC2/ B2r

1 aD3 /C3r

1a D3 /C3t

1aD4 /A 4r

1aD4 /A4t

d3'

c3'

o'

200 ft/s

Acceleration Scale

2

a4'

60˚

A

C

BD

E

F3

2

45

6

e4'

f5'

1aF5 /E5t

1aF5 /E5r

- 130 -

ω33 3

3 3

41 33 75 11 0= = =

vr

D C

D C

/

/

.. . rad / s CW

Also,

vD A4 4 37 19/ .= ft / s

or

ω44 4

4 4

37 194 5 8 264= = =

vr

D A

D ACW/

/

.

. . rad / s

And,

vE5 35 95= . ft / s

Now,

vF5 in horizontal direction

v r v r rF E F E F E F E F E5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω to


vF E5 5 12 21/ .= ft / s

or

ω55 5

5 5

12 2111 1 1 1= = =

vr

F E

F E

/

/

.. . rad / s CCW


a a aC C C B3 2 2 2= = /

a a a a a aD Ar

D At

C Br

C Bt

D Cr

D Ct

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (3)

a aE E5 4=

a a a aF F E F E5 6 5 5 5= = + /

a a a aF E F Er

F Et

5 5 5 5 5 5= + +/ / (4)

Now,

a r a rC Br C B C B

r C B2 2 2 22 2 22 213 614 3 75 695 0/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω ft / s2


- 131 -

a r a rC Bt C B C B

t C B2 2 2 22 2 0 3 75 0/ / / / .= × ⇒ = ⋅ = ⋅ =α α ft / s2

a r a rD Cr D C D C

r D C3 3 3 33 3 32 211 0 3 75 453 8/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω ft / s2

in the direction of - rD C/

a r a r rD Ct D C D C

t D C D C3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rD Ar D A D A

r D A4 4 4 44 4 42 28 264 4 5 307 3/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω ft / s2

in the direction of - rD A/

a r a r rD At D A D A

t D A D A4 4 4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α to

Solve Eq. (3) graphically with an acceleration polygon. From the polygon, using accelerationimage,

aE5 308 0= . ft / s2

Now,

aF5 is in the horizontal direction

a r a rF Er F E F E

r F E5 5 5 55 5 52 21 1 11 1 13 4/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω ft / s2

in the direction of - rF E/

a r a r rF Et F E F E

t F E F E5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥α α to


aF6 83 4= . ft / s2

Also,

aF Et

5 5325 2/ .= ft / s2

or

α55 5

5 5

325 211 1 29 3= = =

ar

F Et

F E

/

/

.. . rad / s CCW2

- 132 -

Problem 2.31

The figure shows the mechanism used in two-cylinder 60-degree V-engine consisting, in part, of anarticulated connecting rod. Crank 2 rotates at 2000 rpm CW. Find the velocities and accelerationof points B, C, and D and the angular acceleration of links 3 and 5.

A

C

2

3

D

5

B

4

E

3

XEA = 1.0"AB = 3.0"BC = 3.0"AC = 1.0"CD = 2.55"

30o

30oY

6

90˚

Position Analysis

Draw the linkage to scale. First locate the two slider lines relative to point E. Then draw link 2 andlocate point A. Next locate points B and C. Next locate point D.

Velocity Analysis:

Find angular velocity of link 2,

ω π π2 30

200030 209 44= ⋅ = ⋅ =n . rad / s

v v v rA A E A A E2 2 2 3 2= = = ×/ /ω

v v v vB B A B A3 4 3 3 3= = + / (1)

v vC C3 5=

- 133 -

10000 in/s

Acceleration Polygon

2

E

C

A

B

D

2

4

5

6

3

100 in/s

Velocity Polygon

a2 a3,

b3 b4b,

c3 , c5

d5 d6,

o

o'

a2’

b3 b4b’,

1aA2 /E2r

1aB3 /A3r

1 aB3

c3’

d5 d6,

1 aC3

1 aD5 /C5r

1 aD5 /C5t

1 aD5

1aB3 /A3t

- 134 -

v v v vD D C D C5 6 5 5 5= = + / (2)

Now,

v r rA A E A E2 2 209 44 1 209 44= = ⋅ = ⊥ω / /. . ( )in / s to

vB3 in the direction of rB E/

v r v r rB A B A B A B A B A3 3 3 33 3/ / / / /( )= × ⇒ = ⊥ω ω to


v vB B3 4 212 7= = . in / s

Also,

vB A3 3 109 3/ .= in / s

or

ω33 3 109 3

3 36 43= = =vrB A

B A

/

/

. . rad / s

To determine the direction of ω3, determine the direction that rB A/ must be rotated to be parallel tovB A3 3/ . This direction is clearly clockwise.

