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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: ( )23
8 6 2 radianst tθ = − −
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 0t − = 0.250 st =
(b) Angular coordinate and angular velocity at t = 0.250 s.
( )( ) ( )( )3 2
8 0.250 6 0.250 2θ = − − 18.25 radiansθ = −
( )( ) ( )( )2
24 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 2.
3
0.5 cos4t
e tπ
θ π−
=
( )3 3
0.5 3 cos4 4 sin 4t td
e t e t
dt
π πθ
ω 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
d
e t e t e t e t
dt
e t e t
π π π π
π π
ω
α 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= − = − 4.71rad/sω = −
( )( )2
0.5 7 34.5α π= − = − 2
34.5 rad/sα = −
( ) 0.125 s, cos4 cos 0, sin 4 sin 1
2 2
b t t t
π π
π π= = = = =
3
0.30786t
e π−
=
( )( )( )0.5 0.30786 0 0θ = = 0θ =
( )( )( )0.5 0.30786 4 1.93437ω π= − = − 1.934 rad/sω = −
( )( )( )2
0.5 0.30786 24 36.461α π= = 2
36.5 rad/sα =
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 3.
7 /6
0 sin 4t
e tπ
θ θ π−
=
7 /6 7 /6
0
7
sin 4 4 cos4
6
t td
e t e t
dt
π πθ π
ω θ π π π− − 
= = − + 
 
2 2 2
7 /6 7 /6 7 /6 2 7 /6
0
7 /6 2 2
0
49 28 28
sin 4 cos4 cos4 16 sin 4
36 6 6
49 28
16 sin 4 cos4
36 3
t t t t
t
d
e t e t e t e t
dt
e t t
π π π π
π
ω π π π
α θ π π π π π
θ π π π π
− − − −
−
 
= = − − −  
 
  
= − − +  
  
( )a 0 0.4 rad,θ = 0.125 st =
7 (0.125)/6
0.63245, 4 , sin 1, cos 0
2 2 2
e tπ π π π
π−
= = = =
( )( )( )0.4 0.63245 1 0.25298 radiansθ = = 0.253 radθ =
( )( ) ( )
7
0.4 0.63245 1 0.92722 rad/s
6
π
ω
 
= − = − 
 
0.927 rad/sω = −
( )( ) ( )2 249
0.4 0.63245 16 1 36.551rad/s
36
α π
 
= − − = − 
 
2
36.6 rad/sα = −
( )b ,t = ∞ 7 /6
0t
e 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 4.
Angular coordinate: 1800 rev = 3600 radiansπθ =
Initial angular velocity: 0 6000 rpm = 200 rad/sω π=
Angular acceleration: constant
d d
dt d
ω ω
α ω
θ
= = =
d dα θ ω ω=
0
0
0
d d
θ
ω
α θ ω ω=∫ ∫
2
0
1
2
αθ ω= −
( )
( )( )
22
20 200
17.4533 rad/s
2 2 3600
πω
α
θ π
= − = − = −
(a) Time required to coast to rest.
0 tω ω α= +
0 0 200
17.4533
t
ω ω π
α
− −
= =
−
36.0 st =
(b) Time to execute the first 900 revolutions.
900 rev = 1800 radiansθ π=
2
0
1
2
t tθ ω α= +
( ) 21
1800 200 17.4533
2
t tπ π= −
2
72 648 0t t− + =
( ) ( )( )2
72 72 4 648
2
t
± −
= 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 5.
( )( )
1 0 1
2400 2
2400 rpm 80 rad/s, 0, 4 s
60
t
π
ω π ω= = = = =
( )a 21
1 0
80
, 20 rad/s
4
t t
t
ω π
ω ω α α α π= + = = = =
( )( )2 2
1 0
1 1 160
0 20 4 160 rad 80 rev
2 2 2
t t
π
θ ω α π π
π
= + = + = = =
1 80 revθ =
(b) 1 2 2 180 rad/s, 0, 40 st tω π ω= = − =
( ) 22 1
2 1 2 1
2 1
0 80
, 2 rad/s
40
t t
t t
ω ω π
ω ω α α π
− −
= + − = = = −
−
( ) ( ) ( )( ) ( )( )2 2
2 1 1 2 1 2 1
1 1
80 40 2 40
2 2
t t t tθ θ ω α π π− = − + − = + −
1600
1600 radians 800 rev
2
π
π
π
= = = 2 1 800 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 6.
Angular acceleration: 0.2
30 t d
e
dt
ω
α −
= =
Angular velocity: 0 0
t
dtω ω α= + ∫
0.2
0
0 30
t t
e dt−
= + ∫
0.2
0
30
0.2
t
t
e− =  −
( )0.2
150 1 t d
e
dt
θ−
= − =
When t = 0.5 s, ( )( )
( )0.2 0.5
150 1 eω −
= −
14.27 rad/sω =
Angular coordinate: 0 0
t
dtθ θ ω= + ∫
( )0.2
0
0 150 1
t t
e dt−
= + −∫
0.2
0
150
150
0.2
t
t
t e− = −  −
( )0.2
150 750 1 t
t e−
= − −
When t = 0.5 s, ( )( ) ( )( )
( )0.2 0.5
150 0.5 750 1 eθ −
= − −
3.63 radiansθ =
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 7.
( ) 0.5 , 0.5 0.5
d
a d d
d
ω
α ω ω ω ω θ
θ
= − = − = −
0
30 0
Integrating, 0.5 30 0.5d d
θ
ω θ θ= − − = −∫ ∫
60
60 radians 9.55 revθ
π
= = =
2
9.55 revθ =
( ) 0.5 2
d d
b dt
dt
ω ω
ω
ω
= − = −
Integrating,
0
0 30
2
t d
dt
ω
ω
= −∫ ∫
0
2 ln
30
t = − = ∞ t = ∞
( )c ( )( )0.02 30 0.6 rad/sω = =
0.6
0 30
2
t d
dt
ω
ω
= −∫ ∫
0.6
2 ln 2 ln 50
30
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 8.
d
k k d k d
d
ω
α θ ω θ ω ω θ θ
θ
= − = − = −
Integrating,
0 6
12 0
d k dω ω θ θ= −∫ ∫
2 2
12 6
0 0
2 2
k
 
− = − −  
 
( )a
2
2
2
12
9 s
6
k −
= = 2
9.00 sk −
=
3
12 0
d k d
ω
ω ω θ θ= −∫ ∫
2 2 2
12 3
9 0
2 2 2
ω  
− = − −  
 
( )b ( )( )22 2 2 2
12 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 9.
( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k
( ) ( )/ 5 in. 15.6 in.B O = +r i j
( ) ( ) ( )2 2 2
5 31.2 12 33.8 in.OAl = + + =
Angular velocity.
( )/
6.76
5 31.2 12
33.8
A O
OAl
ω
= = + +r i j kωωωω
( ) ( ) ( )1.0 rad/s 6.24 rad/s 2.4 rad/s= + +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 2
126.1 in./s 74.3 in./s 246 in./sB = − − +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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 10.
( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k
( ) ( )/ 5 in. 15.6 in.B O = +r i j
( ) ( ) ( )2 2 2
5 31.2 12 33.8 in.OAl = + + =
Angular velocity.
( )/
3.38
5 31.2 12
33.8
A O
OAl
ω
= = + +r i j kωωωω
( ) ( ) ( )0.5 rad/s 3.12 rad/s 1.2 rad/s= + +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.
( )/
5.07
5 31.2 12
33.8
A O
OAl
α −
= = + +r i j kαααα
( ) ( ) ( )2 2 2
0.75 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 2
3.46 in./s 27.6 in./s 73.1in./sB = − − +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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 11.
( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k
2 2 2
0.5 0.225 0.3 0.625 mABl = + + =
Angular velocity vector.
( )/
10
0.5 0.225 0.3
0.625
B A
ABl
ω
= = − +r i j kωωωω
( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j k
( ) ( )/ 300 mm 0.3 mE 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 2
11.52 m/s 5.18 m/s 23.1m/sE = − + +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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 12.
( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k
2 2 2
0.5 0.225 0.3 0.625 mABl = + + =
Angular velocity vector.
( )/
10
0.5 0.225 0.3
0.625
B A
ABl
ω
= = − +r i j kωωωω
( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j k
Angular acceleration vector.
( )/
20
0.5 0.225 0.3
0.625
B A
ABl
α −
= = − +r i j kαααα
( ) ( ) ( )2 2 2
16 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 mC 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 2
18.00 m/s 19.20 m/s 15.60 m/sC = + −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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 13.
( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k
( ) ( ) ( )2 2 2
200 120 90 250 mmDAd = + + =
0.8 + 0.48 + 0.36A/D
DA
DAd
= = −
r
i j kλλλλ
( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −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 2
405 m/s 432 m/s 324 m/sB = − − +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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 14.
( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k
( ) ( ) ( )2 2 2
200 120 90 250 mmADd = + + =
0.8 + 0.48 + 0.36A/D
DA
ADd
= = −
r
i j kλλλλ
( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −i j k i j kωωωω λλλλ
( )( )600 0.8 + 0.48 + 0.36DA
d
dt
ω
= = − − i j kαααα λλλλ
( ) ( ) ( )2 2 2
480 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 2
405 m/s 389 m/s 266 m/sB = − − −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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 15.
9
93,000,000 mi 491.04 10 ft= ×
6
365.24 days 31.557 10 s, 1rev 2 radπ= × =
Angular velocity.
9
6
2
199.11 10 rad/s
31.557 10
π
ω −
= = ×
×
Velocity of the earth.
( )( )9 9 3
491.04 10 199.11 10 97.77 10 ft/sv rω −
= = × × = ×
3
66.7 10 mi/hv = ×
Acceleration of the earth.
( )( )
2
2 3 9 3 2
97.77 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 16.
3
23 h 56min 23.933 h 86.16 10 s, 1rev 2 radπ= = × =
6
3
2
72.925 10 rad/s
86.16 10
π
ω −
= = ×
×
6
6370 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 2
cos cos 33.876 10 cos m/sp R 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 2
33.9 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 =
( )( )3
33.876 10 0.76604−
= − ×a i
( )3 2
0.273 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 17.
300 mm/sB Av v= = 120 mmBr =
( ) 180 mm/sB At
a a= =
( )a
300
, 2.5 rad/s
120
B
B B
B
v
v r
r
ω ω= = = = 2.50 rad/s=ωωωω
( )
( ) 2180
, 1.5 rad/s
120
B t
B Bt
B
a
a r
r
α α= = = = 2
1.500 rad/s=αααα
( )b ( ) ( )( )22 2
120 2.5 750 mm/sB Bn
a r ω= = =
( ) ( ) ( ) ( )2 2 2 2 2
180 750 771mm/sB B Bt n
a a a= + = + =
750
tan , 76.5
180
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 18.
4 rad/sBω = , 120 mmBr =
( ) ( )( )22 2
120 4 1920 mm/sB B Bn
a r ω= = =
2
2400 mm/sBa =
( ) ( )22 2 2 2
2400 1920 1440 mm/sB B Bt n
a a a= − = − = ±
( )
( ) 21440
, 12 rad/s
120
B t
B Bt
B
a
a r
r
α α
±
= = = = ±
2
12.00 rad/s or
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 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
18 rad/s
0.6667
A B
A
A A
v v
r r
ω = = = = 18 rad/sA =ωωωω
296
144 rad/s
0.6667
A B
A
A A
a a
r r
α = = = = 2
144 rad/sA =αααα
Pulley C. 5 in. 0.41667 ftCr = =
12
28.8 rad/s
0.41667
C B
C
C C
v v
r r
ω = = = = 28.8 rad/sC =ωωωω
296
230.4 rad/s
0.41667
C B
C
C C
a a
r r
α = = = = 2
230 rad/sC =αααα
(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = =
( ) 2
96 ft/sP Bt
a a= =
( )
( )22 2
212
345.6 ft/s
0.41667
P B
P n
C C
v v
a
ρ ρ
= = = =
( ) ( ) ( ) ( )2 2 2 2 2
96 345.6 358.7 ft/sP P Pt n
a a a= + = + =
96
tan 15.52
345.6
β 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 20.
1
rev radians, 8 in. 0.6667 ft, 5 in. 0.41667 ft
2
A Cr rπ= = = = =
2
2 2
500
120 0.002 ,
120 0.002 60000
d d d
d
d
ω ω ω ω ω
α ω ω θ
θ ω ω
= − = = =
− −
Integrating and applying initial condition 0 at 0ω θ= = and noting that θ π= radians at the final state,
( )2
20 00
500
250 ln 60000
60000
d
d
ωω πω ω
ω θ π
ω
= − − = =
−
∫ ∫
( )
2
2 60000
250 ln 60000 ln 60000 250 ln
60000
ω
ω π
− − − − = − =
 
2
/25060000
60000
e πω −−
=
2 /250 2 2
60000 1 749.26 rad /se π
ω − = − = 
27.373 rad/sω =
( )( )2
120 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= = = =
( )( ) 2
0.6667 118.50 79.0 ft/sB A Aa a r α= = = =
2
79.0 ft/sBa =
(b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = =
18.249 ft/sP Bv v= =
( )
2 2
218.249
799.3 ft/s
0.41667
B
P n
C
v
ρ
= = =a
( ) 2
79.0 ft/sP Bt
a= =a
( ) ( )2 2 2
799.3 79.0 803.2 ft/sPa = + =
79
tan , 5.64
799.3
β β= = °
2
803 ft/sP =a 5.64°
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 21.
Left pulley.
Inner radius r1 = 50 mm
Outer radius r2 = 100 mm
0.6 m/s = 600 mm/sAv =
1
2
600
6 rad/s
100
Av
r
ω = = =
Speed of intermediate belt.
( )( )1 1 1 50 6 300 mm/sv rω= = =
Right pulley.
Inner radius r3 = 50 mm
Outer radius r4 = 100 mm
1
2
4
300
3 rad/s
100
v
r
ω = = =
(a) Velocity of C.
( )( )3 2 50 3 150 mm/sCv r ω= = =
0.1500 m/sC =v
(b) Acceleration of point B.
( )( )22 2
4 2 100 3 900 mm/sBa r ω= = =
2
0.900 m/sB =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
© 2007 The McGraw-Hill Companies.
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
600
6 rad/s
100
Av
r
ω = = =
( ) 2
1
2
1800
18 rad/s
100
A t
a
r
α = = =
Intermediate belt.
( )( )1 1 1 50 6 300 mm/sv rω= = =
( ) ( )( ) 2
1 1 1 50 18 900 mm/st
a rα= = =
Right pulley.
Inner radius r3 = 50 mm
Outer radius r4 = 100 mm
1
2
4
300
3 rad/s
100
v
r
ω = = =
( )1 2
2
4
900
9 rad/s
100
t
a
r
α = = =
(a) Velocity and acceleration of point C.
( )( )3 2 50 3 150 mm/sCv r ω= = =
0.150 m/sC =v
( ) ( )( )3 2 50 9 450 mm/sC t
a rα= = =
0.450 m/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
© 2007 The McGraw-Hill Companies.
(b) Acceleration of point B.
( ) ( )( )22 2
4 2 100 3 900 mm/sB n
a r ω= = =
( ) 2
0.900 m/sB n
=a
( ) ( )( ) 2
4 2 100 9 900 mm/sB t
a r α= = =
( ) 2
0.900 m/sB t
=a
2
1.273 m/sB =a 45°
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 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
ω
ω = =
1
2
A
B
r
r
ω
ω =
2
1( ) : A Ab On shaft A a rω=
2
1A Arω=a
2
2 1
2 2
2
: A
B B
r
On ring B a r r
r
ω
ω
 
