chapter 2. graphical position velocity

7/17/2019 Chapter 2. Graphical Position Velocity ...

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- 48 -

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, ωω ω ω 3, and ωω ω ω 4.

b) Determine aC4, αα α α 3, and αα α α 4.

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

B

C

2

3

4 A

D

50 mm

250 mm

ω2

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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



- 49 -

v A2 0=

Therefore,

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

Now,

ω 2 2= rad sCCW /

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

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

v r r C D C D C D4 4 / / / ( )= × ⊥ω 44 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,

ω 33 3 156

18286= = =

v

r

C B

C B

/

/ . rad/s

From the directions given in the position and velocity polygons

ω 3 86= . rad / s CW

Also,

ω 44 4 43

10043= = =

v

r

C D

C D

/

/ . rad/s


ω 44 = .43 rad /s CW

Acceleration Analysis:

a a a B 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 r B Ar

B A B Ar

B A mm2 2 2 22 2 2

2 22 75 300 / / / / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2



- 50 -

in the direction of - r B A2 2/

a B At

2 20 / = since link 2 rotates at a constant speed (α 2 0= )

a r a r C Br

C B C Br

C B mm3 3 3 33 3 3

2 286 182 134 6 / / / / . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2

in the direction of - r C B/

a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

C D mm s4 4 4 44 4 4

2 2 243 100 18 5 / / / / . . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω

in the direction of - rC D/

a r a r r C Dt

C D C Dt

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 s

3 319 22 / . / =

2

aC Dt mm s

4 4434 70 / . / =

2

Then,

α 33 3 67 600

2 4227 900= = =

a

r

C Bt

C B

/

/

,.

, rad/s2

α 4 4 4 434 70100 4 347= = =

a

r C D

t

C D / /

. . rad/s2

To determine the direction of αα α α 3, determine the direction that r C B/ must be rotated to be parallel to

aC Bt

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

To determine the direction of α 4, determine the direction that r C D/ must be rotated to be parallel to

aC Dt


From the acceleration polygon,

aC mm s4 435= / 2



- 51 -

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

a) Determinev

C4,ωω ω ω

3, andωω ω ω

4.


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

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 0=

Therefore,



- 52 -

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

Now,

ω 2 500= rad s CCW /

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

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

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


vC B in s3 3 523 5 / . / =

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

Now,

ω 33 3 523 5

2 42216 3= = =

v

r

C B

C B

/

/

..

. rad/s


ω 3 216 3= . rad/ s CCW

Also,

ω 44 4 858

2429= = =

v

r

C D

C D

/

/ rad/s


ω 44 =429 rad s CC / W


a a a B 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 r B Ar

B A B Ar

B A in s2 2 2 22 2 2

2 2500 1 2 300000 / / / / . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a B At




- 53 -

a r a r C Br

C B C Br

C B in3 3 3 33 3 3

2 2216 3 2 42 113 000 / / / / . . ,= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2


a r a r r C Bt

C B C Bt

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

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


a r a r r C Dt

C D C Dt

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


aC Bt in s

3 367561 / / = 2

aC Dt in s

4 4

151437 / / =2

Then,

α 33 3 67561

2 4227 900= = =

a

r

C Bt

C B

/

/ . , rad/s2

α 44 4 151437

275 700= = =

a

r

C 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

a) Determinev

C4,ωω ω ω

3, andωω ω ω

4.


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 v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 0=

Therefore,

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



- 55 -

Now,

ω 2 10= rad s CW /


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

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


vC B mm s3 3 31 3 / . / =

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

Now,

ω 33 3 31 3

260

12= = =v

r

C B

C B

/

/

. . rad/s


ω 3 12=. rad/ s CCW

Also,

ω 44 4 990

1805 5= = =

v

r

C D

C D

/

/ . rad/s


ω 44 =5 5. / rad sC W


a a a B 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 r B Ar

B A B Ar

B A mm s2 2 2 22 2 2

2 210 100 10 000 / / / / , / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a B At


a r a r C Br

C B C Br

C B mm3 3 3 33 3 3

2 212 260 3 744 / / / / . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2



- 56 -


a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

C D mm s4 4 4 44 4 4

2 2 25 5 180 5 445 / / / / . , / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C Dt

C D C Dt

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


aC Bt mm s

3 34784 / / =

2

aC Dt mm s

4 41778 / / =

2

Then,

α 33 3 4785

26018 4= = =

a

r

C Bt

C B

/

/ . rad/s2

α 44 4 1778

1809 88= = =

a

r

C Dt

C D

/

/ . rad/s2


aC Bt



aC Dt



aC mm s

45 700= , / 2



- 57 -

Problem 2.4

In the mechanism shown below, link 2 is rotating CW at the rate of 4 rad/s (constant). In the

position shown, θ is 53˚. Write the appropriate vector equations, solve them using vector polygons,

and

a) Determine vC4, ωω ω ω 3, and ωω ω ω 4.


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

B C

2

3

4

D

ω 2 A

220 mm

160 mm

θ



- 58 -

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 0=

Therefore,

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

Now,

ω 2 4= rad s CW /


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

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


vC B mm s3 3 500 / / =

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

Now,

ω 33 3 500

1603 125= = =

v

r

C B

C B

/

/ . rad/s


ω 3 3 125= . rad/ s CCW

Also,

ω 44 4 300

2001 5= = =

v

r

C D

C D

/

/ . rad/s


ω 44 =1 5. / rad s CC W


a a a B 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 r B Ar

B A B Ar

B A mm s2 2 2 22 2 2

2 24 100 1600 / / / / / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2

in the direction of -r B A2 2/

a B At


a r a r C Br

C B C Br

C B mm3 3 3 33 3 3

2 23 125 160 1560 / / / / .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2


a r a r r C Bt

C B C Bt

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

a r a r C D

r

C D C D

r

C D mm s

4 4 4 44 4 4

2 2 2

1 5 200 450 / / / / . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C Dt

C D C Dt

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


aC Bt mm s

3 3618 5 / . / =

2

aC Dt mm s

4 43 220 / , / =

2

Then,

α 33 3 618 5

1603 87= = =

a

r

C Bt

C B

/

/

. . rad/s2

α 44 4 3220

20016 1= = =

a

r

C Dt

C D

/

/ . rad/s2


aC Bt



aC Dt



aC mm s

43250= / 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

a) Determinev

C4,ωω ω ω

3, andωω ω ω

4.


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 v B C B C 3 3 3 3= + /

v v B B3 2=



- 61 -

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

v A2 0=

Therefore,

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

Now,

ω 2 4= rad s CCW /

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

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

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


vC B in s3 3 6 25 / . / =

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

Now,

ω 33 3 6 25

2 52 5= = =

v

r

C B

C B

/

/

..

. rad/s


ω 3 2 5= . rad/ s CCW

Also,

ω 44 4 3 75

2 51 5= = =

v

r

C D

C D

/

/

..

