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COSMOS: Complete Online Solutions Manual Organization System Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. Chapter 15, Solution 1. Angular coordinate: ( ) 2 3 8 6 2 radians t t θ = Angular velocity: ( ) 2 24 12 2 rad/s d t t dt θ ω = = Angular acceleration: 2 48 12 rad/s d t dt ω α = = (a) When the angular acceleration is zero. 48 12 0 t = 0.250 s t = (b) Angular coordinate and angular velocity at t = 0.250 s. ( )( ) ( )( ) 3 2 8 0.250 6 0.250 2 θ = 18.25 radians θ =− ( )( ) ( )( ) 2 24 0.250 12 0.250 2 ω = 22.5 rad/s ω =

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Transcript of solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 15

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 1.

Angular coordinate: ( )238 6 2 radianst tθ = − −

Angular velocity: ( )224 12 2 rad/s

dt t

dt

θω = = − −

dt

dt

ωα = = −

(a) When the angular acceleration is zero.

48 12 0t − = 0.250 st =

(b) Angular coordinate and angular velocity at t = 0.250 s.

( )( ) ( )( )3 28 0.250 6 0.250 2θ = − − 18.25 radiansθ = −

( )( ) ( )( )224 0.250 12 0.250 2ω = − − 22.5 rad/sω =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 2.

3

0.5 cos4t

e tπθ π−=

( )3 30.5 3 cos4 4 sin 4

t tde t e t

dt

π πθω π π π π− −= = − −

( )( )

2 3 2 3 2 3 2 3

2 3 2 3

0.5 9 cos4 12 sin 4 12 sin 4 16 cos4

0.5 24 sin 4 7 cos4

t t t t

t t

de t e t e t e t

dt

e t e t

π π π π

π π

ωα π π π π π π π π

π π π π

− − − −

− −

= = + + −

= −

(a) 0,t = ( )0.5θ = 0.500 radθ =

( )( )0.5 3 4.71ω π= − = − 4.71 rad/sω = −

( )( )20.5 7 34.5α π= − = − 2

( ) 0.125 s, cos4 cos 0, sin 4 sin 12 2

b t t tπ ππ π= = = = =

30.30786

te

π− =

( )( )( )0.5 0.30786 0 0θ = = 0θ =

( )( )( )0.5 0.30786 4 1.93437ω π= − = − 1.934 rad/sω = −

( )( )( )20.5 0.30786 24 36.461α π= = 2

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 3.

7 /6

0sin 4

te t

πθ θ π−=

7 /6 7 /6

0

7sin 4 4 cos4

6

t tde t e t

dt

π πθ πω θ π π π− − = = − +

2 2 2

7 /6 7 /6 7 /6 2 7 /6

0

7 /6 2 2

0

49 28 28sin 4 cos4 cos4 16 sin 4

36 6 6

49 2816 sin 4 cos4

36 3

t t t t

t

de t e t e t e t

dt

e t t

π π π π

π

ω π π πα θ π π π π π

θ π π π π

− − − −

= = − − −

= − − +

( )a 0

0.4 rad,θ = 0.125 st =

7 (0.125)/60.63245, 4 , sin 1, cos 0

2 2 2e t

π π π ππ− = = = =

( )( )( )0.4 0.63245 1 0.25298 radiansθ = = 0.253 radθ =

( )( ) ( )70.4 0.63245 1 0.92722 rad/s

6

πω = − = −

( )( ) ( )2 2490.4 0.63245 16 1 36.551 rad/s

36α π = − − = −

( )b ,t = ∞

7 /60

te

π− = 0θ =

0ω =

0α =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 4.

Angular coordinate: 1800 rev = 3600 radiansπθ =

Initial angular velocity: 0

6000 rpm = 200 rad/sω π=

Angular acceleration: constantd d

dt d

ω ωα ωθ

= = =

d dα θ ω ω=

0

0

0d d

θωα θ ω ω=∫ ∫

2

0

1

2αθ ω= −

( )

( )( )22

20200

πωαθ π

= − = − = −

(a) Time required to coast to rest.

0

tω ω α= +

00 200

17.4533t

ω ω πα− −= =

− 36.0 st =

(b) Time to execute the first 900 revolutions.

900 rev = 1800 radiansθ π=

2

0

1

2t tθ ω α= +

( ) 211800 200 17.4533

2t tπ π= −

272 648 0t t− + =

( ) ( )( )2

72 72 4 648

2t

± −= 10.54 st =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 5.

( )( )1 0 1

2400 22400 rpm 80 rad/s, 0, 4 s

60t

πω π ω= = = = =

( )a 21

1 0

4t t

t

ω πω ω α α α π= + = = = =

( ) ( )2 2

1 0

1 1 1600 20 4 160 rad 80 rev

2 2 2t t

πθ ω α π ππ

= + = + = = =

180 revθ =

(b) 1 2 2 1

80 rad/s, 0, 40 st tω π ω= = − =

( ) 22 1

2 1 2 1

2 1

40t t

t t

ω ω πω ω α α π− −= + − = = = −−

( ) ( ) ( )( ) ( )( )2 2

2 1 1 2 1 2 1

1 180 40 2 40

2 2t t t tθ θ ω α π π− = − + − = + −

1600

πππ

= = = 2 1800 revθ θ− =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 6.

Angular acceleration: 0.230

t de

dt

ωα −= =

Angular velocity: 0 0

t

dtω ω α= + ∫

0.2

00 30

t te dt

−= + ∫

0.2

0

30

0.2

tt

e− = −

( )0.2150 1

t de

dt

θ−= − =

When t = 0.5 s, ( )( )( )0.2 0.5

150 1 eω −= −

Angular coordinate: 0 0

t

dtθ θ ω= + ∫

( )0.2

00 150 1

t te dt

−= + −∫

0.2

0

150150

0.2

tt

t e− = − −

( )0.2150 750 1

tt e

−= − −

When t = 0.5 s, ( )( ) ( )( )( )0.2 0.5150 0.5 750 1 eθ −= − −

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 7.

( ) 0.5 , 0.5 0.5d

a d dd

ωα ω ω ω ω θθ

= − = − = −

0

30 0Integrating, 0.5 30 0.5d d

θω θ θ= − − = −∫ ∫

60

= = =2 9.55 revθ =

( ) 0.5 2d d

b dtdt

ω ωωω

= − = −

Integrating, 0

0 302

t ddt

ωω

= −∫ ∫

0

2 ln30

t = − = ∞ t = ∞

( )c ( )( )0.02 30 0.6 rad/sω = =

0.6

0 302

t ddt

ωω

= −∫ ∫

0.6

2 ln 2 ln 5030

t = − = 7.82 st =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 8.

dk k d k d

d

ωα θ ω θ ω ω θ θθ

= − = − = −

Integrating, 0 6

12 0d k dω ω θ θ= −∫ ∫

2 2

12 60 0

2 2k

− = − −

( )a

2

2

2

129 s

6k

−= = 2

9.00 sk−=

3

12 0d k d

ω ω ω θ θ= −∫ ∫

2 2 2

12 39 0

2 2 2

ω − = − −

( )b ( )( )22 2 2 212 9 3 63 rad /sω = − = 7.94 rad/sω =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 9.

( ) ( ) ( )/5 in. 31.2 in. 12 in.

AO= + +r i j k

( ) ( )/5 in. 15.6 in.

B O= +r i j

( ) ( ) ( )2 2 25 31.2 12 33.8 in.

OAl = + + =

Angular velocity.

( )/

6.765 31.2 12

33.8AO

OAl

ω= = + +r i j kωωωω

Velocity of point B.

/B B O

= ×v rωωωω

1.0 6.24 2.4 37.44 12 15.6

5 15.6 0

B= = − + −

i j k

v i j k

( ) ( ) ( )37.4 in./s 12.00 in./s 15.60 in./sB

= − + −v i j k

Acceleration of point B.

B B

= ×a vωωωω

1.0 6.24 2.4 126.1 74.26 245.6

37.4 12 15.60

B= = − − +

− −

i j k

a i j k

( ) ( ) ( )2 2 2126.1 in./s 74.3 in./s 246 in./s

B= − − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 10.

( ) ( ) ( )/5 in. 31.2 in. 12 in.

AO= + +r i j k

( ) ( )/5 in. 15.6 in.

B O= +r i j

( ) ( ) ( )2 2 25 31.2 12 33.8 in.

OAl = + + =

Angular velocity.

( )/

3.385 31.2 12

33.8AO

OAl

ω= = + +r i j kωωωω

Velocity of point B.

/B B O

= ×v rωωωω

0.5 3.12 1.2 18.72 6 7.80

5 15.6 0

B= = − + −

i j k

v i j k

( ) ( ) ( )18.72 in./s 6.00 in./s 7.80 in./sB

= − + −v i j k

Angular Acceleration.

( )/

5.075 31.2 12

33.8AO

OAl

α −= = + +r i j kαααα

( ) ( ) ( )2 2 20.75 rad/s 4.68 rad/s 1.8 rad/s= − − −i j kαααα

Acceleration of point B.

/B B O B

= × + ×a r vαααα ωωωω

0.75 4.68 1.8 0.5 3.12 1.2

5 15.6 0 18.72 6 7.8

B= − − − +

− −

i j k i j k

a

28.08 9 11.7 31.536 18.564 61.406= − + − − +i j k i j k

( ) ( ) ( )2 2 23.46 in./s 27.6 in./s 73.1 in./s

B= − − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 11.

( ) ( ) ( ) ( ) ( ) ( )/500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 m

B A= − + = − +r i j k i j k

2 2 20.5 0.225 0.3 0.625 m

ABl = + + =

Angular velocity vector.

( )/

100.5 0.225 0.3

0.625B A

ABl

ω= = − +r i j kωωωω

( ) ( )/300 mm 0.3 m

E B= − = −r k k

Velocity of E.

/8 3.6 4.8 1.08 2.4

0 0 0.3

E E B= × = − = +

i j k

v r i jωωωω

( ) ( )1.080 m/s 2.40 m/sE

= +v i j

Acceleration of E.

8 3.6 4.8 11.52 5.184 23.088

1.08 2.4 0

E E= × = − = − + +

i j k

a v i j kωωωω

( ) ( ) ( )2 2 211.52 m/s 5.18 m/s 23.1m/s

E= − + +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 12.

( ) ( ) ( ) ( ) ( ) ( )/500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 m

B A= − + = − +r i j k i j k

2 2 20.5 0.225 0.3 0.625 m

ABl = + + =

Angular velocity vector.

( )/

100.5 0.225 0.3

0.625B A

ABl

ω= = − +r i j kωωωω

Angular acceleration vector.

( )/

200.5 0.225 0.3

0.625B A

ABl

α −= = − +r i j kαααα

( ) ( ) ( )2 2 216 rad/s 7.2 rad/s 9.6 rad/s= − + −i j k

.Velocity of C /C C B= ×v rωωωω

( ) ( )/500 mm 0.5 m

C B= − = −r i i

8 3.6 4.8 2.4 1.8

0.5 0 0

C= − = − −

i j k

v j k

( ) ( )2.40 m/s 1.800 m/sC

= − −v j k

.Acceleration of C /C C B C= × + ×a r vαααα ωωωω

16 7.2 9.6 8 3.6 4.8

0.5 0 0 0 2.4 1.8

= − − + −− − −

i j k i j k

4.8 3.6 18 14.4 19.2= + + + −j k i j k

18 19.2 15.6= + −i j k

( ) ( ) ( )2 2 218.00 m/s 19.20 m/s 15.60 m/s

C= + −a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 13.

( ) ( ) ( )/200 mm + 120 mm 90 mm

A D= − +r i j k

( ) ( ) ( )2 2 2200 120 90 250 mm

0.8 + 0.48 + 0.36A/D

DA

= = −ri j kλλλλ

( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDA

ω= = − = −i j k i j kωωωω λλλλ

0DA

d

dt

ω= =αααα λλλλ

( ) ( )200 mm = 0.2 mB/A

=r i i

Velocity of corner B.

/60 36 27 5.4 7.2

0.2 0 0

B B A= × = − = −

i j k

v r j kωωωω

( ) ( )5.40 m/s 7.20 m/sB

= −v j k

Acceleration of corner B.

/B B A B

= × + ×a r vαααα ωωωω

0 60 36 27 405 432 324

0 5.4 7.2

= + − = − − +−

i j k

i j k

( ) ( ) ( )2 2 2405 m/s 432 m/s 324 m/s

B= − − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 14.

( ) ( ) ( )/200 mm + 120 mm 90 mm

A D= − +r i j k

( ) ( ) ( )2 2 2200 120 90 250 mm

0.8 + 0.48 + 0.36A/D

DA

= = −ri j kλλλλ

( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDA

ω= = − = −i j k i j kωωωω λλλλ

( )( )600 0.8 + 0.48 + 0.36DA

d

dt

ω= = − − i j kαααα λλλλ

( ) ( ) ( )2 2 2480 rad/s 288 rad/s 216 rad/s= − −i j k

( ) ( )200 mm = 0.2 mB/A

=r i i

Velocity of corner B.

/60 36 27 5.4 7.2

0.2 0 0

B B A= × = − = −

i j k

v r j kωωωω

( ) ( )5.40 m/s 7.20 m/sB

= −v j k

Acceleration of corner B.

/B B A B

= × + ×a r vαααα ωωωω

480 288 216 60 36 27

0.2 0 0 0 5.4 7.2

= − − + −−

i j k i j k

43.2 57.6 405 432 324= − − − −j k i j k

( ) ( ) ( )2 2 2405 m/s 389 m/s 266 m/s

B= − − −a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 15.

993,000,000 mi 491.04 10 ft= ×

6365.24 days 31.557 10 s, 1 rev 2 radπ= × =

Angular velocity.

9

6

31.557 10

πω −= = ××

Velocity of the earth.

( )( )9 9 3491.04 10 199.11 10 97.77 10 ft/sv rω −= = × × = ×

3

66.7 10 mi/hv = ×

Acceleration of the earth.

( )( )22 3 9 3 297.77 10 199.11 10 19.47 10 ft/sa rω − −= = × × = ×

3 2

19.47 10 ft/sa−= ×

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 16.

323 h 56min 23.933 h 86.16 10 s, 1 rev 2 radπ= = × =

6

3

86.16 10

πω −= = ××

66370 km 6.37 10 mR = = ×

, cos sinR Rω φ φ= = +j r i jωωωω

( )( ) ( )6 6

cos

72.925 10 6.37 10 cos 464.53cos m/s

Rω φ

φ φ−

= × = −

= − × × = −

v r k

k k

ωωωω

( ) ( )2 3 2cos cos 33.876 10 cos m/s

pR Rω ω φ ω φ φ−= × = × − = − = − ×a v j i i iωωωω

( ) .a Equator ( )0 cos 1.000φ ϕ= ° =

( )465 m/s= −v k 465 m/sv =

( )3 233.9 10 m/s

−= − ×a i 2

0.0339 m/sa =

( ) .b Philadelphia ( )40 cos 0.76604φ φ= ° =

( )( ) ( )464.52 0.76604 356 m/s= − = −v k k 356 m/sv =

( )( )333.876 10 0.76604

−= − ×a i

( )3 20.273 10 m/s

−= − × i 2

0.0259 m/sa =

( ) .c North Pole ( )90 cos 0φ φ= ° =

0v =

0a =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 17.

300 mm/sB Av v= = 120 mm

Br =

( ) 180 mm/sB At

a a= =

( )a 300

B

B B

B

v

v r

r

ω ω= = = = 2.50 rad/s=ωωωω

( ) ( )2180

B t

B Bt

B

a

a r

r

α α= = = = 2

( )b ( ) ( )( )22 2120 2.5 750 mm/s

B Bna r ω= = =

( ) ( ) ( ) ( )2 2 2 2 2180 750 771mm/s

B B Bt na a a= + = + =

750

tan , 76.5180

β β= = ° 2

771mm/sB

=a 76.5°

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 18.

ω = , 120 mmBr =

( ) ( )( )22 2120 4 1920 mm/s

B B Bna r ω= = =

2

2400 mm/sB

a =

( ) ( )22 2 2 22400 1920 1440 mm/s

B B Bt na a a= − = − = ±

( ) ( )21440

B t

B Bt

B

a

a r

r

α α ±= = = = ±

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 19.

Let Bv and

Ba be the belt speed and acceleration. These are given as 2

12 ft/s and 96 ft/s .B Bv a= =

These are also the speed and tangential acceleration of periphery of each pulley provided no slipping occurs.

(a) Angular velocity and angular acceleration of each pulley.

Pulley A. 8 in. 0.6667 ftAr = =

12

A B

A

A A

v v

r r

ω = = = = 18 rad/sA

=ωωωω

296

A B

A

A A

a a

r r

α = = = = 2

=αααα

Pulley C. 5 in. 0.41667 ftCr = =

12

C B

C

C C

v v

r r

ω = = = = 28.8 rad/sC

=ωωωω

296

C B

C

C C

a a

r r

α = = = = 2

=αααα

(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftC

ρ = =

( ) 296 ft/s

P Bta a= =

( ) ( )22 2

212

345.6 ft/s0.41667

P B

P n

C C

v v

a

ρ ρ= = = =

( ) ( ) ( ) ( )2 2 2 2 296 345.6 358.7 ft/s

P P Pt na a a= + = + =

96tan 15.52

345.6β β= = °

2

359 ft/sP

=a 15.52°

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 20.

1rev radians, 8 in. 0.6667 ft, 5 in. 0.41667 ft

2A Cr rπ= = = = =

2

2 2

500120 0.002 ,

120 0.002 60000

d d dd

d

ω ω ω ω ωα ω ω θθ ω ω

= − = = =− −

Integrating and applying initial condition 0 at 0ω θ= = and noting that θ π= radians at the final state,

( )220 00

500250 ln 60000

60000

dd

ωω πω ω ω θ πω

= − − = =−∫ ∫

( )2

2 60000250 ln 60000 ln 60000 250 ln

60000

ωω π− − − − = − =

2/25060000

60000e

πω −− =

2 /250 2 260000 1 749.26 rad /se

πω − = − =

( )( )2120 0.002 120 0.002 749.26 118.50 rad/sα ω= − = − =

(a) Tangential velocity and acceleration of point B on the belt.

( )( )0.6667 27.373 18.249 ft/sB A Av v r ω= = = =

( )( ) 20.6667 118.50 79.0 ft/s

B A Aa a r α= = = =

2

79.0 ft/sB

a =

(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftC

ρ = =

18.249 ft/sP Bv v= =

( )2 2

218.249799.3 ft/s

0.41667

B

P n

C

v

ρ= = =a

( ) 279.0 ft/s

P Bta= =a

( ) ( )2 2 2799.3 79.0 803.2 ft/s

Pa = + =

79tan , 5.64

799.3β β= = °

2

803 ft/sP

=a 5.64°

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Chapter 15, Solution 21.

Left pulley.

Inner radius r1 = 50 mm

Outer radius r2 = 100 mm

0.6 m/s = 600 mm/sAv =

1

2

100

Av

r

ω = = =

Speed of intermediate belt.

( )( )1 1 150 6 300 mm/sv rω= = =

Right pulley.

Inner radius r3 = 50 mm

Outer radius r4 = 100 mm

1

2

4

100

v

r

ω = = =

(a) Velocity of C.

( )( )3 250 3 150 mm/s

Cv rω= = =

0.1500 m/sC

=v

(b) Acceleration of point B.

( )( )22 2

4 2100 3 900 mm/s

Ba r ω= = =

2

0.900 m/sB

=a

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Chapter 15, Solution 22.

Left pulley.

Inner radius r1 = 50 mm

Outer radius r2 = 100 mm

0.6 m/s = 600 mm/sAv =

( ) 1.8 m/s = 1800 mm/sA t

a = −

1

2

100

Av

r

ω = = =

( )

2

1

2

100

A ta

r

α = = =

Intermediate belt.

( )( )1 1 150 6 300 mm/sv rω= = =

( ) ( )( ) 2

1 1 150 18 900 mm/s

ta rα= = =

Right pulley.

Inner radius r3 = 50 mm

Outer radius r4 = 100 mm

1

2

4

100

v

r

ω = = =

( )1 2

2

4

100

ta

r

α = = =

(a) Velocity and acceleration of point C.

( )( )3 250 3 150 mm/s

Cv rω= = =

0.150 m/sC

=v

( ) ( )( )3 250 9 450 mm/s

C ta rα= = =

0.450 m/sC

=a

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(b) Acceleration of point B.

( ) ( )( )22 2

4 2100 3 900 mm/s

B na r ω= = =

( ) 20.900 m/s

B n=a

( ) ( )( ) 2

4 2100 9 900 mm/s

B ta rα= = =

( ) 20.900 m/s

B t=a

21.273 m/s

B=a 45°

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Chapter 15, Solution 23.

(a) Let point C be the point of contact between the shaft and the ring.

1C Av rω=

1

2 2

C A

B

v r

r r

ωω = =

1

2

A

B

r

r

ωω =

2

1( ) :

A Ab On shaft A a rω=

2

1A Arω=a

2

2 1

2 2

2

:A

B B

rOn ring B a r r

r

ωω

= =

2 2

1

2

A

B

r

r

ω=a

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Chapter 15, Solution 24.

(a) Let point C be the point of contact between the shaft and the ring.

( )( )10.5 25 12.5 in./s

C Av rω= = =

2

2.5

C

B

v

r

ω = = = 5.00 rad/sB

ω =

( ) On shaft :b A ( )( )22

10.5 25

A Aa rω= =

2

312.5 in./s ,= 2

26.0 ft/sA

=a

On ring :B ( )( )22

22.5 5.0

B Ba r ω= =

2

62.5 in./s ,= 2

5.21 ft/sB

=a

( ) At a point on the outside of the ring,c 33.5 in.r r= =

( )( )22 23.5 5.0 87.5 in./s

Ba rω= = =

27.29 ft/sa =

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Chapter 15, Solution 25.

( )a ( )( )600 2

A

πω π= = =

Let points A, B, and C lie at the axles of gears A, B, and C, respectively.

Let D be the contact point between gears A and B.

( )( )/2 20 40 in./s

D D A Av r ω π π= = =

/

40 6010 rad/s 10 300 rpm

4 2

D

B

D B

v

r

πω π ππ

= = = = ⋅ =

300 rpmB

=ωωωω

Let E be the contact point between gears B and C.

( )( )/2 10 20 in./s

E E B Bv r ω π π= = =

( )/

20 603.333 rad/s 3.333 100 rpm

6 2

E

C

E C

v

r

πω π ππ

= = = = =

100 rpmC

=ωωωω

(b) Accelerations at point E.

( )22

2

/

20On gear : 1973.9 in./s

2

E

B

E B

vB a

r

π= = =

2

1974 in./sB

=a

( )22

2

/

20On gear : 658 in./s

6

E

C

E C

vC a

r

π= = =

2

658 in./sC

=a

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Chapter 15, Solution 26.

( ) At timea 2 s,t = ( )( )600 2

A

πω π= = =

A A At

t

ωω α α π= = =

Let D be the contact point between gears A and B.

( ) ( )( ) 2

/2 10 20 in./s

D D A Ata r α π π= = =

( )

2

/

4

D t

B

D B

a

r

πα π= = = 2

=αααα

Let E be the contact point between gears B and C.

( ) ( )( ) 2

/2 5 10 in./s

E E B Bta r α π π= = =

( )

2

/

6

E t

C

E C

a

r

πα π= = = 2

=αααα

( ) Atb 0.5 s.t = ( )( )For gear , 5 0.5 2.5 rad/sB B

B tω α π π= = =

( ) ( )( )22 2

/2 2.5 123.37 in./s

E E B Bnr ω π= = =a

( ) 210 in./s

E tπ=a

231.416 in./s=

( ) ( ) ( ) ( )2 2 2 2 2123.37 31.416 127.3 in./s

E E En ta a a= + = + =

31.416

tan , 14.29123.37

β β= = ° 2

127.3 in./sE

=a 14.29°

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( )( )For gear , 1.6667 0.5 0.83333 rad/sC C

C tω α π π= = =

( ) ( )( )22 2

/6 0.83333 41.123 in./s

E E C Cna r ω π= = =

( ) 231.416 in./s

E ta =

( ) ( ) ( ) ( )2 2 2 2 241.123 31.416 51.75 in./s

E E En ta a a= + = + =

31.416

tan 37.4 ,41.123

β = = ° 2

51.8 in./sE

=a 37.4°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 27.

( ) For the pulley, a ( )1 1, 0.3 0.15 m

2 2A

r d r= = =

( )10.2 0.1 m

2Br = =

( )/

/

,A A B B A B A B A B

A B

A B

v r v r v v v r r

v

r r

ω ω ω

ω

= = = − = −

=−

At 0,t = 0

0.15 0.1ω = =

At 0.25 s,t = 1

0.15 0.1ω = =

21 28 16

ω ωα − −= = = − 2

A Ar α=a ( )( ) 2

0.15 32 4.80 m/s= =

B B

r α=a ( )( ) 20.1 32 3.2 m/s= =

2

/1.6 m/s

A B A B= − =a a a

2

/1.600 m/s

A B=a

( )b ( ) ( ) 2

/ / /0

1

2A B A B A B

t t= +x v a

( )( ) ( )( )210.8 0.25 1.6 0.25 0.15 m

2= − + = −

/150.0 mm

A B=x

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Chapter 15, Solution 28.

For the pulley, ( )1 1, 0.3 0.15 m

2 2A

r d r= = =

( )10.2 0.1m

2Br = =

( )/,

A A B B A B A Bv r v r v r rω ω ω= = = −

/A B

A B

v

r r

ω =−

At 0,θ = 0

0.15 0.1ω = =

2θ π= =

0.15 0.1ω = =

d dd d

dt d

ω ωα ω ω ω α θθ

= = =

( )( )2 2

9 2

18 0

2 2d d

πω ω α θ α π α= − = = −∫ ∫

( ) 2 20.15 38.675 5.8012 m/s or 5.8012 m/s

A Ar α= = − = −a

( ) 2 20.1 38.675 3.8675 m/s or 3.8675 m/s

B Br α= = − = −a

( )a 2 2

/1.9337 m/s or 1.9337 m/s

A B A B= − = −a a a

2

/1.934 m/s

A B=a

( )b ( ) ( ) 2

/ / /0

1

2A B A B A B

t t= +x v a

( )( ) ( )( )210.9 0.3 1.9337 0.3 0.1830 m

2= + − =

/183.0 mm

A B=x

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Chapter 15, Solution 29.

(a) Motion of pulley.

( ) ( )0 0

8 in./sE A

= =v v ( ) 210 in./s

E At= =a a

Fixed axis rotation.

( ) ( )0

0 00

4 in.

E

E A

A

v

v r

r

ω ω= = = =

( ) ( ) 2

4 in.

E t

E At

A

a

a r

r

α α= = = =

Since α is constant,

02 2.5t tω ω α= + = +

2 2

0 0

10 2 1.25

2t t t tθ θ ω α= + + = + +

For t = 3s, ( )( )2 2.5 3 9.5 rad/sω = + =

( )( ) ( )( )20 2 3 1.25 3 17.25 radθ = + + =

In revolutions, 17.25

π= 2.75 revθ =

(b) Motion of load B. t = 3s

( )( )6 9.5B Bv r ω= = 57.0 in./s

B=v

( )( )6 17.25B By r θ∆ = = 103.5 in.

By∆ =

(c) Acceleration of point D. t = 0

0

( ) ( )( ) 26 2.5 15 in./s

D Dtr α= = =a

( ) ( )( )22 26 2 24 in./s

D Dnr ω= = =a

2

28.3 in./sD

=a 32.0°

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Chapter 15, Solution 30.

00ω =

Use equations for constant angular acceleration.

0

2.4t tω ω α= + =

2 2

0 0

11.2

2t t tθ θ ω α= + + =

At t = 4s,

( )( )2.4 4 9.6 rad/sω = =

( )21.2 4 19.2 radθ = =

(a) Load A. at t = 4s, 4 in.Ar =

( ) ( )4 in. 9.6 rad/sA Av r ω= = 38.4 in./s

A=v

( ) ( )4 in. 19.2 radA Ay r θ= = 76.8 in.

A=y

(b) Load B. at t = 4s, 6 in.Br =

( ) ( )6 in. 9.6 rad/sB Bv r ω= = 57.6 in./s

B=v

( ) ( )6 in. 19.2 radB By r θ= = 115.2 in.

B=y

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Chapter 15, Solution 31.

ω π= =

Let C be the contact point between the two gears.

( )( ) 20.15 8 1.2 m/s

C A Av r ω π π= = =

0.2

C

B

B

v

r

πω π= = =

tω π α= =

tω π α= = −

Subtracting, ( )( ) 22 2 rad/sπ α α π= =

( )a 2

( )b 8 8

8 stπ πα π

= = = 8.00 st =

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Chapter 15, Solution 32.

( )0

ω π= = ( ) 118

A Atω π α= −

3 3

2 2 2

1 1 1

1 1 1 0.154

2 2 2 0.2

A

B B A A

B

rt t t

r

θ π α α α = = = =

( )3

2

1

0.15Atα π = =

( )3

1 1 11

0.150.421875

0.2B B A A

t t tω α α α = = =

Let Cv be the velocity at the contact point.

( )( )1 10.15 8 1.2 0.15

C A A A Av r t tω π α π α= = − = −

and ( )( )1 10.2 0.421875 0.084375

C B B A Av r t tω α α= = =

Equating the two expressions for Cv ,

1 1 11.2 0.15 0.084375 or 16.0850 rad/s

A A At t tπ α α α− = =

Then,

2

1

1

1

59.5743.7037 s

16.0850

A

A

tt

t

αα

= = =

( )a 216.0850

Aα = =

Aα =

( )3

0.2B

α = =

2

α =

( ) From above,b 13.70 st =

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Chapter 15, Solution 33.

Motion of disk B. ( )0

ω = =

Assume that the angular acceleration of disk B is constant.

( )0B B B

tω ω α= +

At 60 s,t = 0B

ω =

( )

200 52.360

B B

Bt

ω ωα

− −= = = −

( )( )3 52.360 0.87266 157.08 2.618 in./sB B Bv r t tω= = − = −

Motion of disk A. ( )0

0,A

Aα =

( )0

3A A A

t tω ω α= + =

( )( )2.5 3 7.5 in./sA A Av r t tω= = =

If disks are not to slip, A Bv v=

7.5 157.08 2.618t t= −

(a) 15.52 st =

(b) ( )( )3 15.52 46.6 rad/sA

ω = =

( )( )52.360 0.87266 15.52 38.8 rad/sB

ω = − =

445 rpmA

=ωωωω , 371 rpmB

=ωωωω

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Chapter 15, Solution 34.

Wheel B. ( )0

ω = =

At 12s,t = 75 rpm =7.854 rad/sB

ω =

Angular acceleration.

( )

07.854 31.416

B B

Bt

ω ωα

− −= = = −

Velocity at contact point with disk A at 12 s:t =

( )( )3 7.854 23.562 in./sB B Bv r ω= = =

Wheel A. ( )0

ω =

Assume that slipping ends when 12 s.t =

Then, 23.562 in./sAv =

23.562

A

A

A

v

r

ω = = =

( )

20

A

12

A A

t

ω ωα

− − −= = = −

(a) 2

=αααα

2

=αααα

(b) Time when ωA is zero.

( )0

0A A A

tω ω α= + =

( )0

0 31.416

3.4034

A A

A

t

ω ωα− −= =

− 9.23 st =

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Chapter 15, Solution 35.

Let one layer of tape be wound and let v be the tape speed.

2 andv t r r bπ∆ = ∆ =

2 2

r bv b

t r

ωπ π

∆ = =∆

For the reel: 1 1d d v dv d

vdt dt r r dt dt r

ω = = +

2 2

2

a v dr a v b

r dt rr r

ωπ

= − = −

2

10

2

ba

r

ωπ

= − =

2

0

2

bωπ

=a

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Chapter 15, Solution 36.

Let one layer of paper be unrolled.

2 andv t r r bπ∆ = ∆ = −

2

r bv dr

t r dtπ∆ −= =∆

2

1 10

d d v dv d v drv

dt dt r r dt dt r dtr

ωα = = = + = −

2

2 32 2

v bv bv

rr rπ π− = − =

2

32

bv

rπ=αααα

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Chapter 15, Solution 37.

Velocity analysis.

150 mm/sB

=v 15°

A A=v v

/500

B Aω=v 50°

Plane motion Translation with Rotation about .B B= +

/A B A B= +v v v

Draw velocity vector diagram.

180 50 75 55φ = ° − ° − ° = °

Law of sines.

/

sin 75 sin sin50

A B A Bv v v

φ= =

° °

( )a /

sin 75 150 sin 75189.14 mm/s

sin50 sin50

B

A B

v

v

° °= = =° °

/ 189.14

A B

AB

v

lω = = =

( )b sin 150 sin55

160.4 mm/ssin50 sin50

B

A

v

v

φ °= = =° °

160.4 mm/sA

=v

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Chapter 15, Solution 38.

Velocity analysis.

225 mm/sA

=v

B Bv=v 15°

/ /B A B Av=v 30°

Plane motion Translation with Rotation about .A A= +

/B A B A= +v v v

Draw velocity vector diagram.

180 60 75 45φ = ° − ° − ° = °

Law of sines.

/

sin 75 sin 60 sin

B A B Av v v

φ= =

° °

( )a /

sin 75 225sin 75307.36 mm/s

sin sin 45

A

B A

v

v

φ° °= = =

°

/ 307.36

B A

AB

v

lω = = =

( )b sin 60 225sin 60

276 mm/ssin sin 45

A

B

v

v

φ° °= = =

°

276 mm/sB

=v 15°

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Chapter 15, Solution 39.

Geometry.

12sin , 36.87

20β β= = °

12tan , 67.38

5θ θ= = °

Velocity analysis.

4.2 ft/sA

=v

/

20

12B A AB AB

rω ω= =v β

B Bv=v θ

Plane motion Translation with Rotation about .A A= +

/B A B A= +v v v

Draw velocity vector diagram.

( )180 90 59.49φ θ β= ° − − ° − = °

Law of sines.

( )/

sin sin 90 sin

B A B Av v v

θ β φ= =

° −

/

sin 4.2sin 67.384.5 ft/s

sin sin59.49

A

B A

v

v

θφ

°= = =°

( )a 4.5

ABω = =

=ωωωω

( )b cos 4.2cos36.87

3.90 ft/ssin sin59.49

A

B

v

v

βφ

°= = =°

3.90 ft/sB

=v 67.4°

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Chapter 15, Solution 40.

Geometry.

12sin , 36.87

20β β= = °

12tan , 67.38

5θ θ= = °

Velocity analysis.

=ωωωω

( )/ /

204.2 7.0 ft/s

12B A B A AB

r ω = = =

v β

B Bv=v θ

A Av=v

Plane motion Translation with Rotation about .A A= +

/B A B A= +v v v

Draw velocity vector diagram.

( )180 90 59.49φ θ β= ° − − ° − = °

Law of sines.

( )/

sin sin 90 sin

B AA Bvv v

φ β θ= =

° −

( )a /sin 7sin59.49

6.53 ft/ssin sin 67.38

B A

A

v

v

φθ

°= = =°

6.53 ft/sA

=v

( )b /cos 7cos36.87

6.07 ft/ssin sin 67.38

B A

B

v

v

βθ

°= = =°

6.07 ft/sB

=v 67.4°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 41.

In units of m/s, ( )/ /0.6 0.6 0.6 0.6

B A B Aω ω ω ω= × = × + = − +v k r k i j i j

/ /1.2 1.2

C A C Aω ω ω= × = × =v k r k i j

/A B A= +v v v

B

( ) ( )7.4 7 0.6 0.6B Ay xv v ω ω− + = − − +i j i j i j

Components. ( ): 7.4 0.6A xv ω− = −i (1)

( ): 7 0.6B yv ω= − +j (2)

/A C A= +v v v

C

( ) ( )1.4 7 1.2C A xyv v ω− + = − +i j i j j

Components. ( ): 1.4A xv− =i (3)

( ): 7 1.2C yv ω= − +j (4)

From (3), ( ) 1.4 m/sA xv = −

( ) From (1),a ( )7.4 1.4

ω− − −

From (2), ( ) ( )( )7 0.6 10 1m/sB yv = − + = −

( )b ( ) ( )7.40 m/s 1.000 m/sB

= − −v i j

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 42.

In units of m/s, ( )/ /0.6 0.6 0.6 0.6

B A B Aω ω ω ω= × = × + = − +v k r k i j i j

/ /1.2 1.2

C A C Aω ω ω= × = × =v k r k i j

/B A B A= +v v v

( ) ( )7.4 7 0.6 0.6B Ay xv v ω ω− + = − − +i j i j i j

Components. ( ): 7.4 0.6A xv ω− = −i (1)

( ): 7 0.6B yv ω= − +j (2)

/C A C A

= +v v v

( ) ( )1.4 7 1.2C A xyv v ω− + = − +i i j j

Components. ( ): 1.4A xv− =i (3)

( ): 7 1.2C yv ω= − +j (4)

From (3), ( ) 1.4 m/s, 1.4 7A Axv = − = − −v i j

From (1), ( ) ( )7.4 1.4

ω− − −

= = =−

kωωωω

(a) ( )/ /10 0.6

O A O A A O A A= + = + × = + ×v v v v r v k iωωωω

1.4 7 6 1.4 1= − − + = − −i j j i j

( ) ( )1.400 m/s 1.000 m/sO

= − −v i j

(b) ( )0O

x y= + × +v i jωωωω

( )0 1.4 1 10 1.4 1 10 10x y x y= − − + × + = − − + −i j k i j i j j j

Components. : 0 1.4 10 ,y= − −i 0.14 my = − 140.0 mmy = −

: 0 1 10 ,x= − +j 0.1 mx = 100.0 mmx =

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Chapter 15, Solution 43.

In units of mm/s, ( )/ /125 75 75 125

B A B Aω ω ω ω= × = × + = − +v k r k i j i j

( )/ /50 150 150 50

C A C Aω ω ω ω= × = × + = − +v k r k i j i j

/B A B A

= +v v v

( ) ( )75 100 75 125B x yv v ω ω− = + − +i j i j i j

A

Components. ( ): 100 75B xv ω= −i (1)

( ): 75 125A yv ω− = +j (2)

/C A C A= +v v v

( ) ( )400 100 150 50C A yyv v ω ω+ = + − +i j i j i j

Components. : 400 100 150ω= −i (3)

( ) ( ): 125C A yyv v ω= +j (4)

(a) From (3), 2 rad/sω = − ( )2 rad/s= − kωωωω

(b) From (2), ( ) ( )75 125 75 125 2 175 mm/sA yv ω= − − = − − − =

( ) ( )100.0 mm/s 175.0 mm/sA

= +v i j

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Chapter 15, Solution 44.

In units of mm/s, ( )/ /125 75 75 125

B A B Aω ω ω ω= × = × + = − +v k r k i j i j

( )/ /50 150 150 50

C A C Aω ω ω ω= × = × + = − +v k r k i j i j

/B A B A

= +v v v

( ) ( )75 100 75 125B x yv v ω ω− = + − +i j i j i j

A

Components. ( ): 100 75B xv ω= −i (1)

( ): 75 125A yv ω− = +j (2)

/C A C A

= +v v v

( ) ( )400 100 150 50C A yyv v ω ω+ = + − +i j i j i j

Components. : 400 100 150ω= −i (3)

( ) ( ): 125C A yyv v ω= +j (4)

From (3), ( )2 rad/sω = − k

From (2), ( ) ( )75 125 75 125 2 175 mm/sA yv ω= − − = − − − =

( ) ( )100 mm/s 175 mm/sA

= +v i j

Find the point with zero velocity. Call it D. 0D

=v

( ) ( )/or 0 100 175 2

A D Ax y= + = + + × +v v v i j k i j

D

0 100 175 2 2 0x y= + + − =i j j i

Components. : 0 100 2 , 50 mmy y= − =i

: 0 175 2 , 87 mmx x= + = −j

100 mm2

v

r

ω= = =

Circle of 100.0 mm radius centered at 87.5 mm, 50.0 mmx y= − =

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Chapter 15, Solution 45.