= =  
 
2 2
1
2
A
B
r
r
ω
=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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 24.
(a) Let point C be the point of contact between the shaft and the ring.
( )( )1 0.5 25 12.5 in./sC Av rω= = =
2
12.5
5.0 rad/s
2.5
C
B
v
r
ω = = = 5.00 rad/sBω =
( ) On shaft :b A ( )( )22
1 0.5 25A Aa rω= =
2
312.5 in./s ,= 2
26.0 ft/sA =a
On ring :B ( )( )22
2 2.5 5.0B Ba r ω= =
2
62.5 in./s ,= 2
5.21ft/sB =a
( ) At a point on the outside of the ring,c 3 3.5 in.r r= =
( )( )22 2
3.5 5.0 87.5 in./sBa rω= = = 2
7.29 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 25.
( )a
( )( )600 2
600 rpm 20 rad/s.
60
A
π
ω 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./sD D A Av r ω π π= = =
/
40 60
10 rad/s 10 300 rpm
4 2
D
B
D B
v
r
π
ω π π
π
= = = = ⋅ =
300 rpmB =ωωωω
Let E be the contact point between gears B and C.
( )( )/ 2 10 20 in./sE E B Bv r ω π π= = =
( )
/
20 60
3.333 rad/s 3.333 100 rpm
6 2
E
C
E C
v
r
π
ω π π
π
= = = = =
100 rpmC =ωωωω
(b) Accelerations at point E.
( )22
2
/
20
On gear : 1973.9 in./s
2
E
B
E B
v
B a
r
π
= = =
2
1974 in./sB =a
( )22
2
/
20
On gear : 658 in./s
6
E
C
E C
v
C a
r
π
= = =
2
658 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 26.
( ) At timea 2 s,t =
( )( )600 2
600 rpm 20 rad/s
60
A
π
ω π= = =
2
, 10 rad/sA
A A At
t
ω
ω α α π= = =
Let D be the contact point between gears A and B.
( ) ( )( ) 2
/ 2 10 20 in./sD D A At
a r α π π= = =
( ) 2
/
20
5 rad/s
4
D t
B
D B
a
r
π
α π= = = 2
15.71rad/sB =αααα
Let E be the contact point between gears B and C.
( ) ( )( ) 2
/ 2 5 10 in./sE E B Bt
a r α π π= = =
( ) 2
/
10
1.6667 rad/s
6
E t
C
E C
a
r
π
α π= = = 2
5.24 rad/sC =αααα
( ) Atb 0.5 s.t = ( )( )For gear , 5 0.5 2.5 rad/sB BB tω α π π= = =
( ) ( )( )22 2
/ 2 2.5 123.37 in./sE E B Bn
r ω π= = =a
( ) 2
10 in./sE t
π=a 2
31.416 in./s=
( ) ( ) ( ) ( )2 2 2 2 2
123.37 31.416 127.3 in./sE E En t
a a a= + = + =
31.416
tan , 14.29
123.37
β β= = ° 2
127.3 in./sE =a 14.29°
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.
( )( )For gear , 1.6667 0.5 0.83333 rad/sC CC tω α π π= = =
( ) ( )( )22 2
/ 6 0.83333 41.123 in./sE E C Cn
a r ω π= = =
( ) 2
31.416 in./sE t
a =
( ) ( ) ( ) ( )2 2 2 2 2
41.123 31.416 51.75 in./sE E En t
a a a= + = + =
31.416
tan 37.4 ,
41.123
β = = ° 2
51.8 in./sE =a 37.4°
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 27.
( ) For the pulley,a ( )
1 1
, 0.3 0.15 m
2 2
Ar d r= = =
( )
1
0.2 0.1 m
2
Br = =
( )/
/
,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
0.8
16 rad/s
0.15 0.1
ω = =
−
At 0.25 s,t = 1
0.4
8 rad/s
0.15 0.1
ω = =
−
21 2 8 16
32 rad/s
0.25t
ω ω
α
− −
= = = − 2
32 rad/s=
A Ar α=a ( )( ) 2
0.15 32 4.80 m/s= =
B Br α=a ( )( ) 2
0.1 32 3.2 m/s= =
2
/ 1.6 m/sA B A B= − =a a a 2
/ 1.600 m/sA B =a
( )b ( ) ( ) 2
/ / /0
1
2
A B A B A Bt t= +x v a
( )( ) ( )( )21
0.8 0.25 1.6 0.25 0.15 m
2
= − + = −
/ 150.0 mmA B =x
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 28.
For the pulley, ( )
1 1
, 0.3 0.15 m
2 2
Ar d r= = =
( )
1
0.2 0.1m
2
Br = =
( )/,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
0.9
18 rad/s
0.15 0.1
ω = =
−
At
1
rev radians,
2
θ π= =
0.45
9 rad/s
0.15 0.1
ω = =
−
d d
d d
dt d
ω ω
α ω ω ω α θ
θ
= = =
( )( )
2 2
9 2
18 0
9 18
38.675 rad/s
2 2
d d
π
ω ω α θ α π α= − = = −∫ ∫
2
38.7 rad/sα =
( ) 2 2
0.15 38.675 5.8012 m/s or 5.8012 m/sA Ar α= = − = −a
( ) 2 2
0.1 38.675 3.8675 m/s or 3.8675 m/sB Br α= = − = −a
( )a 2 2
/ 1.9337 m/s or 1.9337 m/sA B A B= − = −a a a
2
/ 1.934 m/sA B =a
( )b ( ) ( ) 2
/ / /0
1
2
A B A B A Bt t= +x v a
( )( ) ( )( )21
0.9 0.3 1.9337 0.3 0.1830 m
2
= + − =
/ 183.0 mmA B =x
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 29.
(a) Motion of pulley.
( ) ( )0 0
8 in./sE A= =v v ( ) 2
10 in./sE At
= =a a
Fixed axis rotation.
( )
( )0
0 00
8 in./s
2 rad/s
4 in.
E
E A
A
v
v r
r
ω ω= = = =
( )
( ) 2
210 in./s
2.5 rad/s
4 in.
E t
E At
A
a
a r
r
α α= = = =
Since α is constant,
0 2 2.5t tω ω α= + = +
2 2
0 0
1
0 2 1.25
2
t t t tθ θ ω α= + + = + +
For t = 3s, ( )( )2 2.5 3 9.5 rad/sω = + =
( )( ) ( )( )2
0 2 3 1.25 3 17.25 radθ = + + =
In revolutions,
17.25
2
θ
π
= 2.75 revθ =
(b) Motion of load B. t = 3s
( )( )6 9.5B Bv r ω= = 57.0 in./sB =v
( )( )6 17.25B By r θ∆ = = 103.5 in.By∆ =
(c) Acceleration of point D. t = 0
0 2 rad/sω ω= = 2
2.5 rad/s=αααα
( ) ( )( ) 2
6 2.5 15 in./sD Dt
r α= = =a
( ) ( )( )22 2
6 2 24 in./sD Dn
r ω= = =a
2
28.3 in./sD =a 32.0°
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 30.
2
2.4 rad/sα = 0 0ω =
Use equations for constant angular acceleration.
0 2.4t tω ω α= + =
2 2
0 0
1
1.2
2
t t tθ θ ω α= + + =
At t = 4s,
( )( )2.4 4 9.6 rad/sω = =
( )2
1.2 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./sA =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./sB =v
( )( )6 in. 19.2 radB By r θ= = 115.2 in.B =y
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 31.
When contact is made, 240 rpm 8 rad/sAω π= =
Let C be the contact point between the two gears.
( )( ) 2
0.15 8 1.2 m/sC A Av r ω π π= = =
1.2
6 rad/s
0.2
C
B
B
v
r
π
ω π= = =
8 rad/sA tω π α= =
( )6 2 rad/sB tω π α= = −
Subtracting, ( )( ) 2
2 2 rad/sπ α α π= =
( )a 2
3.14 rad/sα =
( )b
8 8
8 st
π π
α π
= = = 8.00 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 32.
( )0
240 rpm 8 rad/sAω π= = ( ) 11
8A A tω π α= −
3 3
2 2 2
1 1 1
1 1 1 0.15
4
2 2 2 0.2
A
B B A A
B
r
t t t
r
θ π α α α
   
= = = =   
  
( )
3
2
1
0.2
8 59.574 radians
0.15
A tα π
 
= = 
 
( )
3
1 1 11
0.15
0.421875
0.2
B B A At t tω α α α
 
= = = 
 
Let Cv be the velocity at the contact point.
( )( )1 10.15 8 1.2 0.15C A A A Av r t tω π α π α= = − = −
and ( )( )1 10.2 0.421875 0.084375C 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/sA A At t tπ α α α− = =
Then,
2
1
1
1
59.574
3.7037 s
16.0850
A
A
t
t
t
α
α
= = =
( )a 216.0850
4.3429 rad/s
3.7037
Aα = = 2
4.34 rad/sAα =
( )
3
20.15
4.3429 1.83218 rad/s
0.2
Bα
 
= = 
 
2
1.832 rad/sBα =
( ) From above,b 1 3.70 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 33.
Motion of disk B. ( )0
500 rpm 52.360 rad/sBω = =
Assume that the angular acceleration of disk B is constant.
( )0B B Btω ω α= +
At 60 s,t = 0Bω =
( ) 20 0 52.360
0.87266 rad/s
60
B B
B
t
ω ω
α
− −
= = = −
( )( )3 52.360 0.87266 157.08 2.618 in./sB B Bv r t tω= = − = −
Motion of disk A. ( )0
0,Aω = 2
3 rad/sAα =
( )0
3A A At 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/sAω = =
( )( )52.360 0.87266 15.52 38.8 rad/sBω = − =
445 rpmA =ωωωω , 371 rpmB =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 34.
Wheel B. ( )0
300 rpm 31.416 rad/sBω = =
At 12s,t = 75 rpm =7.854 rad/sBω =
Angular acceleration.
( )0 7.854 31.416
1.9635 rad/s
12
B B
B
t
ω ω
α
− −
= = = −
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
300 rpm = 31.416 rad/sAω =
Assume that slipping ends when 12 s.t =
Then, 23.562 in./sAv =
23.562
9.4248 rad/s
2.5
A
A
A
v
r
ω = = =
9.4248 rad/sAω = 9.4248 rad/s= −
( ) 20
A
9.4248 31.416
3.4034 rad/s
12
A A
t
ω ω
α
− − −
= = = −
(a)
2
3.40 rad/sA =αααα
2
1.963 rad/sB =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 ω
α
− −
= =
−
9.23 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
© 2007 The McGraw-Hill Companies.
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
ω
π π
∆
= =
∆
For the reel:
1 1d d v dv d
v
dt dt r r dt dt r
ω    
= = +   
   
2 2
2
a v dr a v b
r dt rr r
ω
π
= − = −
2
1
0
2
b
a
r
ω
π
 
= − = 
 
2
0
2
bω
π
=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
© 2007 The McGraw-Hill Companies.
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 1
0
d d v dv d v dr
v
dt dt r r dt dt r dtr
ω
α
   