. rad/s


ω 44 =1 5. / rad sC W


a a a B 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 r B Ar

B A B Ar

B A in s2 2 2 22 2 2

2 24 1 25 20 / / / / . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a B At


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


a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

C D in s4 4 4 44 4 4

2 2 21 5 2 5 5 6 / / / / . . . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C D

t

C D C D

t

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


aC Bt in s

3 34 69 / . / =

2

aC Dt in s

4 44 69 / . / =

2

Then,

α

3

3 3 4 69

2 51 87= = =

a

r

C Bt

C B

/

/

.

. . rad/s2

α 44 4 4 69

2 51 87= = =

a

r

C 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 ωω ω ω 3

b) Determine aC4 and αα α α 3

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

B

C

2

3

4

ω 2

A

120 mm

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 0=

Therefore,

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



- 64 -

Now,

ω 2 100= rad s CW /


v r r C 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,

ω 33 3 7500

20037 5= = =

v

r

C B

C B

/

/

. rad/s


ω 3 12=. rad / s CW


a a a B 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 r B Ar

B A B Ar

B A mm s2 2 2 22 2 2

2 2100 60 600 000 / / / / , / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a B At


a r a r C B

r

C B C B

r

C B mm3 3 3 33 3 3

2 2

37 5 200 281 000 / / / / . ,= × ×

( )⇒ = ⋅ = ⋅ =ω ω ω

/ s2


a r a r r C Bt

C B C Bt

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

a aC D C 4 4 4 / = →parallel to ground




- 65 -

aC Bt mm s

3 3211 000 2

/ , / =

a aC D C mm s4 4 4

248 000 2 / , / = =

Then,

α 33 3 211 000

2001060= = =a

r C 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 ωω ω ω 3 and ωω ω ω 4.

b) Determine αα α α 3 and αα α α 4.

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

B

C

23

4

A

8.5 ft

120˚

vC4



- 66 -

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 0=

Therefore,

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

Now,

vC 4 4= f t / s parallel to the ground

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

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


v B C ft s3 3 2 3 / . / =

v B A ft s2 2 2 3 / . / =

or

ω 33 3 2 3

20115= = =

v

r

B C

B C

/

/

. . rad/s




- 67 -

ω 3 115=. rad / s CW

Also,

ω 22 2 2 3

1023= = =

v

r

B A

B A

/

/

. . rad/s


ω 2 23=. rad /s CCW

ω 4 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 r B Ar

B A B Ar

B A ft s2 2 2 22 2 2

2 223 10 529 / / / / . . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2

in the direction of - r B A2 2 /

a r a r r B At

B A B At

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

a r a r C Br

C B C Br

C B ft 3 3 3 33 3 3

2 2115 20 264 / / / / . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2

in the direction of - r C B /

a r a r r C Bt

C B C Bt

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

a C D4 40 / = link 4 is moving at a constant velocity


aC Bt ft s

3 30 045 / . / =

2

a B At ft s

2 20 017 2

/ . / =

Then,

α 33 3 0 45

20023= = =

a

r

C Bt

C B

/

/

. . rad/s2

α 22 2 0 017

100017= = =

a

r

B At

B A

/

/

. . rad/s2



- 68 -

To determine the direction of αα α α 3, determine the direction that r C B / must be rotated to be parallel to

aC Bt


To determine the direction of αα α α 2, determine the direction that r B A / must be rotated to be parallel to

a B 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 ωω ω ω 3 and ωω ω ω 4.

b) Determine αα α α 3 and αα α α 4.

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

2

A

B

C 3

4

7 in

45˚

vC4



- 69 -

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

v A2 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 r B C B C B C 3 3 3 / / / ( )= × ⊥ω to

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


v B C in s3 3 14 1 / . / =

v B A in s2 2 14 1 / . / =

or

ω 33 3 14 1

52 82= = =

v

r

B C

B C

/

/

. . rad/s


ω 3 2 82= . rad /s CCW

Also,

ω 22 2 14 1

52 82= = =

v

r

B A

B A

/

/

. . rad/s


ω 2 2 82= . rad /s CCW

ω 4 0= rad / s since it doesn’t rotate


a a a B 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 r B Ar

B A B Ar

B A in s2 2 2 22 2 2

2 22 82 5 39 8 / / / / . . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a r a r r B At

B A B At

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

a r a r C Br

C B C Br

C B in3 3 3 33 3 3

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


a r a r r C Bt

C B C Bt

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

aC D4 4 0 /

=

link 4 is moving at a constant velocity


aC Bt in s

3 338 8 / . / =

2

a B At in s

2 238 8 2

/ . / =

Then,

α 33 3 38 8

57 76= = =

a

r

C Bt

C B

/

/

. . rad/s2

α 22 2 38 8

57 76= = =

a

r

B At

B A

/

/

. . rad/s2

α 4 0 4= ( )link isnotrotating


aC Bt


To determine the direction of α 22 , determine the direction that r B A / must be rotated to be parallel to

a B At




- 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

4 vC4

135˚

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=

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

Therefore,

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



- 72 -

Now,

vC 4 6= . f t / sparallel to the ground

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

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


v B C ft s3 3 85 / . / =

or

ω 33 3 85

5 122 04= = =

v

r

B C

B C

/

/

.( / )

. rad/s


ω 3 2 04

=

. rad / s CW

Now,

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

Using velocity image,

v A ft s3 1 34= . /


a aC C 4 3

0= =

a a a a a B B B C r

B C t

B C 3 2 3 3 3 3 3 3= = = + / / / (2)

Now,

a r a r B C r

B C B C r

B C ft s3 3 3 33 3 3

2 22 04 5 12 1 73 / / / / . ( / ) . / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2

in the direction of - r B C 3 3 /

a r a r r B C t

B C B C t

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



a B C t ft s

3 31 73 / . / =

2

Then,



- 73 -

α 33 3 1 73

5 124 15= = =

a

r

B C t

B C

/

/

.( / )

. rad/s2

To determine the direction of αα α α 3, determine the direction that r B C / must be rotated to be parallel to

a B C t


Using acceleration image,

a A 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, ωω ω ω 2, and ωω ω ω 3.

b) Determine aB2, aG3, αα α α 2, and αα α α 3.

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,

ovc3 c4,

b2 b3,

25 ft/sec

Velocity Polygon

Velocity Analysis:

v v v B C B C 3 3 3 3= + /

v v B B3 2=



- 75 -

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

v A2 0=

Therefore,

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

Now,

vC 3 75= f t / sin the direction of r C A/

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

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


v B C ft s3 3 69 4 / . / =

or

ω 33 3 69 4

16 1 1252= = =

v

r

B C

B C

/

/

.( / )

rad/s


ω 3 52= rad/ s CCW

Also,

ω 22 2 91 5

4 8 1 12228= = =v

r

B A

B A

/

/

.. ( / )

rad/s


ω 2 228= rad /s CW

To compute the velocity of G3,

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

Using the values computed previously

ω 3 3 3 52 6 0 312× = =r G B / ( . ) in/s

and from the directions given in the velocity and position diagrams

ω 3 3 3 3 3312× = ⊥r r G B G B / / in/s

Now draw vG3 on the velocity diagram

vG3 79 0= . f t / s in the direction shown.