Also,

v vC C3 5 243 3= = . in / s

Now,

v vD D5 6= in the direction of rD E/

v r v r rD C D C D C D C D Cto5 5 5 55 5/ / / / /( )= × ⇒ = ⊥ω ω


v vD D in5 6 189 2= = . / sec

Also,

vD C5 5 135/ = in / s

or

ω55 5 135

2 55 52 9= = =vrD C

D C

/

/ . . rad / s

- 135 -

To determine the direction of ω5, determine the direction that rD C/ must be rotated to be parallel tovD C5 5/ . This direction is clearly clockwise.


a a a aA A A Er

A Et

2 3 2 2 2 2= = +/ /

a a a aB B A B A3 4 3 3 3= = + /

a a a a aB A Er

A Et

B Ar

B At

3 2 2 2 2 3 3 3 3= + + +/ / / / (3)

a aC C3 5=

a a a aD D C D C5 6 5 5 5= = + /

a a a a aD D C D Cr

D Ct

5 6 5 5 5 5 5= = + +/ / (4)

Now,

aB3 in the direction of ± rB E/

a r a rA Er A E A E

r A E2 2 2 22 2 22 2209 44 1 43860/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rA E/

a r a rA Et A E A E

t A E in2 2 2 22 2 20 1 0/ / / / / sec= × ⇒ = ⋅ = ⋅ =α α

a r a rB Ar B A B A

r B A3 3 3 33 3 32 236 43 3 3981/ / / / .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rB A/


t B A B A3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to


aB3 14710= in / s2

Also,

aB At

3 3 38460/ = in / s2

or

α33 3 38460

3 12820= = =arB At

B A

/

/rad / s2

- 136 -

To determine the direction of α3 , determine the direction that rB A/ must be rotated to be parallel to

aB At


Also,

a aC C in s3 5 39 300 2= = , /

Now,

aD5 in the direction of - rD E/

a r a rD Cr D C D C

r D C5 5 5 55 5 52 252 9 2 55 7136/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rD C/

a r a r rD Ct D C D C

t D C D C5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥α α to


a aD D5 6 24 000= = , in / s2

Also,

aD Ct

5 526 300/ ,= in / s2

or

α55 5 26300

2 55 10 300= = =arD Ct

D C

/

/ . , rad / s2

To determine the direction of α5, determine the direction that rD C/ must be rotated to be parallel toaD C


Problem 2.32

In the mechanism shown, ωωωω2 = 4 rad/s CCW (constant). Write the appropriate vector equations,solve them using vector polygons, and

a) Determine vE3, vE4, and ωωωω3.

b) Determine aE3, aE4, and αααα3.

Also find the point in link 3 that has zero acceleration for the position given.

- 137 -

Position Analysis

Locate pivots A and D. Draw link 2 and locate B. Then locate point C. Finally locate point E.

Velocity Analysis

For the velocity analysis, the basic equation is:

v v vB B B A2 3 2 2= = /

v v v v vC B C B C C D3 3 3 3 4 4 4= + = =/ /

Then,

v v vC D C B B A4 4 3 3 2 2/ / /= +

and the vectors are:

v vB A B A B A B A B Ar r r2 2 2 22 2 4 0 5 2/ / / / /. ( )= × ⇒ = = ⋅ = ⊥ω ω m / s to

vC B C B C Br r3 3 3/ / /( )= × ⊥ω to

vC D C D C Dr r4 4 4/ / /( )= × ⊥ω to

The basic equation is used as a guide and the vectors are added accordingly. Each side of theequation starts from the velocity pole. The directions are gotten from a scaled drawing of themechanism.