= = = + = −   
   
2
2 3
2 2
v bv bv
rr rπ π
−  
= − =  
  
2
3
2
bv
rπ
=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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 37.
Velocity analysis.
150 mm/sB =v 15°
A A=v v
/ 500B A ω=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.
/
sin75 sin sin50
A B A B
v v v
φ
= =
° °
( )a /
sin75 150 sin75
189.14 mm/s
sin50 sin50
B
A B
v
v
° °
= = =
° °
/ 189.14
0.378 rad/s
500
A B
AB
v
l
ω = = =
0.378 rad/s=ωωωω
( )b
sin 150 sin55
160.4 mm/s
sin50 sin50
B
A
v
v
φ °
= = =
° °
160.4 mm/sA =v
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 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.
/
sin75 sin60 sin
B A B A
v v v
φ
= =
° °
( )a /
sin75 225sin75
307.36 mm/s
sin sin 45
A
B A
v
v
φ
° °
= = =
°
/ 307.36
0.615 rad/s
500
B A
AB
v
l
ω = = =
0.615 rad/s=ωωωω
( )b
sin 60 225sin60
276 mm/s
sin sin 45
A
B
v
v
φ
° °
= = =
°
276 mm/sB =v 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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 39.
Geometry.
12
sin , 36.87
20
β β= = °
12
tan , 67.38
5
θ θ= = °
Velocity analysis.
4.2 ft/sA =v
/
20
12
B A AB ABrω 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 A
v v v
θ β φ
= =
° −
/
sin 4.2sin67.38
4.5 ft/s
sin sin59.49
A
B A
v
v
θ
φ
°
= = =
°
( )a
4.5
2.7 rad/s
20/12
ABω = =
2.70 rad/sAB =ωωωω
( )b
cos 4.2cos36.87
3.90 ft/s
sin sin59.49
A
B
v
v
β
φ
°
= = =
°
3.90 ft/sB =v 67.4°
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 40.
Geometry.
12
sin , 36.87
20
β β= = °
12
tan , 67.38
5
θ θ= = °
Velocity analysis.
4.2 rad/sAB =ωωωω
( )/ /
20
4.2 7.0 ft/s
12
B A B A ABr 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 B
vv v
φ β θ
= =
° −
( )a
/ sin 7sin59.49
6.53 ft/s
sin sin 67.38
B A
A
v
v
φ
θ
°
= = =
°
6.53 ft/sA =v
( )b
/ cos 7cos36.87
6.07 ft/s
sin sin67.38
B A
B
v
v
β
θ
°
= = =
°
6.07 ft/sB =v 67.4°
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 41.
In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j
/ / 1.2 1.2C A C Aω ω ω= × = × =v k r k i j
/A B A= +v v vB
( ) ( )7.4 7 0.6 0.6B Ay x
v v ω ω− + = − − +i j i j i j
Components. ( ): 7.4 0.6A x
v ω− = −i (1)
( ): 7 0.6B y
v ω= − +j (2)
/A C A= +v v vC
( ) ( )1.4 7 1.2C A xy
v v ω− + = − +i j i j j
Components. ( ): 1.4 A x
v− =i (3)
( ): 7 1.2C y
v ω= − +j (4)
From (3), ( ) 1.4 m/sA x
v = −
( ) From (1),a
( )7.4 1.4
10 rad/s,
0.6
ω
− − −
= =
−
10.00 rad/s=ωωωω
From (2), ( ) ( )( )7 0.6 10 1m/sB y
v = − + = −
( )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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 42.
In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j
/ / 1.2 1.2C 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 x
v v ω ω− + = − − +i j i j i j
Components. ( ): 7.4 0.6A x
v ω− = −i (1)
( ): 7 0.6B y
v ω= − +j (2)
/C A C A= +v v v
( ) ( )1.4 7 1.2C A xy
v v ω− + = − +i i j j
Components. ( ): 1.4 A x
v− =i (3)
( ): 7 1.2C y
v ω= − +j (4)
From (3), ( ) 1.4 m/s, 1.4 7A Ax
v = − = − −v i j
From (1),
( )
( )
7.4 1.4
10 rad/s, 10.00 rad/s
0.6
ω
− − −
= = =
−
kωωωω
(a) ( )/ / 10 0.6O 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) ( )0 O 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 =
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 43.
In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j
( )/ / 50 150 150 50C 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 y
v v ω ω− = + − +i j i j i jA
Components. ( ): 100 75B x
v ω= −i (1)
( ): 75 125A y
v ω− = +j (2)
/C A C A= +v v v
( ) ( )400 100 150 50C A yy
v v ω ω+ = + − +i j i j i j
Components. : 400 100 150ω= −i (3)
( ) ( ): 125C A yy
v 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 y
v ω= − − = − − − =
( ) ( )100.0 mm/s 175.0 mm/sA = +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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 44.
In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j
( )/ / 50 150 150 50C 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 y
v v ω ω− = + − +i j i j i jA
Components. ( ): 100 75B x
v ω= −i (1)
( ): 75 125A y
v ω− = +j (2)
/C A C A= +v v v
( ) ( )400 100 150 50C A yy
v v ω ω+ = + − +i j i j i j
Components. : 400 100 150ω= −i (3)
( ) ( ): 125C A yy
v v ω= +j (4)
From (3), ( )2 rad/sω = − k
From (2), ( ) ( )75 125 75 125 2 175 mm/sA y
v 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 2A D A x y= + = + + × +v v v i j k i jD
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
Radius of locus.
200
100 mm
2
v
r
ω
= = =
Circle of 100.0 mm radius centered at 87.5 mm, 50.0 mmx y= − =
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 45.
7
Slope angle of rod. tan 0.7, 35
10
θ θ= = = °
10
12.2066 in. 20 7.7934 in.
cos
AC CB AC
θ
= = = − =
Velocity analysis.
25 in./sA =v , C Cv=v θ
/C A ABACω=v θ
/C A C A= +v v v
Draw corresponding vector diagram.
/ sin 25sin35 14.34 in./sC A Av v θ= = ° =
( )a
/ 14.34
1.175 rad/s
12.2066
C A
AB
v
AC
ω = = =
1.175 rad/sAB =ωωωω
cos 25cos 20.479 in./sC Av v θ θ= = =
( )( )/ 7.7934 1.175B 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.
/ 9.1551
tan , 24.09
20.479
B C
C
v
v
φ φ= = = °
( )b
20.479
22.4 in./s 1.869 ft/s
cos cos24.09
C
B
v
v
φ
= = = =
°
59.1φ θ+ = °
1.869 ft/sB =v 59.1°
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 46.
Instantaneous geometry. Law of sines:
sin sin120
10 15
φ °
=
10
sin sin120 0.57735
15
φ = ° =
35.264φ = °
Velocity analysis.
1.2 ft/sA =v 14.4 in./s=
/ 10B 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 A
v v v
γ β
= =
°
/
sin 14.4 sin125.264
28.10 in./s
sin sin 24.736
A
B A
v
v
γ
β
°
= = =
°
(a)
28.10
10
ABω = 2.81 rad/sAB =ωωωω
(b)
sin30 14.4 sin30
17.21 in./s
sin sin 24.736
A
B
v
v
β
° °
= = =
1.434 ft/sB =v 35.3°
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
© 2007 The McGraw-Hill Companies.
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 80 Av ω= (1)
Arm AB: 2 120 ABv ω= (2)
Gear B: 1 2 40 Bv v ω= − (3)
3 2 80 Bv v ω= + (4)
Gear C: 3 200 Cv ω= (5)
Data: 0, 5 rad/sA Cω ω= =
From (1), 1 0,v =
From (5), ( )( )3 200 5 1000 mm/sv = =
From (3), 2 40 0Bv ω− = (6)
From (4), 2 80 1000Bv ω+ = (7)
Solving (6) and (7) simultaneously,
(a)
1000
8.333 rad/s
120
Bω = = 8.33 rad/sB =ωωωω
2 (40)(8.333) 333.33 mm/sv = =
(b) From (2),
333.33
2.78 rad/s
120
ABω = = 2.78 rad/sAB =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
© 2007 The McGraw-Hill Companies.
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 80 Av ω= (1)
Arm AB: 2 120 ABv ω= (2)
Gear B: 1 2 40 Bv v ω= − (3)
3 2 80 Bv v ω= + (4)
Gear C: 3 200 Cv ω= (5)
Data: 20 rad/s, 0B Cω ω= =
From (5), 3 0.v =
From (4), ( )( )2 80 80 20Bv ω= − = − 1600 mm/s=
From (3), ( )( )1 1600 40 20 2400v = − − = − 2400 mm/s=
From (1), 1 /80 30 rad/sA vω = = −
(a) 30.0 rad/sA =ωωωω
From (2), 1600 120 ABω=
1600
13.33 rad/s
120
ABω = − = −
(b) 13.33 rad/sAB =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 49.
Data: 3600 rpm 376.99 rad/s, 0A Bω ω= = =
1
1.25 in.
2
A 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
d
ω = =
942.48 rad/s=
/C B C Bv v v= +
1
0 (0.25)(942.48) 235.62 in./s
2
dω= + = =
(a) 236 in./sCv =
(b) Angular velocity of ball.
942.48 rad/sCω = 9000 rpmCω =
(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
l
n
rπ
= = 1500n =
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 50.
Contact point 1 between gears A and B.
Contact point 2 between gears B and C.
Gear B: 6 rad/sBω =
1 (6 rad/s)(10 in.) 60 in./sv = = (1)
2 (6 rad/s)(5 in.) 30 in./sv = = (2)
Arm ABC: ABC ABCω=ωωωω
15A ABCω=v 15C ABCω=v
Gear A: 3 rad/sAω =
1 (5)(3)Av v= + 15 ABCω= 15+ (3)
Matching expressions (1) and (3) for 1,v
(a) : 60 15 15ABCω= + 3.00 rad/sABC =ωωωω
Gear C: C Cω ω=
2 10C Cv v ω= + 15= 10 Cω+ (4)
Matching expressions (2) and (4) for 2,v
(b) : 30 15 10 Cω= + 1.500 rad/sC =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
© 2007 The McGraw-Hill Companies.
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), 2B Ev b a bω ω+ = + (6)
2 1From (1) and (3), B Av b v aω ω− = = (7)
( )
2 2
2
Solving for and ,
2
E A
B
a b a
v v
ω ω
ω
 + + =
( )2
2
E A
B
a b a
b
ω ω
ω
 + − =
From (2),
( )
( )
2
2
2
E A
S S
a b av
a b a b
ω ω
ω ω
 + + = =
+ +
Data: 60 mm, 60 mm, 2 180 mm, 120 mma b a b a b= = + = + =
( )a
( )( )
180 60
1.5 0.5
2 60
E A
B E A
ω ω
ω ω ω
−
= = −
( )( ) ( )( )1.5 120 0.5 150 105 rpm= − =
105.0 rpmB =ωωωω
( )b
( )( )
180 60
0.75 0.25
2 120
E A
S E A
ω ω
ω 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.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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), 2B 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
v
ω ω + + =
( )2
2
E A
B
a b a
b
ω ω
ω
 + − =
From (2),
( )
( )
2
2
2
E A
S S
a b av
a b a b
ω ω
ω ω
 + + = =
+ +
1
Data: 0,
5
E S Aω ω ω= =
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.
( )a
( )( )
( )
2 01
5 2
A
S A
a b a
a b
ω
ω ω
 + + = =
+
( ) ( )
1 1 1
,
2 5 52 1 b
a
a
a b
= =
+ +
2 1 5
b
a
 