- 76 -


a a a B 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 r B Ar

B A B Ar

B A ft s2 2 2 22 2 2

2 2228 4 8 12 20 900 / / / / ( . / ) , / = × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω 2


a r a r r B At

B A B At

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

a r a r C Br

C B C Br

C B ft 3 3 3 33 3 3

2 252 16 12 3605 / / / / ( / )= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω / s2


a r a r r C Bt

C B C Bt

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



aC Bt ft s

3 328 700 / , / =

2

a B At ft s

2 2

20 000 2 / , / =

Then,

α 33 3 28 700

16 1221 500= = =

a

r

C Bt

C B

/

/

,( / )

, rad/s2

α 22 2 20 000

4 8 1250 000= = =

a

r

B At

B A

/

/

,( . / )

, rad/s2

To determine the direction of α 3 , determine the direction that r C B/ must be rotated to be parallel to

aC Bt


To determine the direction of α 22 , determine the direction that r B A / must be rotated to be parallel to

a B C t



a B ft s

228 900= , / 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 ωω ω ω 2 = 50 rad/s CW and αα α α 2 = 1600 rad/s2 CW. Write the

appropriate vector equations, solve them using vector polygons, and

a) Determine vB2, vC3, vE3

, ωω ω ω 3, and ωω ω ω 4.

b) Determine aB2, aC3, aE3

,αα α α 3, and αα α α 4.

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

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



- 78 -

3 3aC / B

t

aC3 / B3

r

aC4 /D4

t

aC 4 /D 4

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 / A2

r

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

v E 3 107 8= . in/s

in the direction shown.

Now

ω 33 3 65 2

5 1512 7= = =

v

r

C B

C B

/

/

..

. rad/s



- 79 -

and

ω 44 4 92 6

2 7533 7= = =

v

r

C D

C D

/

/

..

. rad/s

To determine the direction of ω 3, determine the direction that r C B/ must be rotated to be parallel to

vC B3 3 / . This direction is clearly clockwise.

To determine the direction of ω 4 , determine the direction that r C D/ must be rotated to be parallel to

vC D4 4 / . This direction is clearly clockwise.


a a a B 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 r B Ar

B A B Ar

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


a r a r r B At

B A B At

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

a r a r C Br

C B C Br

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


a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

C D in4 4 4 44 4 4

2 2 233 7 2 75 3123 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C Dt

C D C Dt

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 31563 / = in/s2

aC Dt

4 44881 / = in/s2

Then,



- 80 -

α 33 3 1563

5 15303= = =

a

r

C Bt

C B

/

/ .rad/s2

α 44 4 4881

2 751775= = =

a

r

C Dt

C D

/

/ .rad/s2

To determine the direction ofα

3 , determine the direction thatr C B/ must be rotated to be parallel to

aC Bt



aC Dt


Also

a E 3 5958= in/s2

Problem 2.12

Resolve Problem 2.11 if ωω ω ω 2 = 50 rad/s CCW and αα α α 2 = 0 .

Position Analysis


Velocity Analysis:

The velocity analysis is similar to that in Problem 2.18.

v v v B 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 r B 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

v E 3 116= in/s



- 81 -


Now

ω 33 3 88 8

5 1517 2= = =

v

r

C B

C B

/

/

..

. rad/s

and

ω 44 4 103 1

2 7537 5= = =

v

r

C D

C D

/

/

..

. rad/s


vC B3 3 / . This direction is clearly counterclockwise.


vC D4 4 / . This direction is clearly counterclockwise.



- 82 -

3 3aC /B

t

aC3

/ B3

r

aC4 /D4

t

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/ A2

r

e'3



- 83 -


a a a B 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 r B Ar

B A B Ar

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


a r a r B At

B A B At

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

a r a r C Br

C B C Br

C B3 3 3 3

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


a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

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


a r a r r C Dt

C D C Dt

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

a

C Bt

C Dt

3 3

4 4

2751

1405

/

/

=

=

in/s

in/s

2

2

Then,

α

α

32

4

3 3

4 4

27515 15

534

14052 75

511

= = =

= = =

a

r

a

r

C Bt

C B

C Dt

C D

rad s /

/

/

/

. /

.rad/s2


aC Bt



aC Dt




- 84 -

Also

a E 3 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, ωω ω ω 3, and ωω ω ω 4.

b) Determine aB2, aC3, aE3, αα α α 3, and αα α α 4.

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

v v B B3 2=

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

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

and

v D4 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,

v B3 828= ft / s ( )/⊥ to r B A

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

v r r C 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:

v

v

r C B

C B

C B3 3 3 3583 58312 48 63 /

/

/ .= ⇒ = = =in / s rad / sω


ω 3 48 6= . / rad s CCW

Also,



- 86 -

v v

r C D

C D

C D4 4

4 4475 4759 2

51 64 / /

/ . .= ⇒ = = =rad / s rad / sω


ω 4 51 6= . rad/s CW

To compute the velocity of E3,

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

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

v r r E B E B E Bto3 3 3 / / / ( )= × ⊥ω

and

v r r E C E C E C to3 3 3 / / / ( )= × ⊥ω

Equation (2) can now be solved to give

v E 3 695= in/s


a a a B 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 r B Ar

B A B Ar

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


a r a r B At

B A B At

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

a r a r C Br

C B C Br

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

in the direction of -r C B/

a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

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


a r a r r C Dt

C D C Dt

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

3 396 880 / ,= in/s2

aC Dt

4 49785 / = in/s2

Then,

α 323 3 96876

128073= = =

a

r

C Bt

C Brad s

/

/ /

α 44 4 9785 5

9 21063 6= = =

a

r

C Dt

C D

/

/

..

. rad/s2


aC Bt




- 88 -


aC Dt


Also

a E 3 123 700= , in/s2

and

aC 3 149 780= , in/s2

Problem 2.14

The accelerations of points A and B in the coupler below are as given. Determine the acceleration of the 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 of point G by image.

For the angular acceleration of the body, resolve the acceleration a B At / in terms of components

along and perpendicular to r B A/ . The tangential component is perpendicular to r B A/ .

a r a r B A

t

B A B A

t

B A / / / / = × ⇒ =α α

and

α = = =

a

r

B At

B A

/

/ .31412 8

1122 rad/ s2



- 89 -

o'

2000 in/s

Acceleration Scale

2

A

G

B a'

b'

g'

aB/ At

aG

To determine the direction of α , determine the direction that r B A/ must be rotated to be parallel to

a B At / . 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 ωω ω ω 2

= 60 rad/s (CW). Find the velocity and acceleration of point F and the angular velocity andacceleration of links 3 and 4.