The graphical solution gives:

vC B s3 3 2 400 46 6/ . .= ∠− ° m /

vC D4 4 0 650 1/ . ˚= ∠− m / s

vE3 2 522 93 9= ∠ °. . /m s (by image)

- 138 -

Now,

ω33 3 2 400

0 8 3 0= = =vC B

C Br/

/

.. . rad / s

ω44 4 0 65

0 8 0 81= = =vC D

C Dr/

/

.. . rad / s


To determine the direction of ω4 , determine the direction that rC D/ must be rotated to be parallel tovC D4 4/ . This direction is clearly clockwise.

Find the velocity of E3 and E4 by image. The directions are given on the polygon. The magnitudesare given by,

vE3 2 522= . m / s

vE4 0 797= . m / s

Acceleration Analysis

The graphical acceleration analysis follows the same points as in the velocity analysis. Start at link2.

a a a

a r

a r a r

B A B At

B Ar

B At B A

B Ar B A B A

r B A

ce2 2 2 2 2 2

2 2

2 2 2 2

2 2

2 2 2 2

0 0/ / /

/ /

/ / / /

sin

= +

= × = =

= × ×( )⇒ =

α α

ω ω ω

or

aB Ar

2 2 4 0 0 5 8 02/ ( . ) ( . ) .= = m / s from B to A2

- 139 -

1 m/s

Velocity Scale

C

B

D

2

34

Eb

oo'

1aB 2/ A2

r

1aC3

/ B3

t

c3

3

e3

e4

4 m/s

Acceleration Scale

d'4

1aE4

2

A

e'3

e'4

c'3

1aC3 / B3

r 1aE3

b'3

1aC

4/D

4

r1aC4 /D4

t

1aC3

- 140 -

Now go to Point C and follow the same path as was used with velocities.

a a a r vC D C Dt

C Dr C D C D4 4 4 4 4 4 4 44 4/ / / / /= + = × + ×α ω

Also

a a a a a ar v a

C D C B B A C Bt

C Br B A

C B C B B A

4 4 3 3 2 2 3 3 3 3 2 2

3 3 2 23 3

/ / / / / /

/ / /

= + = + += × + × +α ω

Therefore,

a a a a aC Dt

C Dr

C Bt

C Br B A4 4 4 4 3 3 3 3 2 2/ / / / /+ = + +

and

a vC Dr C D4 4 4 44 0 812 0 650 0 528/ / ( . )( . ) .= × = =[ ]ω from C to D

a r rC Dt C D C D4 4 4/ / /?= × = ⊥α

a vC Br C B3 3 3 33 3 0 2 4 7 2/ / ( . )( . ) .= × = =[ ]ω from C to B

a r rC Bt C B C B3 3 3/ / /?= × = ⊥α

These values permit us to solve for the unknown vectors. We can then find “e” by accelerationimage. From the acceleration polygon,

aC Bt

3 3 16 13/ .= m / s2

Then,

α33 3 16 13

0 8 20 16= = =arC Bt

C B

/

/

.

. . rad / s2


aC Bt


and

aE315 84= . m / s2

aE432 19= . m / s2

- 141 -

Problem 2.33

In the mechanism shown, point A lies on the X axis. Draw the basic velocity and accelerationpolygons and use the image technique to determine the velocity and acceleration of point D4. Thendetermine the velocity and acceleration images of link 4. Draw the images on the velocity andacceleration polygons.

A

B D

E

F (-1.0", -0.75")

2

3

4

5

6

v = 10 in/sA2 (constant)

C X

Y

FE = 1.35"ED = 1.5"BD = CD = 1.0"AB = 3.0"

84˚

90˚

Square

Position Analysis:

Plot the linkage to scale. Start by drawing point D and the rest of link 4. Next draw link B andfinally draw link 3. Links 5 and 6 do not need to be drawn because they do not affect theinformation that is requested.