+ = 
 
1.500
b
a
=
( )b
( )
( )( )
2
0
2 2 2 1.5 3
E A A A A
B
a b a a
b b
ω ω ω ω ω
ω
 + − = = − = − = −
1
3
B Aω=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
© 2007 The McGraw-Hill Companies.
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/sABCω π= =
0Av =
(12)(2.5 ) 30 rad/sBv π π= =
(12 7)(2.5 ) 47.5 rad/sCv π π= + =
Gear A: 10, 0, 0A Av vω = = =
Gear B: 1 4 0B Bv v ω= − =
30 4 0Bπ ω− =
7.5 rad/sBω π=
2 4B Bv v ω= +
30 30 60 in./sπ π π= + =
Gear C: 2 3C Cv v ω= −
60 47.5 3 Cπ π ω= −
4.1667 rad/sCω π= −
Summary: 225 rpmB =ωωωω
125.0 rpmC =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
© 2007 The McGraw-Hill Companies.
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, 0C Cv vω = = =
Gear B: 2 4 56 /3 4 0B B Bv v ω π ω= − = − =
14 /3 rad/sBω π=
1 4 56 /3 56 /3B Bv v ω π π= + = +
112 /3 in./sπ=
Gear A: 1 8A Av v ω= −
112 /3 152 /3 8 Aπ π ω= −
5 /3 rad/sAω π= −
Summary: 50.0 rpmA =ωωωω
140.0 rpmB =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 55.
Geometry.
( ) ( )sin sinOA ABθ β=
( )sin 10 sin30
sin , 1.79
160
OA
AB
θ
β β
°
= = = °
Shaft and eccentric disk. (Rotation about O), 900 rpm 30 rad/sOAω = = π
( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πv
Rod AB. (Plane motion = Translation with A + Rotation about A.)
[/B A B A Bv= +v v v ] [ Av= ] /60 A Bv° +  ]β
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
φ
β
π °
= =
° − °
=
497 mm/sB =v
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 56.
Geometry.
( ) ( ) ( )sin 180 sinOA ABθ β° − =
( ) ( )sin 180 10 sin60
sin , 3.10
160
OA
AB
θ
β β
° − °
= = = °
Shaft and eccentric disk. (Rotation about O) 900 rpm 30 rad/sOAω = = π
( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πv
Rod AB. (Plane motion = Translation with A + Rotation about A.)
[/B A B A Bv= +v v v ] [300= π ] /30 B 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
φ
β
π °
= =
° + °
=
791 mm/sB =v
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 57.
Disk A
Bar :BD
15 rad/sAω = , 2.8 in.AB =
Rotation about a fixed axis.
( ) ( )( )2.8 15 42 in./sB Av AB ω= = =
( )a 0 .θ = ° 42 in./sB =v
2.8
sin , 16.260
10
β β= = °
/D B D B= +v v v
Dv [42= ] /D Bv+  β 
/
42
43.75 in./s
cos
D Bv
β
= =
/ 43.75
,
10
D B
DB
v
DB
ω = = 4.38 rad/sDB =ωωωω
tan ,D Bv v β= 12.25 in./sD =v
( ) 90 .b θ = ° 42 in./sB =v
5.6
sin , 34.06
10
β 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
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.
( )c 180 .θ = ° 42 in./sBv =
2.8
sin , 16.26
10
β β= = °
Bar :BD /D B D B= +v v v
Dv [42= ] /D Bv+  β 
/
42
43.75
cos
D Bv
β
= =
/ 43.75
10
D B
DB
v
DB
ω = =
4.38 rad/sDB =ωωωω
tanD Bv v β= 12.25 in./sD =v
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 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= = +
2.8
sin , 22.9 ,
10 2.8
A
BD A
r
l r
β β= = = °
− −
22.9θ = °
For 180 , sin sin ,θ β θ β= ° + = −
sin sinBD A Al r rβ β= −
2.8
sin , 12.6
10 2.8
A
BD A
r
l r
β β= = = °
+ +
192.6θ = °
( ) For matching magnitudesb /D B Bv v=
( )( )2.8 20
, 5.6 rad/s
10
A A
BD BD A A BD
BD
r
l r
l
ω
ω ω ω= = = =
For 22.9 ,θ = ° 5.60 rad/sBD =ωωωω
For 192.6 ,θ = ° 5.60 rad/sBD =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 59.
Rod BE: 0Ev =
/ (192)(4) 768 mm/sB B E BEr ω= = =v
Rod ABD: D Dv=v
/D B D B= +v v v
Dv 768= 360 ADω+ 60°
Draw diagram for vector addition.
(a) 768 360 sin30ADω= °
4.2667 rad/sADω =
4.27 rad/sAD =ωωωω
(b) 768 tan30Dv= °
1330 mm/sDv =
1.330 m/sD =v
(c) / 240A B ADω=v 60 1024 mm/s° = 60°
/ 768A B A B= + =v v v 1024+ 60 1557 mm/s° = 34.7°
1.557 m/sA =v 34.7°
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 60.
Rod BE: 0E =v B Bv=v
Rod ABD: 1.6 m/sD =v AD ADω=ωωωω
/B D B D= +v v v
Bv 1.6= 0.360 ADω+ 60°
Draw diagram for vector addition.
(a)
1.6
0.360
sin60
ADω =
°
5.13 rad/sAD =ωωωω
(b) 1.6tan30 0.92376Bv = ° =
0.924 m/sB =v
(c) / 0.240A B ADω=v 60 1.2317 m/s° = 60°
/ 0.92376A B B A= + =v v v 1.2317+ 60°
[1.5396= ] [1.0667+ ] 1.873= 34.7°
1.873 m/sA =v 34.7°
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 61.
1000 rpmABω =
( )( )1000 2
104.72 rad/s
60
π
= =
( )a 0 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r
( )( )/ 3 104.72 314.16 in./sB 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./sD D Bv v= =
P D=v v 0P =v
314.16
8
B
BD
v
l
ω = = 39.3 rad/sBD =ωωωω
( )b 90 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r
( )( )/ 3 104.72 314.16 in./sB 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./sD B Dv v= =
/
,
D B
BD
v
l
ω = 0BD =ωωωω
314.16 in./sP D= =v v 26.2 ft/sP =v
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 62.
( )( )1000 2
1000 rpm 104.72 rad/s
60
ABω
π
= = =
60 , . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r 30°
( )( )/ 3 104.72 314.16 in./sB B A ABω= = =v r 60°
Rod BD. (Plane motion Translation with Rotation about .)B B= +
Geometry. sin sinl rβ θ=
3
sin sin sin60
8
18.95
r
l
β θ
β
= = °
= °
/D B D B= +v v v
[ Dv ] [314.16= ] /60 D Bv° +  ]β Draw velocity vector diagram.
( )180 30 90 78.95φ β= ° − ° − ° − = °
Law of sines.
( )
/
sin sin30 sin 90
D BD B
vv v
φ β
= =
° ° −
sin 314.16 sin78.95
326 in./s
cos cos18.95
B
D
v
v
φ
β
°
= = =
°
P Dv v= 27.2 ft/sP =v
/
sin30 314.16sin30
166.08 in./s
cos cos18.95
B
D B
v
v
β
° °
= = =
°
/ 166.08
8
D B
BD
v
l
ω = = 20.8 rad/sBD =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 63.
Bar AB. (Rotation about A)
( ) ( ) ( )/ 4 0.25 1.00 m/sB AB B A= × = − × − = −v r k j iωωωω
Bar ED. (Rotation about E)
( )/ 0.075 0.15 0.15 0.075D 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 0.2D 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 BDω ω ω− = − +i j i j
Components:
: 0.15 1.00, 6.6667 rad/sDE DEω ω= − = −i 6.67 rad/sDE =ωωωω
: 0.075 0.2DE BDω ω− =j
( )( )0.075 6.6667
0.2
BDω
− −
= 2.50 rad/sBD =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 64.
Bar AB. (Rotation about A.) ( ) 0.18B AB ABAB ω ω= =v 30°
Bar DE. (Rotation about E.) ( ) 0.18D DE DEDE ω ω= =v
Bar BGD. (Plane motion = Translation with B + Rotation about .B )
[/D B D B Dv= +v v v ] [ Bv= ]30° /D Bv+  ]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
ω
ω ω= = =
/ /
1
0.09
2
G B GB BD D B ABv l vω ω= = =
/G B G B= +v v v
Draw vector diagram.
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.
Law of cosines
( ) ( ) ( )( )2 22
0.18 0.09 2 0.18 0.09 cos60G AB AB AB ABv ω ω ω ω= + − °
2 2
0.0243G ABv ω=
( )( )6.415 6.416 2.5AB Gvω = =
16.04 rad/sAB =ωωωω
16.04 rad/sBD =ωωωω
0.18D B ABv v ω= =
0.18
0.18
D AB
DE AB
DE
v
l
ω
ω ω= = =
16.04 rad/sDE =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 65.
Bar AB. (Rotation about A.) 25 rad/sAB =ωωωω
( )( )/ 8 25 200 in./sB B A ABr ω= = =v
Bar ED. (Rotation about E.)
D Dv=v 8D DEv ω=
Plate BDHF. (Translation with B + Rotation about .B )
[/D B D B Dv= +v v v ] [ Bv= ] /D Bv+  ]30°
Draw velocity vector diagram.
/
200
230.94 in./s
cos30 cos30
B
D B
v
v = = =
° °
/ 230.94
14.4338 rad/s
16
B D
BDHF
BD
v
l
ω = = =
(a) 14.43 rad/sBDHF =ωωωω
( )( )/ 8 14.4338 115.47 in./sF B BDHFv BFω= = =
/ 115.47 m/sF B =v 30°
[/ 200 in./sF B F B= + =v v v ] [115.47 in./s+ ]30°
(b) [142.265 in./sF =v ] [100 in./s+ ] 173.9 in./sF =v 54.9°
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 66.
Rod DE. (Rotation about E.) 35 rad/sDEω =
( )( )/ 8 35 280 in./sD D E DEr ω= = =v
Rod AB. (Rotation about A.)
B Bv=v 8B ABv ω=
Plate BDHF. (Translation with D + Rotation about .D )
[/B D B D Bv= +v v v ] [ Dv= ] /B Dv+  ]30°
Draw velocity vector diagram.
/
/
280
560 in./s
sin30 sin30
560
35 rad/s
16
D
D B
D B
BDHF
DB
v
v
v
l
ω
= = =
° °
= = =
(a) 35.0 rad/sBDHF =ωωωω
Point of zero velocity lies above point D.
/
280
8 in.
35
D
C D
BDHF
v
y
ω
= = =
(b) 8 in. above point D.
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 67.
(15 rad/s)AB = − kωωωω BD BDω= kωωωω DE DEω=ω k
( )/ 0.2mB A = −r j ( ) ( )/ 0.6 m 0.25 mD B = − −r i j ( )/ 0.2 mD E = −r i
( ) ( ) ( )/ 0 15 0.25 3 m/sB A AB B A= + × = + − × − = −v v ω r k j i
( ) ( )/ / 6 0.2 0.25 0.6D 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 BDω ω= + = − + −v v v i i j (1)
( ) ( )/ /0 0.2 0.2D 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 0BDω− + =i 12 rad/sBDω =
: 0.6 0.2BD DEω ω− = −j 36 rad/sDEω =
( )a Angular velocity of rod BD. 12 rad/sBD =ωωωω
(b) Velocity of the midpoint M of rod BD.
( ) ( )/ /
1
0.3 m 0.125 m
2
M B D B= = − −r r i j
( )/ / 3 12 0.3 0.125M 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°
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 68.
Bar AB. ( ) ( )/ 0.300 m + 0.125 m ,B A =r i j ( )3 rad/sAB = kωωωω
0A =v
/ /0B 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 mD B = −r j BD BDω= kωωωω
( )0.375 + 0.9 + 0.325D B D/B BDω= + = − × −v v v i j k j
0.375 + 0.9 + 0.325 BDω= − 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 DEω= + × − + =v k i j
0.375 0.9 0.325 0.15 0.2 0BD DE DEω ω ω− + + − − =i j i j i
Components:
j: 0.9 0.15 0DEω− = 6 rad/sDEω =
i: 0.375 0.325 0.2 0BD DEω ω− + − =
( )( )0.325 0.375 0.2 6BDω = +
4.85 rad/sBDω = 4.85 rad/sBD =ωωωω
6.00 rad/sDE =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 69.
80 km/h 22.222 m/sA = =v 0C =v
560 mm 280 mm 0.28 m
2
d
d r= = = =
22.222
79.364 rad/s
0.28
Av
r
ω = = =
/ / /B A D A E Av v v rω= = =
( )( )0.28 79.364 22.222 m/s= =
[/ 22.222 m/sB A B A= + =v v v ] [22.222 m/s+ ]
44.4 m/sB =v
[/ 22.222 m/sD A D A= + =v v v ] [22.222 m/s+ ]30°
42.9 m/sD =v 15.0°
[/ 22.222 m/sE A E A= + =v v v ] [22.222 m/s+ ]
31.4 m/sE =v 45.0°
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 70.
(a) 0β = .Wheel AD 0, 45 in./sC D= =v v
45
11.25 rad/s
4
D
AD
v
CD
ω = = =
( ) ( ) 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 Av+  ]ϕ 16.88 in./sB =v
/ 0B Av = 0ABω =
(b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v
2.5
tan , 32.005
4
DA
DC
γ γ= = = °
4.7170 in.
cos
DC
CA
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 , 18.663
12.5
φ φ= = °
Plane motion = Translation with A + Rotation about 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
© 2007 The McGraw-Hill Companies.
[/B A B A Bv= +v v v ] [ Av= ] [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 A
vv v
δ γ φ
= =
° +
( )
( )53.066 sin 39.332sin
sin 90 sin108.663
A
B
v
v
δ
φ
°
= =
° + °
35.5 in./s= 35.5 in./sB =v
( )/
sin (53.066)sin32.005
sin 90 sin108.663
A
B A
v
v
γ
φ
°
= =
° + °
29.686 in./s=
/ 29.686
2.37 rad/s
12.5
B A
AB
v
AB
ω = = = 2.37 rad/sAB =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 71.
0 180 km/h = 50 m/s=v
( )( )180 2
180 rpm = 18.85 rad/s
60
π
ω = =
Top View
0v zω=
0 50
2.65 m
18.85
v
z
ω
= = =
Instantaneous axis is parallel to the y axis and passes through the point
0x =
2.65 mz =
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 72.
1.5 1.0 1
rad/s
1.5 3
E D
ED
v v
l
ω
− −
= = =
1
3
1.0
3 mD
CE
v
l
ω
= = =
(a) 1.5 2 3 0.5 mACl = + − = lies 0.500 m to the right ofC A
(b) ( )
1
0.5 0.1667 m/s
3
A ACv l ω
 
= = = 
 
A 0.1667 m/s=v
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 73.
Contact points:
1 between gears A and B.
2 between gears B and C.
Arm ABC: 4 rad/sABCω =
( )( )15 4 60 in./sAv = =
( )( )15 4 60 in./sCv = =
Gear B: 8 rad/sBω =
( )( )1 10 8 80 in./sv = =
( )( )2 5 8 40 in./sv = =
Gear A:
1 80 60
5 5
A
A
v v
ω
− −
= =
4 rad/sAω =
60
15 in.
4
A
A
A
v
ω
= = =l
COSMOS: Complete Online Solutions Manual Organization System
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© 2007 The McGraw-Hill Companies.
Gear C:
2 60 40
10 10
C
C
v v
ω
− −
= =
2 rad/sCω =
60
30 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.
4.00 rad/sA =ωωωω
2.00 rad/sC =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
© 2007 The McGraw-Hill Companies.
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 Cv r ω=
/
6
6 rad/s
1
B
B C
v
r
ω = = =
6.00 rad/s=ωωωω
(b) Velocity of point A. /A A Cv 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.=
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 75.
12 1.5
35 rad/s
0.3
O A
AO
v v
l
ω
− −
= = =
(a) 31.5
42.86 10 m
35
A
CA
v
l
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 2
0.3 0.34286 0.45558 mCDl = + =
( )( )0.45558 35 15.95 m/sD CDv l ω= = =
0.3
tan , 41.2
0.34286
OD
CO
l
l
θ θ= = = °
15.95 m/sD =v 41.2°
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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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 76.
12 1.5
45 rad/s
0.3
O A
AO
v v
l
ω
− +
= = =
(a) 31.5
33.33 10 m
45
A
AC
v
l
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 2
0.3 0.26667 0.4014 mCEl = + =
( )( )0.4014 45 18.06 m/sE CEv l ω= = =
0.3
tan , 48.4
0.26667
OE
OC
l
l
θ θ= = = °
18.06 m/sE =v 48.4°
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 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= +
1.6667 rad/sω = −
/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./sE Av v− = − =
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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© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 78.
10 in./sD =v , 8 in./sB =v
10 8
4 rad/s
4.5
D Bv v
BD
ω
+ +
= = =
10
2.5 in.
4
Dv
CD
ω
= = =
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.
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 79.
Rod AD.
( )( )/ 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 sin30 0.180 mB Cr = ° =
/
0.768
4.2667
0.180
B
AD
B C
v
r
ω = = =
4.27 rad/sAD =ωωωω "
(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 = ° =
0.600sin30 0.300 mCEl = ° =
0.20785
tan
0.300
β = 34.7β = °
( ) ( )2 2
0.20785 0.300 0.36497 mCAl = + =
( )( )0.36497 4.2667 1.557 m/sA CA ADv l ω= = =
1.557 m/sA =v 34.7° "
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 80.
15 rad/sABω =
( ) ( )( )0.200 15 3 m/sB ABv AB ω= = =
B Bv=v D Dv=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
12 rad/s
0.25
B
BD
v
BC
ω = = = 12.00 rad/sBD =ωωωω
(b) Locate point M, the midpoint of rod BD. Draw CM.
( ) ( )2 2
0.6 0.25 0.65 mBD = + =
0.25
tan 22.62 90 67.38
0.6
β β β= = ° ° − = °
( )
1
0.325 m
2
CM DM MB BD= = = =
( ) ( )( )0.325 12 3.9 m/sMv CM ω= = = 3.90 m/sM =v 67.4°
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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 81.
Bar DC. (rotation about D)
( )( )( ) 18 10C CDv CDω= =
180 in./s=
180 in./sC =v 30°
Bar AB. (rotation about A)
B Bv=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= =
180
18 rad/s
10
C
BC
v
IC
ω = = =
( ) ( )( )10 3 18 311.77 in./sB BCv IB ω= = =
(a)
311.77
31.177 rad/s
10
B
AB
v
AB
ω = = = 31.2 rad/sAB =ωωωω
(b) 18.00 rad/sBC =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/sM =v 30°
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 82.
Bar AB. (rotation about A)
B Bv=v 60°
Bar CD. (rotation about D)
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
9.36 rad/s
10
M
BC
v
IM
ω = = =
(a) ( ) ( )( )10 3 9.36 162.12 in./sB BCv IB ω= = =
162.12
16.21 rad/s
10
B
AB
v
AB
ω = = = 16.21 rad/sAB =ωωωω
(b) 9.36 rad/sBC =ωωωω
(c) ( ) ( )( )10 9.36 93.6 in./sC BCv IC ω= = =
93.6
9.36 rad/s
10
C
CD
v
DC
ω = = = 9.36 rad/sCD =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
© 2007 The McGraw-Hill Companies.
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 Av and BC perpendicular to Bv .
(a)
800
3.0792 rad/s
300cos30
B
ABD
v
BC
ω = = =
°
3.08 rad/sABD =ωωωω
( ) ( )2 2
600cos30 300sin30 540.83 mmCDl = ° + ° =
300sin30
tan 16.10 90 73.9
600cos30
γ γ γ
°
= = ° ° − = °
°
(b) ( )( ) 3
540.83 3.0792 1.665 10 mm/sD CD ABDv l ω= = = ×
1.665 m/sD =v 73.9°
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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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 84.
40 ,θ = ° 0.6 m/sB =v , A Av=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)
0.6
0.46672 rad/s
1.28557
B
ABD
v
BC
ω = = =
0.467 rad/sABD =ωωωω
/ / /D C B C D B= +r r r
[1.28557 m= ] [2 m+ ]40°
2.9930 m= 30.79°
30.79β = °
(b) ( )( )/ 2.9930 0.46672D 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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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 85.
2 2
320 240 400 mmDE = + =
240
tan 0.75
320
β = = 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 mm
tan
AC
β
= =
( )720 320 100 300 mmBC AC AB= − = − + =
Since triangles FCB and BDK are similar,
300
100
b DK
BC BK
= =
( )( )300 300
900 mm.
100
b = =
( )a Distance b. 0.900 mb =
2 2
300 400 500 mm 0.5 mCD = + = =
6
1.2 rad/s
5
D
BDF
v
CD
ω = = =
( )( )0.900 1.2 10.8 m/sF BDFv bω= = =
( )b Velocity of point F. 10.80 m/sF =v
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 86.
Locate the instantaneous center I of rotation of bar ABD as the
intersection of line AI perpendicular to Av and line BI perpendicular to
Bv Triangle IAB is equilateral.
300 mmIA IBl l= =
(a)
900 mm/s
3 rad/s
300 mm
A
ABD
IA
v
l
ω = = =
3.00 rad/sABD =ωωωω
(b) By the law of cosines, ( )600cos30 mmIDl = °
( )( )600cos30 3 1559 mm/sD ID ABDv l ω= = ° =
1.559 m/sD =v
COSMOS: Complete Online Solutions Manual Organization System
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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 87.
A Av=v 45 ,° 7.5 ft/sB =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 Av and
BC perpendicular to .Bv
Let 24 in. 2 ftl AB= = =
Law of sines for triangle ABC.
2.8284 ft
sin75 sin60 sin 45
b a l
= = =
° ° °
2.4495 ft, 2.73205 fta b= =
7.5
2.7452 rad/s
2.73205
Bv
b
ω = = =
(a) ( )( )2.4495 2.7452 6.724 ft/sAv aω= = =
6.72 ft/sA =v 45.0°
(b) 2.75 rad/s=ωωωω
(c) Let M be the midpoint of AB.
Law of cosines for triangle CMB.
( ) ( ) ( )( )( )
2
2 2
2 2
2 cos60
2 2
2.73205 1 2 2.73205 1 cos60
2.3942 ft
l l
m b b
m
 