7Slope angle of rod. tan 0.7, 35

10θ θ= = = °

1012.2066 in. 20 7.7934 in.

cosAC CB AC

θ= = = − =

Velocity analysis.

25 in./sA

=v ,C C

v=v θ

/C A ABACω=v θ

/C A C A= +v v v

Draw corresponding vector diagram.

/sin 25sin35 14.34 in./s

C A Av v θ= = ° =

( )a / 14.34

C A

AB

v

ACω = = =

=ωωωω

cos 25cos 20.479 in./sC Av v θ θ= = =

( )( )/7.7934 1.175

B C ABv CBω= =

9.1551in./s=

/ /has same direction as .

B C C Av v

/B C B C= +v v v

Draw corresponding vector diagram.

/ 9.1551tan , 24.09

20.479

B C

C

v

v

φ φ= = = °

( )b 20.479

22.4 in./s 1.869 ft/scos cos24.09

C

B

v

v

φ= = = =

°

59.1φ θ+ = °

1.869 ft/sB

=v 59.1°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 46.

Instantaneous geometry. Law of sines:

sin sin120

10 15

φ °=

10sin sin120 0.57735

15φ = ° =

35.264φ = °

Velocity analysis.

1.2 ft/sA

=v

14.4 in./s=

/10

B A ABω=v 60°

B Bv=v φ

/B B A Av v= +v

Use the triangle construction to perform the vector addition.

60 24.736β φ= ° − = °

90 125.264γ φ= ° + = °

Law of sines. /

sin sin30 sin

B A B Av v v

γ β= =

°

/

sin 14.4 sin125.26428.10 in./s

sin sin 24.736

A

B A

v

v

γβ

°= = =°

(a) 28.10

10AB

=ωωωω

(b) sin 30 14.4 sin 30

17.21 in./ssin sin 24.736

A

B

v

v

β° °= = =

1.434 ft/sB

=v 35.3°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 47.

Label the contact point between gears A and B as 1, the center of gear B

as 2, and the contact point between gears B and C as 3.

Gear A: 1

80A

v ω= (1)

Arm AB: 2

120AB

v ω= (2)

Gear B: 1 2

40B

v v ω= − (3)

3 2

80B

v v ω= + (4)

Gear C: 3

200C

v ω= (5)

ω ω= =

From (1), 1

0,v =

From (5),

( )( )3200 5 1000 mm/sv = =

From (3), 2

40 0B

v ω− = (6)

From (4), 2

80 1000B

v ω+ = (7)

Solving (6) and (7) simultaneously,

(a)

120B

=ωωωω

2

(40)(8.333) 333.33 mm/sv = =

(b) From (2), 333.33

AB=ωωωω

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 48.

Label the contact point between gears A and B as 1, the center of

gear B as 2, and the contact point between gears B and C as 3.

Gear A: 1

80A

v ω= (1)

Arm AB: 2

120AB

v ω= (2)

Gear B: 1 2

40B

v v ω= − (3)

3 2

80B

v v ω= + (4)

Gear C: 3

200C

v ω= (5)

ω ω= =

From (5), 3

0.v =

From (4),

( )( )280 80 20

Bv ω= − = −

1600 mm/s=

From (3),

( )( )11600 40 20 2400v = − − = −

2400 mm/s=

From (1), 1

vω = = −

=ωωωω

From (2), 1600 120AB

ω= 1600

ABω = − = −

=ωωωω

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Chapter 15, Solution 49.

Data: 3600 rpm 376.99 rad/s, 0A B

ω ω= = =

11.25 in.

2A Ar d= =

diameter of ball 0.5 in.d = =

Velocity of point on inner race in contact with a ball.

(1.25)(376.99) 471.24 in./sA A Av r ω= = =

Consider a ball with its center at point C.

/A B A Bv v v= +

0A Cv dω= +

471.24

0.5

A

C

v

dω = =

/C B C Bv v v= +

10 (0.25)(942.48) 235.62 in./s

2dω= + = =

(a) 236 in./sCv =

(b) Angular velocity of ball.

ω = 9000 rpmC

ω =

(c) Distance traveled by center of ball in 1 minute.

(235.62)(60) 14137.2 in.C Cl v t= = =

Circumference of circle: 2 2 (1.25 0.25)rπ π= +

9.4248 in.=

Number of circles completed in 1 minute:

14137.2

2 9.4248

ln

rπ= = 1500n =

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Chapter 15, Solution 50.

Contact point 1 between gears A and B.

Contact point 2 between gears B and C.

ω =

1(6 rad/s)(10 in.) 60 in./sv = = (1)

2(6 rad/s)(5 in.) 30 in./sv = = (2)

Arm ABC: ABC ABC

ω=ωωωω

15A ABC

ω=v 15C ABC

ω=v

ω =

1(5)(3)

Av v= + 15

ABCω= 15+ (3)

Matching expressions (1) and (3) for 1,v

(a) : 60 15 15ABC

=ωωωω

Gear C: C C

ω ω=

2

10C C

v v ω= + 15= 10C

ω+ (4)

Matching expressions (2) and (4) for 2,v

(b) : 30 15 10C

=ωωωω

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 51.

Let a be the radius of the central gear A, and let b be the radius of the

planetary gears B, C, and D. The radius of the outer gear E is 2 .a b+

Label the contact point between gears A and B as 1, the center of gear B

as 2, and the contact point between gears B and E as 3.

1Gear :

AA v aω= (1)

( )2Spider:

Sv a b ω= + (2)

2 1Gear :

BB v v bω= + (3)

3 2 Bv v bω= + (4)

( )3Gear : 2

EE v a b ω= + (5)

( )2From (4) and (5), 2

B Ev b a bω ω+ = + (6)

2 1From (1) and (3),

B Av b v aω ω− = = (7)

( )2 2

2Solving for and ,

2

E A

B

a b av v

ω ωω

+ + =

( )2

2

E A

B

a b a

b

ω ωω

+ − =

From (2), ( )

( )2

2

2

E A

S S

a b av

a b a b

ω ωω ω

+ + = =+ +

Data: 60 mm, 60 mm, 2 180 mm, 120 mma b a b a b= = + = + =

( )a ( )( )180 60

1.5 0.52 60

E A

B E A

ω ωω ω ω−= = −

( )( ) ( )( )1.5 120 0.5 150 105 rpm= − =

105.0 rpmB

=ωωωω

( )b ( )( )180 60

0.75 0.252 120

E A

S E A

ω ωω ω ω+= = +

( )( ) ( )( )0.75 120 0.25 150 127.5 rpm= + =

127.5 rpmS

=ωωωω

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 52.

Let a be the radius of the central gear A, and let b be the radius of the

planetary gears B, C, and D. The radius of the outer gear E is 2 .a b+

Label the contact point between gears A and B as 1, the center of gear B

as 2, and the contact point between gears B and E as 3.

1Gear :

AA v aω= (1)

( )2Spider:

Sv a b ω= + (2)

2 1Gear :

BB v v bω= + (3)

3 2 Bv v bω= + (4)

( )3Gear : 2

EE v a b ω= + (5)

( )2From (4) and (5), 2

B Ev b a bω ω+ = + (6)

2 1From (1) and (3),

B Av b v aω ω− = = (7)

2Solving for and ,

Bv ω

( )2

2

2

E Aa b a

vω ω + + =

( )2

2

E A

B

a b a

b

ω ωω

+ − =

From (2), ( )

( )2

2

2

E A

S S

a b av

a b a b

ω ωω ω

+ + = =+ +

1Data: 0,

5E S A

ω ω ω= =

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( )a ( )( )

( )2 01

5 2

A

S A

a b a

a b

ωω ω

+ + = =+

( ) ( )1 1 1

, 2 5 52 1 b

a

a

a b= =

+ +

2 1 5b

a

+ =

1.500b

a=

( )b ( )

( )( )2

02 2 2 1.5 3

E A A A A

B

a b a a

b b

ω ω ω ω ωω + − = = − = − = −

1

3B A

ω=ωωωω

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Chapter 15, Solution 53.

Label the contact point between gears A and B as 1 and that between gears B and C as 2.

Rod ABC: 75 rpm 2.5 rad/sABC

ω π= =

0Av =

(12)(2.5 ) 30 rad/sBv π π= =

(12 7)(2.5 ) 47.5 rad/sCv π π= + =

Gear A: 10, 0, 0

A Av vω = = =

Gear B: 14 0

B Bv v ω= − =

30 4 0B

π ω− =

ω π=

24

B Bv v ω= +

30 30 60 in./sπ π π= + =

Gear C: 23

C Cv v ω= −

60 47.5 3C

π π ω= −

ω π= −

Summary: 225 rpmB

=ωωωω

125.0 rpmC

=ωωωω

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 54.

Label the contact point between gears A and B as 1 and that between gears B and C as 2.

Rod ABC: 0, 80 rpm 8 /3 rad/sC ABCv ω π= = =

7 (7)(8 /3) 56 /3 in./sB ABCv ω π π= = =

(7 12) (19)(8 /3) 152 /3 in./sA ABCv ω π π= + = =

Gear C: 20, 0, 0

C Cv vω = = =

Gear B: 24 56 /3 4 0

B B Bv v ω π ω= − = − =

ω π=

14 56 /3 56 /3

B Bv v ω π π= + = +

112 /3 in./sπ=

Gear A: 18

A Av v ω= −

112 /3 152 /3 8A

π π ω= −

ω π= −

Summary: 50.0 rpmA

=ωωωω

140.0 rpmB

=ωωωω

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Chapter 15, Solution 55.

Geometry.

( ) ( )sin sinOA ABθ β=

( )sin 10 sin30

sin , 1.79160

OA

AB

θβ β°= = = °

ω = = π

( ) ( ) ( )10 30 300 mm/sA OA

OA ω= = π = πv

Rod AB. (Plane motion = Translation with A + Rotation about A.)

[/

B A B A Bv= +v v v ] [ A

v=

] /60

A Bv° +

Draw velocity vector diagram. 90 88.21β° − = °

180 60 88.21 31.79φ = ° − ° − ° = °

Law of sines.

( )sin sin 90

B Av v

φ β=

° −

( )( )300 sin 31.79sin

sin 90 sin 88.21

497 mm/s

A

B

v

v

φβ

π °= =

° − °

=

497 mm/sB

=v

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Chapter 15, Solution 56.

Geometry.

( ) ( ) ( )sin 180 sinOA ABθ β° − =

( ) ( )sin 180 10 sin 60

sin , 3.10160

OA

AB

θβ β

° − °= = = °

ω = = π

( ) ( ) ( )10 30 300 mm/sA OA

OA ω= = π = πv

Rod AB. (Plane motion = Translation with A + Rotation about A.)

[/

B A B A Bv= +v v v ] [300= π ] /

30B Av° + ]β

Draw velocity vector diagram. 90 93.10β° + = °

180 30 93.10 56.90φ = ° − ° − ° = °

Law of sines.

( )sin sin 90

B Av v

φ β=

° +

( )( )300 sin 56.90sin

sin 90 sin 93.10

791 mm/s

A

B

v

v

φβ

π °= =

° + °

=

791 mm/sB

=v

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Chapter 15, Solution 57.

Disk A

Bar :BD

ω = , 2.8 in.AB =

( ) ( )( )2.8 15 42 in./sB Av AB ω= = =

( )a 0 . θ = ° 42 in./sB

=v

2.8sin , 16.260

10β β= = °

/D B D B

= +v v v

Dv

[42=

] /D B

v+ β

/

4243.75 in./s

cosD Bv

β= =

/ 43.75

,10

D B

DB

v

DBω = =

=ωωωω

tan ,D Bv v β=

12.25 in./sD

=v

( ) 90 .b θ = °

42 in./sB

=v

5.6sin , 34.06

10β β= = °

Bar :BD /D B D B

= +v v v

Dv

[42=

] /D Bv+

β

Components: : /

0D Bv =

: D Bv v=

0DB

=ωωωω

42.0 in./sD

=v

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( )c 180 . θ = ° 42 in./sBv =

2.8sin , 16.26

10β β= = °

Bar :BD

/D B D B= +v v v

Dv [42= ] /D B

v+ β

/

4243.75

cosD Bv

β= =

/ 43.75

10

D B

DB

v

DBω = =

=ωωωω

tanD Bv v β=

12.25 in./sD

=v

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Chapter 15, Solution 58.

2.8 in., 10 in.A BDr l= =

From geometry, sin sinBD A Al r rβ θ= + (1)

is tangent to the circular path of ,B

Bv

thus B A Ar ω=v θ

For rod BD /B D BD BDl ω=v β

/ /

0B D B D B D

= + = +v v v v

/B D B=v v

( ) For matching directiona or 180θ β θ β= = ° +

For ,β θ=

sin sin so that sin sinBD A Al r rβ θ β β= = +

2.8sin , 22.9 ,

10 2.8

A

BD A

r

l rβ β= = = °

− −

22.9θ = °

For

180 , sin sin ,θ β θ β= ° + = −

sin sinBD A Al r rβ β= −

2.8sin , 12.6

10 2.8

A

BD A

r

l rβ β= = = °

+ +

192.6θ = °

( ) For matching magnitudesb /D B B

v v=

( )( )2.8 20 , 5.6 rad/s

10

A A

BD BD A A BD

BD

rl r

l

ωω ω ω= = = =

For 22.9 ,θ = ° 5.60 rad/sBD

=ωωωω

For 192.6 ,θ = ° 5.60 rad/sBD

=ωωωω

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Chapter 15, Solution 59.

Rod BE: 0Ev =

/

(192)(4) 768 mm/sB B E BE

r ω= = =v

Rod ABD: D D

v=v

/D B D B

= +v v v

Dv 768= 360

ω= °

ω =

=ωωωω

(b) 768 tan30Dv= °

1330 mm/sDv =

1.330 m/sD

=v

(c) /

ω=v 60 1024 mm/s° = 60°

/768

A B A B= + =v v v 1024+ 60 1557 mm/s° = 34.7°

1.557 m/sA

=v 34.7°

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Chapter 15, Solution 60.

Rod BE: 0E

=v B B

v=v

Rod ABD: 1.6 m/sD

/B D B D

= +v v v

Bv

ω+ 60°

(a) 1.6

0.360sin 60

°

=ωωωω

(b) 1.6 tan30 0.92376Bv = ° =

0.924 m/sB

=v

(c) /

ω=v 60 1.2317 m/s° = 60°

/0.92376

A B B A= + =v v v 1.2317+ 60°

[1.5396= ] [1.0667+ ] 1.873=

34.7°

1.873 m/sA

=v 34.7°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 61.

1000 rpmAB

ω =

( ) ( )1000 2104.72 rad/s

60

π= =

( )a 0 . . (Rotation about )Crank AB Aθ = ° /

3 in.B A

=r

( ) ( )/3 104.72 314.16 in./s

B B A ABv ω= = =v

.Rod BD (Plane motion Translation with Rotation about )B B= +

/D B D Bv= +v v

Dv

[314.16= ] /D Bv+ ]

/0, 314.16 in./s

D D Bv v= =

P D=v v

0P

=v

314.16

8

B

BD

v

lω = =

=ωωωω

( )b 90 . . (Rotation about )Crank AB Aθ = °

/3 in.

B A=r

( ) ( )/3 104.72 314.16 in./s

B B A ABr ω= = =v

.Rod BD

(Plane motion Translation with Rotation about .)B B= +

/D B D B= +v v v

Dv 314.16=

/D Bv+ ]β

/0, 314.16 in./s

D B Dv v= =

/,

D B

BD

v

lω =

0BD

=ωωωω

314.16 in./sP D

= =v v

26.2 ft/sP

=v

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 62.

( ) ( )1000 21000 rpm 104.72 rad/s

60AB

ωπ

= = =

60 , . (Rotation about )Crank AB Aθ = ° /

3 in.B A

=r 30°

( ) ( )/3 104.72 314.16 in./s

B B A ABω= = =v r

60°

Rod BD. (Plane motion Translation with Rotation about .)B B= +

Geometry. sin sinl rβ θ=

3sin sin sin 60

8

18.95

r

lβ θ

β

= = °

= °

/D B D B= +v v v

[ Dv ] [314.16= ] /

60D Bv° + ]β Draw velocity vector diagram.

( )180 30 90 78.95φ β= ° − ° − ° − = °

Law of sines.

( )/

sin sin30 sin 90

D BD Bvv v

φ β= =

° ° −

sin 314.16 sin 78.95326 in./s

cos cos18.95

B

D

v

v

φβ

°= = =°

P Dv v= 27.2 ft/s

P=v

/

sin 30 314.16sin 30166.08 in./s

cos cos18.95

B

D B

v

v

β° °= = =

°

/ 166.08

8

D B

BD

v

BD=ωωωω

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Chapter 15, Solution 63.

( ) ( ) ( )/4 0.25 1.00 m/s

B AB B A= × = − × − = −v r k j iωωωω

( )/0.075 0.15 0.15 0.075

D DE D E DE DE DEω ω ω ω= × = × − − = −v k r k i j i j

Bar BD. (Translation with B + Rotation about B.)

/ /0.2 0.2

D B BD D B BD BDω ω ω= × = × =v k r k i j

/D B D B= +v v v

0.15 0.075 1.00 0.2DE DE BD

ω ω ω− = − +i j i j

Components:

: 0.15 1.00, 6.6667 rad/sDE DE

ω ω= − = −i 6.67 rad/sDE

=ωωωω

: 0.075 0.2DE BD

ω ω− =j

( )( )0.075 6.6667

0.2BD

ω− −

=ωωωω

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Chapter 15, Solution 64.

Bar AB. (Rotation about A.) ( ) 0.18B AB AB

AB ω ω= =v 30°

Bar DE. (Rotation about E.) ( ) 0.18D DE DE

DE ω ω= =v

Bar BGD. (Plane motion = Translation with B + Rotation about .B )

[/

D B D B Dv= +v v v ] [ B

v= ]30°/D B

v+ ]30°

Draw the velocity vector diagram.

Equilateral triangle.

/

0.18D B B ABv v ω= =

/ 0.18

0.18

D B AB

BD AB

BD

v

l

ωω ω= = =

/ /

10.09

2G B GB BD D B ABv l vω ω= = =

/G B G B

= +v v v

Draw vector diagram.

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Law of cosines

( ) ( ) ( )( )2 220.18 0.09 2 0.18 0.09 cos60

G AB AB AB ABv ω ω ω ω= + − °

2 2

0.0243G ABv ω=

( )( )6.415 6.416 2.5AB G

vω = =

=ωωωω

=ωωωω

0.18D B ABv v ω= =

0.18

0.18

D AB

DE AB

DE

v

l

ωω ω= = =

=ωωωω

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Chapter 15, Solution 65.

=ωωωω

( )( )/8 25 200 in./s

B B A ABr ω= = =v

D Dv=v

8D DEv ω=

Plate BDHF. (Translation with B + Rotation about .B )

[/

D B D B Dv= +v v v ] [ B

v= ] /D Bv+ ]30°

Draw velocity vector diagram.

/

200230.94 in./s

cos30 cos30

B

D B

v

v = = =° °

/ 230.94

B D

BDHF

BD

v

lω = = =

=ωωωω

( )( )/8 14.4338 115.47 in./s

F B BDHFv BFω= = =

/

115.47 m/sF B

=v 30°

[/200 in./s

F B F B= + =v v v ] [115.47 in./s+ ]30°

(b) [142.265 in./sF=v ] [100 in./s+ ] 173.9 in./s

F=v 54.9°

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Chapter 15, Solution 66.

ω =

( )( )/8 35 280 in./s

D D E DEr ω= = =v

B Bv=v

8B ABv ω=

Plate BDHF. (Translation with D + Rotation about .D )

[/

B D B D Bv= +v v v ] [ D

v= ] /B Dv+ ]30°

Draw velocity vector diagram.

/

/

280560 in./s

sin 30 sin 30

16

D

D B

D B

BDHF

DB

vv

v

= = =° °

= = =

=ωωωω

Point of zero velocity lies above point D.

/

2808 in.

35

D

C D

BDHF

vy

ω= = =

(b) 8 in. above point D.

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Chapter 15, Solution 67.

= − kωωωω BD BDω= kωωωω DE DE

ω=ω k

( )/0.2m

B A= −r j ( ) ( )/

0.6 m 0.25 mD B

= − −r i j ( )/0.2 m

D E= −r i

( ) ( ) ( )/0 15 0.25 3 m/s

B A AB B A= + × = + − × − = −v v ω r k j i

( ) ( )/ /6 0.2 0.25 0.6

D B BD D B BD BD BDω ω ω= × = × − − = −v r k i j i jωωωω

/

3 0.25 0.6D B D B BD BD

ω ω= + = − + −v v v i i j (1)

( ) ( )/ /0 0.2 0.2

D E D E DE D E DE DEω ω= + = + × = × − = −v v v ω r k i j (2)

Equate the expressions (1) and (2) for vD and resolve into components.

: 3 0.25 0BD

ω =

: 0.6 0.2BD DE

ω ω− = −j 36 rad/sDE

ω =

( )a Angular velocity of rod BD. 12 rad/sBD

=ωωωω

(b) Velocity of the midpoint M of rod BD.

( ) ( )/ /

10.3 m 0.125 m

2M B D B

= = − −r r i j

( )/ /3 12 0.3 0.125

M B M B B BD M B= + = + × = − + × − −v v v v ω r i k i j

( ) ( )1.5 m/s 3.6 m/s= − −i j

3.90 m/sM

=v 67.4 °

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Chapter 15, Solution 68.

Bar AB. ( ) ( )/0.300 m + 0.125 m ,

B A=r i j ( )3 rad/s

AB= kωωωω

0A

=v

/ /0

B A B A AB B A= + = + ×v v v rωωωω

( )3 0.3 + 0.125 0.375 + 0.9= × = −k i j i j

Bar BD. ( )/0.325 m

D B= −r j

BD BDω= kωωωω

( )0.375 + 0.9 + 0.325D B D/B BD

ω= + = − × −v v v i j k j

0.375 + 0.9 + 0.325BD

ω= − i j i

Bar DE. ( ) ( )/0.150 m 0.200 m ,

E D= − +r i j

DE DEω= kωωωω

/ /E D E D D DE E D= + = + ×v v v v rωωωω

( )0.150 0.200 0D DE

ω= + × − + =v k i j

0.375 0.9 0.325 0.15 0.2 0BD DE DE

ω ω ω− + + − − =i j i j i

Components:

j: 0.9 0.15 0DE

ω =

i: 0.375 0.325 0.2 0BD DE

ω ω− + − =

( )( )0.325 0.375 0.2 6BD

ω = +

=ωωωω

=ωωωω

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Chapter 15, Solution 69.

80 km/h 22.222 m/s

A= =v 0

C=v

560 mm 280 mm 0.28 m2

dd r= = = =

0.28

Av

r

ω = = =

/ / /B A D A E A

v v v rω= = =

( )( )0.28 79.364 22.222 m/s= =

[/22.222 m/s

B A B A= + =v v v ] [22.222 m/s+ ]

44.4 m/sB

=v

[/22.222 m/s

D A D A= + =v v v ] [22.222 m/s+ ]30°

42.9 m/sD

=v 15.0°

[/22.222 m/s

E A E A= + =v v v ] [22.222 m/s+ ]

31.4 m/sE

=v 45.0°

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Chapter 15, Solution 70.

(a) 0β = .Wheel AD 0, 45 in./sC D

= =v v

45

D

v

CDω = = =

( ) ( ) 4 2.5 1.5 in.CA CD DA= − = − =

( ) ( )( ) 1.5 11.25 16.875 in./sA ADv CA ω= = =

Rod AB. /B A B A= +v v v

[ Bv ] [16.875= ] /B A

v+ ]ϕ 16.88 in./sB

=v

/

0B Av = 0

ABω =

ω= =v

2.5

tan , 32.0054

DA

DCγ γ= = = °

4.7170 in.cos

DCCA

γ= =

( ) ( )( )4.7170 11.25 53.066 in./sA ADv CA ω= = =

[53.066 in./sA

=v ]32.005°

Rod AB. B Bv=v

4

sin , 18.66312.5

φ φ= = °

Plane motion = Translation with A + Rotation about A.

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[/

B A B A Bv= +v v v ] [ A

v= ] [r + vB/A

] φ

Draw velocity vector diagram.

( )180 90δ γ φ= ° − − ° +

90 32.005 18.663 39.332= ° − ° − ° = °

Law of sines.

( )/

sin sin sin 90

B AB Avv v

δ γ φ= =

° +

( )( )53.066 sin 39.332sin

sin 90 sin 108.663

A

B

v

v

δφ

°= =

° + °

35.5 in./s= 35.5 in./sB

=v

( )/

sin (53.066)sin32.005

sin 90 sin108.663

A

B A

v

v

γφ

°= =° + °

29.686 in./s=

/ 29.686

B A

AB

v

ABω = = = 2.37 rad/s

AB=ωωωω

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Chapter 15, Solution 71.

0

180 km/h = 50 m/s=v

( )( )180 2180 rpm = 18.85 rad/s

60

πω = =

Top View

0v zω=

050

2.65 m18.85

v

z

ω= = =

Instantaneous axis is parallel to the y axis and passes through the point

0x =

2.65 mz =

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Chapter 15, Solution 72.

1.5 3

E D

ED

v v

lω − −= = =

1

3

1.03 m

D

CE

vl

ω= = =

(a) 1.5 2 3 0.5 mACl = + − = lies 0.500 m to the right of C A

(b) ( ) 10.5 0.1667 m/s

3A ACv l ω = = =

A0.1667 m/s=v

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Chapter 15, Solution 73.

Contact points:

1 between gears A and B.

2 between gears B and C.

ω =

( )( )15 4 60 in./sAv = =

( )( )15 4 60 in./sCv = =

ω =

( )( )110 8 80 in./sv = =

( )( )25 8 40 in./sv = =

Gear A:

180 60

5 5

A

A

v vω − −= =

ω =

60

15 in.4

A

A

A

v

ω= = =l

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Gear C:

2

60 40

10 10

C

C

v vω − −= =

ω =

6030 in.

2

C

C

C

v

ω= = =l

(a) Instantaneous centers.

Gear A: 15 in. left of A

Gear C: 30 in. left of C

(b) Angular velocities.

=ωωωω

=ωωωω

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Chapter 15, Solution 74.

Since the drum rolls without sliding, the point of contact C with the fixed surface is the instantaneous center.

Let point A be the center of the cylinder and point B the point where the cord breaks contact with the

cylinder.

0 6 in./sC B Dv v v= = =

(a) Angular velocity of cylinder

/B B C

v r ω=

/

1

B

B C

v

r

ω = = =

(b) Velocity of point A. /A AC

v r ω=

( )( )5 6 30= =

30.0 in./sA

=v

(c) Rate of winding of cord. Since ,A Bv v> the cord is wound up at rate of

30 6 24 in./s.A Bv v− = − =

Winding rate 24.0 in./s.=

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Chapter 15, Solution 75.

0.3

O A

AO

v v

lω − −= = =

(a) 31.5

42.86 10 m35

A

CA

vl

ω−= = = ×

42.86 mm=

lies 42.9 mm below .C A

(b) 0.6 0.04286 0.64286 mCBl = + =

( ) ( ) 0.64286 35 22.5 m/sB CBv l ω= = =

22.5 m/sB

=v

(c) 0.3 m, 0.3 0.04286 0.34286 mOD COl l= = + =

( ) ( )2 20.3 0.34286 0.45558 m

CDl = + =

( )( )0.45558 35 15.95 m/sD CDv l ω= = =

0.3tan , 41.2

0.34286

OD

CO

l

lθ θ= = = °

15.95 m/sD

=v 41.2°

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Chapter 15, Solution 76.

12 1.5

O A

AO

v v

lω − += = =

(a) 31.5

33.33 10 m45

A

AC

vl

ω−= = = ×

33.33 mm=

lies 33.3 mm above .C A

(b) ( )0.3 0.3 0.0333 0.56667 mCBl = + − =

( ) ( ) 0.56667 45 25.5 m/sB CBv l ω= = =

25.5 m/sB

=v

(c) 0.3 m, 0.3 0.03333 0.26667 mOE OCl l= = − =

( ) ( )2 20.3 0.26667 0.4014 m

CEl = + =

( ) ( )0.4014 45 18.06 m/sE CEv l ω= = =

0.3tan , 48.4

0.26667

OE

OC

l

lθ θ= = = °

18.06 m/sE

=v 48.4°

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Chapter 15, Solution 77.

(b) Velocity of point B.

(a) Location of instantaneous axis.

8 in./sE D

= =v v

3 in./sA

=v

/E A E A

= + ×v v rωωωω

/E A E Av v r ω= +

8 3 3ω= +

/C A C A= + ×v v rωωωω

0 3 1.6667y= −

1.800 in. above point . y A=

/B A B A= + ×v v rωωωω

( ) ( )3 3 3 3 1.6667 2Bv ω= − = − = −

2.00 in./sB

=v

(c) Since ,D E Av v v= > the paper unwinds.

Rate of unwinding: 8 3 5 in./s E Av v− = − =

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Chapter 15, Solution 78.

10 in./sD

=v , 8 in./sB

=v

4.5

D Bv v

BDω + += = =

10

2.5 in.4

Dv

CDω

= = =

3.0 2.5 0.5 in.CA = − =

(a) C lies 0.500 in. to the right of A.

(b) ( )( )0.5 0.5 4 2 in./sAv ω= = =

2.00 in./sA

=v

(c) 12 in./sD A

− =v v

Cord DE is unwrapped at 12.00 in./s.

6 in./sB A

− =v v

Cord BF is unwrapped at 6.00 in./s.

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Chapter 15, Solution 79.

( ) ( )/ 0.192 4 0.768 m/sB B E BEr ω= = =v

(a) Instantaneous center C is located by noting that CD is perpendicular to vD and CB is perpendicular to vB

/ 0.360 sin 30 0.180 mB Cr = ° =

/

0.7684.2667

0.180B

v

rω = = =

(b) Velocity of D. / 0.360 cos30 0.31177 mD Cr = ° =

( )( )/ 0.31177 4.2667D D Cv r ω= =

1.330 m/sD =v "

(c) Velocity of A.

0.240cos30 0.20785 mAEl = ° =

0.600sin 30 0.300 mCEl = ° =

0.20785

tan0.300

β = 34.7β = °

( ) ( )2 20.20785 0.300 0.36497 mCAl = + =

( ) ( )0.36497 4.2667 1.557 m/sA CA ADv l ω= = =

1.557 m/sA =v 34.7° "

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Chapter 15, Solution 80.

ω =

( ) ( )( )0.200 15 3 m/sB ABv AB ω= = =

B Bv=v D D

v=v

Locate the instantaneous center (point C) of bar BD by noting that

velocity directions at points B and D are known. Draw BC perendicular to

Bv and DC perpendicular to .

Dv

( )a 3

B

BD

v

BCω = = = 12.00 rad/s

BD=ωωωω

(b) Locate point M, the midpoint of rod BD. Draw CM.

( ) ( )2 20.6 0.25 0.65 mBD = + =

0.25tan 22.62 90 67.38

0.6β β β= = ° ° − = °

( )10.325 m

2CM DM MB BD= = = =

( ) ( )( )0.325 12 3.9 m/sMv CM ω= = = 3.90 m/s

M=v 67.4 °

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Chapter 15, Solution 81.

( ) ( )( ) 18 10C CDv CDω= =

180 in./s=

180 in./sC

=v 30°

B B

v=v 30°

Locate the instantaneous center (point I) of bar BC by noting that velocity directions at two points are known.

Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D.

10 in., 10 3 in.IC IB= =

10

C

BC

v

ICω = = =

( ) ( )( )10 3 18 311.77 in./sB BCv IB ω= = =

(a) 311.77

B

AB

v

ABω = = = 31.2 rad/s

AB=ωωωω

=ωωωω

(c) Locate point M, the midpoint of bar BC.

Triangle ICM is an equilateral triangle. 10 in.IM =

( ) ( )( )10 18 180 in./sM BCv IM ω= = = 15.00 ft/s

M=v 30°

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Chapter 15, Solution 82.

B Bv=v 60°

C Cv=v

Bar BC. Locate its instantaneous center (point I) by noting that velocity directions at two points are known.

Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D.

Locate point M, the midpoint of bar BC. From geometry, triangle ICM is an equilateral triangle.

10 in., 10 3 in.IM AB CD IB= = = =

( ) ( )7.8 12

M

BC

v

IMω = = =

(a) ( ) ( )( )10 3 9.36 162.12 in./sB BCv IB ω= = =

162.12

B

AB

v

ABω = = = 16.21 rad/s

AB=ωωωω

=ωωωω

(c) ( ) ( )( )10 9.36 93.6 in./sC BCv IC ω= = =

93.6

C

CD

v

DCω = = = 9.36 rad/s

CD=ωωωω

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 83.

800 mm/sB

=v A Av=v

Locate the instantaneous center of rod ABD by noting that velocity directions at points A and B are known.

Draw AC perpendicular to A

v and BC perpendicular to B

v .

(a) 800

B

ABD

v

BCω = = =

°

=ωωωω

( ) ( )2 2600cos30 300sin 30 540.83 mm

CDl = ° + ° =

300sin 30tan 16.10 90 73.9

600cos30γ γ γ°= = ° ° − = °

°

(b) ( )( ) 3540.83 3.0792 1.665 10 mm/s

D CD ABDv l ω= = = ×

1.665 m/sD

=v 73.9°

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Chapter 15, Solution 84.

40 ,θ = ° 0.6 m/sB

=v , A A

v=v

Locate the instantaneous center (point C) by noting that velocity directions at points A and B are known. Draw AC perpendicular to

Av and BC

perpendicular to .

Bv

( ) sin 2sin 40 1.28557 mBC AB θ= = ° =

(a)

1.28557

B

ABD

v

BCω = = =

=ωωωω

/ / /DC B C D B

= +r r r

[1.28557 m= ] [2 m+ ]40°

2.9930 m= 30.79°

30.79β = °

(b) ( )( )/2.9930 0.46672

D D C ABDv r ω= =

1.397 m/s=

1.397D

=v β

90 59.2β° − = °

1.397 m/sD

=v 59.2°

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Chapter 15, Solution 85.

2 2320 240 400 mmDE = + =

240tan 0.75

320β = = 36.87β = °

( )( ) ( )( )400 15 6000 mm/s 6 m/sD DEv DE ω= = = =

6 m/sD

=v β B Bv=v

Locate point C, the instantaneous of bar DBF, by drawing BC perpendicular to vB and DC perpendicular

to vD.

From the figure: 540

720 mmtan

ACβ

= =

( )720 320 100 300 mmBC AC AB= − = − + =

Since triangles FCB and BDK are similar,

300

100

b DK

BC BK= =

( )( )300 300900 mm.

100b = =

( )a Distance b. 0.900 m b =

2 2

300 400 500 mm 0.5 mCD = + = =

6

D

BDF

v

CDω = = =

( )( )0.900 1.2 10.8 m/sF BDFv bω= = =

( )b Velocity of point F. 10.80 m/s

F=v

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Chapter 15, Solution 86.

Locate the instantaneous center I of rotation of bar ABD as the

intersection of line AI perpendicular to A

v and line BI perpendicular to

Bv Triangle IAB is equilateral.

300 mmIA IBl l= =

(a) 900 mm/s

A

ABD

IA

v

lω = = =

=ωωωω

(b) By the law of cosines, ( )600cos30 mmIDl = °

( )( )600cos30 3 1559 mm/sD ID ABDv l ω= = ° =

1.559 m/sD

=v

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Chapter 15, Solution 87.

A Av=v 45 ,° 7.5 ft/s

B=v

Locate the instantaneous center (point C) of rod AB by noting that velocity

directions at points A and B are known. Draw AC perpendicular to A

v and BC perpendicular to .

Bv

Let 24 in. 2 ftl AB= = =

Law of sines for triangle ABC.

2.8284 ftsin 75 sin 60 sin 45

b a l= = =° ° °

2.4495 ft, 2.73205 fta b= =

2.73205

Bv

bω = = =

(a) ( ) ( )2.4495 2.7452 6.724 ft/sAv aω= = =

6.72 ft/sA

=v 45.0°

(c) Let M be the midpoint of AB.

Law of cosines for triangle CMB.

( ) ( ) ( ) ( ) ( )

2

2 2

2 2

2 cos602 2

2.73205 1 2 2.73205 1 cos60

2.3942 ft

l lm b b

m

= + − °

= + − °

=

Law of sines.

2

sin sin 60 1sin 60, sin , 21.2

2.3942lm

β β β° °= = = °

( ) ( )2.3942 2.7452 6.573 ft/s,Mv mω= = =

6.57 ft/sM

=v 21.2°

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Chapter 15, Solution 88.

Bar DE. ( )( )24 8 192 in./sEDD

v eω= = =

192 in./sD

=v

Bar AB. 8 ABB AB

v aω ω= =

8B AB

ω=v 30°

Locate the instantaneous center (point C) of bar BD by noting that velocity directions at points B and D are known. Draw BC perpendicular to

Bv and

DC perpendicular to .

Dv

Let 24 in.l BD= =

Law of sines for triangle CBD.

2448 in.

sin120 sin 30 sin 30 sin 30

b d l= = = =° ° ° °

41.569 in., 24 in.b d= =

(a)

24

D

BD

v

dω = = = 8.00 rad/s

BD=ωωωω

(b) ( )( )41.569 8 332.55 in./sB BDv bω= = =

8

B

AB

v

a

ω = = = 41.6 rad/sAB

=ωωωω

(c) Law of cosines for triangle CMD.

( ) ( ) ( ) ( )

2

2 2

22

2 cos1202 2

24 12 2 24 12 cos120

31.749 in.

l lm d d

m

= + − °

= + − °

=

Law of sines.

( )2

12 sin120sin sin120, sin , 19.1

31.749lm

β β β°°= = = °

Velocity of M. ( )( )31.749 8 253.99 in./sM BDv mω= = =

21.2 ft/sM

=v 19.1°

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Chapter 15, Solution 89.

( ) , B ABF B Bv AB vω= =v 75°

200 mm/sD

=v

Locate the instantaneous center (point C) of bar DBE by noting that the velocity directions at points B and D

are known. Draw BC perpendicular toB

v and DC perpendicular to .

Dv

Law of sines for triangle . sin150 sin15 sin15

CD BC BDBCD = =

° ° °

180sin150180 mm 347.73 mm

sin15BC BD CD

°= = = =°

347.73

D

DBE

v

CDω = = =

( ) ( )( )180 0.57515 103.528 mm/sB DBEv BC ω= = =

180

B

ABF

v

ABω = = =

( ) ( )( )300 0.57515 172.546 mm/sF ABFv AF ω= = =

Law of cosines for triangle DCE. ( ) ( ) ( ) ( ) ( )2 2 22 cos15CE CD DE CD DE= + − °

( ) ( )( )( )2 2 2347.73 300 2 347.73 300 cos15 , 96.889 mmCE CE= + − ° =

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sin15 300 sin15EH DE= ° = °

300sin15cos 36.7

96.889

EH

CEβ β°= = = °

(a) ( ) ( )( )96.889 0.57515 55.7 mm/s,E BCDv CE ω= = =

55.7 mm/sE

=v 36.7°

(b) 172.5 mm/sF

=v 75.0°

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Chapter 15, Solution 90.

=ωωωω

( ) ( )( )160 3

480 mm/s

D DEv DE ω= =

=

is perpendicular to .D

DEv

B Bv=v

Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are

known. Draw BC perpendicular to B

v and DC perpendicular to .