Velocity Analysis:

v vA A3 2=

v vB B3 4=

v v v v vB A B A B B C3 3 3 3 4 4 4= + = =/ / (1)

Now,

vA2 10= in / s in the horizontal direction

v v r v r rB B B C B B C B C3 4 4 4 44 4= = × ⇒ = ⋅ ⊥ω ω/ / /( )to

v r v r rB A B AB A B A B A3 3 3 3 3 33 3/ // / /( )= × ⇒ = ⊥ω ω to

- 142 -

30 in/s

Acceleration Scale

2

A

B

C

D

5 in/s

Velocity Scale

a3o

b3

c4

b4

d4

1aB3 /A3r

1aB 4/C4r

b3'

o'

a'2a'3c'4

1a B3 /A3t

1a B4 /C 4t

b4'

d4'


vB A3 3 6 64/ .= in / s

or

- 143 -

ω33 3 6 64

3 2 21= = =vrB A

B A

/

/

. . rad / s

To determine the direction of ω3, determine the direction that rB A/ must be rotated to be parallel tovB A3 3/ . This direction is clearly counterclockwise.

Also,

vB4 9 70= . in / s

and

ω44 4 9 70

1 414 6 86= = =vrB C

B C

/

/

.. . rad / s

To determine the direction of ω4 , determine the direction that rB C/ must be rotated to be parallel tovB C4 4/ . This direction is clearly clockwise.

Also,

v rD D C4 6 77= ⊥. ( )/in / s to

Draw the image of link 4 on the velocity polygon. The image is a square.


a aA A2 3=

a a a a aB B A B A B C3 4 3 3 3 4 4= = + =/ /

a a a a aB Cr

B Ct A B A

rB At

4 4 4 4 3 3 3 3 3/ / / /+ = + + (3)

Now,

aA3 0=

a r a rB Cr B C B C

r B C4 4 4 44 4 42 26 86 1 414 66 54/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rB C/


t B C B C4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rB Ar B A B A

r B A3 3 3 33 3 32 22 21 3 14 6/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t B A B A3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

- 144 -


aD4 54 0= . in / s2

The image of link 4 is a square as shown on the acceleration polygon.

Problem 2.34

In the mechanism shown below, the velocity of A2 is 10 in/s to the right and is constant. Draw thevelocity and acceleration polygons for the mechanism, and record values for angular velocity andacceleration of link 6. Use the image technique to determine the velocity of points D3, and E3, andlocate the point in link 3 that has zero velocity.

A C

BD

E

F2

34

5

6v = 10 in/sA2

(constant)

CF = 1.95"FE = 1.45"ED = 1.5"CD = 1.0"BC = 1.45"BD = 1.05"AB = 3.0"

103˚

Position Analysis:

Locate points C and F and the line of action of A. Draw link 6 and locate pont E. Then locate pointD. Next locate point B and finally locate point A.

Velocity Analysis:

The equations required for the velocity analysis are:

v vA A3 2=

v vB B3 4=

v v vB A B A3 3 3 3= + / (1)

v vD D5 4=

v v v vE E D E D5 6 5 5 5= = + / (2)

Now,

vA2 10= in / s in the horizontal direction

- 145 -

v v r v r rB B B C B B C B C3 4 44 4= = × ⇒ = ⋅ ⊥ω ω/ / /( )to

v r v r rB A B A B A B A B A3 3 3 33 3/ / / / /( )= × ⇒ = ⊥ω ω to


vB A3 3 5 97/ .= in / s

ω33 3 5 97

3 1 99= = =vrB A

B A

/

/

. . rad / s

Also,

vB in s4 9 42= . /

or

ω44 4

4 4

9 421 45 6 50= = =

vr

B C

B C

/

/

.

. . rad / s CW

Now,

v rD D Cin s to4 6 50= ⊥. / ( )/

v r v r rE D E D E D E D E D5 5 5 55 5/ / / / /( )= × ⇒ = ⊥ω ω to

v v v r v r rE E E F E F E F E F E F5 6 6 6 6 66 6= = = × ⇒ = ⊥/ / / / /( )ω ω to


vE6 5 76= . in / s

or

ω66 6 5 76

1 45 3 97= = =vrE F

E F

/

/

.