= + − °  
= + − °
=
Law of sines.
2
sin sin60 1sin60
, sin , 21.2
2.3942l
m
β
β β
° °
= = = °
( )( )2.3942 2.7452 6.573 ft/s,Mv mω= = =
6.57 ft/sM =v 21.2°
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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 88.
Bar DE. ( )( )24 8 192 in./sEDDv eω= = =
192 in./sD =v
Bar AB. 8ABB ABv aω ω= =
8B ABω=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.
24
48 in.
sin120 sin30 sin30 sin30
b d l
= = = =
° ° ° °
41.569 in., 24 in.b d= =
(a)
192
8 rad/s
24
D
BD
v
d
ω = = = 8.00 rad/sBD =ωωωω
(b) ( )( )41.569 8 332.55 in./sB BDv bω= = =
332.55
41.6 rad/s
8
B
AB
v
a
ω = = = 41.6 rad/sAB =ωωωω
(c) Law of cosines for triangle CMD.
( ) ( )( )( )
2
2 2
22
2 cos120
2 2
24 12 2 24 12 cos120
31.749 in.
l l
m d d
m
 
= + − °  
= + − °
=
Law of sines.
( )
2
12 sin120sin sin120
, sin , 19.1
31.749l
m
β
β β
°°
= = = °
Velocity of M. ( )( )31.749 8 253.99 in./sM BDv mω= = =
21.2 ft/sM =v 19.1°
COSMOS: Complete Online Solutions Manual Organization System
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Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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 to Bv and DC perpendicular to .Dv
Law of sines for triangle .
sin150 sin15 sin15
CD BC BD
BCD = =
° ° °
180sin150
180 mm 347.73 mm
sin15
BC BD CD
°
= = = =
°
200
0.57515 rad/s
347.73
D
DBE
v
CD
ω = = =
( ) ( )( )180 0.57515 103.528 mm/sB DBEv BC ω= = =
103.528
0.57515 rad/s
180
B
ABF
v
AB
ω = = =
( ) ( )( )300 0.57515 172.546 mm/sF ABFv AF ω= = =
Law of cosines for triangle DCE. ( ) ( ) ( ) ( )( )2 2 2
2 cos15CE CD DE CD DE= + − °
( ) ( )( )( )2 2 2
347.73 300 2 347.73 300 cos15 , 96.889 mmCE CE= + − ° =
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sin15 300 sin15EH DE= ° = °
300sin15
cos 36.7
96.889
EH
CE
β 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°
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 90.
3 rad/sDE =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 Bv and DC perpendicular to .Dv
( )120 mm, cos30 120cos30BD DK BD= = ° = °
120 cos30
cos , 49.495 , 180 30 100.505
160
DK
ED
β β φ β
°
= = = ° = ° − ° − = °
Law of sines for triangle BCD.
sin30 sin sin
CD BC BD
φ β
= =
°
( )
( )
sin30 120sin30
78.911 mm
sin sin
sin 120sin
155.177 mm
sin sin
BD
CD
BD
BC
β β
φ φ
β β
° °
= = =
= = =
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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Law of cosines for triangle ABC. ( ) ( ) ( ) ( )( )2 2 2
2 cos150AC BC AB AB BC= + − °
( ) ( )( )( )2 2 2
155.177 200 2 155.177 200 cos150 , 343.27 mmAC AC= + − ° =
Law of sines.
sin sin150 200 sin150
sin , 16.9
343.27AB AC
γ
γ γ
° °
= = = °
(a)
480
6.0828 rad/s
78.911
D
ABD
v
CD
ω = = =
6.08 rad/sABD =ωωωω
(b) ( ) ( )( )343.27 6.0828 2088 mm/sA ABDv AC ω= = =
2.09 m/sA =v 73.1°
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 91.
AB = 20 in.
Instantaneous centers:
at I for BC.
at J for BD.
Geometry
( )
12
11 8.25 in.
16
IC
 
= = 
 
( )
12
21 15.75 in.
16
JD
 
= = 
 
( )
11
20 13.75 in.
16
AI
 
= =  
20 in. 13.75 in. = 6.25 in.BI AB AI= − = −
6.25 in.BJ BI= =
Member BC. 33 in./sC =v
33
4 rad/s
8.25
C
BC
v
IC
ω = = =
( ) ( )( )6.25 in. 4 rad/s 25 in./sB BCv BI ω= = =
Member BD.
25 in./s
4 rad/s
6.25 in.
B
BD
v
BJ
ω = = =
(a) ( ) ( )( )= = 15.75 in. 4 rad/sD BDJD ωv 63.0 in./sD =v
Member AB.
(b)
25 in./s
20 in.
B
AB
v
AB
ω = = 1.250 rad/sAB =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 92.
6 in./sA =v , B Bv=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 Av and BC perpendicular to Bv .
Triangle ACB. Law of sines.
sin 30 sin 30 sin120
AC CB AB
= =
° ° °
10sin 30
5.7735 in.
sin120
AC CB
°
= = =
°
(a)
6
1.03923 rad/s
5.7735
A
AB
v
AC
ω = = =
1.039 rad/sAB =ωωωω
( ) ( )( )5.7735 1.03923 6 in./sB ABCB 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 to Bv and DI perpendicular to Dv .
Triangle BID. Law of sines.
sin120 sin 30 sin 30
BI DI BD
= =
° ° °
10 sin120
17.3205 in.
sin 30
BI
°
= =
°
, 10 in.DI =
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.
6
0.34641 rad/s
17.3205
B
BD
v
BI
ω = = =
0.346 rad/sBD =ωωωω
(b) ( ) ( )( )10 0.34641 3.46 in./sD BDv DI ω= = =
3.46 in./sD =v
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 93.
Method 1
Assume Dv 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 2
0.27 0.36 0.45 mAD = + =
( ) ( )2 2
0.18 0.135 0.225 mBD = + =
0.27 0.18
sin 0.6, sin 0.8
0.45 0.225
θ 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
cos
c
θ
β
=
0.225sin
cos
d
φ
β
=
0.36 0.27tan 0.135 0.18tana bβ β= − = +
.Kinematics 0.4 m/s, 1 m/sA Bv v= =
,A D
AD
v v
a c
ω = = B D
BD
v v
b d
ω = =
A B
D
cv dv
v
a b
= =
0.45sin cos 0.4
0.6
cos 0.225sin 1
A
B
cv
a b b b
dv
θ β
β φ
= = ⋅ ⋅ =
( )0.36 0.27tan 0.6 0.135 0.18tanβ 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 m
cos cos
c d
β β
= = = =
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.
( )a
0.4
2.4891 rad/s
0.1607
A
AD
v
a
ω = = = 2.49 rad/sAD =ωωωω
( )b
1
3.733 rad/s
0.2679
B
BD
v
b
ω = = = 3.73 rad/sBD =ωωωω
(c) ( )( )0.3356 2.4891 0.835 m/s,D ADv cω= = = 0.835 m/sD =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/sA A B A= =v v
0.27
tan , 36.87
0.36
θ θ= = °
0.36
0.45 m
cos
AD
θ
= =
/ 0.45D A ADω=v θ
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
( )
0.18
0.3 m
sin
cos 0.135 0.375 m
CD
BC CD
θ
θ
= =
= + =
/ 1.4
3.73 rad/s
0.375
B A
BD
v
CB
ω = = =
( ) ( )( )/ 0.3 3.73 1.120 m/sD A BDv CD ω= = =
( )a
/ 1.120
0.45
D A
AD
v
AD
ω = = 2.49 rad/sAD =ωωωω
( )b 3.73 rad/sBD =ωωωω
( )c [/ 0.4 m/sD A D A= + =v v v  [1.120+ ]θ
0.835 m/s= 53.6° 0.835 m/sD =v 53.6°
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 94.
( )( )0.2 0.2 4.5 0.9 m/sB ABω= = =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 b
ω = =
0.9
D BD
a
a
b
ω= =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 bc
ω = =
( )
( )0.9 0.2
0.2 0.7E DE
a d
v d
bc
ω
−
= − = = (1)
0.25 0.15 0.25
, , 0.25tan , 0.15tan
cos cos 0.15
a
a c b d
c
β β
β β
= = = = =
Substituting into (1) ( )
0.25 0.2 0.15tan
0.9 0.7 or 1.2 0.9tan 0.7tan
0.15 0.25tan
β
β β
β
− 
= − = 
 
1.2
tan 0.75 36.87 cos 0.8
1.6
β β β= = = ° =
0.25
0.3125 m, 0.1875 m, 0.1875 m, 0.1125 m
0.8
a b c d= = = = =
( )a
0.9
4.8 rad/s
0.1875
BDω = = 4.80 rad/sBD =ωωωω
( )b
( )( )
( )( )
0.9 0.3125
8 rad/s
0.1875 0.1875
DEω = = 8.00 rad/sDE =ωωωω
( )c ( )( )0.3125 4.8 1.5 m/sDv = = 1.5 m/sD =v 53.1°
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 95.
5 rad/sABCω = , 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 26.565
18
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
5 rad/s
20.125
D
DE
JD
v
l
ω = = =
(a) 5.00 rad/sDE =ωωωω
9 cos 27 in.JE JDl l φ= + =
( )( )27 5E JE DEv l ω= =
(b) 135 in./sE =v
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 96.
12 in./sA =v
B Bv=v
Point C is the instantaneous center of bar AB.
12
20cos30
B
AB
v
AC
ω = =
°
0.69282 rad/s=
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
0.4 rad/s
20cos30
D
DE
v
DI
ω = = =
°
0.400 rad/sDE =ωωωω
( )b ( ) ( )( )20sin30 0.4 4 in./sE DEv EI ω= = ° = 0.333 ft/sE =v 30°
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 97.
Let points A, B, and C move to , , andA B C′ ′ ′ as shown.
Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack.
: lower rackspace 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 gearbody centrode 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
© 2007 The McGraw-Hill Companies.
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 Bv=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 cosCy l β=
( )22 2 2
300 mmC 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 cos2
2
sin
cos sin sin 2
2
C
C
x AC
l
l
y AC
l
l
β
β β
β
β β β
=
= = +
=
= =
( )
2 2
22 2 2
150 150
2 2
C C C C
l l
x y x y
   
− + = = − + =   
   
The body centrode is a semi circle of 150 mm radius centered
midway between A and B.
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 99.
80 km/h 22.222 m/sAv = =
0C =v
1
560 mm, 0.280 mm 0.28 m
2
d r d= = = =
Point C is the instantaneous center.
22.222
79.364 rad/s
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
( )
1
30 15
2
γ = ° = °
( )( )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°
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 100.
900 rpm 30 rad/sOAω π= =
( ) ( )( )10 30 300 mm/sA OAOA 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
( )sin30 10sin30
sin , 1.79
160
OA
AB
β β
° °
= = = °
( ) ( )cos30 cos 10cos30 160cosOB OA AB β β= ° + = ° +
168.582 mm=
168.582
10 184.662 mm
cos30 cos30
OB
AC OA= − = − =
° °
( )tan30 97.377 mmBC OB= ° =
A B
AB
v v
AC BC
ω = =
( )( )97.377 300
497 mm/s,
184.662
B A
BC
v v
AC
π 
= = = 
 