Dv

( )120 mm, cos30 120cos30BD DK BD= = ° = °

120 cos30cos , 49.495 , 180 30 100.505

160

DK

EDβ β φ β°= = = ° = ° − ° − = °

Law of sines for triangle BCD.

sin30 sin sin

CD BC BD

φ β= =

°

( )

( )

sin 30 120sin 3078.911 mm

sin sin

sin 120sin155.177 mm

sin sin

BDCD

BDBC

β βφ φ

β β

° °= = =

= = =

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Law of cosines for triangle ABC. ( ) ( ) ( ) ( ) ( )2 2 22 cos150AC BC AB AB BC= + − °

( ) ( )( )( )2 2 2155.177 200 2 155.177 200 cos150 , 343.27 mmAC AC= + − ° =

Law of sines. sin sin150 200 sin150

sin , 16.9343.27AB AC

γ γ γ° °= = = °

(a) 480

D

ABD

v

CDω = = =

=ωωωω

(b) ( ) ( )( )343.27 6.0828 2088 mm/sA ABDv AC ω= = =

2.09 m/sA

=v 73.1°

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Chapter 15, Solution 91.

AB = 20 in.

Instantaneous centers:

at I for BC.

at J for BD.

Geometry

( )1211 8.25 in.

16IC

= =

( )1221 15.75 in.

16JD

= =

( )1120 13.75 in.

16AI

= =

20 in. 13.75 in. = 6.25 in.BI AB AI= − = −

6.25 in.BJ BI= =

Member BC. 33 in./sC

=v

33

C

BC

v

ICω = = =

( ) ( )( )6.25 in. 4 rad/s 25 in./sB BCv BI ω= = =

Member BD. 25 in./s

B

BD

v

BJω = = =

(a) ( ) ( )( )= = 15.75 in. 4 rad/sD BD

JD ωv 63.0 in./sD

=v

Member AB.

(b) 25 in./s

20 in.

B

AB

v

AB=ωωωω

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 92.

6 in./sA

=v , B B

v=v 30°

Locate the instantaneous center (point C) of rod AB by noting that velocity directions at points A and B are

known. Draw AC perpendicular to A

v and BC perpendicular to B

v .

Triangle ACB. Law of sines. sin 30 sin 30 sin120

AC CB AB= =° ° °

10sin 305.7735 in.

sin120AC CB

°= = =°

(a) 6

A

AB

v

ACω = = =

=ωωωω

( ) ( )( )5.7735 1.03923 6 in./sB AB

CB ω= = =v 30°

D Dv=v

Locate the instantaneous center (point I) of rod BD by noting that velocity directions at points B and D

are known. Draw BI perpendicular toB

v and DI perpendicular to D

v .

Triangle BID. Law of sines. sin120 sin 30 sin 30

BI DI BD= =° ° °

10 sin12017.3205 in.

sin 30BI

°= =°

, 10 in.DI =

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6

B

BD

v

BIω = = =

=ωωωω

(b) ( ) ( )( )10 0.34641 3.46 in./sD BDv DI ω= = =

3.46 in./sD

=v

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Chapter 15, Solution 93.

Method 1

Assume D

v has the direction indicated by the angle β as shown. Draw

CDI perpendicular to .

Dv Then, point C is the instantaneous center of rod

AD and point I is the instantaneous center of rod BD.

.Geometry ( ) ( )2 20.27 0.36 0.45 mAD = + =

( ) ( )2 20.18 0.135 0.225 mBD = + =

0.27 0.18sin 0.6, sin 0.8

0.45 0.225θ φ= = = =

( ) ( )0.45 0.45 0.225 0.225

sin sin 90 cos sin sin 90 cos

c d

θ β β φ β β= = = =

° + ° −

0.45sin

cosc

θβ

= 0.225sin

cosd

φβ

=

0.36 0.27 tan 0.135 0.18 tana bβ β= − = +

.Kinematics 0.4 m/s, 1 m/sA Bv v= =

,A D

v v

a c

ω = = B D

BD

v v

b dω = =

A B

D

cv dvv

a b= =

0.45sin cos 0.40.6

cos 0.225sin 1

A

B

cva b b b

dv

θ ββ φ

= = ⋅ ⋅ =

( )0.36 0.27 tan 0.6 0.135 0.18 tanβ β− = +

0.279 0.378tan , tan 0.73809, 36.43β β β= = = °

( )( )0.36 0.27 0.73809 0.1607 ma = − =

( )( )0.135 0.18 0.73809 0.2679 mb = + =

( )( ) ( )( )0.45 0.6 0.225 0.8

0.3356 m 0.2237 mcos cos

c dβ β

= = = =

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( )a 0.4

A

v

a

=ωωωω

( )b 1

B

BD

v

bω = = = 3.73 rad/s

BD=ωωωω

(c) ( )( )0.3356 2.4891 0.835 m/s,D ADv cω= = = 0.835 m/s

D=v 53.6°

Method 2

Consider the motion using a frame of reference that is translating with

collar A. For motion relative to this frame.

0.4 m/sA

=v

1 m/sB

=v

/ /0, 1.4 m/s

A A B A= =v v

0.27tan , 36.87

0.36θ θ= = °

0.360.45 m

θ= =

/0.45

Locate the instantaneous center (point C) for the relative motion of bar

BD by noting that the relative velocity directions at points B and D are

known. Draw BC perpendicular to B A/

v and DC perpendicular to .

D A/v

( )

0.180.3 m

sin

cos 0.135 0.375 m

CD

BC CD

θθ

= =

= + =

/ 1.4

B A

BD

v

CBω = = =

( ) ( )( )/0.3 3.73 1.120 m/s

D A BDv CD ω= = =

( )a / 1.120

0.45

D A

v

=ωωωω

( )c [/0.4 m/s

D A D A= + =v v v [1.120+ ]θ

0.835 m/s= 53.6° 0.835 m/sD

=v 53.6 °

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Chapter 15, Solution 94.

( )( )0.2 0.2 4.5 0.9 m/sB AB

ω= = =v

Let point C be the instantaneous center of bar BD. Define angle β and

lengths a and b as shown.

0.9B

BD

v

b bω = =

0.9

D BD

aa

bω= =v β

0.7 m/sE

=v

Let point I be the instantaneous center of bar DE. Define lengths c and d

as shown.

0.9D

DE

v a

c bcω = =

( ) ( )0.9 0.20.2 0.7

E DE

a dv d

bcω

−= − = = (1)

0.25 0.15 0.25, , 0.25 tan , 0.15 tan

cos cos 0.15

aa c b d

cβ β

β β= = = = =

Substituting into (1) ( ) 0.25 0.2 0.15 tan0.9 0.7 or 1.2 0.9 tan 0.7 tan

0.15 0.25 tan

β β ββ

− = − =

1.2tan 0.75 36.87 cos 0.8

1.6β β β= = = ° =

0.25

0.3125 m, 0.1875 m, 0.1875 m, 0.1125 m0.8

a b c d= = = = =

( )a 0.9

BD=ωωωω

( )b ( )( )

( )( )0.9 0.3125

DE=ωωωω

( )c ( )( )0.3125 4.8 1.5 m/sDv = = 1.5 m/s

D=v 53.1 °

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Chapter 15, Solution 95.

ω = , vA = vA

B Bv=v ,

E Ev=v

Locate point I, the instantaneous center of rod ABD by drawing IA

perpendicular to vA and IB perpendicular to vB.

9

tan 26.56518

φ φ= = °

18

20.125 in.cos

IDl

φ= =

( )( )5 20.125D ABD IDv lω= =

100.6 in./sD

=v φ

Locate point J, the instantaneous center of rod DE by drawing JD

perpendicular to vD and JE perpendicular to vE.

18

20.125 in.cos

JDl

φ= =

100.6

D

DE

JD

v

lω = = =

=ωωωω

9 cos 27 in.JE JDl l φ= + =

( )( )27 5E JE DEv l ω= =

(b) 135 in./sE

=v

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Chapter 15, Solution 96.

12 in./sA

=v

B Bv=v

Point C is the instantaneous center of bar AB.

12

20cos30

B

AB

v

ACω = =

°

10 in.CD =

( ) ( )( )10 0.69282 6.9282 in./sD ABv CD ω= = =

6.9282 in./sD

=v 30°

E Ev=v 30°

Point I is the instantaneous center of bar DE.

20cos30DI = °

( )a 6.9282

D

DE

v

DIω = = =

=ωωωω

( ) b ( ) ( )( )20sin 30 0.4 4 in./sE DEv EI ω= = ° = 0.333 ft/s

E=v 30 °

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Chapter 15, Solution 97.

Let points A, B, and C move to , , and A B C′ ′ ′ as shown.

Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack.

: lower rack space centrode

Since the point of contact of the gear with the lower rack is always a point on the circumference of the gear, the body centrode is the circumference of the gear.

: circumference of gear body centrode A

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Chapter 15, Solution 98.

Draw x and y axes as shown with origin at the intersection of the two

slots. These axes are fixed in space.

A Av=v , B B

v=v

Locate the space centrode (point C) by noting that velocity directions at points A and B are known. Draw AC perpendicular to

Av and BC

perpendicular to .

Bv

The coordinates of point C are sinCx l β= − and cos

Cy l β=

( )22 2 2300 mm

C Cx y l+ = =

The space centrode is a quarter circle of 300 mm radius centered at O.

Redraw the figure, but use axes x and y that move with the body. Place

origin at A.

( )

( )

( )

2

cos

cos 1 cos22

sin

cos sin sin 22

C

C

x AC

ll

y AC

ll

β

β β

β

β β β

=

= = +

=

= =

( )2 2

22 2 2150 150

2 2C C C C

l lx y x y

− + = = − + =

The body centrode is a semi circle of 150 mm radius centered

midway between A and B.

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Chapter 15, Solution 99.

80 km/h 22.222 m/sAv = =

0 C

=v

1560 mm, 0.280 mm 0.28 m

2d r d= = = =

Point C is the instantaneous center.

0.28

Av

r

ω = = =

2 0.56 mCB r= =

( ) ( )( )0.56 79.364 44.4 m/sBv CB ω= = =

44.4 m/sB

=v

( )130 15

2γ = ° = °

( )( )2 cos15 2 0.28 cos15 0.54092 mCD r= ° = ° =

( ) ( )( )0.54092 79.364 42.9 m/s,Dv CD ω= = =

42.9 m/sD

=v 15.0 °

2 0.28 2 0.39598 mCE r= = =

( ) ( )( )0.39598 79.364 31.4 m/s,Ev CE ω= = =

31.4 m/sE

=v 45.0 °

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Chapter 15, Solution 100.

ω π= =

( ) ( )( )10 30 300 mm/sA OA

OA ω π π= = =v

A Av=v 60 ,°

B Av=v

Locate the instantaneous center (point C of bar BD by noting that

velocity directions at point B and A are known. Draw BC perpendicular to

Bv and AC perpendicular to .

Av

( )sin 30 10sin 30

sin , 1.79160

OA

ABβ β

° °= = = °

( ) ( )cos30 cos 10cos30 160cosOB OA AB β β= ° + = ° +

168.582 mm=

168.582

10 184.662 mmcos30 cos30

OBAC OA= − = − =

° °

( ) tan 30 97.377 mmBC OB= ° =

A B

AB

v v

AC BCω = =

( )( )97.377 300

497 mm/s,184.662

B A

BCv v

AC

π = = =

497 mm/sB

=v

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Chapter 15, Solution 101.

Bar AB.

ω =

( )( )0.25 4B AB ABv l ω= =

1 m/sB

=v

Bar DE.

75

tan 26.565150

φ φ= = °

0.15= 0.167705 m

cosEDl

φ=

0.167705D ED ED EDv l ω ω= =

0.167705D ED

ω=v φ

Bar BD. Locate point I, the instantaneous center of bar BD by drawing IB

perpendicular to vB and ID perpendicular to vD.

0.2

0.4 mtan

IBl

φ= =

0.4

0.44721 mcos

IDl

φ= =

1

B

BD

IB

v

lω = = =

=ωωωω

1.11803 m/sD ID BDv l ω= =

ED

ED

v

ED=ωωωω

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Chapter 15, Solution 102.

(a) 0β = .Wheel AD 0, 45 in./sC D

= =v v

45

D

v

CDω = = =

( ) ( ) 4 2.5 1.5 in.CA CD DA= − = − =

( ) ( )( ) 1.5 11.25 16.875 in./sA ADv CA ω= = =

Rod AB. B Bv=v

Since vA and vB are parallel, the instantaneous center of rod AB lies at infinity.

B A

=v v 16.88 in./sB

=v

0AB

ω =

ω= =v

2.5

tan , 32.0054

DA

DCγ γ= = = °

4.7170 in.cos

DCCA

γ= =

( ) ( )( )4.7170 11.25 53.066 in./sA ADv CA ω= = =

[53.066 in./sA

=v ]32.005°

Rod AB. B Bv=v

4

sin , 18.66312.5

φ φ= = °

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Locate point I, the instantaneous center of bar AB, by drawing IA perpendicular to vA and IB

perpendicular to vB. For triangle ABI,

( )180 90 39.332δ γ φ= ° − − + ° = °

( )12.5

sin sin 90 sin

IA IB

γ φ δ= =

+ °

12.5sin108.66322.345 in.

sin32.005IA

°= =°

12.5sin39.33214.9486 in.

sin32.005IB

°= =°

53.066

A

AB

v

IAω = = =

( ) ( )( )14.9486 2.3748 35.5 in./sB ABv IB ω= = =

35.5 in./sB

=v

=ωωωω

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Chapter 15, Solution 103.

=ωωωω CD CD

ω=ωωωω , BDE BD

ω=ωωωω

( ) ( )( )12 3 36 in./sB AB

AB ω= = =v [, 9D CD

ω=v ]

/D B D B= +v v v

9CD

ω [36= ] [7.5 BDω+ ]

Components: : 7.5 0 0BD BD

ω ω− = =

: 9 36 4 rad/sCD CD

ω ω= =

Acceleration analysis. 0, AB CD CD

α= =αααα αααα , BCD BD

α=αααα

( )B ABAB α= a ( ) 2

ABAB ω +

( )( )12 0=

( )( )2 12 3+

2108 in./s=

( )D CDCD α= a ( ) 2

CDCD ω + [9 CD

α= ( )( )29 4+ [9 CDα =

]

2 144 in./s+

( ) ( )/ /D B D B B Dt n

= + +a a a a

( ) ( )/D B BDt

BD α= a 7.5BD

α =

( ) ( ) 2

/D B BDn

BD ω= a ( )( )27.5 0 = 0=

( ) ( )/ /D B D B D Bt n

= + +a a a a

[9 CDα ] [144+ ] [108= ] [7.5 BD

α+ ] [ ]0+

Components: : 9 0 0CD CD

α α= =

2

: 144 108 7.5 4.8 rad/sBD BD

α α= + =

( )a ( )( )9 0D

= a ] [144+ ] 2144 in./s=

212.00 ft/s

D=a

( )b ( ) ( )/E D ADt

DE α= a ( )( )7.5 4.8 = 2

36 in./s =

( ) ( ) 2

DE ω= a ( )( )20.75 0 = 0 =

( ) ( )/ /E D E D D Et n

= + +a a a a

[144= ] [36+ ] [0+ ] 2180 in./s=

215.00 ft/s

E=a

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Chapter 15, Solution 104.

=ωωωω CD CD

ω=ωωωω , BDE BD

ω=ωωωω

( ) ( )( )12 3 36 in./sB AB

AB ω= = =v [, 9D CD

ω=v ]

/D B D B= +v v v

9CD

ω [36= ] [7.5 BDω+ ]

Components: : 7.5 0 0BD BD

ω ω− = =

: 9 36 4 rad/sCD CD

ω ω= =

.Acceleration analysis

0, AB CD CD

α= =αααα αααα , BCD BD

α=αααα

( )B ABAB α= a ( ) 2

ABAB ω + ( )( )12 0 = ( )( )2 12 3+

2108 in./s=

( )D CDCD α= a ( ) 2

CDCD ω + [9 CD

α = ] ( )( )29 4+ [9 CDα =

]

2 144 in./s+

( ) ( )/ /D B D B B Dt n

= + +a a a a

( ) ( )/D B BDt

BD α= a 7.5BD

α =

( ) ( ) 2

/D B BDn

BD ω= a ( )( )27.5 0 = 0=

( ) ( )/ /D B D B D Bt n

= + +a a a a

[9 CDα ] [144+ ] [108= ] [7.5 BD

α+ ] [ ]0+

Components: : 9 0 0CD CD

α α= =

2: 144 108 7.5 33.6 rad/s

BD BDα α= − + =

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( )a ( )( )9 0D

= a [144+ ] 2144 in./s=

212.00 ft/s

D=a

( )b ( ) ( )/E D ADt

DE α= a ( )( )7.5 33.6= 2

252 in./s =

( ) ( ) 2

DE ω= a ( )( )27.5 0 0 = =

( ) ( )/ /E D E D E Dt n

= + +a a a a

[144= ] [252+ ] 20 396 in./s+ =

233.0 ft/s

E=a

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Chapter 15, Solution 105.

0, ABC ABC

ω α= = =ωωωω αααα , B B

a=a 2

60 , 7.2 m/sD

° =a

/0.45 m

B D=r /

30 , 0.9 mA D

° =r 30°

( ) ( )/ /B D B D D Bt n

= + +a a a a

[ Ba ] [60 7.2° = ] [0.45α+ ] ( )( )260 0.45 0° +

30°

Components.

( ): cos60 7.2 0.45cos60 0B

a α° = − ° + (1)

( ): sin 60 0.45sin 60 0.45B B

a aα α° = ° =

Substituting into (1), ( ) ( )0.45cos60 7.2 0.45cos60α α° = − °

( )a 27.2

α = =°

ABC=αααα

( ) From (2),b ( )( ) 20.45 16 7.2 m/s

Ba = =

27.20 m/s

B=a 60 °

( )c ( ) ( )/ /A D A D A Dt n

= + +a a a a

[7.2= ] [0.9α+ ] ( )( )260 3 0° + 30

°

[7.2= ] [14.4+ ] [ ]60 0° + 2

12.47 m/sA

=a

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Chapter 15, Solution 106.

ABC ABCω= = =ωωωω αααα ,

Collar B slides along a straight rod. ( )B Ba=a 60°

Collar D slides along a straight rod. D Da=a

/0.45 m

B D=r /

30 0.9 mA D

° =r 30°

( ) ( )/ /B D B D B Dt n

= + +a a a a

[ Ba ] [60

Da° = ] ( )( )0.45 12+ ] ( )( )260 1.5 0° + 30

°

Components.

: cos 60 5.4cos60 0B D

a a° = + ° + (1)

2

: sin 60 5.4sin 60 , 5.4 m/s ,B B

a a− ° = ° = −

( ) From (1),a 2

cos60 5.4cos60 10.8cos60 5.4 m/sD B

a a= ° − ° = − ° = −

2

5.40 m/sD

=a

( )b 2

5.40 m/sB

=a 60°

( ) c ( ) ( )/ /A D A D A Dt n

= + +a a a a

[5.4= ] ( )( )0.9 12+ ( )( )260 3 0° + 30°

[5.4= ] [10.8+ ] [ ]60 0° + 2

9.35 m/sA

=a

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Chapter 15, Solution 107.

0ω =

2

2

0.9 ft/sC

=a

Acceleration of A.

/A C A C

= +a a a

2

0.9ft/s= ] [( )AC α+ ]

2

0.9 ft/s= ( )( )210.8 ft 0.8 rad/s+

27.74 ft/s=

2

7.74 ft/sA

=a

Acceleration of B.

/B C B C= +a a a

2

0.9 ft/s= ] [( )BC α+ ]

20.9 ft/s= ] [(1.8) (0.8)+ ] 2

0.54 ft/s=

20.54 ft/s

B=a

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Chapter 15, Solution 108.

Geometry. Let φ be the angle between rod ADE and the vertical.

4

tan7.5

φ = 28.072φ = °

( )2 2 10.5( ) (7.5) (4) 8.5 in. ( ) = 8.5 11.9 in.

Velocity analysis. A Av=v

D Dv=v

Locate point I, the instantaneous center of rod ADB, by

drawing IA perpendicular to vA and ID perpendicular to vD.

0A

AB

v

IAω = =

( ) 0D ABv ID ω= =

0D

DE

v

EDω = =

Acceleration analysis.

2 20.8 ft/s 9.6 in./s

A= =a

0,AB

ω = 0DE

ω =

Rod DE. (Rotation about E) DE DE

α=αααα

( ) ( )D D t D n

= +a a a [6 DEα= ] 2

6DE

ω+ [] 6DE

α= ] (1)

9.6 in./sA

=a AB ABα=αααα

(Plane motion = Translation with A + Rotation about A)

( ) ( )/D A D/A D At n= + +a a a a

[9.6= [] 8.5 AB

α+ ] 28.5

ABφ ω+ ]φ

[9.6= [] 8.5AB

α+ ] 0φ + (2)

Match expressions (1) and (2) for Da and resolve into components.

( ): 0 9.6 8.5 cosAB

φ α= − + 2

α =

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(a) Angular acceleration of rod ADB.

AB=αααα

(b) Acceleration of point B. ( ) ( )/ /B A B A B At n

a= + +a a a

[9.6B

=a [] 11.9AB

α+ ] 2[11.9AB

φ ω+ ]φ

[9.6= ] [15.2319+ ] 0φ +

[3.84= ] [7.1678+ ]

28.1316 in./s= 61.8°

20.678 ft/s

A=a 61.8°

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Chapter 15, Solution 109.

Geometry. Let point C be the center of the center cylinder, B its contact point with

the crate, and D its contact point with the ground. Let r be the radius of

the cylinder. 100 mm.r =

Velocity analysis. Since the contacts at B and D are rolling contacts without slipping,

200 mm/sBv = and 0.

D=v Point D is the instantaneous center of

rotation.

200 mm/s

Bv

r

ω = = =

Point C moves on a horizontal line.

Acceleration analysis. [

C Ca=a ]

/ /( ) ( ) [

D C D C t D C n Ca= + + =a a a a ] [rα+ 2] [rω+ ]

Component : Ca rα0 −==== (1)

/ /( ) ( ) [

B C B C t B C n Ca= + + =a a a a ] [rα+ 2] [rω+ ]

Component : 2

400 mm/sCa rα= + (2)

Solving (1) and (2) simultaneously,

2

200 mm/sCa =

2200 mm/srα =

(a)

2200 mm/s

100 mmα =

/ /( ) ( ) [

A C AC t AC n Ca= + + =a a a a ] [rα+ 2] [rω+ ]

2[200 mm/s= 2] [200 mm/s+ 2] + [(100 mm) (1 rad/s) ]

2[100 mm/s= 2] [200 mm/s+ ]

(b) 223.6 mm/sA

a =

20.224 m/s

A=a 63.4°

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Chapter 15, Solution 110.

Let point C be the contact point.

( )0, 0, C C t

ω= = =v a ωωωω , α=αααα

/0

O C O Cbω= + = +v v v bω=

Point O moves parallel to x-axis.

( ) ( )/ /C O C O C Ot n

= + +a a a a

( )Ct

a ( )Cn

a+ [ Oa= ] [bα+ ] 2

bω+

( )0C

na+ [ O

a= ] [bα+ ] 2bω+

From x-component, , O Oa b bα α= − =a

Acceleration of the point with coordinates ( ),x y

[Oxα= +a a ] [ yα+ ] 2

xω+ 2

yω +

[bα= ] [xα+ ] [ yα+ ] 2xω+

2yω+

Set 0=a and resolve into components.

( ) 2: 0b y xα ω+ + = (1)

2

: 0x yα ω− = (2)

( )2 4

From (2), , From (1), 0y y

x b yω ωαα α

= + + =

4 4

2 2

2 2/

, 1 1

b b yy x

ω ωα α

ω α ωα

−= − = =+ +

Data: 2

12 in., 2 rad/sb ω= = 2

2 4 4

2 2

4 16 25, 1

3 9 9

ω ω ωα α α

= = + =

( )912 4.32 in.,

25y = − = − 5.76 in., 4.32 in. x y= − = −

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Chapter 15, Solution 111.

.Velocity analysis 1.5 in. 10 rad/sr ω= =

Point C is the instantaneous center of the wheel.

( )( )1.5 10 15 in./sA

rω= = =v

.Acceleration analysis

Point moves on a circle of radius .A ρ 6 1.5 7.5 in.R rρ = + = + =

Since the wheel does not slip, C C

a=a

( ) ( )/ /C A C A C At n

a= + +a a a

[ Ca ] ( )A

ta=

2

Av

ρ

+

[rα+

] 2

rω+

( )At

a= ] ( )215

7.5

+

( )( )1.5 30

+

( )( )21.5 10 +

( )A ta= [ 30+ ] [45+ ] [150+ ]

Components. ( ) ( ): 45 0 45 in./sA At t

a a− + = =

: 30 150 120 in./sC Ca a= − + =

(a) Acceleration of point A.

2

45 in./sA

= a 2

30 in./s +

254.1 in./s

A=a 33.7 °

( ) b Acceleration of point B. ( ) ( )/ /B A B A B At n

= + +a a a a

[45B=a ] [30+ ] [rα+ ] 2

rω+

[45= ] [30+ ] ( )( )1.5 30+ ( )( )21.5 10+

2

105 in./s= 2

75 in./s+

2129.0 in./s

B=a 35.5 °

( ) c Acceleration of point C. C Ca=a

2120 in./s

C=a

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Chapter 15, Solution 112.

Velocity analysis.

Let point G be the center of the shaft and point C be the point of contact with the

rails. Point C is the instantaneous center of the wheel and shaft since that point

does not slip on the rails.

24

G

G

v

r

r

ω ω= = = =v

.Acceleration analysis

Since the shaft does not slip on the rails, C C

a=a 20°

Also, 2

10 mm/sG

= a

20 °

( ) ( )/ /C G C G C Gt n

= + +a a a a

[ Ca ] 2

20 10 mm/s° = [20 30α° + ] 220 30ω° + 20 °

Components 2

20 : 10 30 0.33333 rad/sα α° = − =

(a) Acceleration of point A.

( ) ( )/ /A G AG AGt n

= + +a a a a

[10= ] [20 360α° + ] 2360ω+

[9.3969= ] [3.4202+ ] [120+ ] [230.4+ ]

[129.3969= ] [233.4202+ ] 2

267 mm/sA

=a 61.0 °

(b) Acceleration of point B.

( ) ( )/ /B G B G B Gt n

= + +a a a a

[10= ] [20 360α° + ] 2360ω+

[9.3969= ] [3.4202+ ] [120+ ] [230.4+ ]

[110.6031= ] [226.9798+ ] 2

252 mm/sB

=a 64.0 °

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Chapter 15, Solution 113.

.Velocity analysis

160 mm/sD A

= =v v

Instantaneous center is at point .B

( ) ( ), 160 100 60Av AB ω ω= = −

.Acceleration analysis [B B

a=a ] for no slipping α=αααα

2

600 mm/sA

= a

( )A na +

[G Ga=a

]

( ) ( )/ /B A B A B At n

= + +a a a a

[ Ba ] [600= ] ( )A n

a+ ( )100 60 α + − ( ) 2100 60 ω + −

Components 2

: 0 600 40 15 rad/sα α= − + =

( ) ( )/ /B G B G B Gt n

= + +a a a a

[ Ba ] [ G

a= ] [100α+ ] 2100ω+

Components 2

: 0 100 100 1500 mm/sG Ga aα α= − + = =

( )( )2 2

: 100 4 1600 mm/sB

a = = 2

1600 mm/sB

=a

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( ) ( )/ /A G AG AGt n

= + +a a a a

[1500= ] [60α+ ] 260ω+

[1500= ] [900+ ] [960+ ]

2

600 mm/s= 2

960 mm/s +

21132 mm/s

A=a 58.0 °

( ) ( )/ /C G C G C Gt n

= + +a a a a

[1500= ] [100α+ ] 2100ω+

[1500= ] [1500+ ] [1600+ ]

2

3100 mm/s= 2

1500 mm/s +

23440 mm/s

C=a 25.8 °

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Chapter 15, Solution 114.

.Velocity analysis

160 mm/sD B

= =v v

Instantaneous center is at point .A

( ) ( ), 160 100 60Bv AB ω ω= = −

.Acceleration analysis

[A Aa=a ] for no slipping. α=αααα

2

600 mm/sB

= a

( )B na +

[G Ga=a ]

( ) ( )/ /A B A B A Bt n

= + +a a a a

[ Aa ] [600= ] ( )B n

a+ ( )100 60 α + − ( ) 2100 60 ω + −

Components 2

: 0 600 40 15 rad/sα α= − + =

( ) ( )/ /A G AG AGt n

= + +a a a a

A

a [ Ga= ] [60α+ ] 2

60ω+

Components 2

: 0 60 60 900 mm/sG Ga aα α= − = =

( )( )22 2

: 60 60 4 960 mm/sA

a ω= = = 2

960 mm/sA

=a

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( ) ( )/ /B G B G B Gt n

= + +a a a a

[900= ] [100α+ ] 2100ω+

[900= ] [1500+ ] [1600+ ]

2

600 mm/s= 2

1600 mm/s +

2

1709 mm/sB

=a 69.4 °

( ) ( )/ /C G C G C Gt n

= + +a a a a

[900= ] [100α+ ] 2100ω+

[900= ] [1500+ ] [1600+ ]

2

700 mm/s= 2

1500 mm/s +

21655 mm/s

C=a 65.0 °

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Chapter 15, Solution 115.

Velocity analysis. Point A is the instantaneous center of rotation of the cylinder. 12 in./s.C Ev v= =

( )( )12

C

C

v

v r

r

ω ω= = = =

.Acceleration analysis ( )( )22 23 2 12 in./srω = =

[A Aa=a ] ( )

C C ta= a ( )C n

a + [ Ea= ( )C n

a+

( )/C A C At

a=

a a ( )/C At

a+

[ Ea ] ( )C

na+ [ A

a= ] [2rα+ ] 22rω+

[19.2 ] ( )Cn

a+ [ Aa= ] [6α+ ] [24+ ] (1)

From (1), Components 2

: 19.2 6 , 3.2 rad/sα α= =

( ) ( )/ /G A G A G At n

= + +a a a a

[ Ga ] [ A

a= ] [rα+ ] 2rω+

[ Ga ] [ A

a= ] 29.6 in./s+

212 in./s +

From which 2 2 2

9.6 in./s and 12 in./s 9.6 in./sG A Ga a= = =a

From (1), Components ( ) 2: 24 12 24 12 in./s

C Ana a= − + = − + =

Then 2

19.2 in./sC

= a 2

12 in./s + 2

22.6 in./sC

=a 58.0 °

( ) ( )/ /D G DG DGt n

= + +a a a a

[9.6= ] [rα+ ] 2rω+

[9.6= ] [9.6+ ] [12+ ]

[ 221.6 in./s= ] 2

9.6 in./s+

223.6 in./s

D=a 66.0 °

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Chapter 15, Solution 116.

Velocity analysis.

/A B A B= +v v v

[9 in./s ] [6 in./s=

] [(5 in.) ω+ ]

Components

9 6 5ω= − +

.Acceleration analysis

Point A moves on a path parallel to the belt. The path is assumed to be straight.

2

36 in./sA

=a 30°

Since the drum rolls without slipping on the belt, the component of acceleration of point B on the

drum parallel to the belt is the same as the belt acceleration. Since the belt moves at constant

velocity, this component of acceleration is zero. Thus

B Ba=a 60°

Let the angular acceleration of the drum be α .

/ /( ) ( )

B A B A t B A n= + +a a a a

[ Ba ] [36= ] [rα+

2] [rω+ ]

Components : 0 36 5α= −

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.Acceleration of point D

/ /( ) ( )

D A D A t D A n= + +a a a a

[A

a= 30 ] [rα° + 260 ] [rω° + ]30°

[36= 30 ] [(5) (7.2)° + 260 ] [(5) (3)° + 30 ]°

Components: 30 :° 236 45 9 in./s− + =

Components: 60 :° 236 in./s

2 2 2

9 36 37.1 in./sD

a = + =

36

tan 76.09

β β= = °

30 46.0β − °= °

2

37.1 in./sD

=a 46.0°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 117.

.Geometry Define angles and as shown.θ β

0.25cos 0.4 0.25cosβ θ= − (1)

0.25sin 0.25sin 0.2 0.25 sinxβ θ θ= + = + (2)

Dividing (1) by (2) gives 0.2 0.25 sin

tan0.4 0.25 cos

θβθ

+=− (3)

Squaring (1) and (2), adding, and rearranging gives

0.2 cos 0.1 sin 0.2θ θ− = (4)

Let sin .u θ= 2

cos 1 uθ = −

2

0.2 1 0.2 0.1u u− = +

Squaring both sides,

( )2 20.04 1 0.04 0.04 0.01u u u− = + +

2

0.05 0.04 0u u+ =

0 or 0.8u u= = −

Reject the negative root. sin 0 0θ θ= =

From (3), 0.2

tan0.4 0.25

β =− 53.13β = °

.Velocity analysis 1.2 m/sD

= =v

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0.25B AB

ω=v

/D B D B

= +v v v

[1.2 ] [0.25 ABω= ] [0.25 BD

ω+ ]β

Components: : 0.25 sin 0, 0BD BD

ω β ω= =

: 1.2 0.25 4.8 rad/sAB AB

ω ω= =

.Acceleration analysis 0, 0, A D AB AB

α= = =a a αααα

BD BDα=,,,, αααα

( ) ( )/ /B A B A B At n

= + +a a a a

[0 0.25AB

α= + ] 20.25

ABω+ [0.25 AB

α + ] [5.76+ ]

( ) ( )/ /D B D B D Bt n

= + +a a a a

[0 0.25AB

α= ] [5.76+ ] [0.25 BDα+ β

2

0.25BD

ω+ β

[0.25 ABα= ] [5.76+ ] [0.25 BD

α+ ] 0β +

( )a 2

: 0 0 5.76 0.25 sin , 28.800 rad/sBD BD

α β α= + + = −

2

=αααα

( )b ( )( ): 0 0.25 0.25 28.800 cos ,AB

α β= + −

2

=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 118.

.Geometry Define angles and as shown.θ β

0.25cos 0.4 0.25cosβ θ= − (1)

0.25sin 0.25sinxβ θ= + (2)

With 0,x = equation (2) gives .β θ=

From (1), 0.25 cos 0.4 0.25 cosβ β= −

0.4

cos0.5

β = 36.87β = °

36.87θ = °

.Velocity analysis 0.6 m/sD

= =v

[0.25B ABω=v ]θ

/D B D B= +v v v

[0.6 ] [0.25 ABω= ] [0.25 BD

θ ω+ ]β

Components: : 0.6 0.25 cos 0.25 cosAB BD

ω θ ω β= +

: 0 0.25 sin 0.25 sinAB BD

ω θ ω β= −

ω =

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.Acceleration analysis 0, 0, A D AB AB

α α= = =a a , BD BD

α α=

( ) ( )/ /B A B A B At n

= + +a a a a

[0 0.25AB

α= + ] 20.25

ABθ ω+ θ

[0.25 ABα= ] [0.5625θ + ]θ

( ) ( )/ /D B D B D Bt n

= + +a a a a

[0 0.25AB

α= ] [0.5625θ + ] [0.25 BDθ α+ ] 2

0.25BD

β ω+ ]β

[0.25AB

α= ] [0.5625θ + ] [0.25 BDθ α+ ] [0.5625β + ]β

: 0.25 cos 0.5625sin 0.25 cos 0.5625sin 0AB BD

α θ θ α β β+ + − =

: 0.25 sin 0.5625cos 0.25 sin 0.5625cos 0AB BD

α θ θ α β β− − − =

Solving simultaneously, 2 2

α α= = −

( )a 2

=αααα

( )b 2

=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 119.

Law of sines.

sin sin 60, 16.779

2 6

β β°= = °

.Velocity analysis 900 rpm 30 rad/sAB

ω π= =

2 60 in./sB AB

ω π= =v 60°

D D

v=v BD BDω=ωωωω

/

6D B BD

ω=v β

/D B D B

= +v v v

[ Dv ] [60π= ] [60 6

BDω° + ]β

Components

: 0 60 cos60 6 cosBD

π ω β= ° −

6cosBD

πωβ

°= =

.Acceleration analysis 0AB

α =

( )( )22 22 2 30 17765.3 in./s

B ABω π= = =a 30°

D D

a=a BD BDα=αααα

[/6

D B ABα=a ] 2

6BD

β ω+ β

[6 BDα= ] [1615.04β + ]β

( )/ Resolve into components.

D B D B= +a a a

: 0 17765.3cos30 6 cos 1615.04sinBD

α β β= − ° + + 2

α =

( )( ): 17765.3sin 30 6 2597.0 sin 1615.04cosD

a β β= ° − +

2

5931 in./s= P D

=a a 2

494 ft/sP

=a

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 120.

Law of sines.

sin sin120, 16.779

2 6

β β°= = °

.Velocity analysis 900 rpm 30 rad/sAB

ω π= =

2 60 in./sB AB

ω π= =v 60°

D Dv=v BD BD

ω=ωωωω

/6

D B BDω=v β

/D B D B= +v v v

[ Dv ] [60π= ] [60 6

BDω° + ]β

Components

: 0 60 cos60 6 cosB

π ω β= − ° +

6cosBD

πωβ

°= =

.Acceleration analysis 0AB

α =

( )( )22 22 2 30 17765.3 in./s

B ABω π= = =a 30°

D Da=a BD BD

α=αααα

[/6

D B BDα=a ] 2

6BD

β ω+ β

[6 BDα= ] [1615.04β + ]β

/ Resolve into components.

D B D B= +a a a

: 0 17765.3cos30 6 cos 1615.04cosBD

α β β= − ° + +

BDα =

( )( ): 17765.3sin 30 6 2597.0 sin 1615.04cosD

a β β= ° + −

2

11835 in./s= P D

=a a 2

986 ft/sP

=a

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 121.

.Geometry and velocity analysis

0θ =

( ) ( )( )60 16 960 mm/sBv AB ω= = =

960 mm/sB

=v , D

v=v

Instantaneous center of bar lies at .BD C

120

sin 0.6, cos 0.8200

β β= = =

36.9 , 200cos 160 mmCBβ β= ° = =

960

B

BD

v

CBω = = =

.Acceleration analysis 0AB

α =

[60B ABα=a ] 2

60AB

ω+

( )( )20 60 16 = +

215360 mm/s =

Point moves on a straight line.D D Da=a

( ) [/120

D B BDt

α=a

] [160 BDα+ ]

( ) 2

/160

D B BDn

ω= a

2120

BDω + [5760 =

] [4320+ ]

( ) ( )/ /. Resolve into components.

D B D B D Bt n

= + +a a a a

2

: 0 0 160 4320 27 rad/sBD BD

α α= + − =

( )a ( )( ) 2: 15360 120 27 5760 24360 mm/s

Da = − − − = −

224.4 m/s

D=a

( ) [/60

G B BDt

α=a

] [80 BDα+ ] [1620=

] [2160+ ]

( ) 2

/80

G B BDn

ω= a

260

BDω + [2880 =

] [2160+ ]

( )b

( ) ( )/ /G B G B G Bt n

= + +a a a a

[15360= ] [1620+ ] [2160+ ] [2880− ] [2160+ ]

2

19860 mm/s= 2

19.86 m/sG

=a

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 122.

