chpt 2 - part 1

- 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, 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= = =arC 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

- 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



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= = =arC 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= = =arC 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= = =arC 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= = =arC 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

aC 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= = =arC 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

= = =

= = =

arar

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= = =arC 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

aB

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

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

v

vrC BC B

C Bm s3 3

3 312 82 12 820 295

43 453//

/. / .

..= � = = =ω rad / s

Using velocity image,

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= = =ar

tC B

C B

/

/

..

rad / s CW2

Also,

α4

4 4 41 010 301

136= = =artC 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= = =arC 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,

v r v r rB A B A B A B A B A2 2 2 2 100 1 100/ / / / /( )= × ⇒ = ⋅ = ⋅ = ⊥ω ω in / s to



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

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= = =arC Dt

C D

/

/

, rad / s2


aC Bt

3 3/ . This direction is clearly counterclockwise.


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

Solve Eq. (2) graphically with a velocity polygon. From the polygon, using velocity image,

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

Documents

chpt 2 - part 1