497 mm/sB =v
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 101.
Bar AB.
4 rad/sABω =
( )( )0.25 4B AB ABv l ω= =
1 m/sB =v
Bar DE.
75
tan 26.565
150
φ φ= = °
0.15
= 0.167705 m
cos
EDl
φ
=
0.167705D ED ED EDv l ω ω= =
0.167705D EDω=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 m
tan
IBl
φ
= =
0.4
0.44721 m
cos
IDl
φ
= =
1
2.5 rad/s
0.4
B
BD
IB
v
l
ω = = =
2.50 rad/sBD =ωωωω
1.11803 m/sD ID BDv l ω= =
6.67 rad/sD
ED
ED
v
l
ω = = 6.67 rad/sED =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
© 2007 The McGraw-Hill Companies.
Chapter 15, Solution 102.
(a) 0β = .Wheel AD 0, 45 in./sC D= =v v
45
11.25 rad/s
4
D
AD
v
CD
ω = = =
( ) ( ) 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
0ABω =
(b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v
2.5
tan , 32.005
4
DA
DC
γ γ= = = °
4.7170 in.
cos
DC
CA
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 , 18.663
12.5
φ φ= = °
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15
solucionario mecanica vectorial para ingenieros - beer  & johnston (dinamica) 7ma edicion Cap 15