. . rad / s


a aA A2 3=

a a a aB B A B A3 4 3 3 3= = + /

a a a aB Cr

B Ct

B Ar

B At

4 4 4 4 3 3 3 3/ / / /+ = + (3)

a aD D5 4=

a a a a aE E E F D E D5 6 6 6 5 5 5= = = +/ /

- 146 -

a a a a aE Fr

E Ft D E D

rE Dt

6 6 6 6 5 5 5 5 5/ / / /+ = + + (4)

Now,

A C

BD

E

F2

34

6

o'25 in/s

Acceleration Polygon2

1 a B3 /A3r

1 aB3/A3t

1 aB4 /C4r

b3’ b4b’,

d4'

1 aE5/ D 5r

1 a E5/D 5t

1a E6 /F6r

e5'

1a B4/ C4t

1 a E6 /F6t

O2.5 in/s

Velocity Polygon

a2a3

,

b3 b4b,

d4 d5,

e5

3

o

d3

e3

5D

or

a r a rB Cr B C B C

r B C4 4 4 44 4 42 26 50 1 45 61 3/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rB C/


t B C B C4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α to

- 147 -

a r a rB Ar B A B A

r B A3 3 3 33 3 32 21 99 3 11 9/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t B A B A3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to


aB At

3 3 65 3/ .= in / s2

or

α3 23 3 65 33 21 8= = =

arB At

B Arad s CW/

/

. . /

And,

aD3 86 6= . in / s2

Also,

a r a rE Dr E D E D

r E D5 5 5 55 5 52 21 73 1 5 4 49/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rE D/

a r a r rE Dt E D E D

t E D E D5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥α α to

a r a rE Fr E F E F

r E F6 6 6 66 6 62 25 76 1 45 48 1/ / / / . . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2

in the direction of - rE F/

a r a r rE Ft E F E F

t E F E F6 6 6 66 6/ / / / /( )= × ⇒ = ⋅ ⊥α α to

Now solve Eq. (4) using the acceleration polygon. Then,

aE Ft

6 6 38 2/ .= in / s2

or

α66 6 38 2

1 45 26 4= = =arE Ft

E F

/

/

.. . rad / s CCW2

Using the image concept, the velocities of points D3 and E3 are

vD3 10 2= . in / s

- 148 -

vE3 12 7= . in / s

The point in link 3 with zero velocity is shown on position diagram. The point is found by findingthe position image of oa b3 3.

Problem 2.35

The instant center of acceleration of a link can be defined as that point in the link that has zeroacceleration. If the accelerations of Points A and B are as given in the rigid body shown below, findthe Point C in that link at which the acceleration is zero.

B

A

AaaB

70˚

5˚

145˚

Aa = 1500 in/s 2

= 1050 in/saB2

AB = 3.75"


Draw the accelerations of points A and B on an acceleration polygon. Then o' will correspond tothe instant center of acceleration. Find the image of o' on the position diagram, and that will be thelocation of C

- 149 -

500 in/s

Acceleration Scale2

o'

a'

b'

c'

A

B

C

1 in

Problem 2.36

The following are given for the mechanism shown in the figure:

ω α2 26 5 40= =. rad/ s (CCW); rad/ s (CCW)2

Draw the velocity polygon, and locate the velocity of Point E using the image technique.

A

C

B

D (2.2", 1.1")

23

4

E

55˚

X

Y

AB = DE = 1.0 inBC = 2.0 inCD = 1.5 in

Position Analysis

Locate the two pivots A and D. Draw link 2 and locate pivot B. Then find point C and finallylocate E.

- 150 -

Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,

v r v r rB A B A B A B A B Ain to2 2 2 2 2 2 2 2 2 22 2 6 5 1 6 5/ / / / /. . / sec ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω

v r v r rC D C D C D C D C Dto4 4 4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥ω ω

v r v r rC B C B C B C B C Bto3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


vE in4 5 4042= . / sec

c3

b3

o

2 in/sec

Velocity Polygon

e4

A

C

B

DE

- 151 -

Problem 2.37

In the mechanism shown, find ωωωω6 and αααα3. Also, determine the acceleration of D3 by image.

AC

B

D

E

F (-1.0", -0.75")

2

3 4

5

6

v = 10 in/sA2 (constant)

X

Y

81˚

CD = 1.0"BD = 1.05"BC = 1.45"ED = 1.5"FE = 1.4"AB = 3.0"

Velocity Analysis:

v vA A3 2=

v vB B3 4=

v v vB A B A3 3 3 3= + / (1)

v vD D5 4=

v v v vE E D E D5 6 5 5 5= = + / (2)