Y

X A

B

2

3 4

30˚

C

D

E F

AB = 15 cm BC = 29.5 cmCD = 30.1 cm AD = 7.5 cm BE = 14.75 cm EF = 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 r A 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 r C C C D C D3 4 4 4 4= = = · / / ω

Now,

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

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

v r r C C D C D4 4= · ^ωω / / ( )to


v v

r C B

C B

C Bm s3 3

3 312 82 12 82

0 295 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

r C D

C D

C D4 4

4 411 39 11 39

0 301 37 844 /

/

/ .

..

.= = = =m / s rad / sωω


ωω 4 37 84= . rad / s CW


a a a r B B

r B 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 at

C D r

C D r

B A t

B A r

C B t

C B4 4 4 4 2 2 2 2 3 3 3 3 / / / / / / + = + + + (2)

Now,

a r r

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

a r r

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

a r r

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

a r a

r r t

C B C B

t C B

C BC B3 3

3 33 3 / /

/

/ / ( )= · = ^α α to

a r a

r r t

C D C D

t C 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:

α 3 3 3 142 79

0 295 484= = =

a

r

t C B

C B

/

/

..

rad / s CW 2

Also,

α 4 4 4 41 01

0 301 136 = = =

a

r

t C D

C D

/

/

..

rad / s CCW 2

Using acceleration polygon,

aF m s3 256 2= /


Problem 2.16

For the straight-line mechanism shown in the figure, ωω ω ω 2 = 20 rad/s (CW) and αα α α 2 = 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

15o D

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



- 93 -

Velocity Analysis:

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

vC 3 in horizontal direction

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


v B3 77 3= . in/s

Also,

vC A3 3 40 / = in/s

or

ω 33 3

3 3

402

20= = =v

r

C A

C A

/

/

rad/sCCW

Also,

vC 3 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,

aC 3 in horizontal direction

a r a r A Dr

A D A Dr

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

in the direction opposite to r A D/

a r a r r A Dt

A D A Dt

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

a r a r C Ar

C A C Ar

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

in the direction opposite to r C A3 3/



- 94 -

a r a r r C At

C A C At

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,

a B3 955= in /s2

aC A3 3 280 / = in/s2

Also,

α 33 3

3 3

2802

140= = =

a

r

C At

C A

/

/ rad / s CCW2

AC

B

2

4

3

D

o

20 in/s

Velocity Polygon

a2 a3,

c3 , c4

b3

vC3

v A3 vC3 / A3

v B3

v B3 / A3

o'

400 in/s

Acceleration Polygon2

a2' a3',

aA2 / D 2

r

aA2/D 2

t

aC3c3'

aC3 / A3

r

aC3 / A3

t

b3'

aB3



- 95 -

Problem 2.17

For the data given in the figure below, find the velocity and acceleration of points B and C . Assume

vA = 20 ft/s, aA = 400 ft/s2, ωω ω ω 2 = 24 rad/s (CW), and αα α α 2 = 160 rad/s2 (CCW).

Bω 2

15o

vA

aA

α 2

A

C

90˚

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

Position AnalysisDraw the link to scale

o

a2

b2

c2

5 ft/sec

Velocity Polygon

o'

a2'

b2'

2'

80 ft/sec


2

aB2 /A2

r

a B2 /A2

t

c

Velocity Analysis:

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

Now,



- 96 -

v A ft 2 20= /sec in the positive vertical direction

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

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


v B ft 2 11 9= . / sec

Also, from the velocity polygon,

vC ft 2 15 55= . / sec in the direction shown


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

B Ar

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

Now,

a A ft 2 400 2= /sec in the given direction

a B At

B Ar in ft 2 2 2

2 2160 4 05 648 54 / / . / sec / sec= ⋅ = ⋅ = =α

a B Ar

B Ar in ft 2 2 2

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

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

a B ft 2 198 64 2= . / sec

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

aC ft 2 289 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 the

velocity and acceleration polygons for the mechanism, and determine aG3 and αα α α 4.

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 v B 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 r B 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

ω 33 3 11 4

25 7= = =

v

r

C B

C B

/

/

. . rad/s

and

ω 44 4 13 7

34 57= = =

v

r

C D

C D

/

/

. . rad/s





The velocity of point G3

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


a a a B 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 r B Ar

B A B Ar

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


a r a r B At

B A B At

B A2 2 2 22 2 0 / / / / = × ⇒ = =α α

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


a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

C D in4 4 4 44 4 4

2 2 24 57 3 62 66 / / / / . . /sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C D

t

C D C D

t

C D C Dto

4 4 4 44 4 / / / / / ( )= × ⇒ = ⋅ ⊥α α

Solve Eq. (2) graphically for the accelerations.



- 100 -


aC Bt

3 338 / = in/s2

aC Dt

4 4128 / = in/s2

Then,

α 33 3 38

219= = =

a

r

C Bt

C B

/

/ rad/s2

α 44 4 128

342 67= = =

a

r

C Dt

C D

/

/ . rad/s2


aC Bt

3 3 / . This direction is clearly counterclockwise.

To determine the direction ofα

4, determine the direction thatr C D/ must be rotated to be parallel to

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 r G Bt

G B G Bt

G B in3 3 3 33 3

219 1 19 / / / / /sec= × ⇒ = ⋅ = ⋅ =α α

a r a r G Br

G B G Br

G B in3 3 3 33 3 3

2 2 25 7 1 32 49 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


aG3 116= in /s2

Problem 2.19

If ωω ω ω 2 = 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"

ω 2

B

D

E

70˚2

3

4



- 101 -

Position Analysis


Velocity Analysis:

v v v B 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

ω 33 3 77 5

1 7544 29= = =

v

r

C B

C B

/

/

..

. rad/s

and

ω 4 4 4 712 35 5= = =v

r C DC D

/ /

. rad/s





The velocity of point E3

v v v r r E D E D D E D E 4 4 4 4

4 4 35 5 0 8 28 4= + = × = ⋅ = ⋅ = / / / . . .ω ω in /s


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


a r a r B At

B A B At

B A2 2 2 22 2 0 / / / / = × ⇒ = =α α

a r a r C Br

C B C Br

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


a r a r r C Bt

C B C Bt

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

to

a r a r C Dr

C D C Dr

C D in4 4 4 44 4 4

2 2 235 5 2 2520 5 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C Dt

C D C Dt

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

Solve Eq. (2) graphically with acceleration.