.Geometry and velocity analysis

90θ = °

60sin 0.3, 17.458

200β β= = = °

and are parallel, thus the instantaneous center lies at .D B

∞v v

0BD

ω =

.Acceleration analysis 0, 16 rad/s, 0AB AB BD

α ω ω= = =

[60B ABα=a ] 2

60AB

ω+

( )( )20 60 16 = +

215360 mm/s =

Point moves on a straight line.D D Da=a

( ) [/60

D B BDt

α=a

] [200cos BDβα+

( ) 2

/60

D B BDn

ω= a

2200cos

BDβω +

0 =

( ) ( )/ /. Resolve into components.

D B D B D Bt n

= + +a a a a

215360: 0 15360 200cos , 80.508 rad/s

200cosBD BD

βα αβ

= − + = =

( )a ( )( ) 2: 0 60 0 60 80.508 4831 mm/s

D BDa α= − + = − = −

2

4.83 m/sD

=a

( ) [/30

G B BDt

α=a

] [100cos BDβα+ ] [2415=

] [7680+ ]

( ) 2

/30

G B BDn

ω= a 2

100cosBD

βω +

0 =

( )b ( ) ( )/ /G B B G B Gt n

= + +a a a a

[15360= ] [2415+ ] [7680+ ] 0+

2

2415 mm/s=

27680 mm/s +

28.05 m/s

G=a 72.5 °

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 123.

.Disk A

ω = , 0, 2.8 in.A

ABα = =

( ) ( )( )2.8 15 42.0 in./sB Av AB ω= = =

( ) ( )( )22 22.8 15 630 in./s

B Aa AB ω= = =

( ) 0.a θ =

42 in./sB

=v , D D

v=v

2.8sin 16.26

10β β= = °

Instantaneous center of bar BD lies at point C.

10cos

B

BD

v

BCω

β= = =

2630 in./s

B=a ,

D Da a= ,

BD BDα=αααα

( )/D B BDDB α= a ( ) 2

BDDBβ ω + β

[10 BDα= ] [191.406β + ]β

/ Resolve into components.

D B D B= +a a a

( ) 2: 0 0 10cos 53.594, 5.5826 rad/s

BD BDβ α α= − + = −

( )( ) 2: 630 10sin 5.5826 183.75 430.62 in./s

Da β= − − =

2

=αααα , 2431 in./s

D=a

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( ) 90 .b θ = °

42 in./sB

=v , D D

=v v

5.6sin 34.056

10β β= = °

Instantaneous center of bar BD lies at .∞

0BD

ω =

2630 in./s

B=a ,

D Da=a

( )/D B BDDB α= a ( ) 2

BDDBβ ω + 10

BDβ α = β

/ Resolve into components.

D B D B= +a a a

( ) 2: 0 630 10cos 76.042 rad/s

BD BDβ α α= − − = −

( )( ) 2: 0 0 10 76.042 sin 425.8 in./s

D Da aβ= + − − =

2

=αααα , 2426 in./s

D=a

( ) 180 .c θ = °

42 in./sB

=v D Dv=v

2.8sin 16.26

10β β= = °

Instantaneous center of bar BD lies at point C.

cos 10cos

B

BD

v

BDω

β β= = =

2630 in/s

B=a D D

a=a

( )/D B BDBD α= a ( ) 2

BDBDβ ω + β

[10BD

α= ] [191.406β + ]β

/ Resolve into components.

D B D B= +a a a

2

: 0 10 cos 53.594 5.5826 rad/sBD BD

α β α= − + =

( )( ) 2: 630 10 5.5826 sin 183.75 829.4 in./s

Da β= + + =

2

=αααα , 2829 in./s

D=a

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 124.

Crank AB.

0, 0, 1500 rpm 157.08 rad/sA A AB

ω= = = =v a,

0AB

α =

/0 [2

B A B A ABω= + = +v v v 45 ] 314.16 in./s° = 45°

( ) ( )/ /B A B A B At n

= + +a a a a

0 [2AB

α= + 45 ]° 2[2AB

ω+ 45 ]°

2[(2)(157.08)= 245 ] 49348 in./s° = 45°

Rod BD. D D

v=v 45° BD BD

ω=ωωωω

/D B B D

= +v v v

Dv 45 [314.16°= 45 ] [7.5

BDω° + 45 ]°

Components 45 : 0 314.16 7.5 41.888 rad/sBD BD

ω ω° = − =

D Da=a 45°

( ) ( )/ /D B D B D Bt n

= + +a a a a

[D

a 45 ] [49348° = 45 ] [7.5BD

α° + 245 ] [7.5BD

ω° + 45 ]°

Components 45 :° 2 249348 (7.5)(41.888) 62507 in./sD

a = + = 2

5210 ft/sD

=a 45°

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Rod BE. 2

sin , 15.466 , 45 29.5347.5

γ γ β γ= = ° = ° − = °

E E

v=v 45°

Since E

v is parallel to ,B

v 0BE

ω =

E E

a=a 45° ( ) 2

/7.5 0

B E BEn

a ω= =

( ) ( )/E/B E Bt ta=a β

( )/E B B Et

= +a a a

45 15.466γ β= ° − = °

tanE B

a a γ=

49348tan15.466= °

2

13654 in./s=

21138 ft/s

E=a 45°

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 125.

Geometry. Triangle ABD is equilateral. ( ) 180 mmAD =

=ωωωω,

B Bv=v 30 ,

D Dv° =v

Locate point I, the instantaneous center of bar BD, by drawing IB perpendicular to B

v and ID

perpendicular to .

Dv Point I coincides with A.

(0.18)(5) 0.9 m/sBv = =

B

BD

vω = =

0.18 0.9 m/s

D BDv ω= =

0.18

D

DE

vω = =

Acceleration analysis. 0AB

=αααα

( ) ( ) ( ) 20 0.18

B B B ABt nω= + = +a a a

260 4.5 m/s° = 60°

Bar BD. BDα

( Plane motion Translation with Rotation about B B= + )

( ) ( )/ /D B D B D Bt n

= + +a a a a

[4.5= 60 ] [0.18BD

α° + 230 ] [0.18BD

ω° + 60 ]°

[4.5= 60 ] [0.18BD

α° + 30 ] [4.5° + 60 ]° (1)

Bar DE. (Rotation about E) DEα

( ) ( ) [0.18D D D DEt n

α= + =a a a2] [0.18DE

ω+ ]

[0.18DE

α= ] [4.5+ ] (2)

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Match the expressions (1) and (2) for Da and resolve into components.

: 4.5 2.25 0.18 cos30 2.25BD

α= − + ° −

0.18cos30BD

α = =°

: ( )( )0.18 3.8971 0.18 57.735 sin30 3.8971DE

α = − ° −

DEα = − ° = −

(a) Angular acceleration of bar BD. 257.7 rad/s

BD=αααα

(b) Angular acceleration of bar DE. 228.9 rad/s

DE=αααα

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Chapter 15, Solution 126.

60 120tan , 26.565 , 134.164 mm

120 cosDEβ β

β= = ° = =

ω =

( ) ( )( )200 4 800 mm/sB ABv AB ω= = =

B Bv=v ,

D Dv=v β

Point C is the instantaneous center of bar BD.

160320 mm

tan tan

BDBC

β β= = =

357.77 mmcos

BCCD

β= =

320

B

BD

v

BCω = = =

( ) ( )( )357.77 2.5 894.425 mm/sD BDv CD ω= = =

134.164

D

DE

v

DEω = = =

.Acceleration analysis 0, 4 rad/sAB AB

α ω= =

( )B ABAB α= a ( ) 2

ABAB ω+

( )( )20 200 4= + 2

3200 mm/s =

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( )/D B BDBD α= a ( ) 2

BDBD ω +

[160 BDα= ] ( )( )2160 2.5+

[160 BDα= ] 2

1000 mm/s+

( )D DEDE α= a ( ) 2

DEDE ωβ + β

[134.164 DEα= ] [5961.5β + ]β

/ Resolve into components.

D B D B= +a a a

: 134.164 cos 5961.5sin 0 1000DE

α β β+ = −

DEα = −

( )( ): 134.164 30.55 sin 5961.5cos 3200 160BD

β β α− − + = +

( )a 2

=αααα

DE=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 127.

ω =

( ) ( )( )8 19 152 in./sB ABv AB ω= = =

B Bv=v ,

D Dv=v

Instantaneous center of bar BD lies at C.

8

B

BD

v

BCω = = =

( ) ( )( )19.2 19 364.8 in./sD BDv CD ω= = =

15.2

D

DE

v

DEω = = =

.Acceleration analysis 0.AB

α =

( ) 2

B ABAB ω= a ( )( )28 19 = 2888 in./s =

( )D DEDE α= a ( ) 2

DEDE ω +

[15.2 DEα= ] 2

8755.2 in./s+

( ) [/19.2

D B BDt

α=a ] [8 DBα+ ]

( ) 2

/19.2

D B BDn

ω= a 2

8BD

ω + ]

2

6931.2 in./s= 2

2888 in./s +

( ) ( )/ / Resolve into components.

D B D B D Bt n

a= + +a a a

: 8755.2 0 8 6931.2BD

α= + +

( )a 2

=αααα

( )( ): 15.2 2888 19.2 228 2888DE

α = − + −

α = − 2

=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 128.

ω =

( ) ( )( )15.2 18 273.6 in./sD DE

DE ω= = =v

D Dv=v ,

B Bv=v

Point C is the instantaneous center of bar BD.

19.2

D

BD

v

CDω = = =

( ) ( )( )8 14.25 114 in./sB BDv CB ω= = =

8

B

AB

v

ABω = = =

.Acceleration analysis 0DE

α =

( ) 2

D DEDE ω= a ( )( )215.2 18 =

24924.8 in./s =

( )B ABAB α= a ( ) 2

ABAB ω +

[8 ABα= ] 2

1624.5 in./s+

( ) [/19.2

D B BDt

α=a [8 BDα +

( ) 2

/19.2

D B BDn

ω= a 2

8BD

ω +

2

3898.8 in./s= 2

1624.5 in./s +

( ) ( )/ / Resolve into components.

D B D B D Bt n

a= + +a a a

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: 0 1624.5 19.2 1624.5,BD

α= − +

BDα =

( )( ): 4924.8 8 8 169.21875 3898.8AB

α= + +

ABα = −

( )a ( )( )8 40.96875B

= −a 2

1624.5 in./s +

2

327.75 in./s= 2

1624.5 in./s + ,

2

138.1 ft/sB

=a 78.6 °

( )b ( )/ /

1

2G B G B B D B

= + = +a a a a a

( ) ( )1 1

2 2B D B B D

= + − = +a a a a a

327.75 4924.8

2

− +=

1624.5

2

+

22298.5 in./s=

2 812.25 in./s+

2

203 ft/sG

=a 19.5 °

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 129.

Geometry. Lengths are expressed in meters.

/0.150 0.200 , 0.300 , 0.150 0.200

B A D/B E/Dr += − = =r i j i r i j

/0.150 0.200

H B= −r i j

=ωωωω 4 rad/s = − k

=v

/ /0

B A B A AB B A= + = + ×v v v rωωωω

( ) ( ) ( ) ( )4 0.150 0.200 0.8 m/s 0.6 m/s= − × − = − −k i j i j

Cross BD. ( )/ /0.300 0.3

BD BD D B BD D B BD BDω ω ω= = × = × =k v r k i jωωωω ωωωω

/0.8 0.6 0.3

D B D B BDω= + = − − +v v v i j j

( )/ /0.150 0.200 0.2 0.15

E D DE E D DE DE DEω ω ω= × = × + = − +v r k i j i jωωωω

/0.8 0.6 0.3 0.2 0.15

E D E D BD DE DEω ω ω= + = − − + − +v v v i j j i j

Since point E is fixed, 0E

=v

Components. : 0 0.8 0.2 4 rad/sDE DE

ω ω= − − = −i

( )( ): 0 0.6 0.3 0.15 4BD

ω= − + + −j

(a) 4BD

=ωωωω

Acceleration analysis. 0, 4 rad/sAB AB

ω= =αααα

=a

( ) ( ) 2

/ / /0 0

B A B A B A AB B At n

a ω= + + = + −a a a r

( )( ) ( ) ( )2 216 0.150 0.200 2.4 m/s 3.2 m/s+= − − = −i j i j

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Cross BD. , 4 rad/sBD BD BD

α ω= =kαααα

( ) ( ) 2

/ / / / /D B D B D B BD D B BD D Bt n

ω= + = × −a a a r rαααα

( ) ( ) ( )20.300 4 0.300

BDα= × −k i i

0.3 4.8BD

α= −j i

/D B D B= +a a a

7.2 3.2 0.3BD

α= − + +i j j

α ω= = −kαααα

( ) ( ) 2

/ / / / /E D E D E D DE E D DE E Dt n

ω= + = × −a a a r rαααα

( ) ( ) ( )20.150 0.200 4 0.150 0.200

DE+α= × + −k i j i j

0.15 0.2 2.4 3.2DE DE

α α= − − −j i i j

/E D E D= +a a a

0.3 0.15 0.2 9.6BD DE DE

α α α= + − −j j i i

Since point E is fixed, 0E

=a

Components. 2

: 0 0.2 9.6 48 rad/sDE DE

α α= − − = −i

( )( ): 0 0.3 0.15 48BD

α= + −j

BDα = 2

=αααα

(c) Acceleration of point H.

( ) ( )/ /H B H B H Bt n

= + +a a a a

2

/ /B BD H B BD H Bω= + −a r rαααα

( ) ( ) ( )22.4 3.2 24 0.150 0.200 4 0.150 0.200+= − + × − − −i j k i j i j

10= j 2

10.00 m/sH

=a

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Chapter 15, Solution 130.

ω =

( ) ( )( )8 4 32 in./sD DEv DE ω= = =

D Dv=v ,

B Dv v= 45°

Instantaneous center of bar BD lies at point B. 0Bv =

8

D

BD

v

BDω = = =

0B

AB

v

ABω = =

.Acceleration analysis 2

α = ( )22, 16 rad/s

DEω =

( )D DEDE α= a ( ) 2

DEDE ω +

( )( )8 10= ( )( )8 16 +

2

80 in./s= 2

128 in./s +

( )/B D BDDB α= a ( ) 2

BDDB ω +

[8 BDα= ] ( )( )28 4+

[8 BDα= ] 2

128 in./s+

( )B ABAB α= a ( ) 2

45AB

AB ω + ° 45°

[8 ABα= ]45 0° +

/ Resolve into components.

B D D B= +a a a

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2: 8cos 45 128 128, 45.255 rad/s

AB ABα α− ° = + = −

( )( ): 8sin 45 45.255 80 8BD

α° − = − −

( )a 2

=αααα

( )/G D BDDG α= a ( ) 2

BDDG ω +

( )( )4 22= ( )( )24 4+

288 in./s=

264 in./s +

( )b 2

/168 in./s

G D G D= + = a a a

2192 in./s +

2

255.1 in./s= 41.2°

221.3 ft/s

G=a 41.2 °

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 131.

From the solution to Problem 15.68 the angular velocities are:

=ωωωω .

Bar AB. ( ) ( )/0.300 m 0.125 m , 0

B A AB= + =r i j αααα

0A

=a

( ) ( ) 2

/ / / /0

B A B A B A AB B A AB B At n

ω= + + = + × −a a a a r rαααα

( ) ( )20 0 3 0.3 0.125 2.7 1.125= + − + = − −i j i j

Bar BD. ( )/0.325 m ,

D B BD BDα= − =r j kαααα

( ) ( ) ( ) ( ) ( )2

/ / /0.325 4.85 0.325

D B D B D B BDt n

α= + = × − − −a a a k j j

0.325 7.6448BD

+α= i j

/2.7 6.5198 0.325

D B D B BDα= + = − + +a a a i j i

Bar DE. ( ) ( )/0.150 m 0.200 m ,

E D DE DEα= − + =r i j kαααα

( ) ( ) 2

/ / / / /E D E D E D DE E D DE E Dt n

α ω= + = × −a a a k r r

( ) ( ) ( )20.15 0.2 6 0.15 0.2

DE+α= × − + − −k i j i j

0.15 0.2 5.4 7.2DE DE

α α= − − + −j i i j

/E D E D= +a a a

2.7 0.4448 0.325 0.15 0.2BD DE DE

α α α= + + − −i j i j i

Since point E is fixed, 0E

=a

Components.

: 0 0.6802 0.15 4.5347DE DE

α α= − = −j

( )( ): 0 2.7 0.325 0.2 4.534BD

α= + − −i

(a) 11.10BD

α = − 2

=αααα

DE=αααα

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 132.

3tan , 36.87

4β β= = °

45 in.

cosAB

β= =

45 in.

cosDE

β= =

( ) ( )( )5 15

75 in./s

B ABv AB ω= =

=

B Bv=v ,

D Dvβ =v β

Point C is the instantaneous center of bar BD.

cos 6.25

B

BD

vCB

CBω

β= = = = =

( ) ( )( )56.25 in. 6.25 12 75 in./s

cosD BD

CD v CD ωβ

= = = = =

5

D

DE

v

DEω = = =

.Acceleration analysis 0AB

α =

( )B ABAB α= a ( ) 2

ABABβ ω+ β

( )( )20 5 15= + 2

1125 in./sβ = β

( )/D B BDBD α= a ( ) 2

BDBD ω+

[10 BDα= ( )( )210 12 +

[10 BDα=

21440 in./s+

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( )D DEDE α= a ( ) 2

DEDEβ ω+ β

[5 DEα= ] ( )( )25 15β + β

[5 DEα= ] 2

1125 in./sβ + β

/ Resolve into components.

D B D B= +a a a

( )a : 5 sin 1125cos 1125cos 1440DE

α β β β+ = − −

2

α = − 2

=αααα

( )b ( )( )5 1080D

= −a 2

1125 in./sβ + β

2

5400 in./s= 2

1125 in./sβ + β

1125tan 11.77

5400γ γ= = °

2 2 25400 1125 5516 in./s

Da = + =

2

460 ft/s=

90 64.9β γ° − + = °

2

460 ft/sD

=a 64.9 °

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 133.

.Relative position vectors ( ) ( )/ /200 mm , 160 mm

B A D B= − =r j r i

( ) ( )/60 mm 120 mm

D E= − −r i j

= −ω k

( ) ( ) ( )/4 200 800 mm/s

B AB B A= × = − × − = −v r k j iωωωω

/ /160 160

D B BD D B BD BDω ω= × = × =v r k i jωωωω

( )/60 120 120 60

D DE D E DE DE DEω ω ω= + = × − − = −v r k i j i jωωωω

/ Resolve into components.

D B D B= +v v v

:i 120 800 0 6.6667 rad/sDE DE

ω ω= − + = −

:j 60 160 0 2.5 rad/sDE BD BD

ω ω ω− = + =

.Acceleration analysis ( )0, 4 rad/sAB AB

α ω= = − k

( ) ( ) ( )22 2

/ /0 4 200 3200 mm/s

B AB B A AB B Aω= × − = − − =a r r j jαααα

( ) ( ) ( )22

/ / /160 2.5 160

D B BD D B BD B D BDω α= × − = × −a r r k i iαααα

( )2160 1000 mm/sBD

α= −j i

( ) ( ) ( )22

/ /60 120 6.6667 60 120

D DE D E DE D E DEω α= × − = × − − − − −a r r k i j i jαααα

( ) ( )2 2120 60 2666.7 mm/s 5333.3 mm/s

DE DEα α= − + +i j i j

/D B D B= +a a a Resolve into components.

:i 2

120 2666.7 0 1000 30.556 rad/sDE DE

α α+ = − = −

:j 2

60 5333.3 160 3200 24.792 rad/sDE BD BD

α α α− + = + =

( )a 2

=αααα

( )b 2

=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 134.

Relative position vectors. ( ) ( )/4 in. 3 in.

B A= −r i j

( )/10 in.

D B=r i ( ) ( )/

4 in. 3 in.D E

= − −r i j

=ω k

( ) ( ) ( ) ( )/15 4 3 45 in./s 60 in./s

B AB B A= × = × − = +v ω r k i j i j

/ /

10 10D B BD D B BD BD

ω ω= × = × =v ω r k i j

( )/4 3 3 4

D DE D E DE DE DEω ω ω ω= × = × − − = −v k r k i j i j

/D B B D

= +v v v Resolve into components.

:i 3 45 0DE

ω =

:j 4 60 10DE AB

ω ω− = + 12 rad/sBD

ω = −

Acceleration analysis. 0,AB

=ω k

( ) ( ) ( ) ( )22 2 2

/ /0 15 4 3 900 in./s 675 in./s

B AB B A AB B Aω= × − = − − = − +a α r r i j i j

( ) ( ) ( )22

/ / /10 12 10

D B BD D B BD D B BDω α= × − = × −a α r r k i i

( )210 1440 in./sBD

α= −j i

( ) ( ) ( )22

/ /4 3 15 4 3

D DE D E DE D E DErω α= × − = × − − − − −a α r k i j i j

( ) ( )2 23 4 900 in./s 675 in./s

DE DEα α= − + +i j i j

/D B D B

= +a a a Resolve into components.

:i 3 900 900 1440DE

α + = − − 21080 rad/s

DEα = −

DE=αααα

(b) ( )( ) ( )( ) ( ) ( )2 23 1080 4 1080 900 675 2340 in./s 4995 in./s

D= − − − + + = − +a i j i j i j

2

5516 in./sD

=a 64.9° 2

460 ft/sD

=a 64.9°

COSMOS: Complete Online Solutions Manual Organization System

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 135.

( ) At the instantaneous center ,a C 0C

=v

/ /A C AC AC= + × = ×v v r rωωωω ωωωω

( ) 2

/ /A AC ACω× = × × = −v r rωωωω ωωωω ωωωω

/ / /2 2 or

A A

AC C A C Aω ω× ×= − = − =v v

r r r

ωωωω ωωωω

2

A

C A ω×− = v

r r

ωωωω 2

A

C A ω×= + v

r r

ωωωω

( )b / /A C AC AC= + × + ×a a r vαααα ωωωω

( )

( )

2

2

A

C A C

C A A

C A A

αω

αωωαω

×= − × + × −

= − × × + ×

= + + ×

va k v v

a k k v ω v

a v v

ωωωω ωωωω

ωωωω

Set 0.C

=a A A A

αω

= + ×a v vωωωω

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Chapter 15, Solution 136.

Geometry. Let φ be the angle ADB as shown. By the law of cosines for triangle ABD

( ) ( ) ( ) ( ) ( )22 24 2 4 cos 180c b b b b θ= + − ° −

( )2 217 8cosc b θ= + (1)

17 8cosc b θ= +

Law of sines: sin sinc bφ θ= (2)

Also, cos 4 cosc b bφ θ= +

Velocity analysis. Angular velocity of crank AB: =ω θ&

Angular velocity of rod DE: =DEω φ&

Differentiating (1), 2

2 42 8 sin sin

bcc b c

cθθ θθ= − = −& && &

Differentiating (2), cos sin cosc c bφ φ φ θ θ+ =& &&

( )2

44 cos sin sin cos

b bb b b

c cθ φ θ θ θ θ θ

+ + − =

& & &

( )2 2 3 2

2

cos 4 sin4 cos

b c bb b

c

θ θθ φ θ++ =& &

( ) 217 8cos cos 4sin

17 8cosb

θ θ θθ

θ+ +

=+

&

2

4 17cos 4cos

17 8cosb

θ θ θθ

+ +=+

&

( )( )4 cos 1 4cos

17 8cosb

θ θθ

θ+ +

=+

&

1 4cos

17 8cos

θφ θθ

+=+

& &

1 4cos

17 8cosDE

θ ωθ

+=+

ωωωω

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 137.

From horizontal distances,

( )sin 1 sin

cos cos

cos

cos

l b

l b

b

l

φ θ

φφ θθθφ θφ

= +

=

=

& &

& &

From vertical distances,

( )

cos cos

sin sin

cossin sin

cos

y l b

y l b

bl b

l

φ θ

φϕ θθθφ θ θθϕ

= −

= − +

= − +

&& &

& &

[ ]sin tan cosbθ θ φ θ= −& (1)

Now, , D

y v θ ω= =&&

From geometry, ( )sin 1 sinb

lφ θ= +

( )

( )( )

1/22

2

2

1/222 2

cos 1 1 sin

1 sinsintan

cos1 sin

b

l

b

l b

φ θ

θφφφ θ

= − +

+= =

− +

Substituting into (1), ( )

( )1/2

22 2

1 sin cossin

1 sin

D

bv b

l b

θ θω θ

θ

+ = −

− +

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 138.

From horizontal distances, ( )sin 1 sinl bφ θ= +

cos cos

cos

cos

l b

b

l

φφ θθθφ θφ

=

=

& &

& &

Set constant.θ ω= =&

( ) ( )2

2

2

cos

cos

cos sin cos sin

cos

cos sin sin

coscos

cos sin cos sin

cos coscos

b

l

b

l

b

l

b b

l l

ω θφφ

φ θθ θ φφωφφ

ω θ φ θθφφφ

ω θ φ ω θ θωφ φφ

=

− − −=

= −

= ⋅ −

&

& &

&&

&&

2 2

3

cos sin sin

coscos

b b

l l

ω θ φ θφφ

= −

(1)

Now, BDφ α=&&

From geometry, ( )sin 1 sinb

lφ θ= +

( )1/21/2 22 2 21

cos 1 sin 1 sinl bl

φ φ θ = − = − +

Substituting into (1),

( )( ) ( )

22

3/2 1/22 22 2 2 2

cos 1 sin sin

1 sin 1 sin

BD

b b lb l

ll b l b

θ θω θαθ θ

+ = − − + − +

( )( ) ( )

3 2

2

3/2 1/22 22 2 2 2

cos 1 sin sin

1 sin 1 sin

BD

b b

l b l b

θ θ θα ωθ θ

+ = − − + − +

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 139.

, A Ax r y rθ= =

sinP Ax x r θ= −

sinr rθ θ= −

cosP Ay y r θ= −

cosr r θ= −

Ax

r

θ =

, 0,

,

cos 1 cos

A A

A

P x

vx v y

r

vtx vt

r

vt vx v r r r

r r

θ

θ

θ θθ

= = =

= =

= = − = −

&& &

& &&

1 cos x

vtv v

r

= −

sin sinP y

vt vy v r r

r rθθ = = =

&&

sin y

vtv v

r

=

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 140.

tan cotA

A

b xu

x bθ θ= = =

1

cot uθ −=

21

u

u

θ = −+&

&

( )( )2 2

2

2

11

uu u u

uu

θ = −++

& & &&&&

But, and ω θ α θ= =& &&

, , 0A A A Ax x v v

u u ub b b b

= = = − = − =& &

& &&

Then,

( )2 2 2,

1

A

A

v

Ab

xA

b

bv

b xω = =

++

2 2

A

A

bv

b x=

+ωωωω

( )( )( ) ( )

2

2

2 22 2 2

2 20 ,

1

A A

A

x v

b bA A

xA

b

bx v

b x

α = − = ++

( )2

22 2

2A A

A

bx v

b x

=+

αααα

COSMOS: Complete Online Solutions Manual Organization System

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 141.

( ) ( )1/2 1/22 2 2 2

sin , cos A

A A

b x

b x b x

θ θ= =+ +

( )1/22 2

cosB A

A

A

A

x l x

lxx

b x

θ= −

= −+

( )1/22 2

sinB

A

lby l

b x

θ= =+

( ) ( ) ( )2

1/2 3/2 3/22 2 2 2 2 2

A A A A A

B A A

A A A

lx lx x x lb xx x x

b x b x b x

= − − = −+ + +

& & && & &

( )3/22 2

A A

B

A

lbx xy

b x

= −+

&&

But, ( ) ( ), , A A B B B Bx yx v x v y v= − = =& & &

( )( )

2

3/22 2

A

B Ax

A

lb vv v

b x

= −+

( )( )3/22 2

A AB y

A

lbx vv

b x

=+

COSMOS: Complete Online Solutions Manual Organization System

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 142.

cos

tan sinC

b by

θθ θ

= =

( )( ) ( )( )

2 2

sin sin cos cos

sin sin

C

C

dy d b dv b

dt dt dt

θ θ θ θ θ θθ θ

− −= = = −

2 2

0sin sin

Cd v v

dt b b

θ θ θ ω= − = − =

Angular acceleration.

2

0 02 sin cos sind d d v v

dt d dt b b

ω ω θ θ θ θαθ

− − = = =

2

302 sin cos

v

bθ θ =

αααα

COSMOS: Complete Online Solutions Manual Organization System

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 143.

From geometry, sin

ry

θ=

2

cos

sin

dy r d

dt dt

θ θθ

= −

0But, and

dy dv

dt dt

θ ω= − =

0 2

cos

sin

r

v

θ ωθ

=

2

0sin

cos

v

r

θωθ

=

Angular acceleration.

d d d d

dt d dt d

ω ω θ ωα ωθ θ

= = =

( )2 3

2

0 0

2

2cos sin sin sin

coscos

v v

r r

θ θ θ θθθ

+ =

( )2 32

0

3

1 cos sin

cos

v

r

θ θ

θ

+ =

( )2

2 301 cos tan

v

r

θ θ = +

αααα

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 144.

Law of cosines for triangle ABE.

( )2 2 22 cos 180u l b bl θ= + − ° −

2 2

2 cosl b bl θ= + +

coscos

θφ += l b

u

sintan

cos

θφθ

=+b

l b

( ) ( )( ) ( )( )( )

2

2

cos cos sin costan sec

cos

θ θ θ θ θ θφ φφ

θ+ +

= =+

& &&

l b b b bd

dt l b

( ) ( )( )

2 2 2 2

2

cos cos cos sin

cos

φ θ θ θ θφ

θ

+ + =

+

&

&bl b

l b

( )2

2 2 2

coscos

2 cos

b b lbl b

u l b bl

θθ θ θθ

++= =+ +

& &

But, , , and θ ω φ ω= = = −& & &BD E

v u

Hence, ( )

2 2

cos

2 cosBD

b b l

l b bl

θω

θ+

=+ +

ωωωω

Differentiate the expression for 2.u

2 2 sinuu bl θθ= − &&

( )1/22 2

sin

2 cos

E

blv u

l b bl

θ ωθ

= − =+ +

&

( )1/22 2

sin

2 cos

E

bl

l b bl

ω θ

θ=

+ +v

1 sintan

cos

b

l b

θθ

− +

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Chapter 15, Solution 145.

Law of cosines for triangle ABE.

( )2 2 22 cos 180u l b bl θ= + − ° −

2 2

2 cosl b bl θ= + +

coscos

l b

u

θφ +=

sintan

cos

b

l b

θφθ

=+

( ) ( )( ) ( )( )( )

2

2

cos cos sin costan sec

cos

l b b b bd

dt l b

θ θ θ θ θ θφ φφ

θ+ +

= =+

& &&

( ) ( )( )

2 2 2 2

2

cos cos cos sin

cos

bl b

l b

φ θ θ θ θφ

θ

+ + =

+

&

&

( )2

2 2 2

coscos

2 cos

b b lbl b

u l b bl

θθ θ θθ

++= =+ +

& &

( )

2 2

cos

2 cos

b b l

l b bl

θφ θ

θ+

=+ +

&& &&

( )( ) ( )( )

( )2 2

2

22 2

2 cos sin cos 2 sin

2 cos

l b bl bl b b l bl

l b bl

θ θ θ θθ

θ

+ + − − + −+

+ +&

( ) ( )

( )2 2

2

22 22 2

sincos

2 cos 2 cos

bl l bb b l

l b bl l b bl

θθθ θ

θ θ

−+= −

+ + + +&& &

But, , 0, BD

θ ω θ ω φ α= = = =& && &&&

( )2 2

2

2 2

sin

2 cosBD

bl l b

l b bl

θω

θ

−=

+ +αααα

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Chapter 15, Solution 146.

Define angles and θ φ as shown.

, tθ ω θ ω= =&

Since the wheel rolls without slipping, the arc OC is equal to arc PC.

( )r Rφ θ φ+ =

r

R r

θφ =−

r r

R r R r

θ ωφ = =− −

&&

r t

R r

ωφ =−

( )sin sinPx R r rφ θ= − −

( ) ( )cos cosP Pxv x R r rφφ θθ= = − −& &&

( ) ( )( )cos cos

r t rR r r t

R r R r

ω ω ω ω = − − − −

( ) cos cos P x

r tv r t

R r

ωω ω = − −

( )cos cosPy R R r rφ θ= − − −

( ) ( )sin sinP Pyv y R r rφφ θθ= = − +& &&

( ) ( )( )sin sinr t r

R r r tR r R r

ω ω ω ω = − + − −

( ) sin sin P y

r tv r t

R r

ωω ω = + −

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Chapter 15, Solution 147.

Define angles and θ φ as shown.

, , 0tθ ω θ ω θ= = =& &&

Since the wheel rolls without slipping, the arc OC is equal to arc PC.

( ) 2r R rφ θ θ θ+ = =

0φ =

φ θ ω= =& &

0φ θ= =&& &&

( )sin sin sin sin 0Px R r r r rφ θ θ θ= − − = − =

The path is the -axis. y

( )cos cos cos cosPy R R r r R r rφ θ θ θ= − − − = − −

( )1 cosR θ= −

sinP

v y R θθ= = && ( )sin R tω ω=v j

( )2 2cos sin cosa v R Rθθ θθ ω θ= = − =& &&&

( )2cos R tω ω=a j

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Chapter 15, Solution 148.

Geometry.

180 120 20 40APB = ° − ° − ° = °

Law of sines.

sin 40 sin 20 sin120

b AP PB= =° ° °

0.5321 0.15963 m

1.3473 0.40419 m

AP b

BP b

= == =

Let P′ be the point on rod AD that coincides with P at the instant shown.

Let u = u 60° be the velocity of P relative to rod AD.

( )/P P P D AAP ω′ = + = v v v ] [30 u° + ]60°

[1.5963 m/s= ] [30 u° + ]60°

But collar P is attached to rod BP. Β B

ω=ωωωω

( ) 0.40419P B Bv BP ω ω= =

0.40419P B

ω=v 70°

Equating the two vectors for vP gives the vector diagram shown.

From the vector diagram,

1.5968

0.40419cos40

Bω =

°

ω =

1.5968 tan 40 1.340 m/su = ° =

=v 60°

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Chapter 15, Solution 149.

Geometry.

30θ = °

250

288.68 mmcos30

AP = =°

250 tan30 144.34 mmBP = ° =

=ωωωω

Rod AE. Let P′ be the point on rod AE coinciding with the pin P.

( ) ( ) ( )288.68 4 1154.70 mm/sP Av AP ω′ = = =

1154.70 mm/sP′ =v θ

/P/AE P AE

v=v θ

P P P/AE′= +v v v

[1154.70 mm/s= θ ] /P AEv+ ]θ

Rod BD. Let P′′ be the point on rod BD coinciding with the pin P.

( ) ( ) ( )144.34 5 721.69 mm/sP Bv BP ω′′ = = =

721.69 mm/sP′′ =v

/P/BD P BDv=v

P P P/BD′′= +v v v

[721.69 mm/s= ] /P BDv+ ]

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Equate the two expressions for vP and resolve into components. ( )30θ = °

: /

1154.70 sin30 cos30 721.69 0P AEv− ° + = − +

/

166.67 mm/sP AEv = −

/166.67 mm/s

P AE=v 30°

:θ /

1154.70 0 721.69 sin30 cos30P BDv+ = ° + °

/

916.67 mm/sP BDv =

/916.67 mm/s

P BD=v

[1154.70P=v ] [30 166.67° + ]30°

[721.69 mm/s= ] [916.67 mm/s+ ]

1167 mm/sP

=v 51.8°

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Chapter 15, Solution 150.

Coordinates.

( )0

,

0,

, 0

sin

cos

A A A

B B

C A C

P A

P

x x r y r

x y r

x x y

x x e

y r e

θ

θθ

= + =

= == == += +

Data: ( )0

24 in.Ax =

10 in.

7 in.

r

e

==

Velocity analysis. AC AC

ω=ωωωω , BD BDω=ωωωω ,

[/P A P A ACrω= + =v v v ] [ AC

eω+ ]θ

[P P BDx ω′ =v ] ( )cos

BDe θ ω+ ]

/

[ cosP F

u β=v ] [ sinu β+ ]

Use /P P P F′= +v v v and resolve into components.

( ) ( ) ( ): cos cos cosAC BD

r e e uθ ω θ ω β+ = + (1)

( ): sin AC

e θ ω =P BDx ω ( ) sin uβ− (2)

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(a) 0.θ = 24 in. 24 in. 20 rad/sA P ACx x ω= = =

cos 7tan , 16.26

24P

e

x

θβ β= = = °

Substituting into Eqs. (1) and (2),

( ) ( ) ( )10 7 20 7 cos16.26BD

uω+ = + ° (1)

( )0 24 sin16.26BD

uω= − ° (2)

ω =

326.4 in./s,u = /27.2 ft/s

P F=v 16.26 °

( )b 90 .θ = °

( )24 10 7 46.708 in.2

Px

π = + + =

0β =

Substituting into Eqs. (1) and (2),

( ) ( )10 20 u= (1)

200 in./s = 16.67 ft/su =

( ) ( )7 20 46.708BD

ω= (2)

=ωωωω

/

16.67 ft/sP F

=v

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Chapter 15, Solution 151.

Coordinates.

( )0

,

0,

, 0

sin

cos

A A A

B B

C A C

P A

P

x x r y r

x y r

x x y

x x e

y r e

θ

θθ

= + =

= == == += +

Data: ( )0

24 in.Ax =

10 in.

7 in.

r

e

==

Velocity analysis. AC AC

ω=ωωωω , BD BDω=ωωωω ,

[/P A P A ACrω= + =v v v ] [ AC

eω+ ]θ

[P P BDx ω′ =v ] ( )cos

BDe θ ω+ ]

/

[ cosP F

u β=v ] [ sinu β+ ]

Use /P P P F′= +v v v and resolve into components.

( ) ( ) ( ): cos cos cosAC BD

r e e uθ ω θ ω β+ = + (1)

: ( )sinAC

e θ ω = P BDx ω ( ) sin uβ− (2)

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( )30 , 24 10 7sin30 32.736 in.6

Px

πθ = ° = + + ° =

7cos30tan 10.491

32.736β β°= = °

Substituting into Eqs. (1) and (2)

( )( ) ( ) ( )10 7cos30 20 7cos30 cos10.491BD

uω+ ° = ° + ° (1)

( )( ) ( )7sin30 20 32.736 sin10.491BD

uω° = − ° (2)

ω =

303.13 in./su = /

25.3 ft/sP F

=v 10.49 °

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Chapter 15, Solution 152.

ω = , /E E F EFr ω= v ] ( )( )0.375 10 3.75 m/s.= =

[/3.75 m/s.

B E B E′ = + =v v v ] [0.5 DEω+ ]

/B ED

u=v

[/3.75 m/s.

B B B ED′= + =v v v ] [0.5 DEω+ ] [u+ ]

( )B ABAB ω= v ( )0.375

45 15cos45

° = ° [45 5.625 m/s

° =

] [5.625 m/s.+ ]

Equate the two expressions for B

v and resolve into components.

: 3.75 5.625, 1.875 m/s.u u+ = =

: 0.5 5.625, 11.25 rad/sDE DE

ω ω= =

ω =

(b) /

1.875 m/s.B DE

=v

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Chapter 15, Solution 153.

ω = /E E F EFr ω= v ( )( )0.375 15 5.625 m/s = =

[/5.625 m/s

B E B ED′ = + =v v v ] [0.5 DEω+ ] [5.625 m/s= ] [5 m/s+ ]

/B EDu=v

[/5.625 m/s

B B B ED′= + =v v v ] [5 m/s+ ] [u+ ]

( )B AB

AB ω= v 0.375

45cos45

ABω

° = ° [45 0.375

ABω° =

] [0.375 AB

ω+ ]

Equate the two expressions for B

v and resolve into components.