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

  • 1. 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: ( )23 8 6 2 radianst tθ = − − 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 0t − = 0.250 st = (b) Angular coordinate and angular velocity at t = 0.250 s. ( )( ) ( )( )3 2 8 0.250 6 0.250 2θ = − − 18.25 radiansθ = − ( )( ) ( )( )2 24 0.250 12 0.250 2ω = − − 22.5 rad/sω =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 2. 3 0.5 cos4t e tπ θ π− = ( )3 3 0.5 3 cos4 4 sin 4t td e t e t dt π πθ ω 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 d e t e t e t e t dt e t e t π π π π π π ω α 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= − = − 4.71rad/sω = − ( )( )2 0.5 7 34.5α π= − = − 2 34.5 rad/sα = − ( ) 0.125 s, cos4 cos 0, sin 4 sin 1 2 2 b t t t π π π π= = = = = 3 0.30786t e π− = ( )( )( )0.5 0.30786 0 0θ = = 0θ = ( )( )( )0.5 0.30786 4 1.93437ω π= − = − 1.934 rad/sω = − ( )( )( )2 0.5 0.30786 24 36.461α π= = 2 36.5 rad/sα =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 3. 7 /6 0 sin 4t e tπ θ θ π− = 7 /6 7 /6 0 7 sin 4 4 cos4 6 t td e t e t dt π πθ π ω θ π π π− −  = = − +    2 2 2 7 /6 7 /6 7 /6 2 7 /6 0 7 /6 2 2 0 49 28 28 sin 4 cos4 cos4 16 sin 4 36 6 6 49 28 16 sin 4 cos4 36 3 t t t t t d e t e t e t e t dt e t t π π π π π ω π π π α θ π π π π π θ π π π π − − − − −   = = − − −        = − − +      ( )a 0 0.4 rad,θ = 0.125 st = 7 (0.125)/6 0.63245, 4 , sin 1, cos 0 2 2 2 e tπ π π π π− = = = = ( )( )( )0.4 0.63245 1 0.25298 radiansθ = = 0.253 radθ = ( )( ) ( ) 7 0.4 0.63245 1 0.92722 rad/s 6 π ω   = − = −    0.927 rad/sω = − ( )( ) ( )2 249 0.4 0.63245 16 1 36.551rad/s 36 α π   = − − = −    2 36.6 rad/sα = − ( )b ,t = ∞ 7 /6 0t e π− = 0θ = 0ω = 0α =
  • 4. 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 4. Angular coordinate: 1800 rev = 3600 radiansπθ = Initial angular velocity: 0 6000 rpm = 200 rad/sω π= Angular acceleration: constant d d dt d ω ω α ω θ = = = d dα θ ω ω= 0 0 0 d d θ ω α θ ω ω=∫ ∫ 2 0 1 2 αθ ω= − ( ) ( )( ) 22 20 200 17.4533 rad/s 2 2 3600 πω α θ π = − = − = − (a) Time required to coast to rest. 0 tω ω α= + 0 0 200 17.4533 t ω ω π α − − = = − 36.0 st = (b) Time to execute the first 900 revolutions. 900 rev = 1800 radiansθ π= 2 0 1 2 t tθ ω α= + ( ) 21 1800 200 17.4533 2 t tπ π= − 2 72 648 0t t− + = ( ) ( )( )2 72 72 4 648 2 t ± − = 10.54 st =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 5. ( )( ) 1 0 1 2400 2 2400 rpm 80 rad/s, 0, 4 s 60 t π ω π ω= = = = = ( )a 21 1 0 80 , 20 rad/s 4 t t t ω π ω ω α α α π= + = = = = ( )( )2 2 1 0 1 1 160 0 20 4 160 rad 80 rev 2 2 2 t t π θ ω α π π π = + = + = = = 1 80 revθ = (b) 1 2 2 180 rad/s, 0, 40 st tω π ω= = − = ( ) 22 1 2 1 2 1 2 1 0 80 , 2 rad/s 40 t t t t ω ω π ω ω α α π − − = + − = = = − − ( ) ( ) ( )( ) ( )( )2 2 2 1 1 2 1 2 1 1 1 80 40 2 40 2 2 t t t tθ θ ω α π π− = − + − = + − 1600 1600 radians 800 rev 2 π π π = = = 2 1 800 revθ θ− =
  • 6. 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 6. Angular acceleration: 0.2 30 t d e dt ω α − = = Angular velocity: 0 0 t dtω ω α= + ∫ 0.2 0 0 30 t t e dt− = + ∫ 0.2 0 30 0.2 t t e− =  − ( )0.2 150 1 t d e dt θ− = − = When t = 0.5 s, ( )( ) ( )0.2 0.5 150 1 eω − = − 14.27 rad/sω = Angular coordinate: 0 0 t dtθ θ ω= + ∫ ( )0.2 0 0 150 1 t t e dt− = + −∫ 0.2 0 150 150 0.2 t t t e− = −  − ( )0.2 150 750 1 t t e− = − − When t = 0.5 s, ( )( ) ( )( ) ( )0.2 0.5 150 0.5 750 1 eθ − = − − 3.63 radiansθ =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 7. ( ) 0.5 , 0.5 0.5 d a d d d ω α ω ω ω ω θ θ = − = − = − 0 30 0 Integrating, 0.5 30 0.5d d θ ω θ θ= − − = −∫ ∫ 60 60 radians 9.55 revθ π = = = 2 9.55 revθ = ( ) 0.5 2 d d b dt dt ω ω ω ω = − = − Integrating, 0 0 30 2 t d dt ω ω = −∫ ∫ 0 2 ln 30 t = − = ∞ t = ∞ ( )c ( )( )0.02 30 0.6 rad/sω = = 0.6 0 30 2 t d dt ω ω = −∫ ∫ 0.6 2 ln 2 ln 50 30 t = − = 7.82 st =
  • 8. 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 8. d k k d k d d ω α θ ω θ ω ω θ θ θ = − = − = − Integrating, 0 6 12 0 d k dω ω θ θ= −∫ ∫ 2 2 12 6 0 0 2 2 k   − = − −     ( )a 2 2 2 12 9 s 6 k − = = 2 9.00 sk − = 3 12 0 d k d ω ω ω θ θ= −∫ ∫ 2 2 2 12 3 9 0 2 2 2 ω   − = − −     ( )b ( )( )22 2 2 2 12 9 3 63 rad /sω = − = 7.94 rad/sω =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 9. ( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k ( ) ( )/ 5 in. 15.6 in.B O = +r i j ( ) ( ) ( )2 2 2 5 31.2 12 33.8 in.OAl = + + = Angular velocity. ( )/ 6.76 5 31.2 12 33.8 A O OAl ω = = + +r i j kωωωω ( ) ( ) ( )1.0 rad/s 6.24 rad/s 2.4 rad/s= + +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 2 126.1 in./s 74.3 in./s 246 in./sB = − − +a i j k
  • 10. 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 10. ( ) ( ) ( )/ 5 in. 31.2 in. 12 in.A O = + +r i j k ( ) ( )/ 5 in. 15.6 in.B O = +r i j ( ) ( ) ( )2 2 2 5 31.2 12 33.8 in.OAl = + + = Angular velocity. ( )/ 3.38 5 31.2 12 33.8 A O OAl ω = = + +r i j kωωωω ( ) ( ) ( )0.5 rad/s 3.12 rad/s 1.2 rad/s= + +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. ( )/ 5.07 5 31.2 12 33.8 A O OAl α − = = + +r i j kαααα ( ) ( ) ( )2 2 2 0.75 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 2 3.46 in./s 27.6 in./s 73.1in./sB = − − +a i j k
  • 11. 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 11. ( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k 2 2 2 0.5 0.225 0.3 0.625 mABl = + + = Angular velocity vector. ( )/ 10 0.5 0.225 0.3 0.625 B A ABl ω = = − +r i j kωωωω ( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j k ( ) ( )/ 300 mm 0.3 mE 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 2 11.52 m/s 5.18 m/s 23.1m/sE = − + +a i j k
  • 12. 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 12. ( ) ( ) ( ) ( ) ( ) ( )/ 500 mm 225 mm 300 mm 0.5 m 0.225 m 0.3 mB A = − + = − +r i j k i j k 2 2 2 0.5 0.225 0.3 0.625 mABl = + + = Angular velocity vector. ( )/ 10 0.5 0.225 0.3 0.625 B A ABl ω = = − +r i j kωωωω ( ) ( ) ( )8 rad/s 3.6 rad/s 4.8 rad/s= − +i j k Angular acceleration vector. ( )/ 20 0.5 0.225 0.3 0.625 B A ABl α − = = − +r i j kαααα ( ) ( ) ( )2 2 2 16 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 mC 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 2 18.00 m/s 19.20 m/s 15.60 m/sC = + −a i j k
  • 13. 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 13. ( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k ( ) ( ) ( )2 2 2 200 120 90 250 mmDAd = + + = 0.8 + 0.48 + 0.36A/D DA DAd = = − r i j kλλλλ ( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −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 2 405 m/s 432 m/s 324 m/sB = − − +a i j k
  • 14. 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 14. ( ) ( ) ( )/ 200 mm + 120 mm 90 mmA D = − +r i j k ( ) ( ) ( )2 2 2 200 120 90 250 mmADd = + + = 0.8 + 0.48 + 0.36A/D DA ADd = = − r i j kλλλλ ( )( ) ( ) ( ) ( )75 0.8 + 0.48 + 0.36 60 rad/s + 36 rad/s + 27 rad/sDAω= = − = −i j k i j kωωωω λλλλ ( )( )600 0.8 + 0.48 + 0.36DA d dt ω = = − − i j kαααα λλλλ ( ) ( ) ( )2 2 2 480 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 2 405 m/s 389 m/s 266 m/sB = − − −a i j k
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 15. 9 93,000,000 mi 491.04 10 ft= × 6 365.24 days 31.557 10 s, 1rev 2 radπ= × = Angular velocity. 9 6 2 199.11 10 rad/s 31.557 10 π ω − = = × × Velocity of the earth. ( )( )9 9 3 491.04 10 199.11 10 97.77 10 ft/sv rω − = = × × = × 3 66.7 10 mi/hv = × Acceleration of the earth. ( )( ) 2 2 3 9 3 2 97.77 10 199.11 10 19.47 10 ft/sa rω − − = = × × = × 3 2 19.47 10 ft/sa − = ×
  • 16. 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 16. 3 23 h 56min 23.933 h 86.16 10 s, 1rev 2 radπ= = × = 6 3 2 72.925 10 rad/s 86.16 10 π ω − = = × × 6 6370 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 2 cos cos 33.876 10 cos m/sp R 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 2 33.9 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 = ( )( )3 33.876 10 0.76604− = − ×a i ( )3 2 0.273 10 m/s− = − × i 2 0.0259 m/sa = ( ) .c North Pole ( )90 cos 0φ φ= ° = 0v = 0a =
  • 17. 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 17. 300 mm/sB Av v= = 120 mmBr = ( ) 180 mm/sB At a a= = ( )a 300 , 2.5 rad/s 120 B B B B v v r r ω ω= = = = 2.50 rad/s=ωωωω ( ) ( ) 2180 , 1.5 rad/s 120 B t B Bt B a a r r α α= = = = 2 1.500 rad/s=αααα ( )b ( ) ( )( )22 2 120 2.5 750 mm/sB Bn a r ω= = = ( ) ( ) ( ) ( )2 2 2 2 2 180 750 771mm/sB B Bt n a a a= + = + = 750 tan , 76.5 180 β β= = ° 2 771mm/sB =a 76.5°
  • 18. 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 18. 4 rad/sBω = , 120 mmBr = ( ) ( )( )22 2 120 4 1920 mm/sB B Bn a r ω= = = 2 2400 mm/sBa = ( ) ( )22 2 2 2 2400 1920 1440 mm/sB B Bt n a a a= − = − = ± ( ) ( ) 21440 , 12 rad/s 120 B t B Bt B a a r r α α ± = = = = ± 2 12.00 rad/s or
  • 19. 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 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 18 rad/s 0.6667 A B A A A v v r r ω = = = = 18 rad/sA =ωωωω 296 144 rad/s 0.6667 A B A A A a a r r α = = = = 2 144 rad/sA =αααα Pulley C. 5 in. 0.41667 ftCr = = 12 28.8 rad/s 0.41667 C B C C C v v r r ω = = = = 28.8 rad/sC =ωωωω 296 230.4 rad/s 0.41667 C B C C C a a r r α = = = = 2 230 rad/sC =αααα (b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = = ( ) 2 96 ft/sP Bt a a= = ( ) ( )22 2 212 345.6 ft/s 0.41667 P B P n C C v v a ρ ρ = = = = ( ) ( ) ( ) ( )2 2 2 2 2 96 345.6 358.7 ft/sP P Pt n a a a= + = + = 96 tan 15.52 345.6 β β= = ° 2 359 ft/sP =a 15.52°
  • 20. 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 20. 1 rev radians, 8 in. 0.6667 ft, 5 in. 0.41667 ft 2 A Cr rπ= = = = = 2 2 2 500 120 0.002 , 120 0.002 60000 d d d d d ω ω ω ω ω α ω ω θ θ ω ω = − = = = − − Integrating and applying initial condition 0 at 0ω θ= = and noting that θ π= radians at the final state, ( )2 20 00 500 250 ln 60000 60000 d d ωω πω ω ω θ π ω = − − = = − ∫ ∫ ( ) 2 2 60000 250 ln 60000 ln 60000 250 ln 60000 ω ω π − − − − = − =   2 /25060000 60000 e πω −− = 2 /250 2 2 60000 1 749.26 rad /se π ω − = − =  27.373 rad/sω = ( )( )2 120 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= = = = ( )( ) 2 0.6667 118.50 79.0 ft/sB A Aa a r α= = = = 2 79.0 ft/sBa = (b) Acceleration of point P on pulley C. 5 in. 0.41667 ftCρ = = 18.249 ft/sP Bv v= = ( ) 2 2 218.249 799.3 ft/s 0.41667 B P n C v ρ = = =a ( ) 2 79.0 ft/sP Bt a= =a ( ) ( )2 2 2 799.3 79.0 803.2 ft/sPa = + = 79 tan , 5.64 799.3 β β= = ° 2 803 ft/sP =a 5.64°
  • 21. 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 21. Left pulley. Inner radius r1 = 50 mm Outer radius r2 = 100 mm 0.6 m/s = 600 mm/sAv = 1 2 600 6 rad/s 100 Av r ω = = = Speed of intermediate belt. ( )( )1 1 1 50 6 300 mm/sv rω= = = Right pulley. Inner radius r3 = 50 mm Outer radius r4 = 100 mm 1 2 4 300 3 rad/s 100 v r ω = = = (a) Velocity of C. ( )( )3 2 50 3 150 mm/sCv r ω= = = 0.1500 m/sC =v (b) Acceleration of point B. ( )( )22 2 4 2 100 3 900 mm/sBa r ω= = = 2 0.900 m/sB =a
  • 22. 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 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 2 600 6 rad/s 100 Av r ω = = = ( ) 2 1 2 1800 18 rad/s 100 A t a r α = = = Intermediate belt. ( )( )1 1 1 50 6 300 mm/sv rω= = = ( ) ( )( ) 2 1 1 1 50 18 900 mm/st a rα= = = Right pulley. Inner radius r3 = 50 mm Outer radius r4 = 100 mm 1 2 4 300 3 rad/s 100 v r ω = = = ( )1 2 2 4 900 9 rad/s 100 t a r α = = = (a) Velocity and acceleration of point C. ( )( )3 2 50 3 150 mm/sCv r ω= = = 0.150 m/sC =v ( ) ( )( )3 2 50 9 450 mm/sC t a rα= = = 0.450 m/sC =a
  • 23. 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. (b) Acceleration of point B. ( ) ( )( )22 2 4 2 100 3 900 mm/sB n a r ω= = = ( ) 2 0.900 m/sB n =a ( ) ( )( ) 2 4 2 100 9 900 mm/sB t a r α= = = ( ) 2 0.900 m/sB t =a 2 1.273 m/sB =a 45°
  • 24. 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 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 ω ω = = 1 2 A B r r ω ω = 2 1( ) : A Ab On shaft A a rω= 2 1A Arω=a 2 2 1 2 2 2 : A B B r On ring B a r r r ω ω   = =     2 2 1 2 A B r r ω =a
  • 25. 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 24. (a) Let point C be the point of contact between the shaft and the ring. ( )( )1 0.5 25 12.5 in./sC Av rω= = = 2 12.5 5.0 rad/s 2.5 C B v r ω = = = 5.00 rad/sBω = ( ) On shaft :b A ( )( )22 1 0.5 25A Aa rω= = 2 312.5 in./s ,= 2 26.0 ft/sA =a On ring :B ( )( )22 2 2.5 5.0B Ba r ω= = 2 62.5 in./s ,= 2 5.21ft/sB =a ( ) At a point on the outside of the ring,c 3 3.5 in.r r= = ( )( )22 2 3.5 5.0 87.5 in./sBa rω= = = 2 7.29 ft/sa =
  • 26. 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 25. ( )a ( )( )600 2 600 rpm 20 rad/s. 60 A π ω 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./sD D A Av r ω π π= = = / 40 60 10 rad/s 10 300 rpm 4 2 D B D B v r π ω π π π = = = = ⋅ = 300 rpmB =ωωωω Let E be the contact point between gears B and C. ( )( )/ 2 10 20 in./sE E B Bv r ω π π= = = ( ) / 20 60 3.333 rad/s 3.333 100 rpm 6 2 E C E C v r π ω π π π = = = = = 100 rpmC =ωωωω (b) Accelerations at point E. ( )22 2 / 20 On gear : 1973.9 in./s 2 E B E B v B a r π = = = 2 1974 in./sB =a ( )22 2 / 20 On gear : 658 in./s 6 E C E C v C a r π = = = 2 658 in./sC =a
  • 27. 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 26. ( ) At timea 2 s,t = ( )( )600 2 600 rpm 20 rad/s 60 A π ω π= = = 2 , 10 rad/sA A A At t ω ω α α π= = = Let D be the contact point between gears A and B. ( ) ( )( ) 2 / 2 10 20 in./sD D A At a r α π π= = = ( ) 2 / 20 5 rad/s 4 D t B D B a r π α π= = = 2 15.71rad/sB =αααα Let E be the contact point between gears B and C. ( ) ( )( ) 2 / 2 5 10 in./sE E B Bt a r α π π= = = ( ) 2 / 10 1.6667 rad/s 6 E t C E C a r π α π= = = 2 5.24 rad/sC =αααα ( ) Atb 0.5 s.t = ( )( )For gear , 5 0.5 2.5 rad/sB BB tω α π π= = = ( ) ( )( )22 2 / 2 2.5 123.37 in./sE E B Bn r ω π= = =a ( ) 2 10 in./sE t π=a 2 31.416 in./s= ( ) ( ) ( ) ( )2 2 2 2 2 123.37 31.416 127.3 in./sE E En t a a a= + = + = 31.416 tan , 14.29 123.37 β β= = ° 2 127.3 in./sE =a 14.29°
  • 28. 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. ( )( )For gear , 1.6667 0.5 0.83333 rad/sC CC tω α π π= = = ( ) ( )( )22 2 / 6 0.83333 41.123 in./sE E C Cn a r ω π= = = ( ) 2 31.416 in./sE t a = ( ) ( ) ( ) ( )2 2 2 2 2 41.123 31.416 51.75 in./sE E En t a a a= + = + = 31.416 tan 37.4 , 41.123 β = = ° 2 51.8 in./sE =a 37.4°
  • 29. 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 27. ( ) For the pulley,a ( ) 1 1 , 0.3 0.15 m 2 2 Ar d r= = = ( ) 1 0.2 0.1 m 2 Br = = ( )/ / ,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 0.8 16 rad/s 0.15 0.1 ω = = − At 0.25 s,t = 1 0.4 8 rad/s 0.15 0.1 ω = = − 21 2 8 16 32 rad/s 0.25t ω ω α − − = = = − 2 32 rad/s= A Ar α=a ( )( ) 2 0.15 32 4.80 m/s= = B Br α=a ( )( ) 2 0.1 32 3.2 m/s= = 2 / 1.6 m/sA B A B= − =a a a 2 / 1.600 m/sA B =a ( )b ( ) ( ) 2 / / /0 1 2 A B A B A Bt t= +x v a ( )( ) ( )( )21 0.8 0.25 1.6 0.25 0.15 m 2 = − + = − / 150.0 mmA B =x
  • 30. 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 28. For the pulley, ( ) 1 1 , 0.3 0.15 m 2 2 Ar d r= = = ( ) 1 0.2 0.1m 2 Br = = ( )/,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 0.9 18 rad/s 0.15 0.1 ω = = − At 1 rev radians, 2 θ π= = 0.45 9 rad/s 0.15 0.1 ω = = − d d d d dt d ω ω α ω ω ω α θ θ = = = ( )( ) 2 2 9 2 18 0 9 18 38.675 rad/s 2 2 d d π ω ω α θ α π α= − = = −∫ ∫ 2 38.7 rad/sα = ( ) 2 2 0.15 38.675 5.8012 m/s or 5.8012 m/sA Ar α= = − = −a ( ) 2 2 0.1 38.675 3.8675 m/s or 3.8675 m/sB Br α= = − = −a ( )a 2 2 / 1.9337 m/s or 1.9337 m/sA B A B= − = −a a a 2 / 1.934 m/sA B =a ( )b ( ) ( ) 2 / / /0 1 2 A B A B A Bt t= +x v a ( )( ) ( )( )21 0.9 0.3 1.9337 0.3 0.1830 m 2 = + − = / 183.0 mmA B =x