- 103 -


aC Bt

3 33500 / = in/s2

aC Dt

4 410 900 / ,= in/s2

Then,

α 33 3 3500

1 752000= = =

a

r

C Bt

C B

/

/ .rad/s2

α 44 4 10 900

25450= = =

a

r

C Dt

C D

/

/

,rad/s2

To determine the direction of α 3 , determine the direction that r C B/ must be rotated to be parallel toaC B

t 3 3 / . This direction is clearly counterclockwise.


aC Dt


Determine the acceleration of point E4

a a a a a E D E D E Dr

E Dt

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



- 104 -

a r a r E Dt

E D E Dt

E D in4 4 4 44 4

25450 0 8 4360 / / / / . / sec= × ⇒ = ⋅ = ⋅ =α α

a r a r E Dr

E D E Dr

E D in4 4 4 44 4 4

2 2 235 5 0 8 1008 2 / / / / . . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a E 4

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 v B B B A3 2 2 2

= = /

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

v v D D5 3=

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

Now,

v r v r r B 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

vC 3 is in the vertical direction. Then,

v r v r r C 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 s D3 18 7= . /



- 105 -

2

4

5

A

B

C

D

E

3

6

b3

c3

d3

o

10 in/sec

Velocity Polygon

e5

d5

Now,

v E 5 in horizontal direction

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


v v in s E E 5 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

ω 2

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 v A A A E 3 2 2 2= = /

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

Find vC 3 by image.

v vC C 5 3=

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

Now,

v B3 in horizontal direction

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

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




- 107 -

E

A

B

C

D

F

2

3

4

6

Slider Line

Velocity Polygon

2.5 in/s

a3

o

f 6b3

c3c5

d5

v B3 3 29=

. in/s


v vC C 5 3 6 78= = . in/ s

Now,

v r v r r D F D F D F D F D F to6 6 6 66 6 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

v r v r r D C D C D C D C D C to5 5 5 55 5 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω


v v v D D D F 5 6 6 6 6 78= = = / . in/s

or

ω 66 6

6 6

6 781 5

4 52= = =v

r D F

D F

/

/

..

. rad/sCCW



- 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,

ωω ω ω

2 = 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"

B A

E

C

D

F

G

H

I

J H

F C D

K L

2

3

4

6

7

85

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 r C 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 C 3 2=

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

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

v vF F 5 3=

v v E E 6 2=

So,



- 109 -

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

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

v v H H 8 3=

So,

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

c3

d3

f 3

h3

e2

g6

i8

j8

AB

CD

E

G

FH

I

J

2 4

3

6

5

7

8

10 in/s

Velocity Polygon

o



- 110 -

Now,

v r C C A3 30 74= ⊥. ( ) / in / s to

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

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


v D4 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 of point E.

vF 5 30 5= . in /s

v H 3 30 3= . in /s

and

v E 6 3 80= . in/s

Now,

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

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


vG6 37 8= . in/s

Now,

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

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

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

v J 8 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

5

6

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 and

of 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 in

radius 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 r B B A B A2 2 2 2= = × / / ω

v v B 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 E 3 4 4 4 4 4 4= = + = / /

where,

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

v r C B C B to CB3 3 3 / / ( )= × ⊥ω

v r C E C E to CE 4 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 v D D5 3=

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

and

v vF F 5 6=

where

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

and vF 6 is along the slide direction.. Then the velocity of F5 (and F6) can be found using thevelocity polygon. From the polygon,

vF 6 = 43.33 in/sec

vC B in3 3 26 6 / . / sec=

vF D in5 5 18 54 / . / sec=

ω 33 3 26 6

22 41 187= = =

v

r C B

C Brad

/

/

.

. . / sec

ω 55 5 18 54

21 50 862= = =

v

r F D

F Drad

/

/

..

. / sec

To determine the direction for ω 3, determine the direction that rC B/ must be rotated to be in the

direction of vC B3 3 / . From the polygon, this direction is CCW.

To determine the direction for ω 5, determine the direction that rF D/ must be rotated to be in the

direction 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 of point 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 v A A3 2=

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

v v B B3 5=

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

Now,

v A m3 15= /sec in vertical direction

vC 3 in horizontal direction

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


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

Now,



- 115 -

3

4

5

C

B

D

2A

6

a3

c3o

10 m/sec

Velocity Polygon

b3b5b

d5

v D5 along the inclined path

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


v D m6 12 31= . / sec

Also,

v D B m5 5 16 61 / . / sec=

or

ω 55 5 16 605

3 74 488= = =

vr D 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

ω 2

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 v A A A E 3 2 2 2= = /

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

v vC C 5 3=

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

Now,

v B3 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 r A E A E A E A E A E in to2 2 2 2 2 22 2 42 0 55 23 1 / / / / / . . / sec ( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω

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


v B in4 17 76= . / sec


vC in3 18 615= . / sec

Now,

v D5 in vertical direction

v r v r r D C D C D C D C D C to5 5 5 5 5 55 5 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω



- 118 -


v D in5 5 63= . / sec

Problem 2.26

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

.

A

B

C

D

2

4

5 E

6

3 X

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 and

D.

Velocity Analysis:

v v A A4 3=

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

v v D D3 2=

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



- 119 -

v v v B C B C 5 5 5 5= + /

Now,

v A ft 4 1 0= . / sec in horizontal direction

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

v r v r r D E D E D E D E D E to2 2 2 22 2 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω


From the polygon, using velocity image,

vC A ft 3 3 1 28 / . / sec=

and,

ω 33 3 1 28

2 30 56= = =

v

r C A

C A

/

/

..

. rad/s

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

Now,

v B6 is horizontal direction

v r v r r B C B C B C B C B C to5 5 5 55 5 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω



- 120 -


v B ft 6 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 v A A3 2=

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

v v B B4 3=

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

v v E E 5 4=

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



- 121 -

v vG G6 5=

Now,

v A cm2 10= /sec ( ) / ⊥ to A C r

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

v r v r r B F B F B F B F B F to4 4 4 44 4 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

From the polygon,

v B cm4 6 6= . / sec


v E cm4 3 12= . / sec

Now,

vG6 is horizontal direction

v r v r r G E G E G E G E G E to5 5 5 55 5 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

For the velocity image

draw a line ⊥ to C E r / at e

draw a line ⊥ to C E r / at g

and find the point “c”

From the velocity polygon

v B 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

v 2A

27˚

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

B

C 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 v A A3 2=

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

v v D D4 3=

v vC C 5 3=

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

v v E E 6 5=

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

Now,

v A in2 10= /sec



- 123 -

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

From the polygon,

v D A in3 3 9 1 / . / sec=


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

vC A in3 3 4 32 / . / sec=

Now,

v r v r r E C E C E C E C E C to5 5 5 55 5 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

v r v r r E F E F E F E F E F to6 6 6 66 6 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

from the velocity polygon

v E in6 2 75= . / sec

and

ω 6 6 6 2 752 1 375= = =vr E F E F

/ /

. . rad/s

To determine the direction of ω 6 , determine the direction that r E F / must be rotated to be parallel to

v E F 6 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 pointF

6.

D 2

3

4

5

6

C

B

E

F

ω2

α2 ,

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 v B 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 v E E 5 3=

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

Now,

v r v r r B 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 r C 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 r C 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

ω 33 3

3 3

6 592

3 29= = =v

r C B

C B

/

/

. . rad/sCCW

Also,

vC D4 4 8 19 / .= in/s

or

ω 44 4

4 4

8 190 95

8 62= = =v

r C D

C D

/

/

.