: 5.625 0.375AB

u ω− + = − (1)

: 5

AB ABω ω= = =

ω =

From (1), ( )( )5.625 0.375 13.333 0.625 m/su = − =

(b) /

0.625 m/sB ED

=v

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Chapter 15, Solution 154.

ω =

( )( )12 8 96 in./sB

= =v

ω =

( )( )12 6 72 in./sD′ = =v

Bar BD. Assume angular velocity is BDω

.

Plane motion = Translation with B + Rotation about B.

[96D B D/B= + =v v v ] [24 BD

ω+ ] [12 BDω+ ]

(1)

Collar D. Sliding on rotating rod EF with relative velocity u .

[72D D D/EF′= + =v v v ] [u+ ] (2)

Matching the expressions (1) and (2) for v ,D

Components : 96 12 72BD

ω+ = − 14BD

ω = −

ω =

Components : 24BD

uω = ( )( )24 14 336 in./su = − = −

(b) = 28.0 ft/su

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Chapter 15, Solution 155.

ω =

( )( )1 ft 4 rad/s 4 ft/sB

= =v

Bar BD. Angular velocity is BDω .

Plane motion = Translation with B + Rotation about B.

[4D B D/B= + =v v v ] [2 BD

ω+ ] [1 BDω+ ]

Magnitude of : 20 ft/sD D

v =v

( ) ( ) ( )2 2 224 2 20

D BD BDv ω ω= + + =

2

5 8 384 0BD BD

ω ω+ − =

8 88

10BD

ω − ±= Positive root 8 rad/sBD

ω =

[4D=v ] ( )( )2 8+ ] ( )( )1 8+ ] [12= ] [16+ ] (1)

Rod EF. (Rotation about E) Angular velocity = EF

ω

( )1D' EFω= v ]

Collar D. Slides on rotating rod EF with relative velocity u .

[1D D' D/EF EFω= + =v v v ] [u+ ] (2)

Matching the expressions (1) and (2) for ,D

v

(a) Component : 12 1EF

ω =

(b) Component : 16 = u 16 ft/su =

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Chapter 15, Solution 156.

Body ACD rotates about point A with angular velocity

If the rod EF were not moving relative to body ACD, the velocity of point H ′ currently at point H, would be

( )/1.6 600 0.96 mm/s

H A H A′ = × = × =v r k i jωωωω

The velocity of point H relative to H ′ is

/

300 mm/sH H ′ =v ( )300 mm/s= i

(a) Velocity of tip H.

( ) ( )/960 mm/s + 300 mm/s

H H H H′ ′= + =v v v j i

H1006 mm/s=v 72.6°

Angular acceleration of body ACD: 0A

=αααα

( ) ( )22

/0 1.6 600

H A H/A A H Aω′ = × − = −a r r iαααα

( )21536 mm/s= − i

Since the sliding motion of H relative to H ′ occurs at constant velocity,

/

0H H ′ =a

Coriolis acceleration. 2c A H/H ′= ×a vωωωω

( ) ( ) ( ) ( )22 1.6 300 960 mm/sca = × =k i j

(b) Acceleration of tip H.

/H H H H c′ ′= + +a a a a

( ) ( )2 21536 mm/s + 0 + 960 mm/s= − i j

21811 mm/s

H=a 32.0°

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Chapter 15, Solution 157.

Body ACD rotates about point A with angular velocity

If the rod EF were not moving relative to body ACD, the velocity of point ,G′ currently at point G, would be

( )1

/1.6 600 400

G A G A= × = × +v r k i jωωωω

( ) ( )640 mm/s 960 mm/s= − +i j

The velocity of point G relative to G′ is

/

300 mm/sGG′ =v ( )300 mm/s= i

(a) Velocity of tip G.

/

640 960 300G G G G′ ′= + = − + +v v v i j i

( ) ( )340 mm/s 960 mm/s= − +i j

1018 mm/sG

=v 70.5°

Angular acceleration of body ACD is 0.A

=αααα

2

/ /G A G A A G Aω′ = × −a r rαααα

( ) ( )20 1.6 600 + 400= − i j

( ) ( )2 21536 mm/s 1024 mm/s= − −i j

Since the sliding motion of G relative to G′ occurs at constant velocity,

/

0G G′ =a

Coriolis acceleration. /

2c A G G′= ×a vωωωω

( ) ( ) ( ) ( )22 1.6 300 960 mm/sca = × =k i j

(b) Acceleration of tip G.

/G G G G c′ ′= + +a a a a

1536 1024 0 960= − − + +i j j

( ) ( )2 21536 mm/s 64 mm/s= − −i j

21537 mm/s

G=a 2.4°

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Chapter 15, Solution 158.

For each pin: /P P P F c′= + +a a a a

Acceleration of the coinciding point P of the plate.′

2For each pin, towards the center .P

r Oω′ =a

Acceleration of the pin relative to the plate.

1 2 4For pins , , and ,P P P /

0P F

=a

3For pin ,P

2

/P F

u

r

=a

.

cCoriolis acceleration a

For each pin 2ca uω= with

ca in a direction obtained by rotating u through 90° in the sense of ,ωωωω i.e. .

Then, 2

1rω= a [2 uω + ]

2

12 r uω ω= −a i j

2

2rω= a [2 uω + ]

2

22 u rω ω= −a i j

2

3rω= a

2u

r

+

[2 uω

+

]

2

2

32

u

r u

r

ω ω

= − + +

a i

2

4rω= a [2 uω + ] ( )2

42 r uω ω= +a j

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Chapter 15, Solution 159.

For each pin: /P P P F c′= + +a a a a

Acceleration of the coinciding point P of the plate.′

2For each pin towards the center .P

r Oω′ =a

Acceleration of the pin relative to the plate.

1 2 4For pins , , and ,P P P /

0P F

=a

3For pin ,P

2

/P F

u

r

=a

.

cCoriolis acceleration a

For each pin 2ca uω= with

ca in a direction obtained by rotating u through 90° in the sense of .ωωωω

Then, 2

1rω= a [2 uω + ]

2

12 r uω ω= +a i j

2

2rω= a [2 uω + ]

2

22 u rω ω= − −a i j

2

3rω= a

2u

r

+

[2 uω

+

]

2

2

32

u

u r

r

ω ω

= − −

a i

2

4rω= a [2 uω + ] ( )2

42r uω ω= −a j

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Chapter 15, Solution 160.

16 in. 1.3333 ft, 16 2 in. 1.3333 2 ftAP BP= = = =

=ωωωω , /0, 8 ft/s

AE P AE= =vαααα , /

0.P AE

=a

Find: /

and .BD P BD

aαααα

Velocity of coinciding point P′ on rod AE.

( ) ( )( )1.3333 6 8 ft/sP AE

AP ω′ = = =v = ( )8 ft/s i

Velocity of P relative to rod AE. ( )/8 ft/s

P AE=v j

Velocity of point P. ( ) ( )/8 ft/s 8 ft/s

P P P AE′= + = +v v v i j

Velocity of coinciding point P′′ on rod BD.

( )P BDBP ω′′ =v 45 1.3333 2

BDω° = 45 1.3333 1.3333

BD BDω ω° = − +i j

Velocity of P relative to rod BD. ( ) ( )/cos45 sin 45

P BDu u= ° + °v i j

Velocity of point P. /P P P BD′′= +v v v

( ) ( )1.3333 1.3333 cos45 sin 45P BD BD

u uω ω= − + + ° + °v i j i j

Equating the two expressions for P

v and resolving into components.

( ): 8 1.3333 cos 45BD

uω= − + °i (1)

( ): 8 1.3333 sin 45BD

uω= + °j (2)

Solving (1) and (2), ( ) ( )/0, 8 2 ft/s, 8 ft/s 8 ft/s

BD P BDuω = = = +v i j

Acceleration of coinciding point P′ on rod AE.

( ) ( ) ( )( ) ( )22 20 1.3333 6 48 ft/s

P AE AEAP APα ω′ = − = − = −a i j j j

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Acceleration of P relative to rod AE. /

0P AE

=a

Coriolis acceleration. ( )( ) ( )2/

2 2 6 8 96 ft/sAE P AE

× = − × =v k j iωωωω

Acceleration of point P.

( ) ( )2 2

/ /2 96 ft/s 48 ft/s

P P P AE AE P AE′= + + × = −a a a v i jωωωω

Acceleration of coinciding point P′′ on rod BD.

2

/ /1.3333 1.3333 0

P BD P B BD P B BD BDα ω α α′′ = × − = − + +a k r r i j

Acceleration of P relative to rod BD. ( ) ( )/cos45 sin 45

P BD r ra a= ° + °a i j

Coriolis acceleration. /

2 0BD P BD

× =vωωωω

Acceleration of point P.

/ /2

P P P AE BD P BD′′= + + ×a a a vωωωω

( ) ( )1.3333 1.3333 cos45 sin 45BD BD r r

a aα α= − + + ° + °i j i j

Equating the two expressions for P

a and resolving into components.

( ): 96 1.3333 cos 45BD r

aα= − + °i (3)

( ): 48 1.3333 sin 45BD r

aα− = + °j (4)

Solving (3) and (4), 254 rad/s , 24 2 ft/s

BD raα = − =

BD=α

(b) 2

/33.9 ft/s

P BD=a 45 °

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Chapter 15, Solution 161.

16 in. 1.3333 ft, 16 2 in. 1.3333 2 ftAP BP= = = =

=ωωωω , /0, 10.5 ft/s

BD P BD= =vαααα /

45 , 0.P BD

° =a

Find: /

and .AE P AE

α a

Velocity of coinciding point P′ on rod AE.

( ) 1.3333P AE AE

AP ω ω′ = =v 1.3333AE

ω= − i

Velocity of P relative to rod AE. /P AE

u=v j

Velocity of point P. /

1.3333P P P AE AE

uω′= + = − +v v v i j

Velocity of coinciding point P′′ on rod BD.

( )P BDBP ω′′ =v ( )( )45 1.3333 2 6° = ( ) ( )45 8 ft/s 8 ft/s° = − +i j

Velocity of P relative to rod BD. ( ) ( )/

5.25 2 ft/s 5.25 2 ft/sP BD

= − −v i j

Velocity of point P. ( ) ( )/

15.4246 ft/s 0.57538 ft/sP P P BD′′= + = − +v v v i j

Equating the two expressions for P

v and resolving into components.

: 1.3333 15.4246 11.5685 rad/sAE AE

ω− = − =i ω

: 0.57538u = −j /

0.57538 ft/sP AE

=v

Acceleration of coinciding point P′ on rod AE.

( )( )22

/ /1.3333 1.3333 11.5685

P AE P A AE P A AEω α′ = × − = − −a k r r i jαααα

1.3333 178.440AE

α= − −i j

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Acceleration of P relative to rod AE. /P AE r

a=a j

Coriolis acceleration. ( )( )( ) ( )2/

2 2 11.5685 0.57538 13.3126 ft/sAE P AE

ω × = − = −k v i i

Acceleration of point P. / /

2P P P AE AE P AE′= + + ×a a a vωωωω

( )1.3333 13.3126 178.440P AE r

aα= − + + − −a i j i j

Acceleration of coinciding point P′′ on rod BD.

( ) ( ) ( ) ( )22 2 2

/ /0 6 1.3333 1.3333 48 ft/s 48 ft/s

P BD P B BD B Dα ω′′ = × − = − + = − −a k r r i j i j

Acceleration of P relative to rod BD. /

0P BD

=a

Coriolis acceleration. ( )( )( )/2 2 6 10.5 126

BD P BDvω = = 45°

( ) ( )2 2126cos 45 ft/s 126sin 45 ft/s= ° − °i j

Acceleration of point P. / /

2P P P AE BD P BD

vω′′= + +a a a

( ) ( )2 248 48 126cos 45 126sin 45 41.0955 ft/s 137.0955 ft/s

P= − − + ° − ° = −a i j i j i j

Equating the two expressions P

a and resolving into components.

: 1.3333 13.3126 41.0955AE

α− − =i 2

α = −

: 178.440 137.0955ra − = −j

241.345 ft/s

ra =

AE=αααα

(b) 2

/41.3 ft/s

P AE=a

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Chapter 15, Solution 162.

Earth makes one revolution ( )2 radiansπ in 23.933 h = 86160 s.

86160

π −= = ×Ω j

Speed of sled. 600 mi/hr 880 ft/su = =

Velocity of sled relative to the Earth.

( )/earth880 sin cosφ φ= − +

Pv i j

Coriolis acceleration. /earth

2c P

= ×a vΩΩΩΩ

( )( ) ( )62 72.926 10 880 sin cos

0.12835sin

cφ φ

φ

− = × × − +

=

a j i j

k

At latitude , 40 ,φ = °

0.12835sin 40c

= °a k

20.0825 ft/s= k

20.0825 ft/s west

c=a

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Chapter 15, Solution 163.

Earth makes one revolution ( )2 radiansπ in 23.933 h (86160 s).

86160

π −= = ×j jΩΩΩΩ

Velocity relative to the Earth at latitude angle .φ

( )/earth12.2 cos sin

Pφ φ= − −v i j

Coriolis acceleration .

ca

( )( ) ( )

( )

/earth

6

3

2

2 72.926 10 12.2 cos sin

1.7794 10 cos

c P

φ φ

φ

= ×

= × × − −

= ×

a v

j i j

k

ΩΩΩΩ

(a) 0 , cos 1.000φ φ= ° =

3 21.779 10 m/s west

c

−= ×a

(b) 40 , cos 0.76604φ φ= ° =

3 21.363 10 m/s west

c

−= ×a

(c) 40 , cos 0.76604φ φ= − ° =

3 21.363 10 m/s west

c

−= ×a

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Chapter 15, Solution 164.

Geometry: ( ) ( )/16 3 in. 16 in.

P A= +r i j

.Motion of coinciding point P on rod AB′

( ) ( ) ( ) ( )( ) ( )

( ) ( )

/

22

/ /

2 2

5 16 3 16 80 in./s 80 3 in./s

0 5 16 3 16

400 3 in./s 400 in./s

P AB P A

P AB P A AB P Aω

= × = × + = − +

= × − = − +

= − −

v v k i j i j

a r r i j

i j

ωωωω

αααα

Motion of collar P relative to rod AB. 6 ft/s 72 in./s=

( ) ( )/ /72cos30 72sin 30 36 3 in./s 36 in./s , 0

P AB P AB= − ° − ° = − − =v i j i j a

Coriolis acceleration. /2

AB P AB× vωωωω

( )( ) ( ) ( ) ( )2 22 5 72cos30 72sin30 360 in./s 360 3 in./s× − ° − ° = −k i j i j

(a) Motion of collar P.

/80 80 3 36 3 36

P P P AB′= + = − + − −v v v i j i j

( ) ( )142.35 in./s 102.56 in./s ,= − +i j 14.62 ft/sP

=v 35.8 °

/ /2 400 3 400 0 360 360 3

P P P AB AB P AB′= + + × = − − + + −a a a v i j i jωωωω

( ) ( )2 2332.82 in./s 1023.54 in./s ,= − −i j

289.7 ft/s

P=a 72.0 °

(b) Motion of point D.

/142.35 102.56 8

D P D P PDω= + = − + +v v v i j j

16D DE

ω= −v i

Equating the two expressions for D

v and resolving into components,

: 142.35 16DE

ω =

: 0 102.56 8 12.820 rad/sPD PD

ω ω= + + = −j

( )( ) ( )16 8.897 142.35 in./sD

= − = −v i i 11.86 ft/sD

=v

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( ) ( )

( )

2

/

2

8 8

332.82 1023.54 8 1314.82

16 16 16 1266.5

D P P D P PD PD

PD

D DE DE DE

α ω

α

α ω α

= + = + × −

= − − + −

= × − = − −

a a a a k i i

i j j i

a k j j i j

Equating the two expressions for and resolving into components,D

a

: 1647.64 16 ,DE

α− = −i

DEα =

2: 1023.54 8 1266.5, 30.375 rad/s

PD PDα α− + = − = −j

( )( ) ( ) ( )2 216 102.98 1266.5 1647.7 in./s 1266.5 in./s

D= − − = − −a i j i j

2173.2 ft/s

D=a 37.5 °

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Chapter 15, Solution 165.

Geometry. ( ) ( )/16 3 in. 16 in.

P A= +r i j

Motion of coinciding point P on rod AB.′

( ) ( ) ( ) ( )/4 16 3 16 64 in./s 64 3 in./s

P AB P A′ = × = − × + = −v r k i j i jωωωω

( ) ( )22

/ /0 4 16 3 16

P AB P A AB P Aω′ = × − = − +a r r i jαααα

( ) ( )2 2256 3 in./s 256 in./s= − −i j

Motion of collar P relative to rod AB. 9.6 ft/s 115.2 in./s=

( ) ( )/ /115.2cos30 115.2sin 30 57.6 3 in./s 57.6 in./s , 0

P AB P AB= − ° − ° = − − =v i j i j a

Coriolis acceleration. /

2AB P AB

× vωωωω

( )( ) ( ) ( ) ( )2 22 4 115.2cos30 115.2sin 30 460.8 in./s 460.8 3 m/s − × − ° − ° = − +k i j i j

Motion of collar P.

/64 64 3 57.6 3 57.6

P P P AB′= + = − − −v v v i j i j

( ) ( )35.7661in./s 168.451in./s= − −i j

/ /2 256 3 256 0 460.8 460.8 3

P P P AB AB P AB′= + + × = − − + − +a a a v i j i jωωωω

( ) ( )2 2904.205 in./s 542.129 in./s = − +i j

/35.7661 168.451 8

D P D P PDω= + = − − +v v v i j j

16D DE

ω= −v i

Equating the two expressions for D

v and resolving into components,

: 35.7661 16 ,DE

ω =

: 0 168.451 8 ,PD

ω =

( ) ( )2

/8 8

D P P D P PD PDα ω= + = + × −a a a a k i i

904.205 542.129 8 3546.98PD

α= − + + −i j j i

( )216 16 16 79.952

D DE DE DEα ω α= × − = − −a k j j i j

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Equating the two expressions for D

a and resolving into components,

: 4451.18 16 ,DE

α− = −i

DEα =

: 542.129 8 79.952,PD

α+ = −j

PDα = −

=ωωωω

PD=αααα ,

DE=αααα

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 166.

For each configuration ( ) ( )/275 mm 100 mm

D A= +r i j

Acceleration of coinciding point .D′ 2

/ /D D A D Aα ω′ = × −a k r r

( ) ( ) ( ) ( )2 2 20 2.4 275 100 1584 mm/s 576 mm/s

D′ = − + = − −a i j i j

Acceleration of point D relative to arm AB. /0

D AB=a

Length .CD 2 2

75 100 125 mmCD = + =

Velocity of point D relative to the arm AB.

Case (a) ( )

/250 mm/s

D AB=v i

Case (b) ( ) ( ) ( )/

25075 100 150 mm/s 200 mm/s

125D AB

= + = +v i j i j

Coriolis acceleration. /2

D Bω ×k v

Case (a) ( )( ) ( )22 2.4 250 1200 mm/s× =k i j

Case (b) ( )( ) ( ) ( ) ( )2 22 2.4 150 200 960 mm/s 720 mm/s× + = − +k i j i j

Acceleration of nozzle D.

/ /2

D D D AB D AB′= + + ×a a a k vωωωω

(a) ( ) ( )2 21584 576 1200 1584 mm/s 624 mm/s

D= − − + = − +a i j j i j

( ) ( )2 2 21584 624 1702 mm/s

D= + =a

624tan , 21.5

1584β β= = °

21.702 m/s

D=a 21.5 °

(b) ( ) ( )1584 576 960 720 2544 mm/s 144 mm/sD

= − − − + = − +a i j i j i j

( ) ( )2 2 22544 144 2548 mm/s

D= + =a

144tan , 3.24

2544β β= = °

22.55 m/s

D=a 3.24 °

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Chapter 15, Solution 167.

For each configuration ( ) ( )/275 mm 100 mm

D A= +r i j

.Acceleration of coinciding point D′ 2

/ /D D A D Aα ω′ = × −a k r r

( ) ( ) ( ) ( )2 2 20 2.4 275 100 1584 mm/s 576 mm/s

D′ = − + = − −a i j i j

. Acceleration of point D relative to arm AB /0

D AB=a

Length .CD 2 2

75 100 125 mmCD = + =

Velocity of point D relative to the arm AB.

Case (a) ( )/250 mm/s

D AB= −v i

Case (b) ( ) ( ) ( )/

25075 100 150 mm/s 200 mm/s

125D AB

= + = − −v i j i j

Coriolis acceleration. /2

D Bω ×k v

Case (a) ( )( ) ( ) ( )22 2.4 250 1200 mm/s× − = −k i j

Case (b) ( )( ) ( ) ( ) ( )2 22 2.4 150 200 960 mm/s 720 mm/s× − − = −k i j i j

Acceleration of nozzle D.

/ /2

D D D AB D AB′= + + ×a a a k vωωωω

(a) ( ) ( )1584 576 1200 1584 mm/s 1776 mm/sD

= − − − = − −a i j j i j

( ) ( )2 2 21584 1776 2380 mm/s

D= + =a

1776tan , 48.3 ,

1584β β= = °

22.38 m/s

D=a 48.3 °

(b) ( ) ( )2 21584 576 960 720 624 mm/s 1296 mm/s

D= − − + − = − −a i j i j i j

( ) ( )2 2 2624 1296 1438 mm/s

D= + =a

1296tan , 64.3 ,

624β β= = °

21.438 m/s

D=a 64.3 °

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Chapter 15, Solution 168.

=ωωωω , 0AB

α =

( )( )0.4 sin 45 15 4.2426 m/sE Ev ′ ′= ° =v

( ) ( )( ) ( )0.4sin 45 0 0E Et t

a ′ ′= ° =a

( ) ( )( ) ( )2 20.4sin 45 15 63.64 m/s

E En na ′ ′= ° =a

Collar E. (Slides on rotating rod AB) 3 m/su = , 0u =&

/[4.2426

E E E AB′= + =v v v ] [3+ ]

Coriolis acceleration: ( )( )( )2 2 15 3 90 m/sABuω = =

2

90 m/sc

=a

/E E E AB c′= + +a a a a

[63.64= ] [u+ & ] [90+ ] [63.64= ] [90+ ]

Rod DE. Angular velocity and acceleration: DEωωωω ,

DEαααα

D Dv=v ,

D Da=a

Plane motion Translation with Rotation about E E.= +

/D E D E

= +v v v

[Dv ] [4.2426= ] [3+ ] [0.4

DEω+ 45 ]°

Components : 0 3 0.4 sin 45 10.6066DE DE

ω ω= + ° = −

( ) ( )/ / /D E D E E D E D Et n

= + = + +a a a a a a

[D

a ] [63.64= ] [90+ ] 0.4DE

α+ 245 ] [0.4DE

ω° + 45 ]°

Components : 0 63.64 0.4 sin 45 45sin 45DE

α= + ° − °

α = −

(a) Angular velocity of rod DE. 10.61 rad/sDE

=ωωωω

(b) Angular acceleration of rod DE. 2112.5 rad/s

DE=αααα

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Chapter 15, Solution 169.

ω = , 0AB

α =

( )( )0.4 sin 45 10 2.8284 m/sE Ev ′ ′= ° =v

( ) ( )( ) ( )0.4sin 45 0 0E Et t

a ′ ′= ° =a

( ) ( )( ) ( )2 20.4sin 45 10 28.284 m/s

E En na ′ ′= ° =a

Collar E. (Slides on rotating rod AB) 2 m/su = , 0u =&

/[2.8224

E E E AB′= + =v v v ] [2+ ]

Coriolis acceleration: ( )( )( ) 22 2 10 2 40 m/s

ABuω = =

2

40 m/sc

=a

/E E E AB c′= + +a a a a

[28.284= ] [u+ & ] [40+ ] [28.284= ] [40+ ]

Rod DE. Angular velocity and acceleration: DEωωωω ,

DEαααα

D Dv=v ,

D Da=a

Plane motion Translation with Rotation about E E.= +

/D E D E

= +v v v

[Dv ] [2.8284= ] [2+ ] [0.4

DEω+ 45 ]°

Components : 0 2 0.4 sin 45 7.0711DE DE

ω ω= − + ° =

Components : ( )( )2.8284 0.4 7.07011 cos45 4.8284D Dv v= + ° =

( ) ( )/ / /D E D E E D E D Et n

= + = + +a a a a a a

[D

a ] [28.284= ] [40+ [] 0.4DE

α+ 245 ] [0.4DE

ω° + 45 ]°

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Components : 0 28.284 0.4 sin 45 20sin 45DE

α= + ° − °

α = −

Components : ( )( )40 0.4 50 cos45 20cos45D

a = + − ° + °

40 D

α =

(a) Velocity of collar D. 4.83 m/sD

=v

(b) Acceleration of collar D. 240.0 m/s

D=a

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Chapter 15, Solution 170.

6 rad/sω = , 0, 180 mm/s 0.18 m/s, 0u uα = = = =&

( )22

2 2180

200 mm, 162 mm/s 0.162 m/s200

uρρ

= = = =

( )( )( )2 2 2 2 236 rad /s , 2 2 6 180 2160 mm/s 2.16 m/suω ω= = = =

(a) Point A. /0, 0.18 m/s

A A F= =r v

2

/0,

A A F

u

ρ′ = =a a 2

0.162 m/s=

Coriolis acceleration. 2 uω 2

2.16 m/s=

[/2

A A A Fuω′= + +a a a ] 2

2.322 m/s=

22.32 m/s

A=a

(b) Point B. 0.2 2 mB

=r /45 , 0.18 m/s

B F° =v

( )( )236 0.2 2

B Bω′ = − = −a r

245 7.2 2 m/s° = 45°

2

2

/0.162 m/s

B F

u

ρ= =a

Coriolis acceleration. 2

2 2.16 m/suω =

[/2

B B B Fuω′= + +a a a ] 2

9.522 m/s=

27.2 m/s +

211.94 m/s

B=a 37.1°

(c) Point C. 0.4 mC

=r /, 0.18 m/s

C F=v

( )(236 0.4

C Cω′ = − = −a r ) 2

14.4 m/s=

2

2

/0.162 m/s

C F

u

ρ= =a

Coriolis acceleration. 2

2 2.16 m/suω =

[/2

C C C Fuω′= + +a a a ] 2

16.722 m/s=

216.72 m/s

C=a

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Chapter 15, Solution 171.

60 3 2

π π πω α θ= = = = = ° =

.Uniform rotational motion 0tθ θ ω= +

0 2

2

3

0.75 st

π

πθ θ

ω−= = =

.Uniform motion along rod 0r r ut= +

0/

20 10 40 40 in./s, in./s

0.75 3 3P AB

r ru

t

− −= = = =v

Acceleration of coinciding point P on the rod.′ ( )20 in.r =

( )2 2

2 22 8020 in./s

3 9P

r

π πω′ = = =

a

287.730 in./s=

.Acceleration of collar P relative to the rod /0

P AB=a

Coriolis acceleration. ( ) 2

/

2 402 2 2 55.851 in./s

3 3P AB

u

πω × = = =

vωωωω

.Acceleration of collar P

/ /2

P P P AB P AB′= + + ×a a a vωωωω

287.730 in./s

P= a

2 55.851 in./s +

2104.0 in./s

P=a 57.5°

2104.0 in./s

Pa =

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Chapter 15, Solution 172.

Given:

75 mm/sB

=v

0B

=a

150 mmb =

Definitions:

ω=ωωωω , α=αααα

902

θφ = ° −

2 cosAB b φ=

Point B′ is point on rod AD that currently

coincides with pin B.

Velocity analysis /B B B AD′= +v v v

B B

v=v

( ) ( )2 cosB

AB bω φ ω′ = =v

φ

[Bv ( )] [ 2 cosb φ ω=

/] [

Components.

( ): cos 2 cos sin 0Bv bφ φ φ ω φ = +

2 sin

Bv

b φ=ω (1)

( ) /: 0 2 cos cos

B ADb vφ ω φ= −

/2

sin

B

vbω

φ= =v

φ (2)

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Acceleration analysis. /B B B AD c′= + +a a a a

0B

=a

[( )B

AB α′ =a2] ( )[AB ω+ ] [(2 cos )b φ α= 2] [(2 cos )b φ ω+ ]

/ /[( )

/] [( )

/] [( )

2

/]

φρ

+

φ

/[( )

2

2]

sin

Bv

φ

+

/[2

2

2]

sin

Bv

φ

=

φ

Components.

2: 0 (2 cos ) sin (2 cos ) cosb bφ φ α φ φ ω φ= −2 2

2 20

sin sin

B Bv v

b bφ φ+ − +

2 cos

sin

φα ωφ

= (3)

Data: 75 mm/s, 150 mm, 110Bv b θ= = = °

90 55 35φ = ° − ° = °

(a) From (1), ( )( )75

ω = =°

(b) From (3), ( )2 2cos350.43586 0.271 rad/s

sin35α °= =

°

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Chapter 15, Solution 173.

Given:

75 mm/sB

=v

0B

=a

150 mmb =

Definitions:

ω=ωωωω , α=αααα

902

θφ = ° −

2 cosAB b φ=

Point B′ is point on rod AD that currently

coincides with pin B.

Velocity analysis /B B B AD′= +v v v

B B

v=v

( ) ( )2 cosB

AB bω φ ω′ = =v

φ

[Bv ( )] [ 2 cosb φ ω=

/] [

Components.

( ): cos 2 cos sin 0Bv bφ φ φ ω φ = +

2 sin

Bv

b φ=ω (1)

( ) /: 0 2 cos cos

B ADb vφ ω φ= −

/2

sin

B

vbω

φ= =v

φ (2)

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Acceleration analysis. /B B B AD c′= + +a a a a

0B

=a

[( )B

AB α′ =a2] ( )[AB ω+ ] [(2 cos )b φ α= 2] [(2 cos )b φ ω+ ]

/ /[( )

/] [( )

/] [( )

2

/]

φρ

+

φ

/[( )

2

2]

sin

Bv

φ

+

/[2

2

2]

sin

Bv

φ

=

φ

Components.

2: 0 (2 cos ) sin (2 cos ) cosb bφ φ α φ φ ω φ= −2 2

2 20

sin sin

B Bv v

b bφ φ+ − +

2 cos

sin

φα ωφ

= (3)

Data: 75 mm/s, 150 mm, 90Bv b θ= = = °

90 45 45φ = ° − ° = °

(a) From (1), ( )( )75

ω = =°

(b) From (3), ( )2 2cos450.35355 0.1250 rad/s

sin 45α °= =

°

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Chapter 15, Solution 174.

Coordinates.

( )0,

0,

, 0

sin , cos

A A A

B B

C A C

P A P

x x r y r

x y r

x x y

x x e y r e

θ

θ θ

= + =

= == == + = +

Data: ( )024 in.Ax =

10 in.r =

7 in.e =

0 24 in.Pxθ = =

Velocity analysis.

20 rad/sAC =ωωωω , BD BDω=ωωωω

( )P ACr e ω= +v

( )( )10 7 20= +

340 in./s=

[P P BDx ω′ =v ] [ BDeω+ ]

[/ cosP F u β=v ] [ sinu β+ ]

7tan

24

16.260P

e

β

= =

= °

Use /P P P F′= +v v v and resolve into components.

: 340 7 cosBD uω β= + (1)

: 0 24 sinBD uω β= − (2)

Solving (1) and (2), 3.8080 rad/s, 326.4 in./sBD u= =ωωωω

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Acceleration analysis. 0, AC BD BDα= =α αα αα αα α

( )( )22 2/0 7 20 2800 in./sA P A ABrω= = = =a a

2/ 2800 in./sP A P A= + =a a a

P P BDx α′ = a [ Beα + [ 2P BDx ω +

2BDeω +

24 BDα= [7 BDα + ( )( )224 3.8080 + ( )( )2

7 3.8080+

24 BDα= [7 BDα + [ 2348.021 in./s + [ 2101.506 in./s+

[/ cosP F u β=a & [ sinu β + &

Coriolis acceleration.

( )( )( ) [ 22 2 3.8080 326.4 2485.86 in./sBDuω = = ]β

Use / 2P P P F BD uω′ = + + a a a ]β and resolve into components.

: 0 7 348.021 cos 2485.86sinBD uα β β= − + +&

or 7 cos 348.02BD uα β+ = −& (3)

: 2800 24 101.506 sin 2485.86cosBD uα β β= + − +&

or 24 sin 312.07BD uα β− =& (4)

Solving (3) and (4), 8.09 rad/s, 421.48 in./sBD uα = = −&

( )a 28.09 rad/sBD =αααα !

( )b 2

/ 35.1 ft/sP F =a 16.26° !

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Chapter 15, Solution 175.

Coordinates.

( )0,

0,

, 0

sin , cos

A A A

B B

C A C

P A P

x x r y r

x y r

x x y

x x e y r e

θ

θ θ

= + =

= == == + = +

Data: ( )024 in.Ax =

10 in.r =

7 in.e =

( ) ( )90 24 10 7 46.708 in., 02Pxπθ β = ° = + + = =

Velocity analysis.

20 rad/sAC =ωωωω , BD BDω=ωωωω

/P A P A= +v v v

[ ACrω= ] [ ACeω+ ]

( )( )10 20= ( )( )7 20+

[200 in./s= [140 in./s ] + ]

p BDP x ω′ = v 46.708 BDω = , /P F u=v

Use /P P P F′= +v v v and resolve into components.

: 200 u= 200 in./s.u =

: 140 46.708 BDω= 2.9973 rad/sBDω =

Acceleration analysis. 0, AC BD BDα= =α αα αα αα α

( )( )22 2/0, 7 20 2800 in./sA P A ABrω= = = =a a

2/ 2800 in./sP A P A= + =a a a

P P BDx α′ = a 2

P BDx ω + [46.708 BDα = ( )( )246.708 2.9973+

46.708 BDα= 2419.616 in./s +

/P F u=a &

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Coriolis acceleration ( )( )( ) 22 2 2.9973 200 1198.92 in./sBDuω = =

Use / 2P P P F Buω′= + +a a a and resolve into components.

2: 2800 419.616 , 2380.4 in./s ,u u− = − + = −& &

: 0 46.708 1198.92BDα= + 225.7 rad/sBD = −αααα

(b) 2/ 2380 in./sP F =a 2

/ 198.4 ft/sP F =a !

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Chapter 15, Solution 176.

Geometry. 5

tan , 26.56510

β β= = °

1011.1803 in.

cosAEl

β= =

=ω ,BD BD

ω=ω

( ) ( )( )5 12 60 in./sB AB

AB ω= = =v

( )E B BDBE ω′ = +v v

β

[60=

] [11.1803 BDω+

[/E BDu=v

], 0E

β =v

Use /E E E BD′= +v v v and resolve into components.

+

: 0 60sin 11.1803 , 2.400 rad/sBD BD

β β ω ω= − + =

+

: 0 60cos , 53.666 m/su uβ β= − =

[60E′ =v ] ( )( )11.1803 2.400+ 53.7 in./sβ = 63.4°

Acceleration analysis.

( ) ( )( )22 25 12 720 in./s

B ABAB ω= = =a

( )E B BDBE α′ = + a a ( ) 2

BDBEβ ω + β

[720= ] [11.1803 BDα+ ] [64.399β + ]β

[/E BDu=a & 0

Eβ = a

Coriolis acceleration.

( )( )( ) [2 2 2.400 53.666 257.60BDuω = = ]β

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Use [/2

E E E BD BDuω′= + +a a a ]β and resolve into components.

+

: 0 720cos 11.1803 257.60BD

β β α= − + +

BDα =

+

2: 0 720sin 64.399 , 257.59 in./su uβ β= − + − = −& &

[720E′ =a ] ( )( )11.1803 34.56+ ] [64.399β + ]β

[720= ] [386.39+ ] [64.399β + ]β

2365 in./s= 18.4°

Summary:

BD=αααα

(b) 53.7 in./sE′ =v 63.4 ,° 2

365 in./sE′ =a 18.4 °

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Chapter 15, Solution 177.

.Geometry 18

18 in., 30 , 20.7846 in.cos30

R ABθ= = ° = =°

:Pin B B Bv=v ,

B Ba=a

Let rod be a rotating frame of reference.AC 2

( ) [20.7846B

AB ω ω′ = =v ]30°

[20.7846Bα′ =a ] 2

30 20.7847ω° + 30°

Motion of B relative to the frame.

[/B ACu=v ] [/

30 , B AC

u° =a & 30 °

B B B ACω′= + =v v v

Bv [ 20.7846ω= ] [30 u° + ]30°

( )( ): 0 20.7846 6 sin30 cos30u= − ° + °

72.000 in./s,u =

( )( ): 20.7846 6 cos30 72.000sin30Bv = ° + °

144.000 in./sBv = 144.0 in./s

B=v

.Coriolis acceleration ( )( )( ) 22 2 6 72.000 864 in./suω = = 30

°

.Acceleration analysis [/2

B B B ACuω′= + +a a a ] 2

[ Ba ] [20.7846α= ] 2

30 20.7847ω° + ] [30 u° + & [30 1728° + ]30°

[83.1384= ] [30 748.246° + [30 u° + & [30 864° + ]30°

[947.14= ] [30 748.246° + ] [30 u° + & 30°

: 0 947.14sin30 748.246cos30 cos30u= − ° − ° + °&

21295.51 in./su =&

: 947.14cos30 748.246sin30 1295.51sin30B

a = ° − ° + °

2

1094 in./s= 2

1094 in./sB

=a

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Chapter 15, Solution 178.

Geometry.

Law of cosines.

( ) ( ) ( )2 2 2

1.25 2.50 2 1.25 2.50 cos 30

1.54914 in.

r

r

= + − °

=

Law of sines. sin sin 30

1.25 r

β °=

23.794β = °

Let disk S be a rotating frame of reference. S

ω=ΩΩΩΩ ,S

α=&ΩΩΩΩ

Motion of coinciding point P′ on the disk.

1.54914P S S

rω ω′ = =v β

[2

/ /1.54914

P S P O S P O Sα ω α′ = − × − =a k r r ] 2

1.54914S

β ω+ β

Motion relative to the frame.

/P Su=v /

P Suβ =a & β

Coriolis acceleration. 2Suω β

[/1.54914

P P P S Sω′= + =v v v ] [uβ + ]β

/2

P P P S Suω= + +a a a

[1.54914 Sα= ] 2

1.54914S

β ω+ [uβ +& [2 S

uω + ]β

Motion of disk D. (Rotation about B)

( ) ( )( )1.25 8 10 in./sP D

BP ω= = =v 30°

( )P DBP α= a ( ) 2

60S

BP ω° + ( )( )230 0 1.25 8° = + 30 °

280 in./s= 30°

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Equate the two expressions for P

v and resolve into components.

( ): 1.54914 10cos 30S

β ω β= ° +

1.54914S

ω °= =

=ωωωω

( ): 10sin 30 10sin 53.794 8.0690 in./suβ β= ° + = ° =

Equate the two expressions for P

a and resolve into components.

( ): 1.54914 2 80sin 30S S

uβ α ω β− = ° +

( )( )( ) 280sin 53.794 2 3.8130 8.0690

° += =

S=αααα

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Chapter 15, Solution 179.

Geometry.

Law of cosines.

( ) ( ) ( )2 2 2

1.25 2.50 2 1.25 2.50 cos 45

1.84203 in.

r

r

= + − °

=

Law of sines. sin sin 45

1.25 r

β °=

28.675β = °

Let disk S be a rotating frame of reference. S

ω=ΩΩΩΩ ,S

α=&ΩΩΩΩ

Motion of coinciding point P′ on the disk.

1.84203P S S

rω ω′ = =v β

[2

/ /1.84203

P S P O S P O Sα ω α′ = − × − =a k r r ] 2

1.84203S

β ω+ β

Motion relative to the frame.