Now,

vA in2 10= / sec in the horizontal direction

v v r v r rB B B C B B C B Cto3 4 4 4 4 4 4 4 44 4= = × ⇒ = ⋅ ⊥ω ω/ / /( )

v r v r rB A B AB A B A B Ato3 3 3 3 3 3 3 3 3 33 3/ // / /( )= × ⇒ = ⊥ω ω


vB A in3 3 6 282/ . / sec=

or

ω33 3

3 3

6 2823 2 094= = =

vrB A

B Arad/

/

. . / sec

- 152 -

A C

B

D

E

F

34

5

62

d4 d5,

2.5 in/sec

Velocity Polygon

o

b3 b4b,

a2 a3,

e5

25 in/sec

Acceleration Polygon2o'1a B3 /A3

r

1a B3 / A3t

1aB4/C4

t

1aB4/C4r

b3’ b4b’,

d3'

Also,

vB in4 9 551= . / sec

or

ω44 4

4 4

9 5511 45 6 587= = =

vr

B C

B Crad CW/

/

.. . / sec

Now,

vD in4 6 587= . / sec ( )/⊥ to D Cr 4 4

- 153 -

v r v r rE D E D E D E D E Dto5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⊥ω ω

v v v r v r rE E E F E F E F E F E Fto5 6 6 6 6 6 6 6 6 6 6 66 6= = = × ⇒ = ⊥/ / / / /( )ω ω


vE in6 6 132= . / sec

or

ω66 6

6 6

6 1321 45 4 229= = =

vrE F

E Frad CW/

/

.. . / sec


a aA A2 3=

a a a aB B A B A3 4 3 3 3= = + /

a a a aB Cr

B Ct

B Ar

B At

4 4 4 4 3 3 3 3/ / / /+ = + (3)

Now,

a r a rB Cr B C B C

r B C in4 4 4 4 4 4 4 44 4 42 2 26 587 1 45 62 913/ / / / . . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω

in the direction of rB C4 4/

a r a rB Ct B C B C

t B C4 4 4 4 4 4 4 44 4/ / / /= × ⇒ = ⋅α α ( )/⊥ to B Cr 4 4

a r a rB Ar B A B A

r B A in3 3 3 3 3 33 3 32 2 22 094 3 13 155/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω

in the direction of rB A/


t B A B Ato3 3 3 3 3 3 3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α


aB At in3 3 68 568 2

/ . / sec=

or

α3 23 3

3 3

68 5683 22 856= = =

ar

B At

B Arad CW/

/

. . / sec

Also,

aD in3 91 390 2= . / sec

- 154 -

Problem 2.38

In the mechanism shown, ωωωω2 = 1 rad/s (CCW) and αααα2 = 0 rad/s2. Find ωωωω5, αααα5, vE6, aE6 for theposition given. Also find the point in link 5 that has zero acceleration for the position given.

A

C

B

D

2

3

4E 5

6

30˚

AD = 1 mAB = 0.5 mBC = 0.8 mCD = 0.8 mBE = 0.67 m

0.52 m

Velocity Analysis

v v v vB B B B A2 3 5 2 2= = = /

v v v vE E B E B5 6 5 5 5= = + / (1)

Now,

v r v r rB A B A B A B A B Am to2 2 2 2 2 2 2 2 2 22 2 1 0 5 0 5/ / / / /. . / sec ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω

15vE in the horizontal direction

v r v r rE B E B E B E B E Bto5 5 5 5 5 5 5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥ω ω


vE B m5 5 0 47313/ . / sec=

or

ω55 5

5 5

0 473130 67 0 706= = =

vrE B

E Brad CCW/

/

.. . / sec

Also,

- 155 -

A

C

B

D

2

3

4E

5

6

o'

0.1 m/secAcceleration Polygon

2

1 aE5

1aB2/A2r

1a E5/B5r

1aE5 /B5tb3b'

e5'

2.2°

27.8°

O'

0.1 m/secVelocity Polygon

b5

o

e 5

vE m6 0 441= . / sec


a a a aB B B B A2 3 5 2 2= = = /

a a a aE E B E B5 6 5 5 5= = + /

a a a a aE B Ar

B At

E Br

E Bt

5 2 2 2 2 5 5 5 5= + + +/ / / / (2)