. . rad/sCW

And,

v E 5 5 09= . in/s

Now,

vF 5 in horizontal direction

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


vF E 5 5 3 97 / .= in /s

or

ω 55 5

5 5

3 972 5

1 59= = =v

r F E

F E

/

/

..

. rad/s CCW

Also,

vF 6 3 79= . in/s


a a a B 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/s


f'5

1 aF5 /E5

t

1 aF5

1a E5

D

2

3

4

5

C

B

EF

A

1a F5 /E5

r

1aC3 /B3

t

1aC3 /B3r

1 aC4/D4t

1aC4 /D4r

1aB2 /A2r

1aB2 /A2t

c3

a a E E 5 3=

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

a a a aF E F E r

F E t

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

Now,

a r a r B Ar

B A B Ar

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



- 127 -


a r a r r B At

B A B At

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 r C Br

C B C Br

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 - r C B3 3/

a r a r r C Bt

C B C Bt

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

a r a r C Dr

C D C Dr

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 - r C D4 4/

a r a r r C Dt

C D C Dt

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,

a E 5 252 0= . in /s2

Now,

aF 5 in horizontal direction

a r a r F E r

F E F E r

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

in the direction of - r F E 5 5/

a r a r r F E t

F E F E t

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


aF 6 152 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

B

D

E

F

3

2

4

5

6

Velocity Analysis:

ω π

2 130 130260

13 614= = =rpm rad / s.

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

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

v v E E 5 4=

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

Now,

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

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


v D C 3 3 41 3 / .= f t / s

or



- 129 -

20 ft/s

Velocity Scale

c3

d3

o

e4

f 5

1 aC2/ B2r

1 aD3 /C3

r

1aD3 /C3

t1aD4 /A4

r

1aD4 /A4

t

d3'

c3'

o'

200 ft/s

Acceleration Scale

2

a4'

60˚

A

C

BD

E

F

3

2

4

5

6

e4'

f 5'

1aF5 /E5t

1aF5 /E5r



- 130 -

ω 33 3

3 3

41 33 75

11 0= = =v

r D C

D C

/

/

..

. rad/sCW

Also,

v D A4 4 37 19 / .= f t / s

or

ω 44 4

4 4

37 194 5

8 264= = =v

r D A

D ACW

/

/

.

. . rad/s

And,

v E 5 35 95= . f t / s

Now,

vF 5 in horizontal direction

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


vF E 5 5 12 21 / .= f t / s

or

ω 55 5

5 5

12 2111 1

1 1= = =v

r F E

F E

/

/

..

. rad/sCCW


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

a a a a a a D Ar

D At

C Br

C Bt

D C r

D C t

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

a a E E 5 4=

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

a a a aF E F E r F E t

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

Now,

a r a r C Br

C B C Br

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

in the direction of - rC B/



- 131 -

a r a r C Bt

C B C Bt

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

a r a r D C r

D C D C r

D C 3 3 3 33 3 32 211 0 3 75 453 8 / / / / . . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω f t / s2

in the direction of - rD C/

a r a r r D C t

D C D C t

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

a r a r D Ar

D A D Ar

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

in the direction of - rD A/

a r a r r D At

D A D At

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,

a E 5 308 0= . f t / s2

Now,

aF 5 is in the horizontal direction

a r a r F E r

F E F E r

F E 5 5 5 55 5 52 21 1 11 1 13 4 / / / / . . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω f t / s2

in the direction of - rF E/

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


aF 6 83 4= . f t / s2

Also,

aF E t

5 5325 2 / .= f t / s2

or

α 55 5

5 5

325 211 1

29 3= = =a

r F E t

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

X EA = 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 π

2 30200030 209 44=

⋅

=

⋅

=

n. rad/s

v v v r A A E A A E 2 2 2 3 2= = = × / / ω

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

v vC C 3 5=



- 133 -

10000 in/s


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 /C 5r

1 aD5 /C 5

t

1 aD5

1aB3 /A3t



- 134 -

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

Now,

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

v B3 in the direction of r B E /

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


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

Also,

v B A3 3 109 3 / .= in/s

or

ω 33 3 109 3

336 43= = =

v

r B A

B A

/

/

. . rad/s

To determine the direction of ω 3, determine the direction that r B A/ must be rotated to be parallel to

v B A3 3 / . This direction is clearly clockwise.

Also,

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

Now,

v v D D5 6= in the direction of r D E /

v r v r r D C D C D C D C D C to5 5 5 55 5 / / / / / ( )= × ⇒ = ⊥ω ω


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

Also,

v D C 5 5 135 / = in/s

or

ω 55 5 135

2 5552 9= = =

v

r D C

D C

/

/ . . rad/s



- 135 -

To determine the direction of ω 5, determine the direction that r D C / must be rotated to be parallel to

v D C 5 5 / . This direction is clearly clockwise.


a a a a A A A E

r

A E

t

2 3 2 2 2 2= = +

/ /

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

a a a a a B A E r

A E t

B Ar

B At

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

a aC C 3 5=

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

a a a a a D D C D C r

D C t

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

Now,

a B3 in the direction of ± r B E /

a r a r A E r

A E A E r

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

in the direction of - rA E/

a r a r A E t

A E A E t

A E in2 2 2 22 2

20 1 0 / / / / /sec= × ⇒ = ⋅ = ⋅ =α α

a r a r B Ar B A B Ar B A3 3 3 33 3 3 2 236 43 3 3981 / / / / .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω in /s2

in the direction of - rB A/

a r a r r B At

B A B At

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


a B3 14710= in/s2

Also,

a B At

3 338460 / = in/s2

or

α 33 3 38460

312820= = =

a

r B At

B A

/

/ rad/s2



- 136 -

To determine the direction of α 3 , determine the direction that r B A/ must be rotated to be parallel to

a B At


Also,

a aC C in s3 5 39 3002

= = , /

Now,

a D5 in the direction of - r D E /

a r a r D C r

D C D C r

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

in the direction of - r D C /

a r a r r D C t

D C D C t

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


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

Also,

a D C t

5 526 300 / ,= in /s2

or

α 5 5 5 263002 55 10 300= = =

a

r D C t

D C

/

/ . , rad/s2

To determine the direction of α 5 , determine the direction that rD C/ must be rotated to be parallel to

a D C t


Problem 2.32

In the mechanism shown, ωω ω ω 2 = 4 rad/s CCW (constant). Write the appropriate vector equations,

solve them using vector polygons, and

a) Determine vE3, vE4, and ωω ω ω 3.

b) Determine aE3, aE4, and αα α α 3.

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 v B 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 v B 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

v E 3 2 522 93 9= ∠ °. . / m s (by image)



- 138 -

Now,

ω 33 3 2 400

0 83 0= = =

vC B

C Br /

/

..