/P Su=v /

P Suβ =a & β

Coriolis acceleration. 2Suω β

[/1.84203

P P P S Sω′= + =v v v ] [uβ + ]β

/2

P P P S Suω= + +a a a

[1.84203 Sα= ] 2

1.84203S

β ω+ [uβ + & [2 Suω + ]β

Motion of disk D. (Rotation about B)

( ) ( )( )1.25 8 10 in./sP D

BP ω= = =v 30°

( )P DBP α= a ( ) 2

45S

BP ω ° + ] ( )( )245 0 1.25 8° = + 45°

280 in./s= 45°

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Equate the two expressions for P

v and resolve into components.

( ): 1.84203 10cos 45S

β ω β= ° +

1.84203S

ω °= =

=ωωωω

( ): 10sin 45 10sin 73.675 9.5968 in./suβ β= ° + = ° =

Equate the two expressions for P

a and resolve into components.

( ): 1.84203 2 80sin 45S S

uβ α ω β− = ° +

( )( )( ) 280sin 73.675 2 1.52595 9.5968

° += =

S=αααα

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Chapter 15, Solution 180.

Geometry. 30θ = °

250

288.68 mmcos30

AP = =°

250 tan30 144.34 mmBP = ° =

See Solution of Problem 15.149, which obtains

/166.67 mm/s

P AE=v 30°

/916.67 mm/s

P BD=v

Acceleration analysis. 0, 0A B

= =α α

Rod AE. Let P′ be the point on rod AE coinciding with the pin P.

( ) ( ) 0P At

a AP α′ = =

( ) ( ) ( )( )22 2288.68 4 4618.9 mm/s

P Ana AP ω′ = = =

2

4618.9 mm/sP

=a 30°

/ /P AE P AEa=a 30°

Coriolis acceleration: ( ) /2

c A P AEAEa vω=

( )( )( ) 22 4 166.67 1333.3 mm/s= =

( ) 21333.3 mm/s

c AE=a 60°

( )/P P P AE c AE′= + +a a a a

= 24618.9 mm/s ] /

30P AE

a° + ] 230 1333.3 mm/s° + 60°

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Rod BD. Let P′′ be the point on rod BD coinciding with the pin P.

( ) ( ) 0P Bt

a BP α′′ = =

( ) ( ) ( )( )22 2144.34 5 3608.5 mm/s

P Bna BP ω′′ = = =

( ) 23608.5 mm/s

P′′ =a

/ /P BD P BDa=a

Coriolis acceleration. ( ) /2

c B P BDBDa vω=

( )( )( ) 22 5 916.67 9166.7 mm/s= =

( ) 29166.7 mm/s

c BD=a

( )/P P P BD c BD′′= + +a a a a

[3608.5= ] /P BDa+ ] 2

9166.7 mm/s+ ]

Equate the two expressions for Pa and resolve into components.

/: 4618.9cos30 cos30 1333.3cos60

P AEa− °+ °+ °

0 0 9166.7= + −

2 2

/ /6735.7 mm/s 6735.7 mm/s

P AE P AEa = − =a 30°

/

60 : 0 0 1333.3 3608.5sin 60 sin 60 9166.7cos60P BD

a° + + = °− °− °

/ /

3223.4 mm/s 3223.4 mm/sP BD P BD

a = − =a

Using Pa from rod AE.

4618.9P

=a 30 6735.7°+ 30 1333.3°+ 60°

29166.7 mm/s= ] 2

6832.0 mm/s+ ]

Check using Pa from rod BD.

[3608.5P=a ] [3223.4+ ] [9166.7+ ]

29166.7 mm/s= ] 2

6831.9 mm/s+ ]

2

11430 mm/sP

=a 36.7° 2

11.43 m/sP

=a 36.7°

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Chapter 15, Solution 181.

Position of A. ( ) ( ) ( )100 mm + 250 mm 300 mmA

= − −r i j k

Velocity of A. ( ) ( ) ( )10 mm/s + + 80 mm/sA A yv=v i j k

A A

= ×v rωωωω

Since vA is perpendicular to rA, the scalar product

0A A

⋅ =r v

( ) ( ) ( ) ( ) ( ) ( )100 10 250 300 80 0A A A yv⋅ = − + − =r v

( ) 100 mm/sA yv =

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Chapter 15, Solution 182.

Radius of ball: 4.3 in. = 0.35833 ft

At the given instant, the origin is not moving.

: 14.4 14.4 10.8

0.35833 0.35833 0

A A x y zω ω ω= × − + =i j k

v r i j kωωωω

( )14.4 14.4 10.8 0.35833 0.35833 0.35833z z x y

ω ω ω ω− + = − + + −i j k i j k

( )

z z

z z

x y x y

ω ωω ω

ω ω ω ω

− = = −= − = −

− = − =

i

j

k

: 28.8 21.6

0 0.71667 0

D D x y zω ω ω= × + =i j k

v r i kωωωω

28.8 21.6 0.71667 0.71667z x

ω ω+ = − +i k i k

z z

x x

ω ωω ω

− = = −= =

i

k

30.140 0y x

ω ω= − =

( ) .a Angular velocity ( ) ( )30.1 rad/s 40.2 rad/s = −i kωωωω

( ) b Velocity of point C.

( )30.140 40.186 0.35833

14.4 10.8

C C= × = − ×

= +

v r i k j

i k

ωωωω

( ) ( )14.4 ft/s 10.8 ft/s C

= +v i k

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Chapter 15, Solution 183.

At the given instant, the origin is not moving.

: 10.8 14.4 14.4

0 0.35833 0.35833

B B x y zω ω ω= × − + =i j k

v r i j kωωωω

( )10.8 14.4 14.4 0.35833 0.35833 0.35833y z x x

ω ω ω ω− + = − − +i j k i j k

( ): 0.35833 10.8 30.140 rad/s

y z y z

x x

x x

ω ω ω ω

ω ωω ω

− = − =

− = − == =

i

j

k

: 21.6 28.8

0 0.71667 0

D D x y zω ω ω= × + =i j k

v r i kωωωω

21.6 28.8 0.71667 0.71667z x

ω ω+ = − +i k i k

z z

x x

ω ωω ω

− = = −= =

i

k

30.140 0y z

ω ω= + =

( ) .a Angular velocity ( ) ( )40.2 rad/s 30.1 rad/s = −i kωωωω

( ) b Velocity of point C.

( ) ( )40.186 30.140 0.35833

10.8 14.4

C C= × = − ×

= +

v r i k j

i k

ωωωω

( ) ( )10.8 ft/s 14.4 ft/s C

= +v i k

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Chapter 15, Solution 184.

.Total angular velocity 2 1ω= +j kωωωω ωωωω

.Angular acceleration

1Frame is rotating with angular velocity .Oxyz ω=Ω k

( )( )( )

1 2 1 1 20

5 4 20

Oxyz

ω ω ω ω ω

= = + ×

= + × + = −

= − = −

k j k i

i i

& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

αααα

( )220.0 rad/s = − i αααα

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Chapter 15, Solution 185.

( ) ( )1450 rpm 15 rad/sπ= − = −i iωωωω

( ) ( )23 rpm 0.100 rad/sπ= − = −j jωωωω

Let the frame rotate with the motor housing.Oxyz

Rate of rotation of frame :Oxyz ( )20.100 rad/sω π= = − jΩΩΩΩ

.Angular acceleration

( ) ( )1 2 1 2 1 2Oxyz= + = + + × +& & & &αααα ωωωω ωωωω ωωωω ωωωω ΩΩΩΩ ωωωω ωωωω

( ) ( ) ( )0 0 0.100 15 0.100π π π= + + − × − −j i j

2

1.500π= − k ( )214.80 rad/s = − kαααα

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Chapter 15, Solution 186.

Let be the spin of gear about the axle .s

. Total angular velocity ( )1sin cos

sω ω θ θ= + − +j i jωωωω (1)

( ) ( )sin cos cos sinC

L r L rθ θ θ θ= − − +r i j

Since gear is fixed, B 0.C

=v

( )( ) ( )

1sin cos 0 0

sin cos cos sin 0

C C s s

L r L r

ω θ ω ω θθ θ θ θ

= × = − + =− − +

i j k

v rωωωω

( ) ( )sin cos sin cos sin coss s

L r L rω θ θ θ ω θ θ θ + − − k

( )1sin cos 0L rω θ θ− − =k

( ) ( )2 2

1sin cos sin cos

s sr r L rω θ θ ω ω θ θ+ = = −

1sin cos

s

L

rω ω θ θ = −

( ) . a Angular velocity ( )1 1sin cos sin cos

L

rω ω θ θ θ θ = + − − +

j i jωωωω

1sin cos sin sin cos

L L

r rω θ θ θ θ θ = − + +

i j ωωωω

( ) .b Angular acceleration

1Frame is rotating with angular velocity .Oxyz ω= jΩΩΩΩ

Oxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

1 10 sin cos sin cos sin

L L

r rω ω θ θ θ θ θ

= + × − + +

j i j

2

1sin sin cos

L

rω θ θ θ = −

k αααα

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Chapter 15, Solution 187.

Let be the spin of gear about the axle .s

.Total angular velocity ( )1sin cos

sω θ θ= + − +j i jωωωω ωωωω

( ) ( )sin cos cos sinC

L r L rθ θ θ θ= − − +r i j

2Since gear is rotating with angular velocity , on gear ,B Bω j

( )2 / 2sin cos

C C BL rω ω θ θ= × = − −v j r k

On gear A ( )( ) ( )

1sin cos 0

sin cos cos sin 0

C C s s

L r L r

ω θ ω ω θθ θ θ θ

= × = − +− − +

i j k

v rωωωω

( ) ( )( )1

sin cos sin cos sin cos

sin cos

C s sL r L r

L r

ω θ θ θ ω θ θ θ

ω θ θ

= + − −

− −

v k

k

Equating the two expressions for and solving for ,C s

ωv

( )1 2sin cos

s

L

rω ω ω θ θ = − −

( ) . a Angular velocity ( ) ( )1 1 2sin cos sin cos

L

rω ω ω θ θ θ θ = + − − − +

j i jωωωω

( )1 2sin cos sin sin cos cos sin sin cos

L L L

r r rω θ θ θ θ θ ω θ θ θ θ = − + + + − − +

i j i jωωωω

( ) .b Angular acceleration

1Frame is rotating with angular velocity .Oxyz ω= jΩΩΩΩ

Oxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

( )1 1 1 20 sin cos

L

rω ω ω ω θ θ = + × = − −

j kωωωω

( )1 1 2sin cos

L

rω ω ω θ θ = − −

k αααα

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Chapter 15, Solution 188.

Angular velocity. 1 2ω ω= +i kωωωω

( ) a Angular acceleration.

1Frame is rotating with angular velocity .Oxyz = iΩΩΩΩ ωωωω

Oxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

( )1 1 2 1 20 ω ω ω ω ω= + × + = −i i k j

( )( )4 5 20= − = −j j ( )220.0 rad/s = − jαααα

( ) 0. b Acceleration of point P.θ =

( ) ( )( )

( )

60 mm 0.06 m

5 4 0.06 0.24

20 0.06 5 4 0.24

1.2 1.2 0.96 0.96 2.4

P

P P

P P P

= =

= × = + × =

= × + × = − × + + ×

= + − = − +

r i i

v r i k i j

a r v j i i k j

k k i i k

ωωωω

αααα ωωωω

( ) ( )2 20.960 m/s 2.40 m/s

P= − +a i k

( ) 90 . c Acceleration of point P.θ = °

( )( )

( ) ( )

0.06 m

5 4 0.06 0.24 0.3

20 0.06 5 4 0.24 0.3

0 0 1.5 0.96 0 2.46

P

P P

P P P

=

= × = + × = − +

= × + ×

= − × + + × − +

= + − − + = −

r j

v r i k j i k

a r v

j j i k i k

j j j

ωωωω

αααα ωωωω

( )22.46 m/s P

= −a j

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Chapter 15, Solution 189.

Angular velocity. 1 2ω ω= +i kωωωω

Angular acceleration.

1Frame is rotating with angular velocity .Oxyz = iΩΩΩΩ ωωωω

Oxyz= = + ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

( )1 1 2 1 20 ω ω ω ω ω= + × + = −i i k j

( )( )4 5 20= − = −j j ( )220.0 rad/s= − jαααα

( )( )( )( )

30 , 60 mm cos30 sin30

0.06 m cos30 sin30

Pθ = ° = ° + °

= ° + °

r i j

i j

( ) ( ) ( )

5 0 4

0.06cos30 0.06sin30 0

0.12 m/s 0.20785 m/s 0.15 m/s

P P= × =

° °

= − + +

i j k

v r

i j k

ωωωω

Acceleration of point P.

0 20 0 5 0 4

0.06cos30 0.06sin 30 0 0.12 0.20785 0.15

1.03923 0.8314 1.23 1.03925

P P P= × + ×

= − +° ° −

= − − +

a r v

i j k i j k

k i j k

αααα ωωωω

( ) ( ) ( )2 2 20.831 m/s 1.230 m/s 2.08 m/s

P= − − +a i j k

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Chapter 15, Solution 190.

1

2

180 6

180 18

d

dt

d

dt

β π π

γ π π

= − = − = −

= − = − = −

j j j

i i i

ωωωω

ωωωω

( ) a Angular velocity. 1 2= +ωωωω ωωωω ωωωω

( ) .b Angular acceleration

1Frame is rotating with angular velocity Oxyz =ΩΩΩΩ ωωωω

( )1 1 1 2 1 20Oxyz= = + × = + × + = ×& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω ωωωω ωωωω ωωωω ωωωω ωωωω

2

6 18 108

π π π = − × − = −

j i k ( )20.0914 rad/s = − k αααα

( ) c Velocity and acceleration of point P.

For 90 and 30 ,β γ= ° = ° ( )( )12 ft sin30 cos30P

= ° + °r j k

018 6

0 12sin30 12cos30

5.4414 1.81380 1.04720

P C

π π = × = − −

° °

= − + −

i j k

v r

i j k

ωωωω

( ) ( ) ( )5.44 ft/s 1.814 ft/s 1.047 ft/s P

= − + −v i j k

( )2

6 10.3923 0108 18 6

5.4414 1.81380 1.04720

0.54831 0 0.54831 0.18277 3.1657

P P P

π π π

= × + ×

= − × + + − −

− −

= + + − −

a r v

i j k

k j k

i j i j k

αααα ωωωω

( ) ( ) ( )2 2 21.097 ft/s 0.1828 ft/s 3.17 ft/s

P= − −a i j k

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Chapter 15, Solution 191.

Contact points. ( ) ( ) ( ) ( )1 26 in. 9 in. 4 in. 3 in.= − + = +r i j r i j

= iωωωω

( ) ( )1 115 6 9 135 in./s

C= × = × − + =v r i i j kωωωω

= iωωωω

( ) ( )2 230 4 3 90 in./s

D= × = × + =v r i i j kωωωω

Gears A and B. x y z

ω ω ω= + +i j kωωωω

( )1 19 6 9 6

6 9 0

x y z z z x yω ω ω ω ω ω ω= × = = − + + +−

i j k

v r i j kωωωω

Matching expressions for 1,v : 9 0, : 6 0

z zω ω− = =i j

: 9 6 135x y

ω ω+ =k (1)

( )2 23 4 3 4

4 3 0

x y z z z x yω ω ω ω ω ω ω= × = = − + + −i j k

v r i j kωωωω

Matching expressions for 2.v : 3 0, : 4 0

z zω ω− = =i j

: 3 4 90x y

ω ω− =k (2)

ω= = −ωωωω

(a) Angular velocity. ( ) ( )20.0 rad/s 7.50 rad/s = −i jωωωω

Shaft HFG and thus the frame Fxyz rotates with angular velocity

ω= =i iΩΩΩΩ

( )0 20 20 7.5Fxyz= = + × = + × −i i j& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

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(b) Angular acceleration. ( )2150.0 rad/s = − k αααα

(c) Acceleration of tooth of gear B at point 2.

2 2 2

0 0 150 20 7.5 0

4 3 0 0 0 90

450 600 0 675 1800 0

= × + ×

= − + −

= − + − − +

a r v

i j k i j k

i j k i j k

αααα ωωωω

( ) ( )2 2225 in./s 2400 in./s= − −i j ( ) ( )2 2

218.75 ft/s 200 ft/s = − −a i j

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Chapter 15, Solution 192.

Results from the solution to Prob. 15.186.

Position vector of contact point C: ( ) ( )sin cos cos sinC

L r L rθ θ θ θ= − − +r i j

Velocity of contact point C on gear A: 0C

=v

Angular velocity of gear A: 1sin cos sin sin cos

L L

r rω θ θ θ θ θ = − + +

i jωωωω

Angular acceleration of gear A: 2

1sin sin cos

L

rα ω θ θ θ = −

k

Acceleration of point C on gear A.

C C C C

r= × + × = ×a r vαααα ωωωω αααα

( ) ( )2

1sin sin cos sin cos cos sin

C

LL r L r

rω θ θ θ θ θ θ θ

= − × − − +

a k i j

2

1sin sin cos cos sin sin cos

C

L L Lr

r r rω θ θ θ θ θ θ θ = − + + −

a i j

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Chapter 15, Solution 193.

Since the cone rolls on the xz plane, 0

0A

= =v v

The instantaneous axis of rotation lies along the x-axis.

Total angular velocity. ω= iωωωω

(a) Rate of spin. ( )cos sins s

ω β β= +i jωωωω

1 spinω= +jωωωω ωωωω

( ) ( ) 1cos sin

s sω ω β ω β ω= + +i i j j

Components.

1

1

1 1

s

: 0 sinsin

cos: cos

sin tan

s s

ωω β ω ωβ

ω β ωω ω ββ β

= + = −

= = − = −

j

i

1

sins

ωωβ

=

(b) Angular velocity. 1

tan

ωβ

= − iωωωω

(c) Angular acceleration. The vector ωωωω rotates about the y-axis at a rate ω1.

1

ω= jΩΩΩΩ

1

1tan

ωωβ

= × = × −

j iαααα ΩΩΩΩ ωωωω

2

1

tan

ωβ

= kαααα

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Chapter 15, Solution 194.

Geometry.

plane. Law of cosines.yz

( )2 2 211.2 8 2 8 cos120d d= + − °

By solving the quadratic equation 4.8 in.d =

Let point be the midpoint of rod .H AB ( ) ( )/4.8 in. sin 60 10.4 in.

H D= ° −r j k

( ) ( ) ( ) ( ) ( ) ( )/ /4 in. 4.8 in. sin 60 10.4 in. , 4 in. 4.8 in. sin 60 10.4 in.

B D A D= + ° − = − + ° −r i j k r i j k

Let be a unit vector normal to plane .ABDλλλλ sin30 cos30= ° + °j kλλλλ

The projection of onto the normal is zero.H

v λλλλ

/

0 sin 30 cos30

8.8 0, 0

0 4.8sin 60 10.4

H H D x y z x xω ω ω ω ω ω° °

⋅ = ⋅ × = = = =° −

λ v λ r

/Also, the projection of onto the normal is zero.

B Av λλλλ

( )/ /

0 sin 30 cos30

0 8 sin 30 cos30

8 0 0

0 3

B A B A y z z y

z y

ω ω ω ω

ω ω

° °⋅ = ⋅ × = = ° − °

= =

λ v λ rωωωω

Then, 3y y

ω ω= +j kωωωω

( ) ( ) ( )

/0 3

4 4.8sin 60 10.4

17.6 6.9282 4

B B D y y

y y y

ω ω

ω ω ω

= × =° −

= − + −

i j k

v r

i j k

ωωωω

( ) ( ) ( ) ( )2 2 22 2 2 2 2 217.6 6.9282 4 4B B B B yx y z

v v v v ω= + + = + + =

y yω ω= = −

ω = − = −

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( ) a Angular velocity. 0.20690 0.35836= − −j kωωωω

( ) ( )0.207 rad/s 0.358 rad/s = − −j k ωωωω

( ) . b Velocity of point A /A A D= ×v rωωωω

0 0.20690 0.35836

4 4.8sin 60 10.4

A= − −

− ° −

i j k

v

( ) ( ) ( )3.64 in./s 1.433 in./s 0.828 in./s A

= + −v i j k

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Chapter 15, Solution 195.

From the solution to Problem 15.194, ( ) ( )0.20690 rad/s 0.35836 rad/s= − −j kωωωω

( ) ( ) ( ) ( ) ( )/ /4.8 in. sin 60 10.4 in. , 4 in. 4.8 in. sin 60 10.4 in.

H D B D= ° − = + ° −r j k r i j k

sin 30 cos30 , 4 in./sBv= ° + ° =j kλλλλ

Note that ωωωω is parallel to .λλλλ

The projection of H

a onto the direction λλλλ is zero.

/ /0

H H D H H D⋅ = ⋅ × + ⋅ × = ⋅ × +a r v rλλλλ λλλλ αααα λλλλ ωωωω λλλλ αααα

0 sin 30 cos30

8.8 0 0

0 4.8sin 60 10.8

x y z x xα α α α α

° °= = =

° −

Also, the projection of /B A

a onto the direction λλλλ is zero.

/ / / /0

B A B A B A B A⋅ = ⋅ × + ⋅ × = ⋅ × +a r r rλλλλ λλλλ αααα λλλλ ωωωω λλλλ αααα

( )0 sin 30 cos30

0 8 sin 30 cos30 0 3

8 0 0

y z z y z yα α α α α α

° °= ° − ° = =

Velocity at B. /B B D

= ×v rωωωω

( ) ( ) ( )0 0.20690 0.35836 3.6414 in./s 1.4334 in./s 0.8276 in./s

4 4.8sin 60 10.4

B= − − = − +

° −

i j k

v i j k

Unit vector tangent to the path of point B. B

t

Bv

= v

e

0.91036 0.35836 0.20690t

= − +e i j k

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Component of acceleration tangent to the path

( ) / /0

0.91036 0.35836 0.20690

0 3 19.333

4 4.8sin 60 10.4

B t B t B D t B t B Dt

y y y

a

α α α

= ⋅ = ⋅ × + ⋅ × = ⋅ × +

= = −° −

e a e r e v e rαααα ωωωω αααα

But ( )B ta is given as 2

8 in./s , thus 219.333 8 in./s

yα− =

y zα α= − = − = −

(a) Angular acceleration. ( ) ( )2 20.414 rad/s 0.717 rad/s = − −j kαααα

Normal component of acceleration. ( )B Bn= ×a vωωωω

( )

( ) ( ) ( )( ) ( ) ( ) ( )

2 2 2

2 2 2 2

0 0.20690 0.35836

3.6414 1.4334 0.8276

0.6849 in./s 1.3049 in./s 0.7534 in./s

0.6849 1.3049 0.7534 1.6551in./s

B n

B na

= − −−

= − − +

= + + =

i j k

a

i j k

But ( ) ( )( )22 2 4

9.67 in.1.6551

B B

B n

B n

v v

a

a

ρρ

= = = =

(b) Radius of curvature of path. 9.67 in. ρ =

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Chapter 15, Solution 196.

Geometry. ( ) ( )2 22 2 2 2 2 2

/ / /: 325 195 100 , 240 mm

AB A B A B A Bl x y z c c= + + = − + + − =

( ) ( ) ( )/195 mm 240 mm 100 mm

A B= − + −r i j k

Velocity of collar B. ( )1m/sB

= −v k ( )1000 mm/s= − k

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )195 240 100 195 240 100 1000Av− + − ⋅ = − + − ⋅ −i j k j i j k k

240 100000 or 416.67 mm/sA Av v= = ( )0.417 m/s

A=v j

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Chapter 15, Solution 197.

Geometry. ( ) ( )/

2 22 2 2 2 2 2

/ /: 325 195 156 , 208 mm

A BAB A B A Bl x y z c c= + + = − + + − =

( ) ( ) ( )/195 mm 208 mm 156 mm

A B= − + −r i j k

Velocity of collar B. ( ) ( )1.6 m/s 1600 mm/sB

= =v k k

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )195 208 156 195 208 156 1600Av− + − ⋅ = − + − ⋅i j k j i j k k

208 249600 or 1200 mm/sA Av v= − = − ( )1.200 m/s

A= −v j

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Chapter 15, Solution 198.

Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.

B/A C B= − = + +r k r i j k

( ) ( )/12 in. 8 in.

C D= +r i j 2 2

12 8 208 in.CDl = + =

Velocity at B. ( ) ( )024 2 48 in./s

B B/Aω= × = × − = −v j r j k i

Velocity of collar C. / 12 8

208

D C

CD

CDl

+= =r i jλλλλ

( )0.83205 0.55470C C CD C

v v= = +v i jλλλλ

/ / / where

C B C B C B BC C B= + = ×v v v v rωωωω

Noting that /C B

v is perpendicular to /,

C Br we get

/ /0.

C B C B⋅ =r v

Forming /

,C B C

⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /C B C C B B⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Cv+ + ⋅ = + ⋅i j k i j i j k i++++ −−−−

17.7504 768Cv = − 43.267 in./s

Cv = −

( )( )43.267 0.83205 0.55470C

= −v i j++++

( ) ( )36.0 in./s 24.0 in./s C

= − −v i j

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Chapter 15, Solution 199.

Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.

B/A C B= − = + +r k r i j k

( ) ( )/12 in. 8 in.

C D= +r i j 2 2

12 8 208 in.CDl = + =

Velocity at B. ( ) ( )024 2 48 in./s

B B/Aω= × = × − =v i r i k j

Velocity of collar C. / 12 8

208

DC

CD

CDl

+= =r i jλλλλ

( )0.83205 0.55470C C CD C

v= = +v v i jλλλλ

/ / / where

C B C B C B BC C B= + = ×v v v v rωωωω

Noting that /C B

v is perpendicular to /,

C Br we get

/ /0.

C B C B⋅ =r v

Forming /

,C B C

⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /C B C C B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Ev+ + ⋅ = + ⋅i j k i j i j k j++++

17.7504 384Cv = 21.633 in./s

Cv =

( )( )21.633 0.83205 0.55470C

=v i j++++

( ) ( )18.00 in./s 12.00 in./s C

= +v i j

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Chapter 15, Solution 200.

Geometry. ( ) ( ), 4.5 in. 9 in.A D C

y= = =r j r i r k

( ) ( ) ( ) ( )2 2

/4.5 in. 9 in. 4.5 9 10.0623 in.

D C D C CDl= − = − = + − =r r r i k

( ) ( ) ( ) ( )/

/

4 4.5 92 in. 4 in.

9 9

D C

B C

c −= = = −

r i kr i k

( ) ( )/9 2 4 2 in. 5 in.

B C B C= + = + − = +r r r k i k i k

( ) ( ) ( )/2 in. in. 5 in.

A B A By= − = − + −r r r i j k

( ) ( )2 22 2 2 2 2 2

/ /: 15 2 5

AB A B A Bl x y z y= + + = − + + −

( ) ( ) ( )/14 in., 2 in. 14 in. 5 in.

A By = = − + −r i j k

Velocity of collar B. /D C

B B

CD

vl

=r

v

( )( ) ( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s

10.0623B

−= = −

i kv i k

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v ω r

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )2 14 5 2 14 5 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k

( )( ) ( )( )14 2 1.11803 5 2.23607 or 0.63888 in./sA Av v= − + − − =

( )0.639 in./s A

=v j

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Chapter 15, Solution 201.

Geometry. ( ) ( ), 4.5 in. 9 in.A D C

y= = =r j r i r k

( ) ( ) ( ) ( )2 2

/4.5 in. 9 in. 4.5 9 10.0623 in.

D C D C CDl= − = − = + − =r r r i k

( ) ( ) ( ) ( )/

/

6 4.5 93 in. 6 in.

9 9

DC

B C

c −= = = −

r i kr i k

( ) ( )/9 3 6 3 in. 3 in.

B C B C= + = + − = +r r r k i k i k

/3 3

A B A By= − = − + −r r r i j k

2 2 2 2 2 2 2

/ /: 15 3 3

AB A B A Bl x y z y= + + = + +2

( ) ( ) ( )/14.3875 in., 3 in. 14.3875 in. 3 in.

A By = = − + −r i j k

Velocity of Collar B. /D C

B B

CD

vl

=r

v

( )( ) ( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s

10.0623B

−= = −

i kv i k

Velocity of Collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )3 14.3875 3 3 14.3875 3 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k

( )( ) ( )( )14.3875 3 1.11803 3 2.23607 or 0.23313 in./sA Av v= − + − − =

( )0.233 in./s A

=v j

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Chapter 15, Solution 202.

Geometry. ( ) ( )/

2 22 2 2 2 2 2

/ / /: 0.5 0.24 0.4

A BAB A B A B A Bl x y z y= + + = − + + −

( ) ( ) ( )/ /0.18 m, 0.24 m 0.18 m 0.4 m

A B A By = = − + −r i j k

( ) ( ) ( ) ( )2 2

/0.24 m 0.18 m , 0.24 0.18 0.3 m

D C CDl= − = + − =r i j

Velocity of collar B. /D C

B B

CD

rv

l=v

( ) ( ) ( ) ( )0.24 0.180.200 0.16 m/s 0.12 m/s

0.3B

−= = −

i jv i j

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )0.24 0.18 0.4 0.24 0.18 0.4 0.16 0.12Av− + − ⋅ = − + − ⋅ −i j k j i j k i j

( ) ( ) ( ) ( )0.18 0.24 0.16 0.18 0.12 or 0.333 m/sA Av v= − + − = −

( )0.333 m/s A

= −v j

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Chapter 15, Solution 203.

Geometry. ( )/

22 2 2 2 2 2

/ / /: 0.5 0 0.4

A BAB A B A B A Bl x y z y= + + = + + −

( ) ( )/ /0.3 m, 0.3 m 0.4 m

A B A By = = −r j k

( ) ( ) ( ) ( )2 2

/0.24 m 0.18 m , 0.24 0.18 0.3 m

D C CDl= − = + − =r i j

Velocity of collar B. /D C

B B

CD

vl

=r

v

( ) ( ) ( ) ( )0.24 0.180.200 0.16 m/s 0.12 m/s

0.3B

−= = −

i jv i j

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )0.3 0.4 0.3 0.4 0.16 0.12Av− ⋅ = − ⋅ −j k j j k i j

( )( )0.3 0.3 0.12 0.12 m/sA Av v= − = −

( )0.1200 m/s A

= −v j

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Chapter 15, Solution 204.

Angular velocity of shaft AC. 1AC

ω= kωωωω

Let 3

ω j be the angular velocity of body D relative to shaft AD.

Angular velocity of body D. 1 3D

ω ω= +k jωωωω

Angular velocity of shaft EG. ( )2cos20 sin 20

EGω= ° − °k jωωωω

Let 4

ω i be the angular velocity of body D relative to shaft EG.

Angular velocity of body D. ( )2 4cos20 sin 20

Dω ω= ° − ° +k j iωωωω

Equate the two expressions for D

ω and resolve into components.

4

: 0 ω=i (1)

3 2

: sin 20ω ω= − °j (2)

1 2

: cos20ω ω= °k (3)

From (3), 1

2

cos 20

ωω =°

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Chapter 15, Solution 205.

Angular velocity of shaft AC. 1AC

ω= kωωωω

Let 3

ω i be the angular velocity of body D relative to shaft AD.

Angular velocity of body D. 1 3D

ω ω= +k iωωωω

Angular velocity of shaft EG. ( )2cos20 sin 20

EGω= ° − °k jωωωω

Let 4

ω λ be the angular velocity of body D relative to shaft EG.

Where λ is a unit vector along the axis the clevis axle attached to shaft EG.

cos20 sin 20= ° + °j kλλλλ

4 4 4

cos20 sin 20ω ω ω= ° + °j kλλλλ

Angular velocity of body D. 4D EG

ω= +ωωωω ωωωω λλλλ

( ) ( )4 2 4 2cos20 sin 20 sin 20 cos20

Dω ω ω ω= ° − ° + ° + °j kωωωω

Equate the two expressions for D

ωωωω and resolve into components.

3

: 0ω =i (1)

4 2

: 0 cos20 sin 20ω ω= ° − °j (2)

1 4 2

: sin 20 cos20ω ω ω= ° + °k (3)

From (2) and (3), 2 1cos20 ω ω= °

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Chapter 15, Solution 206.

Geometry. ( ) ( ) ( )640 mm , 24 in. 10 in. C C B

x= + = +r i j r j k

2 224 10 26 in.

ABl = + =

( ) ( )/8 in. 10 in.

C B Cx= + −r i j k

Length of rod BC. 2 2 2 2 242 8 10

BC Cl x= = + +

Solving for ,Cx 40 in.

Cx =

( ) ( ) ( )/40 in. 8 in. 10 in.

C B= + −r i j k

Velocity. ( ) ( ) ( )19.524 10 18 in./s 7.5 in./s

26B

= − − = − −v j k j k

C C

v=v i

Angular velocity of collar C. C C

ω= iωωωω

The axle of the clevis at C is perpendicular to the x-axis and to the rod BC.

A vector along this axle is /C B

= ×p i r

( ) ( ) ( )2 2

40 8 10 10 in. 8 in.

10 8 12.806 in.p

= × + − = +

= + =

p i i j k j k

Let λλλλ be a unit vector along the axle. 0.78087 0.62470p

= = +pj kλλλλ

Let s s

ω=ωωωω λλλλ be the angular velocity of rod BC relative to collar C.

0.78087 0.62470s s s

ω ω= +j kωωωω

Angular velocity of rod BC. BC C s

= +ωωωω ωωωω ωωωω

0.78087 0.62470BC C s s

ω ω ω= + +i j kωωωω

/C B BC C B= + ×v v rωωωω

18 7.5 0.78087 0.62470

40 8 10

C C s sv ω ω ω= − − +

i j k

i j k

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Resolving into components,

: 12.806C sv ω= −i (1)

: 0 18 10 24.988C s

ω ω= − + +j (2)

: 0 7.5 8 31.235C s

ω ω= − + −k (3)

Solving the simultaneous equations (1), (2), and (3),

vω ω= = = −

(a) Angular velocity of rod BC.

( )( ) ( )( )1.4634 0.78087 0.13470 0.62470 0.13470BC

= + +i j kωωωω

= + +i j kωωωω

(b) Velocity of collar C. ( )1.725 in./s C

= −v i

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Chapter 15, Solution 207.

Geometry. Determine the position of collar A.

( ) ( )/

, 6 in. 2 in.

6 2

A A B B B

A B A

z x y

z

= = + = +

= − −

r k r i j i j

r k i j

Length of rod AB: 2 2 2 2 211 6 2

AB Al z= = − −

Solving for ,Az 9 in.

Az =

( ) ( ) ( )/6 in. 2 in. 9 in.

A B= − − +r i j k

Velocity. ( ) ( )4.5 ft/s 54 in./s ,B A A

v= − − =v j j v k====

Angular velocity of collar B. B B

ω= jωωωω

The axle of the clevis at B is perpendicular to both the y-axis and the rod AB.

A vector along this axle is /A B

= ×p j r

( ) ( ) ( )/

2 2

6 2 9 9 in. + 6 in.

9 6 10.8167 in.

A Br= × = × − − + =

= + =

p j j i j k i k

p

Let λ be a unit vector along the axle. 0.83205 0.55470p

= = +pλ i k

Let s s

ω=ω λ be the angular velocity of rod AB relative to collar B.

0.83205 0.55470s s s

ω ω= +i kωωωω

Angular velocity of rod AB. AB B s

= +ωωωω ωωωω ωωωω

0.83205 0.55470AB s B s

ω ω ω= + +ω i j k

/A B AB A B= + ×v v rωωωω

54 0.83205 0.55470

6 2 9

A s B sv ω ω ω= − +

− −

i j k

k j

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Resolving into components,

: 0 0 9 1.1094B s

ω ω= + +i (1)

: 0 54 10.8167s

ω= − −j (2)

: 0 6 1.6641A B sv ω ω= + −k (3)

ω = −

ω =

(a) Angular velocity of rod AB.

( )( ) ( )( )0.83205 4.9923 0.61539 0.55470 4.9923AB

= − + + −i j kωωωω

= − + −i j k ωωωω

(b) Velocity of collar A.

From (3), ( )( ) ( )( )6 0.61539 1.6641 4.9923 12.00 in./sAv = − − =

( )1.000 ft/s A

=v k

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Chapter 15, Solution 208.

Geometry. ( ) ( )2 22 2 2 2 2 2

/ / /: 325 195 100 , 240 mm.

AB A B A B A Bl x y z c c= + + = − + + − =

( ) ( ) ( )/195 mm 240 mm 100 mm

A B= − + −r i j k

Velocity of collar B. ( ) ( )1m/s 1000 mm/sB

= − −v k k====

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )195 240 100 195 240 100 1000Av− + − ⋅ = − + − ⋅ −i j k j i j k k

240 100000 or 416.67 mm/sA Av v= =

Relative velocity. /A B A B

= −v v v

( ) ( ) ( ) ( )2 22 2 2 2

/ /416.67 mm/s 1000 mm/s 416.67 1000 1083.33 mm /s

A B A Bv= + = + =v j k

Acceleration of collar B. 0B

=a

Acceleration of collar A. A A

a=a j

/ / / /, where

A B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωω

Noting that /AB A B

× rαααα is perpendicular to /,

A Br we get

/ /0

A B AB A B⋅ × =r rαααα

We note also that / / / / /A B AB A B A B A B A B

⋅ × = ⋅ ×r v v rωωωω ωωωω

( )2/ / /A B A B A Bv= − ⋅ = −v v

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Then, ( ) ( )2 2

/ / / /0

A B A B A B A Bv v⋅ = − = −r a

Forming /

,A B A

⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)

From (2), ( ) ( ) ( )2195 240 100 0 1083.33A

a− + − ⋅ = −i j k j

( )2 2240 1083.33 4890 mm/s

A Aa a= − = − ( )24.89 m/s

A= −a j

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Chapter 15, Solution 209.

Geometry. ( ) ( )2 22 2 2 2 2 2

/ / /: 325 195 156 , 208 mm

AB A B A B A Bl x y z c c= + + = − + + − =

( ) ( ) ( )/195 mm 208 mm 156 mm

A B= − + −r i j k

Velocity of collar B. ( ) ( )1.6 m/s 1600 mm/sB

=v k k====

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )195 208 156 195 208 156 1600Av− + − ⋅ = − + − ⋅i j k j i j k k

208 249600 or 1200 mm/sA Av v= − = −

Relative velocity. /A B A B

= −v v v

( ) ( ) ( )2 2 2 2 2

/ /1.2 m/s 1.6 m/s 1.2 1.6 4 m /s

A B A Bv= − − = + =v j k

Acceleration of collar B. 0B

=a

Acceleration of collar A. A A

a=a j

/ / / /, where

A B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωω

Noting that /AB A B

× rαααα is perpendicular to /,

A Br we get

/ /0

A B AB A B⋅ × =r rαααα

We note also that / / / / /A B AB A B A B A B A B

⋅ × = ⋅ ×r v v rωωωω ωωωω

( )2/ / /A B A B A Bv= − ⋅ = −v v

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Then, ( ) ( )2 2

/ / / /0

A B A B A B A Bv v⋅ = − = −r a

Forming /

,A B A

⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)

From (2), ( ) ( )0.195 0.208 0.156 0 4A

a− + − ⋅ = −i j k j

0.208 4A

a = −

( )219.23 m/s A

= −a j

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Chapter 15, Solution 210.

Geometry. ( ) ( ) ( ) ( )/2 in. 16 in. 8 in. 2 in.

B/A C B= − = + +r k r i j k

( ) ( )/12 in. 8 in.

C D= +r i j

2 212 8 208 in.

CDl = + =

Velocity at B. ( ) ( )024 2 48 in./s

B B/Aω= × = × − = −v j r j k i

Velocity of collar C. / 12 8

208

D C

CD

CDl

+= =r i jλλλλ

( )0.83205 0.55470C C CD C

v v= = +v i jλλλλ

/ / / where

C B C B C B BC C B= + = ×v v v v rωωωω

Noting that /C B

v is perpendicular to /,

C Br we get

/ /0.