Now,

- 156 -

aE5 in horizontal direction

a r a rB Ar B A B A

r B A m2 2 2 22 2 22 2 21 0 5 0 5/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω

in the direction opposite to rB A/

a r a rB At B A B A

t B A m2 2 2 22 2 20 1 0/ / / / / sec= × ⇒ = ⋅ = ⋅ =α α

a r a rE Br E B E B

r E B m5 5 5 55 5 5

2 2 20 706 0 67 0 334/ / / / . . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω

in the direction opposite to rE B/

a r a r rE Bt E B E B

t E B E Bto5 5 5 55 5/ / / / /( )= × ⇒ = ⋅ ⊥α α


aE m6 0 042 2= . / sec

Also,

aE Bt m

5 50 42 2

/ . / sec=

or

α5 25 5

5 5

0 420 67 0 626= = =

ar

E Bt

E Brad CW/

/

.

. . / sec

Using acceleration image, point O' is in the location in which the acceleration of link five is zero.

- 157 -

Problem 2.39

Part of an eight-link mechanism is shown in the figure. Links 7 and 8 are drawn to scale, and thevelocity and acceleration of point D7 are given. Find ωωωω7 and αααα7 for the position given. Also findthe velocity of G7 by image.

D

E

7

6

G

8

X

Y

1.25"1.6"

DE = 1.5"DG = 0.7"GE = 1.65"

= 5.0 320˚ in/sec = 40 260˚ in/s 2vD7 aD7

Velocity Analysis

Compute the velocity of Points E7 and E8.

v vv v v

E E

E D E D

7 8

7 7 7 7

== + /

and because points on the same link are involved,

v v v r vD E D D E D E7 7 7 7 87+ = + × =/ /ω

From the velocity polygon:

vE in8 2 884= . / sec

in the direction shown, and

vE D in7 7 3 405/ . / sec=


ω77 7 3 405

1 50 2 27= = =vrE D

E Drad/

/

.. . / sec

- 158 -

o'

D

E

7

7d'

6

G 8

1a E7 /D7r

1a E7 /D7t

e'72 in/sec

Velocity Scale

220 in/sec

Acceleration Scale

7d

7e 8, e

7g

o

The direction can be found by rotating rE D/ 90˚ in the direction of ω7 to get vE D7 7/ . From thepolygon, the direction must be counter clockwise. Therefore,

ω7 2 27= . / secrad CCW

The velocity of G7 is found by image. The magnitude of the velocity is:

vG in7 6 016= . / sec



Use the same points as were used in the velocity analysis.

a a a a a r aD E D F D E Dr E D F7 7 7 8 7 7 7 7 7 87+ = = + + × =/ / /α

where

a vrE D

r E D

E Din7 7

7 72 2

23 4051 50 7 729/

/

/

.. . / sec= = =

- 159 -

in the direction opposite rE D/ .

From the polygon,

a rE Dt E D7 7 7 77 0/ /= × =α

Therefore,

α7 27 7 42 591 50 28 40= = =

arE Dt

E Drad/

/

.. . / sec

The direction can be found by rotating rE D/ 90˚ in the direction of ω7 to get aE Dt

7 7/ . From thepolygon, the direction must be counter clockwise. Therefore,

a7 228 40= . / secrad CCW

Problem 2.40

In the mechanism shown below, link 2 is rotating CW at the rate of 3 rad/s (constant). In theposition shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vectorpolygons, and

a) Determine vC4, vE4, ωωωω3, and ωωωω4.

b) Determine aC4, aE4, αααα3, and αααα4.

Link lengths: AB = 3 in, BC = BE = CE = 5 in, CD = 3 in

B

C

2

3

4

D

ω 2

A

7"

E

Position Analysis

Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B andthen C. Then locate G.

- 160 -

Velocity Analysis:

v v vB B B A3 2 2 2= = /

v v v v vC C C D B C B3 4 4 4 3 3 3= = = +/ / (1)

Now,





From the polygon,

vC B3 3 11 16/ .= in / s

vC D4 4 6 57/ .= in / s


- 161 -

Now

ω33 3 11 16

5 2 232= = =vrC B

C B

/

/

. . rad / s

and

ω44 4 6 57

3 2 19= = =vrC D

C D

/

/

. . rad / s



The velocity of point E3

v v v v rE B E B B E B3 3 3 3 3 3= + = + ×/ /ω

from the polygon

vE3 13 5= . in / s

vC4 6 57= . in / s


a a aB B B A3 2 2 2= = /

a a a a aC C C D B C B3 4 4 4 3 3 3= = = +/ /

a a a a a aC Dr

C Dt

B Ar

B At

C Br

C Bt

4 4 4 4 2 2 2 2 3 3 3 3/ / / / / /+ = + + + (2)