. rad/s

ω 4

4 4 0 650 8 0 81= = =

vC D

C Dr

/

/

.. . rad/s

To determine the direction of ω 3, determine the direction that rC B/ must be rotated to be parallel to


To determine the direction of ω 4 , determine the direction that rC D/ must be rotated to be parallel to

vC 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,

v E 3 2 522=

. m / s

v E 4 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 A

r B A

B A

r B A

ce

2 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

a B Ar

2 24 0 0 5 8 02

/ ( . ) ( . ) .= = m / s from B to A2



- 139 -

1 m/s

Velocity Scale

C

B

D

2

34

E

b

oo'

1aB 2 / A2

r

1aC

3/ B

3

t

c3

3

e3

e4

4 m/s

AccelerationScale

d'4

1aE4

2

A

e'3

e'4

c'3

1aC3 / B3

r 1aE3

b'3

1aC

4/D

4

r1a

C4 /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 a

r 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 D

r

C D4 4 4 44 0 812 0 650 0 528 / / ( . )( . ) .= × = =

[ ]ω

from C to D

a r r C 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 r C 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 B

t 3 3 16 13 / .

=m / s

2

Then,

α 33 3 16 13

0 820 16= = =

a

r C Bt

C B

/

/

.

. . rad/s2


aC Bt


and

a E 315 84= . m / s2

a E 432 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 v A A3 2=

v v B B3 4=

v v v v v B A B A B B C 3 3 3 3 4 4 4= + = = / / (1)

Now,

v A2 10= in / s in the horizontal direction

v v r v r r B B B C B B C B C 3 4 4 4 44 4= = × ⇒ = ⋅ ⊥ω ω / / / ( )to

v r v r r B A B A B 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/C4

r

b3'

o'

a'2a'3c'4

1a B3/A3t

1a B4 /C 4

t

b4'

d4'


v B A3 3 6 64 / .= in/s

or



- 143 -

ω 33 3 6 64

32 21= = =

v

r B A

B A

/

/

. . rad/s

To determine the direction of ω 3, determine the direction that r B A/ must be rotated to be parallel to

v B A3 3 / . This direction is clearly counterclockwise.

Also,

v B4 9 70= . in/s

and

ω 44 4 9 70

1 4146 86= = =

v

r B C

B C

/

/

..

. rad/s

To determine the direction of ω 4 , determine the direction that r B C / must be rotated to be parallel to

v B C 4 4 / . This direction is clearly clockwise.

Also,

v r D D C 4 6 77= ⊥. ( ) / in / s to

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


a a A A2 3=

a a a a a B B A B A B C 3 4 3 3 3 4 4= = + = / /

a a a a a B C r

B C t

A B Ar

B At

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

Now,

a A3 0=

a r a r B C r

B C B C r

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

in the direction of - r B C /

a r a r r B C t

B C B C t

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

a r a r B Ar

B A B Ar

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

in the direction of - r B A/

a r a r r B At

B A B At

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



- 144 -


a D4 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 E 3, andlocate the point in link 3 that has zero velocity.

AC

B D

E

F 2

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 v A A3 2=

v v B B3 4=

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

v v D D5 4=

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

Now,

v A2 10= in/s in the horizontal direction



- 145 -

v v r v r r B B B C B B C B C 3 4 44 4= = × ⇒ = ⋅ ⊥ω ω / / / ( )to

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


v B A3 3 5 97 / .= in/s

ω 33 3 5 97

31 99= = =

v

r B A

B A

/

/

. . rad/s

Also,

v B in s4 9 42= . /

or

ω 4

4 4

4 4

9 421 45 6 50= = =

v

r

B C

B C

/

/

.

. . rad/sCW

Now,

v r D D C in s to4 6 50= ⊥. / ( ) /

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

v v v r v r r E E E F E F E F E F E F 5 6 6 6 6 66 6= = = × ⇒ = ⊥ / / / / / ( )ω ω to


v E 6 5 76= . in/s

or

ω 66 6 5 76

1 453 97= = =

v

r E F

E F

/

/

.

. . rad/s


a a A A2 3=

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

a a a a B C r

B C t

B Ar

B At

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

a a D D5 4=

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



- 146 -

a a a a a E F r

E F t

D E Dr

E Dt

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

Now,

A C

B

D

E

F2

34

6

o'

25 in/s


2

1aB3 /A3

r

1aB3/A3

t

1 aB4 /C 4

r

b3’ b4b’,

d4'

1aE5/ D 5r

1aE5/D 5

t

1aE6 /F6

r

e5'

1aB4/ C4t

1 aE6 /F6

t

O2.5 in/s

Velocity Polygon

a2

a3

,

b3 b4b,

d4 d5,

e5

3

o

d3

e3

5D

or

a r a r B C r

B C B C r

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

in the direction of - r B C /

a r a r r B C t

B C B C t

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



- 147 -

a r a r B Ar

B A B Ar

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

in the direction of - rB A/

a r a r r B At

B A B At

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


a B At

3 365 3 / .= in/s2

or

α 323 3 65 3

321 8= = =

a

r B At

B Arad s CW

/

/

. . /

And,

a D3 86 6= . in/s2

Also,

a r a r E Dr

E D E Dr

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

in the direction of - r E D/

a r a r r E Dt

E D E Dt

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

a r a r E F r

E F E F r

E F 6 6 6 66 6 62

25 76 1 45 48 1 / / / / . . .= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω in/s2

in the direction of - r E F /

a r a r r E F t

E F E F t

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

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

a E F t

6 638 2 / .= in/s2

or

α 66 6 38 2

1 4526 4= = =

a

r E F t

E F

/

/

..

. rad / s CCW2

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

v D3 10 2= . in/s



- 148 -

v E 3 12 7= . in/s

The point in link 3 with zero velocity is shown on position diagram. The point is found by finding

the 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

Aa

aB

70˚

5˚

145˚

Aa = 1500 in/s2

= 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 Scale

2

o'

a'

b'

c'

A

B

C

1 in

Problem 2.36

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

ω α 2 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")

2

34

E

55˚

X

Y

AB = DE = 1.0 in BC = 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 v B 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 r B 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 r C D C D C D C D C Dto4 4 4 4 4 44 4 / / / / / ( )= × ⇒ = ⋅ ⊥ω ω

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


v E in4 5 4042= . / sec

c3

b3

o

2 in/sec

Velocity Polygon

e4

A

C

B

D

E



- 151 -

Problem 2.37

In the mechanism shown, find ωω ω ω 6 and αα α α 3. 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 v A A3 2=

v v B B3 4=

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

v v D D5 4=

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

Now,

v A in2 10= /sec in the horizontal direction

v v r v r r B B B C B B C B C to3 4 4 4 4 4 4 4 44 4= = × ⇒ = ⋅ ⊥ω ω / / / ( )

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


v B A in3 3 6 282 / . / sec=

or

ω 33 3

3 3

6 2823

2 094= = =v

r

B A

B A

rad /

/

. . / sec



- 152 -

AC

B

D

E

F

3

45

62

d4 d5,

2.5 in/sec

Velocity Polygon

o

b3 b4b,

a2 a3,

e5

25 in/sec


2o'1a B3 /A3

r

1aB3 / A3

t

1aB4/C4

t

1aB4/C4r

b3’ b4b’,

d3'

Also,

v B in4 9 551= . / sec

or

ω 44 4

4 4

9 5511 45

6 587= = =v

r B C

B C rad CW

/

/

..