C B C B⋅ =r v

Forming /

,C B C

⋅r v we get ( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /C B C C B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 + 0.55470 16 8 2 48Cv+ + ⋅ = + ⋅ −i j k i j i j k i++++

17.7504 768Cv = − 43.267 in./s

Cv = −

( )( )43.267 0.83205 0.55470C

= −v i j++++

(36.0 in./s) (24.0 in./s)= − −i j

Relative velocity. /C B C B

= −v v v

/

36 24 ( 48 ) (12 in./s) (24 in./s)C B

= − − − − = −v i j i i j

2 2 2 2 2

/( ) (12) (24) 720 in. /s

C Bv = + =

Acceleration at B. 2

/ /B AB B A D B Aω= × −a r rαααα

2 20 (24) ( 2 ) (1152 in./s )B

= − − =a k k

Acceleration of collar C. (0.83205 0.55470 )C C CD C

a a= = +a i jλλλλ

/ / /

whereC B C B C/B BC C B BC C B

= + = × + ×a a a a α r vωωωω

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Noting that /BC C B

×α r is perpendicular to /,

B Cr we get

/ /( ) 0.

C B BC C B⋅ × =r α r

We also note that / / / /C B BC C B C B C B BC

⋅ × = ⋅ ×r ω v v r ω

2

/ / /( )

C B C B C B= − ⋅ = −v v v

Then, 2 2

/ / / /0 ( ) ( )

C B C B C B C Bv v⋅ = − = −r a

Forming /C B C

⋅r a we get

/ / / / / /

( )C B C C B B C B C B B C B C B

⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or 2

/ / /( )

C B C C B B C Bv⋅ = ⋅ −r a r a (2)

From (2), (16 8 2 ) (0.83205 0.55470 ) (16 8 2 ) (1152 ) 720Ca+ + ⋅ + = + + ⋅ −i j k i j i j k k

217.7504 1584 89.237 in./s

C Ca a= =

(89.237)(0.83205 0.55470 )C

= +a i j

2 2(74.3 in./s ) (49.5 in./s )C

= +a i j

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Chapter 15, Solution 211.

Geometry. ( ) ( ) ( ) ( )/ /2 in. 16 in. 8 in. 2 in.

B A C B= − = + +r k r i j k

( ) ( ) 2 2

/12 in. 8 in. 12 8 208 in.

C D CDl= + = + =r i j

Velocity at B. ( ) ( )0 /24 2 48 in./s

B B Aω= × = × − =v i r i k j

Velocity of collar C. / 12 8

208

DC

CD

CDl

+= =r i j

λ

( )0.83205 0.55470C C CD C

v v= = +v λ i j

/ / /where

C B C B C B BC C B= + = ×v v v v ω r

Noting that /C Bv is perpendicular to /

,C Br we get / /

0C B C B

⋅ =r v

Forming /

,

C B C⋅r v we get ( )

/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ = ⋅r v r v v r v r v

or / /C B C C B B⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )16 8 2 0.83205 0.55470 16 8 2 48Cv+ + ⋅ + = + + ⋅i j k i j i j k j

17.7504 384 21.633 in./sC Cv v= =

( )( ) ( ) ( )21.633 0.83205 0.55470 18 in./s 12 in./sC

= + = +v i j i j

Relative velocity. /C B C B= −v v v

( ) ( )/18 12 48 18 in./s 36 in./s

C B= + − = −v i j j i j

( ) ( ) ( )2 2 2 2 2

/18 36 1620 in. /s

C Bv = + =

Acceleration at B. 2

/ 0 /B AB B A B Aω= × −a α r r

( ) ( ) ( )2 20 24 2 1152 in./s

B= − − =a k k

Acceleration of collar C. ( )0.83205 0.55470C C CD C

a aλ= = +a i j

/ / / /where

C C C B C B BC C B BC B C= + = × + ×a a a a r ω vαααα

Noting that /BC C B×α r is perpendicular to /

,C Br we get / /

0C B BC C B

⋅ × =r α r

We also note that ( )2/ / / / / / /C B BC C B C B C B BC C B C B C Bv⋅ × = ⋅ × = − ⋅ = −r ω v v r ω v v

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Then, ( ) ( )2 2

/ / / /0

C B C B C B C Bv v⋅ = − = −r a

Forming /

we getC B C

⋅r a

( )/ / / / / /C B C C B B C B C B B C B C B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or ( )2/ / /C B C B C B C Bv⋅ = ⋅ −r a r a (2)

From (2) ( ) ( )16 8 2 0.83205 0.55470C

+ + ⋅ +i j k i j a

( ) ( )16 8 2 1152 1620= + + ⋅ −i j k k

2

17.7504 684 38.534 in./sC Ca a= =

( )( )38.534 0.83205 0.55470C

= +a i j

( ) ( )2 232.1 in./s 21.4 in./s

C= +a i j

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Chapter 15, Solution 212.

Geometry. ( ) ( ), 4.5 in. 9 in.A D C

y= = =r j r i r k

( ) ( ) ( ) ( )2 2

/4.5 in. 9 in. 4.5 9 10.0623 in.

D C D C CDl= − = − = + − =r r r i k

( ) ( ) ( ) ( )/

/

4 4.5 92 in. 4 in.

9 9

D C

B C

c −= = = −

r i kr i k

( ) ( )/9 2 4 2 in. 5 in.

B C B C= + = + − = +r r r k i k i k

/

2 5A B A B

y= − = − + −r r r i j k

( ) ( )2 22 2 2 2 2 2

/ /: 15 2 5

AB A B A Bl x y z y= + + = − + + −

( ) ( ) ( )/14 in., 2 in. 14 in. 5 in.

A By r= = − + −i j k

Velocity of collar B. /DC

B B

CD

vl

=r

v

( )( ) ( ) ( )2.5 4.5 91.11803 in. 2.23607 in./s

10.0623B

−= = −

i kv i k

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )2 14 5 2 14 5 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k

( )( ) ( )( )14 2 1.11803 5 2.23607 or 0.63888 in./sA Av v= − + − − =

Relative velocity. /A B A B

= −v v v

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )/

2 2 2 2 2

/

0.63888 in./s 1.11803 in./s 2.23607 in./s

0.63888 1.11803 2.23607 6.6582 in./s

A B

A Bv

= − +

= + − + =

v j i k

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Acceleration of collar B. 0B

=a

Acceleration of collar A. A A

a=a j

/ / / /, where

A B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωω

Noting that /AB A B

× rαααα is perpendicular to /,

A Br we get

/ /0

A B AB A B⋅ × =r α r

We note also that / / / / /A B AB A B A B A B A B

⋅ × = ⋅ ×r v v rωωωω ωωωω

( )2/ / /A B A B A Bv= − ⋅ = −v v

Then, ( ) ( )2 2

/ / / /0

A B A B A B A Bv v⋅ = − = −r a

Forming /

,A B A

⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)

From (2), ( ) ( )2 14 5 0 6.6582A

a− + − ⋅ = −i j k j

14 6.6582A

a = − ( )20.476 in./s A

= −a j

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 213.

Geometry. ( ) ( ), 4.5 in. 9 in.A D C

y= = =r j r i r k

( ) ( ) ( ) ( )2 2

/4.5 in. 9 in. 4.5 9 10.0623 in.

D C D C CDl= − = − = + − =r r r i k

( ) ( ) ( ) ( )/

/

6 4.5 93 in. 6 in.

9 9

DC

B C

c −= = = −

r i kr i k

( ) ( )/9 3 6 3 in. 3 in.

B C B C= + = + − = +r r r k i k i k

/

3 3A B A B

y= − = − + −r r r i j k

2 2 2 2 2 2 2 2

/ /: 15 3 3

AB A B A Bl x y z y= + + = + +

( ) ( ) ( )/14.3875 in. 3 in. 14.3875 in. 3 in.

A By = = − + −r i j k

Velocity of collar B. /DC

B B

CD

vl

=r

v

( )( ) ( ) ( )2.5 4.5 91.11803 in./s 2.23607 in./s

10.0623B

−= = −

i kv i k

Velocity of collar A. A A

v=v j

/ / /, where

A B A B A B AB A B= + = ×v v v v rωωωω

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )3 14.3875 3 3 14.3875 3 1.11803 2.23607Av− + − ⋅ = − + − ⋅ −i j k j i j k i k

( ) ( ) ( ) ( )14.3875 3 1.11803 3 2.23607 or 0.23313 in./sA Av v= − + − − =

Relative velocity. /A B A B

= −v v v

( ) ( ) ( )/0.23313 in./s 1.11803 in./s 2.23607 in./s

A B= − +v j i k

( ) ( ) ( ) ( ) ( )2 2 2 2 2

/0.23313 1.11803 2.23607 6.30435 in./s

A Bv = + − + =

Acceleration of collar B. 0B

=a

Acceleration of collar A. A A

a=a j

/ / / /, where

A B A B A B AB A B AB A B= + = × + ×a a a a r vαααα ωωωω

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Noting that /AB A B

× rαααα is perpendicular to /,

A Br we get

/ /0

A B AB A B⋅ × =r rαααα

We note also that / / / / /A B AB A B A B A B A B

⋅ × = ⋅ ×r v v rωωωω ωωωω

( )2/ / /A B A B A Bv= − ⋅ = −v v

Then, ( ) ( )2 2

/ / / /0

A B A B A B A Bv v⋅ = − = −r a

Forming /

,A B A

⋅r a we get ( )/ / / / / /A B A A B A A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r a r a a r a r a

or ( )2/ / /A B A A B B A Bv⋅ = ⋅ −r a r a (2)

From (2), ( ) ( )3 14.3875 3 0 6.30435A

a− + − ⋅ = −i j k j

14.3875 6.30435A

a = −

( )20.438 in./s A

= −a j

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 214.

Geometry. / / /

20(24 in.) (10 in.)

24B A D A B A

= + =r i j r r

( ) ( )/ / /8 in. 8.3333 in.

D C D A C A= − = +r r r i j

2 224 10 26 in.

ABl = + =

Unit vector along AB. / 12 5

13 12

B A

AB

ABl

= = +r

λ i j

Let Oxyz be a frame of reference currently coinciding with OXYZ but rotating with angular velocity

( )16 rad/sω= =Ω j j

(a) Velocity of D. /D D D AB′= +v v v

( ) ( )/6 8 8.3333 48 in./s

D D C′ = × = × + = −v Ω r j i j k

( ) ( )/

12 578 72 in./s 30 in./s

13 13D AB AB

u

= = + = +

v λ i j i j

( ) ( ) ( )72.0 in./s 30.0 in./s 48 in./sD

= + −v i j k

(b) Acceleration of D. / /

2D D D AB D AB′= + + ×a a a Ω v

( )( ) ( )22 2

18 6 288 in./s

Drω′ = − = − = −a i i i

/

0D AB

=a

( )( ) ( ) ( )2/2 2 6 72 30 864 in./s

D F× = × + = −Ω v j i j k

( ) ( )2 2288 in./s 864 in./s

D= − −a i k

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Chapter 15, Solution 215.

Geometry. ( )( )/10 in. cos30 sin30

D C= ° − °r i j

( ) ( )5 3 in. 5 in.= −i j

Let frame Cxyz, which at the instant shown coincides with CXYZ, rotate with angular velocity

( )110 rad/s .ω= =Ω j j

Motion of coinciding point D′ in the frame.

( ) ( )/10 5 3 5 50 3 in./s

D DC′ = × = × + = −v Ω r j i j k

( ) ( )2 2 210 10cos30 500 3 in./s

Dr′ = −Ω = − ° = −a i i

Motion of point D relative to the frame 4.5 ft/s 54 in./su = =

( ) ( ) ( )/sin30 cos30 27 in./s 27 3 in./s

D Fu= ° + ° = +v i j i j

( )2

/cos30 sin30

D F

u

ρ= ⋅ − ° + °a i j

( )2

54cos30 sin30

10= − ° + °i j

( ) ( )2 2145.8 3 in./s 145.8 in./s= − +i j

(a) Velocity of point D. /D D D F′= +v v v

( ) ( ) ( )27 in./s 27 3 in./s 50 3 in./sD

= + −v i j k

( ) ( ) ( )2.25 ft/s 3.90 ft/s 7.22 ft/sD

= + −v i j k

Coriolis acceleration. /

2D F

×Ω v

( )( ) ( ) ( )2/2 2 10 27 27 3 540 in./s

D F× = × + = −Ω v j i j k

(b) Acceleration of point D. / /

2D D D F D F′= + + ×a a a Ω v

( ) ( ) ( )2 2 2645.8 3 in./s 145.8 in./s 540 in./s

D= − + −a i j k

( ) ( ) ( )2 293.2 ft/s 12.15 ft/s 45.0 ft/s

D= + −a i j k

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Chapter 15, Solution 216.

With the origin at point A, ( ) ( ) ( )0.2 m 0.3 m 0.15 mD

= + −r i j k

( ) ( ) 2 2

/0.2 m 0.15 m , 0.2 0.15 0.25 m

C B BCl= − = + =r i k

Let the frame Axyz rotate with angular velocity ( )110 rad/sω= =i iΩΩΩΩ

(a) ( ) ( ) ( )10 0.2 0.3 0.15 1.5 m/s 3 m/sD D′ = × = × + − = +v r i i j k j kΩΩΩΩ

( ) ( ) ( )/

0.60.6 m/s, 0.2 0.15 0.48 m/s 0.36 m/s

0.25D F

u = = − = −v i k i k

( ) ( ) ( )/0.48 m/s 1.5 m/s 2.64 m/s

D D D F′= + = + +v v v i j k

Velocity of D. ( ) ( ) ( )0.480 m/s 1.500 m/s 2.64 m/sD

= + +v i j k

(b) ( ) ( ) ( )2 210 1.5 3 30 m/s 15 m/s

D D′ ′= × = × + = − +a v i j k j kΩΩΩΩ

/0

D F=a

( )( ) ( ) ( )2/2 2 10 0.48 0.36 7.2 m/s

D F× = × − =v i i k jΩΩΩΩ

( ) ( )2 2

/ /2 22.8 m/s 15 m/s

D D D F D F′= + + × = − +a a a v j kΩΩΩΩ

Acceleration of D. ( ) ( )2 222.8 m/s 15.00 m/s

D= − +a j k

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Chapter 15, Solution 217.

Geometry. ( ) ( ) ( )0.36 m , 0.15 m 0.52 mB D

= = +r j r i k

( ) ( ) ( ) 2 2 2

/0.15 m 0.36 m 0.52 m 0.15 0.36 0.52 0.65 m

B D DBl= − + − = + + =r i j k

( ) ( ) ( ) ( )10.075 m 0.18 m 0.26 m

2C B D

= + = + +r r r i j k

Let frame Axyz rotate with angular velocity ( )15 rad/sω= =k kΩΩΩΩ

Velocity Analysis. 975 mm/s 0.975 m/su = =

( ) ( ) ( )5 0.075 0.18 0.26 0.9 m/s 0.375 m/sC C′ = × = × + + = − +v r k i j k i jΩΩΩΩ

( ) ( ) ( ) ( )/ /

0.9750.15 0.36 0.52 0.225 m/s 0.54 m/s 0.78 m/s

0.65C F D B

DB

u

l= = − + − = − + −v r i j k i j k

/C C C F′= +v v v

( ) ( ) ( )1.125 m/s 0.915 m/s 0.780 m/sC

= − + −v i j k

Acceleration Analysis. /

0C F

=a

( ) ( ) ( )2 25 0.9 0.375 1.875 m/s 4.5 m/s

C C′ ′= × = × − + = − −a v k i j i jΩΩΩΩ

( )( ) ( ) ( ) ( )2 2

/2 2 5 0.225 0.54 0.78 5.4 m/s 2.25 m/s

C F× = × − + − = − −v k i j k i jΩΩΩΩ

/ /2

C C C F C F′= + + ×a a a vΩΩΩΩ

( ) ( )2 27.28 m/s 6.75 m/s

C= − −a i j

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Chapter 15, Solution 218.

Geometry. With the origin at A, ( )6.75 in.B

=r j

Let frame Axyz rotate about the Y axis with constant angular velocity ( )19 rad/s .ω= =k kΩΩΩΩ Then, the

motion relative to the frame consists of rotation about the x axis with constant angular velocity

( )2 212 rad/s .ω= =i iωωωω

Motion of coinciding point .B′

( )9 6.75 60.75 in./sB B′ = × = × = −v r k j iΩΩΩΩ

B B B′ ′= × + ×a r vαααα ΩΩΩΩ

( ) ( )20 9 60.75 546.75 in./s= + × − = −k i j

Motion relative to the frame.

( )/ 212 6.75 81 in./s

B F B= × = × =v r i j kωωωω

( )/ 2 2 /

2 0 12 81 972 in./s

B F B B F= × + ×

= + × = −

a r v

i k j

ωωωωαααα

(a) Velocity of point B. /B B B F′= +v v v

( ) ( )60.8 in./s 81.0 in./sB

= − +v i k

Coriolis acceleration. /

2B F

× vΩΩΩΩ

( )( )/2 2 9 81 0

B F× = × =v k kΩΩΩΩ

(b) Acceleration of point B. / /

2B B B F B F′= + + ×a a a vΩΩΩΩ

( )21518.75 in./sB

= −a j

( )2126.6 ft/sB

= −a j

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Chapter 15, Solution 219.

Geometry. With the origin at A, ( ) ( )6.75 in. 4.5 in.C

= +r j k

Let frame Axyz rotate about the Y axis with constant angular velocity ( )19 rad/s .ω= =k kΩΩΩΩ Then, the motion

relative to the frame consists of rotation about the x axis with constant angular velocity

( )2 212 rad/s .ω= =i iωωωω

Motion of coinciding pointC ′ in the frame.

( ) ( )9 6.75 4.5 60.75 in./sC C′ = × = × + = −v r k j k iΩΩΩΩ

C C C′ ′= × + ×a r vαααα ΩΩΩΩ

( ) ( )20 9 60.75 546.75 in./s= + × − = −k i j

Motion relative to the frame.

( ) ( ) ( )

( ) ( ) ( )

/ 2

/ 2 2 /

2 2

12 6.75 4.5 54 in./s 81 in./s

0 12 54 81 972 in./s 648 in./s

C F C

C F C C F

= × = × + = − +

= × + ×

= + × − + = − −

v r i j k j k

a r v

i j k j k

ωωωω

αααα ωωωω

(a) Velocity of point C. /C C C F′= +v v v

( ) ( ) ( )60.8 in./s 54.0 in./s 81.0 in./sC

= − − +v i j k

Coriolis acceleration. /

2C F

× vΩΩΩΩ

( )( ) ( ) ( )2/2 2 9 54 81 972 in./s

C F× = × − + =v k j k iΩΩΩΩ

(b) Acceleration of point C. / /

2C C C F C F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 2972 in./s 1518.75 in./s 648 in./s

C= − −a i j k

( ) ( ) ( )2 2 281.0 ft/s 126.6 ft/s 54.0 ft/s

C= − −a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 220.

Geometry. ( )( )0.36 m cos 20 sin 20C

= ° − °r i j

Let frame Oxyz rotate about the Y axis with angular velocity 1

ω= jΩΩΩΩ and angular acceleration 0.=&ΩΩΩΩ

Then, the motion relative to the frame consists of rotation with angular velocity 2

ω=2

kωωωω and angular

acceleration 2

(a) ( )3 0.36cos 20 0.36sin 20 1.08cos 20C C′ = × = × ° − ° = − °v r j i j kΩΩΩΩ

( )/ 24 0.36cos 20 0.36sin 20

1.44sin 20 1.44cos 20

C F C= × = × ° − °

= ° + °

v r k i j

i j

ωωωω

/1.44sin 20 1.44cos 20 1.08cos 20

C C C F′= + = ° + ° − °v v v i j k

( ) ( ) ( )0.493 m/s 1.353 m/s 1.015 m/sC

= + −v i j k

(b) ( )3 1.08cos 20 3.24cos 20C C′ ′= × = × − ° = − °a v j k iΩΩΩΩ

( )/ 2 /4 1.44sin 20 1.44cos 20

5.76cos 20 5.76sin 20

C F C F= × = × ° + °

= − ° + °

a v k i j

i j

ωωωω

( ) ( ) ( )/2 2 3 1.44sin 20 1.44cos 20 8.64sin 20

C F× = × ° + ° = − °v j i j kΩΩΩΩ

( )/ /

2

3.24 5.76 cos 20 5.76sin 20 8.64sin 20

C C C F C F′= + + ×

= − + ° + ° − °

a a a v

i j k

ΩΩΩΩ

( ) ( ) ( )2 2 28.46 m/s 1.970 m/s 2.96 m/s

C= − + −a i j k

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Chapter 15, Solution 221.

Geometry. ( ) ( ) ( )0.36 m cos 20 sin 20 0.180 mD

= ° − ° +r i j k

Let frame Oxyz rotate about the Y axis with angular velocity 1

ω= jΩΩΩΩ and angular acceleration 0.=&ΩΩΩΩ

Then, the motion relative to the frame consists of rotation with angular velocity 2 2

ω= kωωωω and angular

acceleration 2

(a) ( )3 0.36cos20 0.36sin 20 0.180D D′ = × = × ° − ° +v r j i j kΩΩΩΩ

0.540 1.08cos 20= − °i k

( )

)/ 2

4 0.36cos20 0.36sin 20 0.180

1.44sin 20 1.44cos20

D F D= × = × ° − ° +

= ° + °

v r k i j k

i j

ωωωω

( )/0.540 1.44sin 20 1.44cos 20 1.08cos 20

D D D F′= + = + ° + ° − °v v v i j k

( ) ( ) ( )1.033 m/s 1.353 m/s 1.015 m/sD

= + −v i j k

(b) ( )3 0.540 1.08cos20D D′ ′= × = × − °a v j i kΩΩΩΩ

3.24 cos 20 1.62= − ° −i k

( )/ 2 /4 1.44sin 20 1.44cos20

D F D F= × = × ° + °a v k i jωωωω

5.76cos 20 5.76sin 20= − ° + °i j

( ) ( ) ( )/2 2 3 1.44sin 20 1.44cos 20 8.64sin 20

D F× = × ° + ° = − °v j i j kΩΩΩΩ

( ) ( )/ /

2

3.24 5.76 cos 20 5.76sin 20 1.62 8.64sin 20

D D D F C F′= + + ×

= − + ° + ° − + °

a a a v

i j k

ΩΩΩΩ

( ) ( ) ( )2 2 28.46 m/s 1.970 m/s 4.58 m/s

D= − + −a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 222.

With the origin at point A, ( ) ( ) ( )0.2 m 0.3 m 0.15 mD

= + −r i j k

( ) ( ) 2 2

/0.2 m 0.15 m , 0.2 0.15 0.25 m

C B BCl= − = + =r i k

Let the frame Axyz rotate with angular velocity ( )110 rad/sω= =i iΩΩΩΩ and angular acceleration

( )2115 rad/s .ω= = −i i& &ΩΩΩΩ

(a) ( ) ( ) ( )10 0.2 0.3 0.15 1.5 m/s 3 m/sD D′ = × = × + − = +v r i i j k j kΩΩΩΩ

( ) ( ) ( )/

0.60.6 m/s, 0.2 0.15 0.48 m/s 0.36 m/s

0.25D F

u = = − = −v i k i k

( ) ( ) ( )/0.48 m/s 1.5 m/s 2.64 m/s

D D D F′= + = + +v v v i j k

Velocity of D. ( ) ( ) ( )0.480 m/s 1.500 m/s 2.64 m/sD

= + +v i j k

(b) D D D′ ′= × + ×a r v

&ΩΩΩΩ ΩΩΩΩ

( ) ( )15 0.2 0.3 0.15 10 1.5 3= − × + − + × +i i j k i j k

( ) ( )2 22.25 4.5 30 15 32.25 m/s 10.5 m/s= − − − + = − +j k j k j k

2

rel3m/sa =

( ) ( ) ( )2 2

rel

30.2 0.15 2.4 m/s 1.8 m/s

0.25= − = −a i k i k

( )( ) ( ) ( )2/2 2 10 0.48 0.36 7.2 m/s

D F× = × − =v i i k jΩΩΩΩ

( ) ( ) ( )rel /2 2.4 m/s 25.05 m/s 8.70 m/s

D D D F′= + + × = − +a a a Ω v i j k

Acceleration of D. ( ) ( ) ( )2 2 22.40 m/s 25.1m/s 8.70 m/s

D= − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

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Chapter 15, Solution 223.

Geometry. ( ) ( ) ( )0.36 m , 0.15 m 0.52 mB D

= = +r j r i k

( ) ( ) ( ) 2 2 2

/0.15 m 0.36 m 0.52 m 0.15 0.36 0.52 0.65 m

B D DBl= − + − = + + =r i j k

( ) ( ) ( ) ( )10.075 m 0.18 m 0.26 m

2C B D

= + = + +r r r i j k

Let frame Axyz rotate with angular velocity ( )15 rad/s .ω= =k kΩΩΩΩ

Velocity analysis. 2 2975 mm/s 0.975 m/su = =

( ) ( ) ( )5 0.075 0.18 0.26 0.9 m/s 0.375 m/sC C′ = × = × + + = − +v r k i j k i jΩΩΩΩ

( ) ( ) ( ) ( )/ /

0.9750.15 0.36 0.52 0.225 m/s 0.54 m/s 0.78 m/s

0.65C F D B

DB

u

l= = − + − = − + −v r i j k i j k

/C C C F′= +v v v

( ) ( ) ( )1.125 m/s 0.915 m/s 0.780 m/sC

= − + −v i j k

Acceleration analysis. 210 rad/s, 6.5 m/suα = = −k &

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

2 2

/

2 2 2

5 0.9 0.375 1.875 m/s 4.5 m/s

6.510 0.075 0.18 0.26 0.15 0.36 0.52

0.65

1.8 m/s 0.75 m/s 1.5 m/s 3.6 m/s 5.2 m/s

C C

C F C

′ ′= × = × − + = − −

= × +

−= × + + + − + −

= − + + − +

a v k i j i j

a r u

k i j k i j k

i j i j k

&

ΩΩΩΩ

αααα

( )( ) ( ) ( ) ( )2 2

/2 2 5 0.225 0.54 0.78 5.4 m/s 2.25 m/s

C F× = × − + − = − −v k i j k i jΩΩΩΩ

/ /2

C C C F C F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 27.58 m/s 9.60 m/s 5.20 m/s

C= − − +a i j k

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Chapter 15, Solution 224

Geometry. With the origin at A, ( ) ( )6.75 in. 4.5 in.C

= +r j k

Let frame Axyz rotate about the y axis with angular velocity ( )19 rad/sω= =k kΩΩΩΩ and angular acceleration

( )2145 rad/s .α= = −k kαααα Then, the motion relative to the frame consists of rotation about the x axis with

angular velocity ( )2 212 rad/sω= =i iωωωω and angular acceleration ( )2 2

60 rad/s .α= = −i iαααα

Motion of coinciding pointC ′ in the frame.

( ) ( )

( ) ( ) ( )( ) ( )2 2

9 6.75 4.5 60.75 in./s

45 6.75 4.5 9 60.75

303.75 in./s 546.75 in./s

C C

C C C

′ ′

= × = × + = −

= × + ×

= − × + + × −

= −

v r k j k i

a r v

k j k k i

i j

ΩΩΩΩ

αααα ΩΩΩΩ

Motion relative to the frame.

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

/ 2

/ 2 2 /

2 2

12 6.75 4.5 54 in./s 81 in./s

60 6.75 4.5 12 54 81

270 405 972 648

702 in./s 1053 in./s

C F C

C F C C F

= × = × + = − +

= × + ×

= − × + + × − +

= + − − −

= − −

v r i j k j k

a r v

i j k i j k

j k j k

j k

ωωωω

αααα ωωωω

(a) Velocity of point C. /C C C F′= +v v v

( ) ( ) ( )60.8 in./s 54.0 in./s 81.0 in./sC

= − − +v i j k

Coriolis acceleration. /

2C F

× vΩΩΩΩ

( )( ) ( ) ( )2/2 2 9 54 81 972 in./s

C F× = × − + =v k j k iΩΩΩΩ

(b) Acceleration of point C. / /

2C C C F C F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 21275.75 in./s 1248.75 in./s 1053 in./s

C= − −a i j k

( ) ( ) ( )2 2 2106.3 ft/s 104.1 ft/s 87.8 ft/s

C= − −a i j k

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Chapter 15, Solution 225.

Let frame Oxyz rotate with angular velocity1.ω= jΩΩΩΩ

Let s s DA

ω=ωωωω λλλλ be the spin of gear A about the axle AD, where sin cosDA

θ θ= − +λ i j is a unit vector along

the axle.

( ) ( )sin cos cos sinC

L r L rθ θ θ θ= − − +r i j

Due to motion of the frame, ( )1cos sin

C Cr Lω θ θ′ = × = −v r kΩΩΩΩ

Due to spin ,s

ωωωω / /C F s C A s

rω= × =v r kωωωω

Then, ( )/ 1cos sin

C C C F sr L rω θ θ ω′ = + = − + v v v k

Since gear B is fixed, 0C

=v

1sin cos

s

L

rω ω θ θ = −

(a) Angular velocity. s

= +ωωωω ΩΩΩΩ ωωωω

( )1 1sin cos sin cos

L

rω ω θ θ θ θ = + − − +

j i jωωωω

1sin cos sin sin cos

L L

r rω θ θ θ θ θ = − + +

i jωωωω

(b) Angular acceleration.

1 10 sin cos sin cos sin

Oxyz

L L

r r

ω

ω ω θ θ θ θ θ

= = + ×

= + × − + +

j i j

& &αααα ωωωω ΩΩΩΩ ωωωω

2

1sin sin cos

L

rω θ θ θ = −

kαααα

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Chapter 15, Solution 226.

Let frame Oxyz rotate with angular velocity 1.ω= jΩΩΩΩ

Let s s DA

ω=ωωωω λλλλ be the spin of gear A about the axle AD, where sin cosDA

θ θ= − +λ i j is a unit vector along

the axle.

( ) ( )sin cos cos sinC

L r L rθ θ θ θ= − − +r i j

Due to motion of the frame, ( )1cos sin

C Cr Lω θ θ′ = × = −v r kΩΩΩΩ

Due to spin ,s

ωωωω / /C F s C A s

rω= × =v r kωωωω

Then, ( )/ 1cos sin

C C C F sr L rω θ θ ω′ = + = − + v v v k

Since gear B is rotating with angular velocity2,ω j on gear B

( )2 / 2sin cos

C C BL rω ω θ θ= × = − −v j r k

Equating the two expressions for C

v and solving for ,s

ω

( )1 2sin cos

s

L

rω ω ω θ θ = − −

(a) Angular velocity. ( ) ( )1 1 2sin cos sin cos

L

rω ω ω θ θ θ θ = + − − − +

j i jωωωω

1sin cos sin cos sin

L L

r rω θ θ θ θ θ = − + +

i jωωωω

( )2cos sin sin cos

L

rω θ θ θ θ + − − +

i j

(b) Angular acceleration.

Frame Oxyz is rotating with angular velocity 1.ω= jΩΩΩΩ

( )1 1 1 20 sin cos

Oxyz

L

rω ω ω ω θ θ

= = + ×

= + × = − −

j k

& &αααα ωωωω ωωωω ΩΩΩΩ ωωωω

ωωωω

( )1 1 2sin cos

L

rω ω ω θ θ = − −

kαααα

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 227.

Let frame Oxyz rotate with angular velocity ( )15 rad/s .ω= =i iΩΩΩΩ The motion relative to the frame is the spin

( )24 rad/s .ω =k k

(a) Angular acceleration.

2Oxyz ω= = + × k& &αααα ωωωω ωωωω ΩΩΩΩ

0 5 4 20= + × = −i k j ( )220.0 rad/s= − jαααα

(b) 0. .Acceleration at point Pθ =

( ) ( )

( )

( ) ( )( )( ) ( ) ( )

( ) ( )

1

/ 2

1 1

2

/ 2 2 /

2

/

2 2

/

60 mm 0.06 m

5 0.06 0

4 0.06 0.24 m/s

0

0 4 0.24 0.96 m/s

2 2 5 0.24 2.4 m/s

0.96 m/s 2.4 m/s

P

P P

P F P

P P P

P F P P F

c P F

P P P F c

v

ωω

ω ω

ω ω

′ ′

= =

= × = × =

= × = × =

= × + × =

= × + × = + × = −

= × = × =

= + + = − +

r i i

v i r i i

v k r k i j

a i r i v

a k r k v k j i

a i j k

a a a a i k

&

&

ΩΩΩΩ

( ) ( )2 20.960 m/s 2.40 m/s

P= − +a i k

(c) 90 . .Acceleration at point Pθ = °

( ) ( )( )

( )( )

( ) ( )( )( ) ( )

1

/ 2

2

1 1

2

/ 2 2 /

/

/

60 mm 0.06 m

5 0.06 0.3 m/s

4 0.06 0.24 m/s

0 5 0.3 1.5 m/s

0 4 0.24 0.96 m/s

2 2 5 0.24 0

P

P P

P F P

P P P

P F p P F

c P F

P P P F c

ω

ω

ω ω

ω ω

′ ′

= =

= × = × =

= × = × = −

= × + × = + × = −

= × + × = + × − = −

= × = × − =

= + +

r j j

v i r i j k

v k r k j i

a i r i v i k j

a k r k v k i j

a v i i

a a a a

&

&

ΩΩΩΩ

( )22.46 m/sP

= −a j

COSMOS: Complete Online Solutions Manual Organization System

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Chapter 15, Solution 228.

Let frame Oxyz rotate with angular velocity ( )15 rad/sω= =i iΩΩΩΩ

The motion relative to the frame is the spin ( )24 rad/sω =k k

( )( )30 , 60 mm cos30 sin 30P

θ = ° = ° + °r i j

( )( )0.06 m cos30 sin 30P

= ° + °r i j

( ) ( )15 0.06cos30 0.06sin 30 0.15 m/s

P Pω′ = × = × ° + ° =v i r i i j k

( )/ 24 0.06cos30 0.06sin30

P F Pω= × = × ° + °v k r k i j

( ) ( )0.12 m/s 0.20785 m/s= − +i j

( )21 10 5 0.15 0.75 m/s

P P Pω ω′ ′= × + × = + × = −a i r i v i k j&

( )/ 2 2 /0 4 0.12 0.20785

P F P P Fω ω= × = × = + × − +a k r k v k i j&

( ) ( )2 20.8314 m/s 0.48 m/s= − −i j

( )( ) ( ) ( )2/2 2 5 0.12 0.20785 2.0785 m/s

c P F= × = × − + =a v i i j kΩΩΩΩ

Acceleration at point P.

/P P P F c′= + +a a a a

( ) ( ) ( )2 2 20.831m/s 1.230 m/s 2.08 m/s

P= − − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 229.

Geometry. ( )( ) ( ) ( )/6 m sin 30 cos30 3 m 3 3 m

B A B= = ° + ° = +r r j k j k

Method 1

Let the unextending portion of the boom AB be a rotating frame of reference.

Its angular velocity is ( ) ( )2 10.40 rad/s 0.25 rad/s .ω ω= + = +i j i jΩΩΩΩ

Its angular acceleration is ( )21 2 1 20.10 rad/s .ω ω ω ω= × = − = −j i k kαααα

Motion of the coinciding point B′ in the frame.

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )2 2 2

0.40 0.25 3 3 3 0.75 3 m/s 1.2 3 m/s 1.2 m/s

0 0 0.10 0.40 0.25 0

0 3 3 3 0.75 3 1.2 3 1.2

0.30 0.30 0.48 1.15614 0.60 m/s 0.48 m/s 1.15614 m/s

B B

B B B

= × = + × + = − +

= × + × = − +

= + − − = − −

v r i j j k i j k

i j k i j k

a r v

i i j k i j k

ΩΩΩΩ

αααα ΩΩΩΩ

Motion relative to the frame.

( ) ( ) ( )/

/

sin 30 cos30 0.45 m/s sin 30 0.45 m/s cos30

0

B F

B F

u= ° + ° = ° + °

=

v j k j k

a

Velocity of point B. /B B B F′= +v v v

0.75 3 1.2 3 1.2 0.45sin30 0.45cos30B

= − + + ° + °v i j k j k

( ) ( ) ( )1.299 m/s 1.853 m/s 1.590 m/sB

= − +v i j k

Coriolis acceleration. /

2B F

× vΩΩΩΩ

( )( ) ( )( ) ( ) ( )

/

2 2 2

2 2 0.40 0.25 0.45sin 30 0.45cos30

0.194856 m/s 0.31177 m/s 0.18 m/s

B F× = + × ° + °

= − +

v i j j k

i j k

ΩΩΩΩ

Acceleration of point B. / /

2B B B F B F′= + + Ω ×a a a v

( ) ( ) ( )0.60 0.194856 0.48 0.31177 1.15614 0.18B

= + − + + − +a i j k

( ) ( ) ( )2 2 20.795 m/s 0.792 m/s 0.976 m/s

B= − −a i j k

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Method 2

Let frame ,Axyz which at the instant shown coincides with AXYZ, rotate with an angular

velocity ( )1 10.25 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to this frame consists of turning

the boom relative to the cab and extending the boom.

Motion of the coinciding point B′ in the frame.

( ) ( )( ) ( )2

0.25 3 3 3 0.75 3 m/s

0.25 0.75 3 0.1875 3 m/s

B B

B B

′ ′

= = × + =×

= × = × = −

v j j k ir

a v j i k

ΩΩΩΩ

ΩΩΩΩ

Motion of point B relative to the frame.

Let the unextending portion of the boom be a rotating frame with constant angular velocity

( )2 20.40 rad/s .ω= =i iΩΩΩΩ The motion relative to this frame is the extensional motion with speed u.

( ) ( ) ( )

( ) ( ) ( )2

2 2

2

0.40 3 3 3 1.2 3 m/s 1.2 m/s

0.40 1.2 3 1.2 0.48 m/s 0.48 3 m/s

B B

B B

′′

′′ ′′

= × = × + = − +

= × = × − + = − −

v r i j k j k

a v i j k j k

ΩΩΩΩ

ΩΩΩΩ

( ) ( ) ( )/boom

/boom

sin 30 cos30 0.45 m/s sin 30 0.45 m/s cos30

0

B

B

u= ° + ° = ° + °

=

v j k j k

a

( )( ) ( )( ) ( )

2 /boom

2 2

2 2 0.40 0.45sin 30 0.45cos30

0.31177 m/s 0.18 m/s

B× = × ° + °

= − +

v i j k

j k

ΩΩΩΩ

/ /boom

1.2 3 1.2 0.45sin30 0.45cos30B F B B′′= + = − + + ° + °v v v j k j k

( ) ( )1.85346 m/s 1.58971 m/s= − +j k

( ) ( )/ /boom 2 /boom

2 2

2

0.48 0.48 3 0 0.31177 0.18 0.79177 m/s 0.65138 m/s

B F B B B′′= + + ×

= − − + − + = − −

a a a v

j k j k j k

ΩΩΩΩ

Velocity of point B. /B B B F′= +v v v

0.75 3 1.85346 1.58971B

= − +v i j k

( ) ( ) ( )1.299 m/s 1.853 m/s 1.590 m/sB

= − +v i j k

Coriolis acceleration. 1 /

2B F

× vΩΩΩΩ

( )( ) ( )1 /2 2 0.25 1.85346 1.58971 0.79486

B F× = × − + =v j j k iΩΩΩΩ

Acceleration of point B. / 1 /

2B B B F B F′= + + ×a a a vΩΩΩΩ

0.1875 3 0.79177 0.65138 0.79486B

= − − − +a k j k i

( ) ( ) ( )2 2 20.795 m/s 0.792 m/s 0.976 m/s

B= − −a i j k

COSMOS: Complete Online Solutions Manual Organization System

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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 230.