Now,

a r a rB Ar B A B A

r B A2 2 2 22 2 22 23 3 27/ / / /= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2


a r a rB At B A B A

tB A2 2 2 22 2 0/ / / /= × ⇒ = =α α

a r a rC Br C B C B

r C B3 3 3 33 3 32 22 232 5 24 9/ / / / . .= × ×( )⇒ = ⋅ = ⋅ =ω ω ω in / s2



t C B C B3 3 3 33 3/ / / / /( )= × ⇒ = ⋅ ⊥α α to

- 162 -

a r a rC Dr C D C D

r C D in4 4 4 44 4 42 2 22 19 3 14 39/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω



t C D C Dto4 4 4 44 4/ / / / /( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically with acceleration.


aC Bt

3 3 0/ = in / s2

aC Dt

4 4 46 71/ .= in / s2

Then,

α33 3 0

2 0= = =arC Bt

C B

/

/rad / s2

α44 4 46 71

3 15 57= = =a

rC Dt

C D

/

/

. . rad / s2


aC Dt


Determine the acceleration of point E3

a a a a a aE B E B B E Br

E Bt

3 3 3 3 2 3 3 3 3= + = + +/ / /

- 163 -

a r a rE Bt E B E B

t E B in3 3 3 33 3 20 1 0/ / / / / sec= × ⇒ = ⋅ = ⋅ =α α

a r a rE Br E B E B

r E B in3 3 3 33 3 32 2 22 232 5 24 9/ / / / . . / sec= × ×( )⇒ = ⋅ = ⋅ =ω ω ω


aE3 34 29= . in / s2

aC4 48 87= . in / s2

Problem 2.41

Part of a 10-link mechanism is shown in the figure. Links 7 and 8 are drawn to scale, and thevelocity and acceleration of points D7 and F8 are given. Find ωωωω8 and αααα7 for the position given.Also find the velocity of G7 by image.

D

E

F (1.8", -1.05")

7

8

9

6

G

X

Y

DE = 1.5"EF = 1.45"DG = 0.7"EG = 1.65"

= 6.0 353˚ in/s

= 7.5 54˚ in/s

= 40 235˚ in/s 2

= 30 305˚ in/s 2

v D

vF

7

8

aD

a F

7

8

Velocity Analysis

Compute the velocity of Points E7 and E8.

v vv v vv v v

E E

E D E D

E F E F

7 8

7 7 7 7

8 8 8 8

== += +

/

/

Therefore,

v v v vD E D F E F7 7 7 8 8 8+ = +/ /

and because points on the same link are involved,

- 164 -

v v v v v r v rD E D F E F D E D F E F7 7 7 8 8 8 7 7 7 8 8 87 8+ = + = + × = + ×/ / / /ω ω

From the velocity polygon:

vE F8 8 0/ =

so

ω8 0=

The velocity of G7 is found by image. The magnitude of the velocity is

vG7 9 1= . in / s

o'

o

7d

E

7d'

8f

8f'

D

F

7

8

9

6

G

8e,

7g

7e,

8e' 7e',

5 in/s

Velocity Polygon

220 in/s

Velocity Polygon

vG7 9 1= . in / s


Use the same points as were used in the velocity analysis for the acceleration analysis.

a a a aa a r a a r

D E D F E F

D E Dr E D F E F

r E F

7 7 7 8 8 8

7 7 7 7 7 8 8 8 8 87 8

+ = += + + × = + + ×

/ /

/ / / /α α

- 165 -

avr

avr

E Dr E D

E D

E Fr E F

E F

7 77 7

7 7

8 88 8

8 8

2 2

2 2

6 921 50

32 0

01 44

0

//

/

//

/

..

.

.

= = =

= = =

in / s

in / s

2

2

From the polygon,

a rE Dt E D7 7 7 77 0/ /= × =α

Therefore,

α7 0=

solutions to chapter 2 exercise problems problem 2 · solutions to chapter 2 exercise problems...

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