. / sec

Now,

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



- 153 -

v r v r r E 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 r E E E F E F E F E F E F to5 6 6 6 6 6 6 6 6 6 6 66 6= = = × ⇒ = ⊥ / / / / / ( )ω ω


v E in6 6 132= . / sec

or

ω 66 6

6 6

6 1321 45

4 229= = =v

r E F

E F rad CW

/

/

..

. / sec


a a A A2 3=

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

a a a a B C r

B C t

B Ar

B At

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

Now,

a r a r B C r

B C B C r

B C in4 4 4 4 4 4 4 44 4 4

2 2 26 587 1 45 62 913 / / / / . . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω

in the direction ofrB C4 4/

a r a r B C t

B C B C t

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

a r a r B Ar

B A B Ar

B A in3 3 3 3 3 33 3 3

2 2 22 094 3 13 155 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω

in the direction of rB A/

a r a r r B At

B A B At

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


a B At in

3 368 568 2

/ . / sec=

or

α 323 3

3 3

68 5683

22 856= = =a

r B At

B Arad CW

/

/

. . / sec

Also,

a D in3 91 390 2= . / sec



- 154 -

Problem 2.38

In the mechanism shown, ωω ω ω 2 = 1 rad/s (CCW) and αα α α 2 = 0 rad/s2. Find ωω ω ω 5, αα α α 5, vE6, aE6 for the

position given. Also find the point in link 5 that has zero acceleration for the position given.

A

C

B

D

2

3

4 E 5

6

30˚

AD = 1 m AB = 0.5 m BC = 0.8 mCD = 0.8 m BE = 0.67 m

0.52 m

Velocity Analysis

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

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

Now,

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


v E B m5 5 0 47313 / . / sec=

or

ω 55 5

5 5

0 473130 67

0 706= = =v

r E B

E Brad CCW

/

/

..

. / sec

Also,



- 155 -

A

C

B

D

2

3

4

E5

6

o'

0.1 m/sec


1 aE5

1aB2/A2

r

1a E5/B5

r

1aE5 /B 5

tb3b'

e5'

2.2°

27.8°

O'

0.1 m/sec

Velocity Polygon

b5

o

e 5

v E m6 0 441= . / sec


a a a a B B B B A2 3 5 2 2= = =

/

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

a a a a a E B Ar

B At

E Br

E Bt

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

Now,



- 156 -

a E 5 in horizontal direction

a r a r B Ar

B A B Ar

B A m2 2 2 22 2 2

2 2 21 0 5 0 5 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω

in the direction opposite to rB A/

a r a r B At

B A B At

B A m2 2 2 22 2

20 1 0 / / / / /sec= × ⇒ = ⋅ = ⋅ =α α

a r a r E Br

E B E Br

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 r E Bt

E B E Bt

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


a E m6 0 042 2= . / sec

Also,

a E Bt m

5 50 42 2

/ . / sec=

or

α 525 5

5 5

0 420 67

0 626= = =a

r 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 the

velocity and acceleration of point D7 are given. Find ωω ω ω 7 and αα α α 7 for the position given. Also find

the 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 v

v v v E E E D E D

7 87 7 7 7

=

= + /

and because points on the same link are involved,

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

From the velocity polygon:

v E in8 2 884= . / sec

in the direction shown, and

v E D in7 7 3 405 / . / sec=


ω 77 7 3 405

1 502 27= = =

v

r E D

E Drad

/

/

..

. / sec



- 158 -

o'

D

E

7

7d'

6

G 8

1a E7 /D7r

1a E7 /D7

t

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 ω 7 to get v E D7 7 / . From the

polygon, the direction must be counter clockwise. Therefore,

ω 7 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 a D E D F D E Dr

E D F 7 7 7 8 7 7 7 7 7 87+ = = + + × = / / / α

where

a v

r E Dr E D

E Din

7 77 7

2 223 405

1 507 729 /

/

/

..

. / sec= = =



- 159 -

in the direction opposite rE D/ .

From the polygon,

a r E Dt

E D7 7 7 77 0 / / = × =α

Therefore,

α 727 7 42 59

1 5028 40= = =

a

r E Dt

E Drad

/

/

..

. / sec

The direction can be found by rotating rE D/ 90˚ in the direction of ω 7 to get a E Dt

7 7 / . From the

polygon, 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

, ωω ω ω 3, and ωω ω ω 4.

b) Determine aC4, aE4

, αα α α 3, and αα α α 4.

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 v B 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

ω 33 3 11 16

52 232= = =

v

r C B

C B

/

/

. . rad/s

and

ω 44 4 6 57

32 19= = =

v

r C D

C D

/

/

. . rad/s





The velocity of point E3

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

from the polygon

v E 3 13 5= . in/s

vC 4 6 57= . in/s


a a a B 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 r B Ar

B A B Ar

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


a r a r B At B A B At B A2 2 2 22 2 0 / / / / = × ⇒ = =α α

a r a r C Br

C B C Br

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


a r a r r C Bt

C B C Bt

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



- 162 -

a r a r C Dr

C D C Dr

C D in4 4 4 44 4 4

2 2 22 19 3 14 39 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a r a r r C Dt

C D C Dt

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

4 446 71 / .= in/s2

Then,

α 33 3 0

20= = =

a

r C Bt

C B

/

/ rad/s2

α 44 4 46 71

315 57= = =

a

r C Dt

C D

/

/

. . rad/s2


aC Dt


Determine the acceleration of point E3

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

E Bt

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



- 163 -

a r a r E Bt

E B E Bt

E B in3 3 3 33 3

20 1 0 / / / / /sec= × ⇒ = ⋅ = ⋅ =α α

a r a r E Br

E B E Br

E B in3 3 3 33 3 3

2 2 22 232 5 24 9 / / / / . . / sec= × ×( ) ⇒ = ⋅ = ⋅ =ω ω ω


a E 3 34 29= . in/s2

aC 4 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 the

velocity and acceleration of points D7 and F 8 are given. Find ωω ω ω 8 and αα α α 7 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 v

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

and because points on the same link are involved,



- 164 -

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

From the velocity polygon:

v E F 8 8 0 / =

so

ω 8 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 a

a 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



a v

r

a v

r

E Dr E D

E D

E F r E F

E F

7 7

7 7

7 7

8 8

8 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 r E Dt

E D7 7 7 77 0 / / = × =α

Therefore,

α 7 0=

chapter 2. graphical position velocity

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