Geometry. ( )( ) ( )( )160 mm sin 45 cos 45 0.16 m sin 45 cos 45A

= ° + ° = ° + °r j k j k

Let frame ,Oxyz which coincides with the fixed frame OXYZ at the instant shown, be rotating about the

y axis with constant angular velocity ( )10.8 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consists

of a rotation about the x axis with constant angular velocity

( ) ( )2300 rpm 10 rad/s .

d

dt

θ π= − = − = −i i iωωωω

Motion of the coinciding point A′ in the frame.

( ) ( )( )2

0.8 0.16sin 45 0.16 cos 45 0.128 m/s cos 45

0.8 0.128cos 45 0.1024 m/s cos 45

A A

A A

′ ′

= × = × ° + ° = °

= × = × ° = − °

v r j j k i

a v j i k

ΩΩΩΩ

ΩΩΩΩ

Motion relative to the frame.

( )( ) ( )

( )( ) ( )

/ 2

/ 2 /

2 2 2 2

10 0.16sin 45 0.16cos 45

1.6 m/s cos 45 1.6 m/s sin 45

10 1.6 cos 45 1.6 sin 45

16 m/s sin 45 16 m/s cos 45

A F A

A F A F

π

π π

π π π

π π

= × = − × ° + °

= ° − °

= × = − × ° − °

= − ° − °

v r i j k

j k

a v i j k

j k

ωωωω

ωωωω

Velocity of point A. /A A A F′= +v v v

( ) ( ) ( )0.128 m/s cos 45 1.6 m/s cos 45 1.6 m/s sin 45A

π π= ° + ° − °v i j k

( ) ( ) ( )0.0905 m/s 3.55 m/s 3.55 m/sA

= + −v i j k

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( ) ( ) ( ) ( )2/2 2 0.8 1.6 cos 45 1.6 sin 45 2.56 m/s sin 45

A Fπ π π× = × ° − ° = − °v j j k iΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

( )2 22.56 sin 45 16 sin 45 16 cos 45 0.1024cos 45

Aπ π π= − ° − ° − ° + °a i j k

( ) ( ) ( )2 2 25.68 m/s 111.7 m/s 111.7 m/s

A= − − −a i j k

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Chapter 15, Solution 231.

Geometry. ( )( ) ( ) ( )60 , 8 in. cos60 sin 60 4 in. 4 3 in.A

θ = ° = ° + ° = +r i j i j

Let the frame Oxyz rotate about the Y axis with constant angular velocity ( )112 rad/s .ω= =j jΩΩΩΩ Then, the

motion relative to the frame consists of rotation about the z axis with constant angular velocity

( )2 28 rad/s .ω= =k kωωωω

Motion of the coinciding point A′ in the frame.

( ) ( )

( ) ( )

1

1

1

2

12 4 4 3 48 in./s

0 12 48 576 in./s

A A

A A A

d

dt

ω

ω ω

′ ′

= × = × + = −

= × + ×

= + × − = −

v j r j i j k

a j r j v

j k i

Motion relative to the frame.

( ) ( ) ( )

( ) ( ) ( )

/ 2

2/ 2 /

2 2

8 4 4 3 32 3 m/s 32 m/s

0 8 32 3 32 256 in./s 256 3 in./s

A F A

A F A A F

d

dt

ω ω

= × = × + = − +

= × + ×

= + × − + = − −

v k r k i j i j

a k r k v

k i j i j

ωωωω

Velocity of point A. /A A A F′= +v v v

( ) ( ) ( )32 3 in./s 32 in./s 48 in./sA

= − + −v i j k

( ) ( ) ( )4.62 ft/s 2.67 ft/s 4.00 ft/sA

= − + −v i j k

Coriolis acceleration.

( )( ) ( ) ( )2/2 2 12 32 3 32 768 3 in./s

A F× = × − + =v j i j kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 2832 in./s 256 3 in./s 768 3 in./s

A= − − +a i j k

( ) ( ) ( )2 2 269.3 ft/s 37.0 ft/s 110.9 ft/s

A= − − +a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 232.

Geometry. ( )( ) ( ) ( )60 , 8 in. cos60 sin 60 4 in. 4 3 in.A

θ = ° = ° + ° = +r i j i j

Let the frame Oxyz rotate about the Y axis with angular velocity ( )112 rad/sω= =j jΩΩΩΩ and angular

d

dt

ω = = −

j jαααα Then, the motion relative to the frame consists of rotation about

the z axis with constant angular velocity ( )2 28 rad/sω= =k kωωωω and angular acceleration

( )22

d

dt

ω = =

k kαααα

Motion of the coinciding point A′ in the frame.

( ) ( )

( ) ( ) ( ) ( ) ( )

1

1

1

2 2

12 4 4 3 48 in./s

16 4 4 3 12 48 64 in./s 576 in./s

A A

A A A

d

dt

ω

ω ω

′ ′

= × = × + = −

= × + ×

= − × + + × − = −

v j r j i j k

a j r j v

j i j j k k i

Motion relative to the frame.

( ) ( ) ( )

( ) ( )( ) ( )

/ 2

2

/ 2 /

8 4 4 3 32 3 in./s 32 in./s

10 4 4 3 8 32 3 32

40 3 40 256 256 3 256 40 3 256 3 40

A F A

A F A A F

d

dt

ω

ω ω

= × = × + = − +

= × + ×

= × + + × − +

= − + − − = − + − −

v k r k i j i j

a k r k v

k i j k i j

i j i j i j

Velocity of point A. /A A A F′= +v v v

( ) ( ) ( )32 3 in./s 32 in./s 48 in./sA

= − + −v i j k

( ) ( ) ( )4.62 ft/s 2.67 ft/s 4.00 ft/sA

= − + −v i j k

Coriolis acceleration.

( )( ) ( ) ( )2/2 2 12 32 3 32 768 3 in./s

A F× = × − + =v j i j kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )832 40 3 256 3 40 768 3 64A

= − + − − + +a i j k

( ) ( ) ( )2 2 275.1 ft/s 33.6 ft/s 116.2 ft/s

A= − − +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 233.

. Let moving frame rotate about the axis with angular velocityFrame of reference Axyz Y

( )10.15 rad/s .ω= =j jΩΩΩΩ

.Geometry ( ) ( )/5cos 20 5sin 20 4.6985 m 1.7101 m

B A= − ° + ° = − +r i j i j

Place point O on Y axis at same level as point A.

( ) ( ) ( )/ / / /0.8 m 3.8985 m 1.7101 m

B O B A AO B A= + = + = − +r r r r i i j

Motion of corresponding point B′ in the frame.

( ) ( ) ( )/0.15 3.8985 1.7101 0.58477 m/s

B B O′ = × = × − + =v r j i j kΩΩΩΩ

( ) ( ) ( )20.15 0.58477 0.087715 m/sB B′ ′= × = × =a v j k iΩΩΩΩ

.Motion of point B relative to the frame ( )20.25 rad/s

d

dt

θ= − =k kωωωω

( ) ( )/ 2 /0.25 4.6985 1.7101

B F B A= × = − × − +v r k i jωωωω

( ) ( )0.42753 m/s 1.17462 m/s= − +i j

( ) ( )/ 2 /0.25 0.42753 1.17462

B F B F= × = − × − +a v k i jωωωω

( ) ( )2 20.29365 m/s 0.106881 m/s= +i j

.Velocity of point B /B B B F′= +v v v

( ) ( ) ( )0.428 m/s 1.175 m/s 0.585 m/s B

= − + +v i j k

.Coriolis acceleration ( )( ) ( )/2 2 0.15 0.42753 1.17462

B F× = − × − +v j i jΩΩΩΩ

( )20.128259 m/s= − k

.Acceleration of point B / /2

B B B F B F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 20.381 m/s 0.1069 m/s 0.1283 m/s

B= + −a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 234.

Geometry. Dimensions in meters.

/

0.1C B

=r k /

0.070711 0.070711DC

= − +r j k

/

0.070711 0.170711D B

= − +r j k

Method 1

Let the rigid body ABC be a rotating frame of reference.

Its angular velocity and angular acceleration are

( )18 rad/s ,ω= = = 0j j &ΩΩΩΩ ΩΩΩΩ

Motion of the coinciding point D′ in the frame.

( )/8 0.070711 0.170711 1.36569

D D B′ = × = × − + =v r j j k iΩΩΩΩ

/

0 8 0.136569 10.9255D D B D′ ′= × + × = + × = −a r v j i k&ΩΩΩΩ ΩΩΩΩ

Motion of point D relative to the frame.

( )/ 2 /4 0.070711 0.070711

D ABC D Cω= × = × − +v i r i j k

0.28284 0.28284= − −j k

/ 2 / 2 /D ABC D C D ABC

ω ω= × + ×a i r i v&

( )5 0.070711 0.070711= × − +i j k

( )+ 4 0.28284 0.28284× − −i j k

0.77781 1.48492= −j k

Velocity of point D. /D D D ABC′= +v v v

( ) ( ) ( )1.366 m/s 0.283 m/s 0.283 m/sD

= − −v i j k

Coriolis acceleration. /

2D ABC

× vΩΩΩΩ

( )( ) ( )/2 2 8 0.28284 0.28284 4.5255

D ABC× = × − − = −v j j k iΩΩΩΩ

Acceleration of point D. / /

2D D D ABC D ABC′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 24.53 m/s 0.778 m/s 12.41 m/s

D= − + −a i j k

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Method 2

Bent rod ABC. (Rotation about the fixed y-axis)

1

8ABC

ω= =j jωωωω 1

0ABC

ω= =j&αααα

/

8 0.1 0.8C ABC C B

= × = × =v r j k iωωωω

/

0 8 0.8 6.4C ABC C B ABC C

= × + × = + × = −a r v j i kαααα ωωωω

Rod CD. 1 2

8 4CD

ω ω= + = +j i j iωωωω

1 2 1 2

0 5 8 4 5 32CD

ω ω ω ω= + + × = + + × = −j i j i i j i i k& &αααα

/ /

4 8 0

0 0.070711 0.070711

0.56569 0.28284 0.28284

D C CD D C= × =

= − −

i j k

v r

i j k

ωωωω

/ / /

5 0 4 8 0

0 0 070711 0 070711 0.56569 0 28284 0 28284

2.26276 0.35356 0.35356 2.26272 1.13136 5.65688

DC CD D C CD C D= × + ×

= −32 +− . . − . − .

= − − − − + −

a r v

i j k i j k

i j k i j k

αααα ωωωω

Velocity of point D. /D C D C

= +v v v

( ) ( ) ( )1.366 m/s 0.283 m/s 0.283 m/sD

= − −v i j k

Acceleration of point D. /D C D C

= +a a a

( ) ( ) ( )4.53 m/s 0.778 m/s 12.41 m/sD

= − + −a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 235.

Geometry. ( ) ( ) ( ) ( )/ /7.5 in. 9 in. 18 in. , 9 in.

A D AC= + − =r i j k r j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the Y axis

with constant angular velocity ( )18rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consists of a

rotation of the disk AB about the bent axle CD with constant angular velocity ( )2 212 rad/s .ω= =k kωωωω

Motion of the coinciding point A′ in the frame.

( ) ( ) ( )/8 7.5 9 18 144 in./s 60 in./s

A A D′ = × = × + − = − −v r j i j k i kΩΩΩΩ

( ) ( ) ( )2 28 144 60 480 in./s 1152 in./s

A A′ ′= × = × − − = − +a v j i k i kΩΩΩΩ

Motion of point A relative to the frame.

( )/ 2 /12 9 108 in./s

A F A D= × = × = −v r k j iωωωω

( ) ( )2/ 2 /12 108 1296 in./s

A F A F= × = × − = −a v k i jωωωω

Velocity of point A. /A A A F′= +v v v

( ) ( )144 60 108 252 in./s 60 in./sA

= − − − = − −v i k i i k

( ) ( )21.0 ft/s 5.00 ft/sA

= − −v i k

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( )( ) ( ) ( )/2 2 8 108 1728 in./s

A F× = × − =v j i kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 2480 1152 1296 1728 480 in./s 1296 in./s 2880 in./s

A= − + − + = − − +a i k j k i j k

( ) ( ) ( )2 2 240.0 ft/s 108.0 ft/s 240 ft/s

A= − − +a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 236.

Geometry. ( ) ( ) ( ) ( )/ /7.5 in. 9 in. 18 in. , 9 in.

B D B C= − − = −r i j k r j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the Y axis

with constant angular velocity ( )18 rad/s .ω= =j jΩΩΩΩ Then, the motion relative to the frame consists of a

rotation of the disk AB about the bent axle CD with constant angular velocity ( )2 212 rad/s .ω= =k kωωωω

Motion of the coinciding point B′ in the frame.

( ) ( ) ( )( ) ( ) ( )

/

2 2

8 7.5 9 18 144 in./s 60 in./s

8 144 60 480 in./s 1152 in./s

B B D

B B

′ ′

= × = × − − = − −

= × = × − − = − +

v r j i j k i k

a v j i k i k

ΩΩΩΩ

ΩΩΩΩ

Motion of point B relative to the frame.

( ) ( )( )

/ 2 /

2

/ 2 /

12 9 108 in./s

12 108 1296 in./s

B F B D

B F B F

= × = × − =

= × = × =

v r k j i

a v k i j

ωωωω

ωωωω

Velocity of point B.

/B B B F′= +v v v

( ) ( )144 60 108 36 in./s 60 in./sB

= − − + = − −v i k i i k

( ) ( )3.00 ft/s 5.00 ft/sB

= − −v i k

Coriolis acceleration.

/2

B F× vΩΩΩΩ

( )( ) ( ) ( )2/2 2 8 108 1728 in./s

B F× = × = −v j i kΩΩΩΩ

Acceleration of point B.

/ /2

B B B F B F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 2480 1152 1296 1728 480 in./s 1296 in./s 576 in./s

B= − + + − = − + −a i k j k i j k

( ) ( ) ( )2 2 240.0 ft/s 108.0 ft/s 48.0 ft/s

B= − + −a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 237.

Geometry. ( ) ( )120 mm 180 mmA

= +r i k

Method 1

Let the rigid body DCB be a rotating frame of reference.

Its angular velocity is ( ) ( )1 21.2 rad/s 1.5 rad/s .ω ω= + = +i k i kΩΩΩΩ

Its angular acceleration is ( )21 2 1 21.8 rad/s .ω ω ω ω= × = − = −i k j jαααα

Motion of the coinciding point A′ in the frame.

( ) ( )( )

1.2 1.5 120 180

216 180 36 mm/s

A A′ = × = + × +

= − + = −

v r i k i k

j j j

ΩΩΩΩ

( ) ( ) ( ) ( )( ) ( )2 2

1.8 120 180 1.2 1.5 36

216 324 43.2 54 270 mm/s 172.8 mm/s

A A A′ ′= × + ×

= − × + + + × −

= − − + = − +

a r v

j i k i k j

k i k i i k

αααα ΩΩΩΩ

Motion of point A relative to the frame.

( )/ /60 mm/s , 0

A F A Fu= = =v i i a

Velocity of point A. /A A A F′= +v v v

( ) ( )60.0 mm/s 36.0 mm/sA

= −v i j

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( )( ) ( )2/2 2 1.2 1.5 60 180 mm/s

A F× = + × =v i k i jΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

( ) ( ) ( )2 2 2270 mm/s 180.0 mm/s 172.8 mm/s

A= − + +a i j k

Method 2

Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity

11.2 rad/s.ω= =i iΩΩΩΩ Then the motion relative to the frame consists of the rotation of body DCB about the z

axis with angular velocity ( )21.5 rad/sω =k k plus the sliding motion ( )60 mmu= =u i i of the rod AB

relative to the body DCB.

Motion of the coinciding point A′ in the frame.

( ) ( )( ) ( )2

1.2 120 180 216 mm/s

1.2 216 259.2 mm/s

A A

A A

′ ′

= × = × + = −

= × = × − = −

v r i i k j

a v i j k

ΩΩΩΩ

ΩΩΩΩ

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Motion of point A relative to the frame.

( ) ( ) ( )( ) ( )

( )( ) ( )( ) ( )

/ 2

/ 2 2 2 2

2 2

1.5 120 180 60 180 mm/s 60 mm/s

2

0 1.5 180 0 2 1.5 60

180 mm/s 270 mm/s

A F A

A F A A

u

u u

ω

ω ω ω

= × + = × + + = +

= × + × × + + ×

= + × + + ×

= −

v k r i k i k i j i

a k r k k r i j i

k j k i

j i

&αααα

Velocity of point A. /A A A F′= +v v v

216 180 60A

= − + +v j j i

( ) ( )60 mm/s 36 mm/sA

= −v i j

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( )( ) ( ) ( )2/2 2 1.2 180 60 432 mm/s

A F× = × + =v i j i kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

259.2 180 270 432A

= − + − +a k j i k

( ) ( ) ( )2 2 2270 mm/s 180.0 mm/s 172.8 mm/s

A= − + +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 238.

Geometry. ( ) ( )120 mm 180 mmA

= +r j k

Method 1

Let the rigid body DCB be a rotating frame of reference.

Its angular velocity is ( ) ( )1 21.2 rad/s 1.5 rad/s .ω ω= + = −i k i kΩΩΩΩ

Its angular acceleration is ( )21 2 1 21.8 rad/s .ω ω ω ω= × = − =i k j jαααα

Motion of the coinciding point A′ in the frame.

( ) ( )( ) ( ) ( )

1.2 1.5 120 180

144 216 180 180 mm/s 216 mm/s 144 mm/s

A A′ = × = − × +

= − + = − +

v r i k j k

k j i i j k

ΩΩΩΩ

( ) ( )2 2

0 1.8 0 1.2 0 1.5

0 120 180 180 216 144

324 324 442.8 259.2 442.8 mm/s 259.2 mm/s

A A A′ ′= × + ×

= + −− +

= − + − = −

a r v

i j k i j k

i i j k j k

αααα ΩΩΩΩ

Motion of point A relative to the frame.

( )/ /60 mm/s , 0

A F A Fu= = =v j j a

Velocity of point A. /A A A F′= +v v v

180 216 144 60A

= − + +v i j k j

( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA

= − +v i j k

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( )( ) ( ) ( )2 2

/2 2 1.2 1.5 60 180 mm/s 144 mm/s

A F× = − × = +v i k j i kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

442.8 259.2 180 144A

= − − + +a j k i k

( ) ( ) ( )2 2 2180 mm/s 443 mm/s 115.2 mm/s

A= − −a i j k

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Method 2

Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity

11.2 rad/s.ω= =i iΩΩΩΩ Then the motion relative to the frame consists of the rotation of body DCB

about the z axis with angular velocity ( )21.5 rad/sω = −k k plus the sliding motion ( )60 mmu= =u i j

of the rod AB relative to the body DCB.

Motion of the coinciding point A′ in the frame.

( ) ( ) ( )( ) ( ) ( )2 2

1.2 120 180 216 mm/s 144 mm/s

1.2 216 144 172.8 mm/s 259.2 mm/s

A A

A A

′ ′

= × = × + = − +

= × = × − + = − −

v r i j k j k

a v i j k j k

ΩΩΩΩ

ΩΩΩΩ

Motion of point A relative to the frame.

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( )( ) ( )

/ 2

/ 2 2 2 2

2 2

1.5 120 180 60 180 mm/s 60 mm/s

2

0 1.5 180 0 2 1.5 60

270 180 180 mm/s 270 mm/s

A F A

A F A A

u

u u

ω

ω ω ω

= × + = − × + + = +

= × + × × + + ×

= + − × + + − ×

= − + = −

v k r j k j k j i j

a k r k k r j k j

k i k j

j i i j

&αααα

Velocity of point A. /A A A F′= +v v v

216 144 180 60A

= − + − +v j k i j

( ) ( ) ( )180.0 mm/s 156.0 mm/s 144.0 mm/sA

= − +v i j k

Coriolis acceleration. /

2A F

× vΩΩΩΩ

( )( ) ( ) ( )2/2 2 1.2 180 60 144 mm/s

A F× = × + =v i i j kΩΩΩΩ

Acceleration of point A. / /

2A A A F A F′= + + ×a a a vΩΩΩΩ

172.8 259.2 180 270 144A

= − − + − +a j k i j k

( ) ( ) ( )2 2 2180.0 mm/s 443mm/s 115.2 mm/s

A= − −a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 239.

Geometry. ( ) ( )6 in. 9 in.A

= −r i j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the x axis

with constant angular velocity ( )16 rad/s .ω= =i iΩΩΩΩ Then, the motion relative to the frame is a spin about the

axle CD of angular velocity ( )28 rad/s .ω− = −j j

Motion of the coinciding point A′ in the frame.

( ) ( )( ) ( )2

6 6 9 54 in./s

6 54 324 in./s

A A

A A

′ ′

= × = × − = −

= × = × − =

v r i i j k

a v i k j

ΩΩΩΩ

ΩΩΩΩ

Motion of point A relative to the frame.

( ) ( )( )

/ 2

/ 2 /

8 6 9 48 in./s

8 48 384 in./s

A F A

A F A F

= − × = − × − =

= − × = − × = −

v j r j i j k

a j v j k i

ωωωω

ωωωω

Velocity of point A.

/A A A F′= +v v v

( )54 48 6 in./sA

= − + = −v k k k

( )0.500 ft/sA

= −v k

Coriolis acceleration.

/2

A F× vΩΩΩΩ

( )( ) ( )2/2 2 6 48 576 in./s

A F× = × = −v i k jΩΩΩΩ

Acceleration of point A.

/ /2

A A A F A F′= + + ×a a a vΩΩΩΩ

( ) ( )2 2324 384 576 384 in./s 252 in./s

A= − − = − −a j i j i j

( ) ( )2 232.0 ft/s 21.0 ft/s

A= − −a i j

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

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Chapter 15, Solution 240.

.Geometry ( ) ( )6 in. 14 in.B

= −r i j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the x axis

with constant angular velocity ( )16 rad/s .ω= =i iΩΩΩΩ Then, the motion relative to the frame is a spin about the

axle CD of angular velocity ( )28 rad/s .ω− = −j j

Motion of the coinciding point B′ in the frame.

( ) ( )( ) ( )2

6 6 14 84 in./s

6 84 504 in./s

B B

B B

′ ′

= × = × − = −

= × = × − =

v r i i j k

a v i k j

ΩΩΩΩ

ΩΩΩΩ

Motion of point B relative to the frame.

( ) ( )( )

/ 2

/ 2 /

8 6 14 48 in./s

8 48 384 in./s

B F B

B F B F

= − × = − × − =

= − × = − × = −

v j r j i j k

a j v j k i

ωωωω

ωωωω

Velocity of point B.

/B B B F′= +v v v

( ) ( ) ( )84 in./s 48 in./s 36 in./sB

= − + = −v k k k

( )3.00 ft/sB

= −v k

Coriolis acceleration.

/2

B F× vΩΩΩΩ

( )( ) ( )2/2 2 6 48 576 in./s

B F× = × = −v i k jΩΩΩΩ

Acceleration of point B.

/ /2

B B B F B F′= + + ×a a a vΩΩΩΩ

( ) ( )2 2504 384 576 384 in./s 72 in./s

B= − − = − −a j i j i j

( ) ( )2 232.0 ft/s 6.00 ft/s

B= − −a i j

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 241.

Let the moving frame of reference be the unit less the pulleys and belt. It rotates about the Y axis with constant

angular velocity ( )1.6 rad/s .ω =1

= j jΩΩΩΩ The relative motion is that of the pulleys and belt with speed

90 mm/s.u =

(a) Acceleration at point A.

( ) ( )100 mm 380 mmA

= − +r i j

( ) ( )1.6 100 380 160 mm/sA A′ = × = × − + =v r j i j kΩΩΩΩ

( )21.6 160 256 mm/sA A′ ′= × = × =a v j k iΩΩΩΩ

( )/90 mm/s

A Fu= =v k k

( )2 2

2

/

90135 mm/s

60A F

u

ρ

= − = − = −

a j j j

( )( ) ( ) ( )2/2 2 1.6 90 288 mm/s

A F× = × =v j k iΩΩΩΩ

/ /2 256 135 288

A A A F A F′= + + × = − +a a a v i j iΩΩΩΩ

( ) ( )2 2544 mm/s 135.0 mm/s

A= −a i j

(b) Acceleration of point B.

( ) ( ) ( )100 mm 200 mm 60 mmB

= − + +r i j k

( ) ( ) ( )1.6 100 200 60 96 mm/s 160 mm/sB B′ = × = × − + + = +v r j i j k i kΩΩΩΩ

( ) ( ) ( )2 21.6 96 160 256 mm/s 153.6 mm/s

B B′ ′= × = × + = −a v j i k i kΩΩΩΩ

( )/90 mm/s

B Fu= − = −v j j

/

0B F

=a

( )( ) ( )/2 2 1.6 90 0

B F× = × − =v j jΩΩΩΩ

/ /2 256 153.6 0 0

B B B F B F′= + + × = − + +a a a v i kΩΩΩΩ

( ) ( )2 2256 mm/s 153.6 mm/s

B= −a i k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 242.

Let the moving frame of reference be the unit less the pulleys and belt. It rotates about the Y axis with constant

angular velocity ( )11.6 rad/s .ω= =j jΩΩΩΩ The relative motion is that of the pulleys and belt with speed

90 mm/s.u =

(a) Acceleration at point C.

( ) ( )100 mm 80 mmC

= − +r i j

( ) ( )1.6 100 80 160 mm/sC A′ = × = × − + =v r j i j kΩΩΩΩ

( )21.6 160 256 mm/sC A′ ′= × = × =a v j k iΩΩΩΩ

( )/90 mm/s

C Fu= − = −v k k

( )2 2

2

/

90135 mm/s

60C F

u

ρ

= = =

a j j j

( )( ) ( ) ( )2/2 2 1.6 90 288 mm/s

C F× = × − = −Ω v j k i

/ /2 256 135 288

C C C F C F′= + + × = + −a a a v i j iΩΩΩΩ

( ) ( )2 232.0 mm/s 135.0 mm/s

C= − +a i j

(b) Acceleration at point D.

( ) ( ) ( )100 mm 200 mm 60 mmD

= − + −r i j k

( ) ( ) ( )1.6 100 200 60 96 mm/s 160 mm/sD B′ = × = × − + − = − +v r j i j k i kΩΩΩΩ

( ) ( ) ( )2 21.6 96 160 256 mm/s 153.6 mm/s

D B′ ′= × = × − + = +a v j i k i kΩΩΩΩ

( )/90 mm/s

D Fu= =v j j

/

0D F

=a

( )( ) ( )/2 2 1.6 90 0

D F× = × − =v j jΩΩΩΩ

/ /2 256 153.6 0 0

D D D F D F′= + + × = + + +a a a v i kΩΩΩΩ

( ) ( )2 2256 mm/s 153.6 mm/s

D= +a i k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 243.

( ) ( ) ( ) 2 2 2

/16 in. 16 in. 8 in. 16 16 8 24 in.

A E AEl= − + + = + + =r i j k

Angular velocity.

( )/

1216 16 8

24A E

AEl

ω= = − + +r i j kωωωω

( ) ( )/16 in. 6 in.

C E= − +r i j

Velocity of C.

/8 8 4 24 64 80

16 6 0

C C E= × = − = − − +

i j k

v r i j kωωωω

( ) ( ) ( )24.0 in./s 64.0 in./s 80.0 in./sC

= − − +v i j k

Angular Acceleration. 0=αααα

Acceleration of C.

/0 8 8 4

24 64 80

C C E C= × + × = + −

− −

i j k

a r vαααα ωωωω

896 544 704= + +i j k

( ) ( ) ( )2 2 2896 in./s 544 in./s 704 in./s

C= + +a i j k

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 244.

( )( )0

2400 22400 rpm 80 rad/s, 1.00 in.

60r

πω π= = = =

1 10, 10 stω = =

(a) Before power is turned off, 0α =

( )( )01.00 80 251.33 in./s,

Cv rω π= = = 20.9 ft/s

Cv =

2 2

2251.3363165 in./s

1.00

C

n

v

a

r

= = =

2

0, 63165 in./s ,t Ca r aα= = =

25260 ft/s

Ca =

(b) Uniformly decelerated motion. 1 0 1

tω ω α= +

21 0

1

10t

ω ω πα π− −= = = −

At 9 s, ( )( )080 8 9 8 rad/stω ω α π π π= + = − =

( )( )1.00 8 25.133 in./s,Cv rω π= = = 2.09 ft/s

Cv =

( )2 2

225.133631.65 in./s

1.00

C

C nC

v

a

r

= = =

( ) ( )( ) 21.00 8 25.133 in./s

C Cta r α π= = − = −

( )22 2 2 2631.65 25.133 632.15 in./s

C n ta a a= + = + =

2

52.7 ft/sCa =

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 245.

Arm . Fixed axis rotation.ACB

/ /

6 mm, AC A AC ABr r ω= =v

( ) ( )6 40= 240 mm/s=

/ /36 mm,

B C B B C ABr r ω= =v

( ) ( )36 40= 1440 mm/s=

Disk B: Plane motion = Translation with B + Rotation about B.

/12 mm,

B D B B Ar = = −v v v

0 1440= 12B

ω+

12B

ω = =

/E B E B= +v v v

1440= ( ) ( ) 6 120+ 2880 mm/s=

Disk A: Plane motion = Translation with A + Rotation about A.

/30 mm,

A E A E Ar = = −v v v

2880 240=

30A

ω+

30A

ω += =

=ωωωω

=ωωωω

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 246.

Rod AB. (Rotation about A) ( ) 0.12B AB AB

AB ω ω= =v 45°

Rod DE. (Rotation about E) ( ) 0.15D DE DE

ED ω ω= =v 45°

Rod BGD. Plane motion = Translation with B + Rotation about .B

[/

D B D B Dv= +v v v ] [45

Bv° = ]45° + [ /D B

v ]

Draw velocity vector diagram.

/

/

/ /

/

2 0.12 2

0.12 2 12

0.24 2

10.06 2

2

Draw vector diagram.

sin 45 .12 sin 45

D B B AB

D B AB

BD AB

BD

G B BG BD D B AB

G B G B

G B AB

v v

v

l

l v

v v

ω

ωω ω

ω ω

ω

= =

= = =

= = =

= +

= ° = °

v

v v v

3.6

,0.12sin 45 0.12sin 45

G

AB

vω = =° °

=ωωωω

( )12 42.4 ,

2BD

ω =

=ωωωω

0.12 5.0912 m/sD B ABv v ω= = =

5.0912

0.15

D

DE

ED

v

DE=ωωωω

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 247.

Since the drum rolls without sliding, its instantaneous center lies at B.

120 mm/sE D

= =v v

/ /,

A A B D D Bv r v rω ω= =

(a) /

100 60

D

D B

v

r

ω = = =−

(b) ( ) ( )60 3.00 180 mm/sAv = =

180 mm/sA

=v

Since Av is to the right and

Dv is to the left, cord is being unwound.

180 120 300 mm/sA Ev v− = + =

(c) Cord unwound per second = 300 mm

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 248.

16 in./sB

=v 5

tan12

EF

DFγ = = D D

v=v γ

Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are

known. Draw BC perpendicular to B

v and DC perpendicular to .

Dv

( ) ( ) 5tan 6.4 2.6667 in.,

12CJ DJ γ = = =

12.8 2.6667 10.1333 in.CB JB CJ= − = − =

(a) 16

B

ABD

v

CBω = = = 1.579 rad/s

ABD=ωωωω

10.1333 7.2 17.3333 in.CK CB BK= + = + =

3.6tan , 11.733 , 90 78.3

17.3333

KA

CKβ β β= = = ° ° − = °

17.333317.7032 in.

cos cos 11.733

CKAC

β= = =

°

(b) ( ) ( )( )17.7032 1.57895 28.0 in./s,A ABDv AC ω= = = 28.0 in./s

A=v 78.3°

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 249.

.Velocity analysis

30 in./sD

=v , 12 in./sC

α=v , ω

[/ or 12

C D C D= +v v v ] [30= ] [6ω+ ]

4212 30 6 , 7 rad/s

6ω ω− = − = =

.Acceleration analysis

236 in./s

D=a ,

C Ca=a

, α

( ) ( )/ /C D C D C Dt n

= + +a a a a

[ Ca ] [36= ] [6α+ ] 2

6ω+

Components: 2

: 0 36 6 6 rad/sα α= − =

( )( )2 2: 6 7 294 in./s

Ca = =

224.5 ft/s

C=a

( ) ( )/ /A D A D A Dt n

= + +a a a a

[36= ] [6α+ ] 26ω+

[36= ] [36+ ] 294+

2

72 in./s= 2

294 in./s +

225.2 ft/s

A=a 76.2 °

( ) ( )/ /B D B D B Dt n

= + +a a a a

[36= ] [6α+ ] 26ω+

[36= ] [36+ ] [294+ ]

2

258 in./s= 2

36 in./s +

221.7 ft/s

B=a 7.9 °

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 250.

ω =

( ) ( )( )12 8 96 in./sB ABv AB ω= = =

B B

v=v , D D

v=v

Instantaneous center of bar BD lies at .∞

0, 96 in./sBD D B

ω = = =v v

96

D

DE

v

DEω = = =

.Acceleration analysis ( )220, 64 rad/s

AB ABα ω= =

( )B ABAB α= a ] ( ) 2

ABAB ω+

( )( )0 12 64= + 2

768 in./s =

[/2

D B BDlα=a ] [ BD

lα+ ] 22

BDlω+

2

BDlω +

[24 BDα= ] [12 BD

α+ ] 0+ [24 BDα= ] [12 BD

α+ ]

( )D DEDE α= a ( ) 2

DEDE ω+

[12 DEα= ] ( )( )12 64+

[12 DEα= ] 2

/ Resolve into components.

D B D B= +a a a

2: 768 768 24 64 rad/s

BD BDα α= − + =

[/ /

112

2C B D B BD

α= =a a ] [6 BDα+ ]

2

768 in./s= 2

384 in./s +

[/768

C B C B= + =a a a ] [768+ ] [384+ ]

2

384 in./sC

=a

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 251.

0.5500 mm 0.5 m, 0.5 tan 30 ,

cos30AB AP BP= = = ° =

°

=ωωωω

Let P′ be the coinciding point on AE and 1u be the outward velocity of the collar along the rod AE.

( )/P P P AE AEAP ω′ = + = v v v ] [ 1

u+ ]

Let P′′ be the coinciding point on BD and 2

u be the outward speed along the slot in rod BD.

( )/P P P BD BDBP ω′′ = + = v v v ] [ 2

30 u° + ]60°

Equate the two expressions for P

v and resolve into components.

: ( )( )1 2

0.53 cos30 cos60

cos30u u

= ° + ° °

or 1 2

1.5 0.5u u= + (1)

: ( )( ) ( ) 2

0.50.5 tan30 8 3 sin30 sin 60

cos30u

− ° = − ° + ° °

[ ]2

11.5 tan 30 4 tan 30 1.66667 m/s

sin 60u = ° − ° = −

°

From (1), ( )( )11.5 0.5 1.66667 0.66667 m/su = + − =

( )( )0.5 tan30 8P

= °v [0.66667+ ] 2.3094 m/s= [0.66667 m/s+ ]

2 22.3094 0.66667 2.4037 m/s

Pv = − + =

2.3094tan 73.9

0.66667β β= = °

2.40 m/sP

=v 73.9 °

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 252.

Let the arm AB be a rotating frame of reference. 0.75 rad/s=ΩΩΩΩ ( )0.75 rad/s= − k

Link 1: ( )1 1/2 in. ,

ABu= − =r i v

( )4 in./s= j

( ) ( ) ( )22

1 10.75 2 1.125 in./s′ = −Ω = − − =a r i i

2 2

2

1/

48 in./s

2AB

u

ρ= = =a ( )28 in./s= i

( )( ) ( ) ( )/2 2 0.75 4 6 in./s

P AB× = − × =v k j iΩΩΩΩ

( )21 1 1/ 1/2 15.125 in./s

AB AB′= + + × =a a a v iΩΩΩΩ

2

115.13 in./s=a

Link 2: ( ) ( )2 2/8 in. 2 in.

ABu= + =r i j v

( )4 in./s= i

( ) ( ) ( ) ( )22 2 2

2 20.75 8 2 4.5 in./s 1.125 in./s′ = −Ω = − + = − −a r i j i j

2/0

AB=a

( )( ) ( ) ( )22/2 2 0.75 4 6 in./s

AB× = − × = −v k i jΩΩΩΩ

( ) ( )2 2 2/ 2/

2 2

2

4.5 1.125 6 4.5 in./s 7.125 in./s

AB AB′= + + ×

= − − − = − −

a a a Ω v

i j j i j

( ) ( )2 2 2

24.5 7.125 8.43 in./sa = + =

7.125

tan , 57.74.5

β β= = °

2

28.43 in./s=a 57.7 °

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 253.

Geometry. ( ) ( ) ( ) ( )/ /0.16 m , 0.5 m 0.4 m 0.16 m

B C A B= − = − + −r k r i j k

Velocity at B. ( ) ( )0 /3 0.16 0.48 m/s

B B Cω= × = × − =v j r j k i

Velocity of collar A. A A

v=v j

/ / /

, where A B A B A B AB A B

= + = ×v v v v ω r

Noting that /A B

v is perpendicular to /,

A Br we get

/ /0.

A B A B⋅ =r v

Forming /

,A B A

⋅r v we get ( )/ / / / / /A B A A B B A B A B B A B A B⋅ = ⋅ + = ⋅ + ⋅r v r v v r v r v

or / /A B A A B B

⋅ = ⋅r v r v (1)

From (1), ( ) ( ) ( ) ( )0.5 0.4 0.16 0.5 0.4 0.16 48Av− + − ⋅ = − + − ⋅i j k j i j k i

0.4 0.24 or 0.6 m/sA Av v= − = − ( )0.600 m/s

A= −v j

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 15, Solution 254.

Let frame Oxyz rotate with angular velocity ( )18 rad/s .ω= =i iΩΩΩΩ

Note that point B does not move.

Geometry. ( )/sin 1 cos , 120 mm, 60

A Bρ θ ρ θ ρ θ= + − = = °r i j

( ) ( )/103.923 mm 60 mm

A B= +r i j

Motion of coinciding point A′

( ) ( )/8 103.923 60 480 mm/s

A A B′ = × = × + =v r i i j kωωωω

( )2/0 8 480 3840 mm/s

A A B A′ ′= × + × = + × = −a r v i k jαααα ωωωω

Motion relative to the frame.

( )/cos sin , 600 mm/s

P Fu uθ θ= + =v i j

( ) ( )/300 mm/s 519.62 mm/s

P F= +v i j

( )2

/sin cos

A F

u θ θρ

= − +a i j

( ) ( )2 2

/2598.1 mm/s 1500 mm/s

A F= − +a i j

Coriolis acceleration. /

2c P F

= ×a vΩΩΩΩ

( )( ) ( ) ( )22 8 300 519.62 8313.9 mm/sc

= × + =a i i j k

(a) Velocity of A. /A A A F′= +v v v

( ) ( ) ( )300 mm/s 519.62 mm/s 480 mm/sA

= + +v i j k

( ) ( ) ( )0.300 m/s 0.520 m/s 0.480 m/sA

= + +v i j k

(b) Acceleration of A. /A A A F c′= + +a a a a

( ) ( ) ( )2 2 22598.1 mm/s 2340 mm/s 8313.9 mm/s

A= − − +a i j k

( ) ( ) ( )2 2 22.60 m/s 2.34 m/s 8.31m/s

A= − − +a i j k