  • 31. 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 29. (a) Motion of pulley. ( ) ( )0 0 8 in./sE A= =v v ( ) 2 10 in./sE At = =a a Fixed axis rotation. ( ) ( )0 0 00 8 in./s 2 rad/s 4 in. E E A A v v r r ω ω= = = = ( ) ( ) 2 210 in./s 2.5 rad/s 4 in. E t E At A a a r r α α= = = = Since α is constant, 0 2 2.5t tω ω α= + = + 2 2 0 0 1 0 2 1.25 2 t t t tθ θ ω α= + + = + + For t = 3s, ( )( )2 2.5 3 9.5 rad/sω = + = ( )( ) ( )( )2 0 2 3 1.25 3 17.25 radθ = + + = In revolutions, 17.25 2 θ π = 2.75 revθ = (b) Motion of load B. t = 3s ( )( )6 9.5B Bv r ω= = 57.0 in./sB =v ( )( )6 17.25B By r θ∆ = = 103.5 in.By∆ = (c) Acceleration of point D. t = 0 0 2 rad/sω ω= = 2 2.5 rad/s=αααα ( ) ( )( ) 2 6 2.5 15 in./sD Dt r α= = =a ( ) ( )( )22 2 6 2 24 in./sD Dn r ω= = =a 2 28.3 in./sD =a 32.0°
  • 32. 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 30. 2 2.4 rad/sα = 0 0ω = Use equations for constant angular acceleration. 0 2.4t tω ω α= + = 2 2 0 0 1 1.2 2 t t tθ θ ω α= + + = At t = 4s, ( )( )2.4 4 9.6 rad/sω = = ( )2 1.2 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./sA =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./sB =v ( )( )6 in. 19.2 radB By r θ= = 115.2 in.B =y
  • 33. 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 31. When contact is made, 240 rpm 8 rad/sAω π= = Let C be the contact point between the two gears. ( )( ) 2 0.15 8 1.2 m/sC A Av r ω π π= = = 1.2 6 rad/s 0.2 C B B v r π ω π= = = 8 rad/sA tω π α= = ( )6 2 rad/sB tω π α= = − Subtracting, ( )( ) 2 2 2 rad/sπ α α π= = ( )a 2 3.14 rad/sα = ( )b 8 8 8 st π π α π = = = 8.00 st =
  • 34. 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 32. ( )0 240 rpm 8 rad/sAω π= = ( ) 11 8A A tω π α= − 3 3 2 2 2 1 1 1 1 1 1 0.15 4 2 2 2 0.2 A B B A A B r t t t r θ π α α α     = = = =       ( ) 3 2 1 0.2 8 59.574 radians 0.15 A tα π   = =    ( ) 3 1 1 11 0.15 0.421875 0.2 B B A At t tω α α α   = = =    Let Cv be the velocity at the contact point. ( )( )1 10.15 8 1.2 0.15C A A A Av r t tω π α π α= = − = − and ( )( )1 10.2 0.421875 0.084375C 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/sA A At t tπ α α α− = = Then, 2 1 1 1 59.574 3.7037 s 16.0850 A A t t t α α = = = ( )a 216.0850 4.3429 rad/s 3.7037 Aα = = 2 4.34 rad/sAα = ( ) 3 20.15 4.3429 1.83218 rad/s 0.2 Bα   = =    2 1.832 rad/sBα = ( ) From above,b 1 3.70 st =
  • 35. 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 33. Motion of disk B. ( )0 500 rpm 52.360 rad/sBω = = Assume that the angular acceleration of disk B is constant. ( )0B B Btω ω α= + At 60 s,t = 0Bω = ( ) 20 0 52.360 0.87266 rad/s 60 B B B t ω ω α − − = = = − ( )( )3 52.360 0.87266 157.08 2.618 in./sB B Bv r t tω= = − = − Motion of disk A. ( )0 0,Aω = 2 3 rad/sAα = ( )0 3A A At 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/sAω = = ( )( )52.360 0.87266 15.52 38.8 rad/sBω = − = 445 rpmA =ωωωω , 371 rpmB =ωωωω
  • 36. 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 34. Wheel B. ( )0 300 rpm 31.416 rad/sBω = = At 12s,t = 75 rpm =7.854 rad/sBω = Angular acceleration. ( )0 7.854 31.416 1.9635 rad/s 12 B B B t ω ω α − − = = = − Velocity at contact point with disk A at 12 s:t = ( )( )3 7.854 23.562 in./sB B Bv r ω= = = Wheel A. ( )0 300 rpm = 31.416 rad/sAω = Assume that slipping ends when 12 s.t = Then, 23.562 in./sAv = 23.562 9.4248 rad/s 2.5 A A A v r ω = = = 9.4248 rad/sAω = 9.4248 rad/s= − ( ) 20 A 9.4248 31.416 3.4034 rad/s 12 A A t ω ω α − − − = = = − (a) 2 3.40 rad/sA =αααα 2 1.963 rad/sB =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 ω α − − = = − 9.23 st =
  • 37. 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 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 ω π π ∆ = = ∆ For the reel: 1 1d d v dv d v dt dt r r dt dt r ω     = = +        2 2 2 a v dr a v b r dt rr r ω π = − = − 2 1 0 2 b a r ω π   = − =    2 0 2 bω π =a
  • 38. 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 36. Let one layer of paper be unrolled. 2 andv t r r bπ∆ = ∆ = − 2 r bv dr t r dtπ ∆ − = = ∆ 2 1 1 0 d d v dv d v dr v dt dt r r dt dt r dtr ω α     = = = + = −        2 2 3 2 2 v bv bv rr rπ π −   = − =      2 3 2 bv rπ =αααα
  • 39. 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 37. Velocity analysis. 150 mm/sB =v 15° A A=v v / 500B A ω=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. / sin75 sin sin50 A B A B v v v φ = = ° ° ( )a / sin75 150 sin75 189.14 mm/s sin50 sin50 B A B v v ° ° = = = ° ° / 189.14 0.378 rad/s 500 A B AB v l ω = = = 0.378 rad/s=ωωωω ( )b sin 150 sin55 160.4 mm/s sin50 sin50 B A v v φ ° = = = ° ° 160.4 mm/sA =v
  • 40. 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 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. / sin75 sin60 sin B A B A v v v φ = = ° ° ( )a / sin75 225sin75 307.36 mm/s sin sin 45 A B A v v φ ° ° = = = ° / 307.36 0.615 rad/s 500 B A AB v l ω = = = 0.615 rad/s=ωωωω ( )b sin 60 225sin60 276 mm/s sin sin 45 A B v v φ ° ° = = = ° 276 mm/sB =v 15°
  • 41. 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 39. Geometry. 12 sin , 36.87 20 β β= = ° 12 tan , 67.38 5 θ θ= = ° Velocity analysis. 4.2 ft/sA =v / 20 12 B A AB ABrω 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 A v v v θ β φ = = ° − / sin 4.2sin67.38 4.5 ft/s sin sin59.49 A B A v v θ φ ° = = = ° ( )a 4.5 2.7 rad/s 20/12 ABω = = 2.70 rad/sAB =ωωωω ( )b cos 4.2cos36.87 3.90 ft/s sin sin59.49 A B v v β φ ° = = = ° 3.90 ft/sB =v 67.4°
  • 42. 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 40. Geometry. 12 sin , 36.87 20 β β= = ° 12 tan , 67.38 5 θ θ= = ° Velocity analysis. 4.2 rad/sAB =ωωωω ( )/ / 20 4.2 7.0 ft/s 12 B A B A ABr 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 B vv v φ β θ = = ° − ( )a / sin 7sin59.49 6.53 ft/s sin sin 67.38 B A A v v φ θ ° = = = ° 6.53 ft/sA =v ( )b / cos 7cos36.87 6.07 ft/s sin sin67.38 B A B v v β θ ° = = = ° 6.07 ft/sB =v 67.4°
  • 43. 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 41. In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j / / 1.2 1.2C A C Aω ω ω= × = × =v k r k i j /A B A= +v v vB ( ) ( )7.4 7 0.6 0.6B Ay x v v ω ω− + = − − +i j i j i j Components. ( ): 7.4 0.6A x v ω− = −i (1) ( ): 7 0.6B y v ω= − +j (2) /A C A= +v v vC ( ) ( )1.4 7 1.2C A xy v v ω− + = − +i j i j j Components. ( ): 1.4 A x v− =i (3) ( ): 7 1.2C y v ω= − +j (4) From (3), ( ) 1.4 m/sA x v = − ( ) From (1),a ( )7.4 1.4 10 rad/s, 0.6 ω − − − = = − 10.00 rad/s=ωωωω From (2), ( ) ( )( )7 0.6 10 1m/sB y v = − + = − ( )b ( ) ( )7.40 m/s 1.000 m/sB = − −v i j
  • 44. 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 42. In units of m/s, ( )/ / 0.6 0.6 0.6 0.6B A B Aω ω ω ω= × = × + = − +v k r k i j i j / / 1.2 1.2C 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 x v v ω ω− + = − − +i j i j i j Components. ( ): 7.4 0.6A x v ω− = −i (1) ( ): 7 0.6B y v ω= − +j (2) /C A C A= +v v v ( ) ( )1.4 7 1.2C A xy v v ω− + = − +i i j j Components. ( ): 1.4 A x v− =i (3) ( ): 7 1.2C y v ω= − +j (4) From (3), ( ) 1.4 m/s, 1.4 7A Ax v = − = − −v i j From (1), ( ) ( ) 7.4 1.4 10 rad/s, 10.00 rad/s 0.6 ω − − − = = = − kωωωω (a) ( )/ / 10 0.6O 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) ( )0 O 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 =
  • 45. 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 43. In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j ( )/ / 50 150 150 50C 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 y v v ω ω− = + − +i j i j i jA Components. ( ): 100 75B x v ω= −i (1) ( ): 75 125A y v ω− = +j (2) /C A C A= +v v v ( ) ( )400 100 150 50C A yy v v ω ω+ = + − +i j i j i j Components. : 400 100 150ω= −i (3) ( ) ( ): 125C A yy v 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 y v ω= − − = − − − = ( ) ( )100.0 mm/s 175.0 mm/sA = +v i j
  • 46. 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 44. In units of mm/s, ( )/ / 125 75 75 125B A B Aω ω ω ω= × = × + = − +v k r k i j i j ( )/ / 50 150 150 50C 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 y v v ω ω− = + − +i j i j i jA Components. ( ): 100 75B x v ω= −i (1) ( ): 75 125A y v ω− = +j (2) /C A C A= +v v v ( ) ( )400 100 150 50C A yy v v ω ω+ = + − +i j i j i j Components. : 400 100 150ω= −i (3) ( ) ( ): 125C A yy v v ω= +j (4) From (3), ( )2 rad/sω = − k From (2), ( ) ( )75 125 75 125 2 175 mm/sA y v 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 2A D A x y= + = + + × +v v v i j k i jD 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 Radius of locus. 200 100 mm 2 v r ω = = = Circle of 100.0 mm radius centered at 87.5 mm, 50.0 mmx y= − =
  • 47. 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 45. 7 Slope angle of rod. tan 0.7, 35 10 θ θ= = = ° 10 12.2066 in. 20 7.7934 in. cos AC CB AC θ = = = − = Velocity analysis. 25 in./sA =v , C Cv=v θ /C A ABACω=v θ /C A C A= +v v v Draw corresponding vector diagram. / sin 25sin35 14.34 in./sC A Av v θ= = ° = ( )a / 14.34 1.175 rad/s 12.2066 C A AB v AC ω = = = 1.175 rad/sAB =ωωωω cos 25cos 20.479 in./sC Av v θ θ= = = ( )( )/ 7.7934 1.175B 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. / 9.1551 tan , 24.09 20.479 B C C v v φ φ= = = ° ( )b 20.479 22.4 in./s 1.869 ft/s cos cos24.09 C B v v φ = = = = ° 59.1φ θ+ = ° 1.869 ft/sB =v 59.1°
  • 48. 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 46. Instantaneous geometry. Law of sines: sin sin120 10 15 φ ° = 10 sin sin120 0.57735 15 φ = ° = 35.264φ = ° Velocity analysis. 1.2 ft/sA =v 14.4 in./s= / 10B 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 A v v v γ β = = ° / sin 14.4 sin125.264 28.10 in./s sin sin 24.736 A B A v v γ β ° = = = ° (a) 28.10 10 ABω = 2.81 rad/sAB =ωωωω (b) sin30 14.4 sin30 17.21 in./s sin sin 24.736 A B v v β ° ° = = = 1.434 ft/sB =v 35.3°
  • 49. 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 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 80 Av ω= (1) Arm AB: 2 120 ABv ω= (2) Gear B: 1 2 40 Bv v ω= − (3) 3 2 80 Bv v ω= + (4) Gear C: 3 200 Cv ω= (5) Data: 0, 5 rad/sA Cω ω= = From (1), 1 0,v = From (5), ( )( )3 200 5 1000 mm/sv = = From (3), 2 40 0Bv ω− = (6) From (4), 2 80 1000Bv ω+ = (7) Solving (6) and (7) simultaneously, (a) 1000 8.333 rad/s 120 Bω = = 8.33 rad/sB =ωωωω 2 (40)(8.333) 333.33 mm/sv = = (b) From (2), 333.33 2.78 rad/s 120 ABω = = 2.78 rad/sAB =ωωωω
  • 50. 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 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 80 Av ω= (1) Arm AB: 2 120 ABv ω= (2) Gear B: 1 2 40 Bv v ω= − (3) 3 2 80 Bv v ω= + (4) Gear C: 3 200 Cv ω= (5) Data: 20 rad/s, 0B Cω ω= = From (5), 3 0.v = From (4), ( )( )2 80 80 20Bv ω= − = − 1600 mm/s= From (3), ( )( )1 1600 40 20 2400v = − − = − 2400 mm/s= From (1), 1 /80 30 rad/sA vω = = − (a) 30.0 rad/sA =ωωωω From (2), 1600 120 ABω= 1600 13.33 rad/s 120 ABω = − = − (b) 13.33 rad/sAB =ωωωω
  • 51. 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 49. Data: 3600 rpm 376.99 rad/s, 0A Bω ω= = = 1 1.25 in. 2 A 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 d ω = = 942.48 rad/s= /C B C Bv v v= + 1 0 (0.25)(942.48) 235.62 in./s 2 dω= + = = (a) 236 in./sCv = (b) Angular velocity of ball. 942.48 rad/sCω = 9000 rpmCω = (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 l n rπ = = 1500n =
  • 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 © 2007 The McGraw-Hill Companies. Chapter 15, Solution 50. Contact point 1 between gears A and B. Contact point 2 between gears B and C. Gear B: 6 rad/sBω = 1 (6 rad/s)(10 in.) 60 in./sv = = (1) 2 (6 rad/s)(5 in.) 30 in./sv = = (2) Arm ABC: ABC ABCω=ωωωω 15A ABCω=v 15C ABCω=v Gear A: 3 rad/sAω = 1 (5)(3)Av v= + 15 ABCω= 15+ (3) Matching expressions (1) and (3) for 1,v (a) : 60 15 15ABCω= + 3.00 rad/sABC =ωωωω Gear C: C Cω ω= 2 10C Cv v ω= + 15= 10 Cω+ (4) Matching expressions (2) and (4) for 2,v (b) : 30 15 10 Cω= + 1.500 rad/sC =ωωωω
  • 53. 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 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), 2B Ev b a bω ω+ = + (6) 2 1From (1) and (3), B Av b v aω ω− = = (7) ( ) 2 2 2 Solving for and , 2 E A B a b a v v ω ω ω  + + = ( )2 2 E A B a b a b ω ω ω  + − = From (2), ( ) ( ) 2 2 2 E A S S a b av a b a b ω ω ω ω  + + = = + + Data: 60 mm, 60 mm, 2 180 mm, 120 mma b a b a b= = + = + = ( )a ( )( ) 180 60 1.5 0.5 2 60 E A B E A ω ω ω ω ω − = = − ( )( ) ( )( )1.5 120 0.5 150 105 rpm= − = 105.0 rpmB =ωωωω ( )b ( )( ) 180 60 0.75 0.25 2 120 E A S E A ω ω ω ω ω + = = + ( )( ) ( )( )0.75 120 0.25 150 127.5 rpm= + = 127.5 rpmS =ωωωω
  • 54. 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 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), 2B 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 v ω ω + + = ( )2 2 E A B a b a b ω ω ω  + − = From (2), ( ) ( ) 2 2 2 E A S S a b av a b a b ω ω ω ω  + + = = + + 1 Data: 0, 5 E S Aω ω ω= =
  • 55. 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. ( )a ( )( ) ( ) 2 01 5 2 A S A a b a a b ω ω ω  + + = = + ( ) ( ) 1 1 1 , 2 5 52 1 b a a a b = = + + 2 1 5 b a   + =    1.500 b a = ( )b ( ) ( )( ) 2 0 2 2 2 1.5 3 E A A A A B a b a a b b ω ω ω ω ω ω  + − = = − = − = − 1 3 B Aω=ωωωω
  • 56. 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 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/sABCω π= = 0Av = (12)(2.5 ) 30 rad/sBv π π= = (12 7)(2.5 ) 47.5 rad/sCv π π= + = Gear A: 10, 0, 0A Av vω = = = Gear B: 1 4 0B Bv v ω= − = 30 4 0Bπ ω− = 7.5 rad/sBω π= 2 4B Bv v ω= + 30 30 60 in./sπ π π= + = Gear C: 2 3C Cv v ω= − 60 47.5 3 Cπ π ω= − 4.1667 rad/sCω π= − Summary: 225 rpmB =ωωωω 125.0 rpmC =ωωωω
  • 57. 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 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, 0C Cv vω = = = Gear B: 2 4 56 /3 4 0B B Bv v ω π ω= − = − = 14 /3 rad/sBω π= 1 4 56 /3 56 /3B Bv v ω π π= + = + 112 /3 in./sπ= Gear A: 1 8A Av v ω= − 112 /3 152 /3 8 Aπ π ω= − 5 /3 rad/sAω π= − Summary: 50.0 rpmA =ωωωω 140.0 rpmB =ωωωω
  • 58. 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 55. Geometry. ( ) ( )sin sinOA ABθ β= ( )sin 10 sin30 sin , 1.79 160 OA AB θ β β ° = = = ° Shaft and eccentric disk. (Rotation about O), 900 rpm 30 rad/sOAω = = π ( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πv Rod AB. (Plane motion = Translation with A + Rotation about A.) [/B A B A Bv= +v v v ] [ Av= ] /60 A Bv° +  ]β 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 φ β π ° = = ° − ° = 497 mm/sB =v
  • 59. 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 56. Geometry. ( ) ( ) ( )sin 180 sinOA ABθ β° − = ( ) ( )sin 180 10 sin60 sin , 3.10 160 OA AB θ β β ° − ° = = = ° Shaft and eccentric disk. (Rotation about O) 900 rpm 30 rad/sOAω = = π ( ) ( )( )10 30 300 mm/sA OAOA ω= = π = πv Rod AB. (Plane motion = Translation with A + Rotation about A.) [/B A B A Bv= +v v v ] [300= π ] /30 B 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 φ β π ° = = ° + ° = 791 mm/sB =v
  • 60. 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 57. Disk A Bar :BD 15 rad/sAω = , 2.8 in.AB = Rotation about a fixed axis. ( ) ( )( )2.8 15 42 in./sB Av AB ω= = = ( )a 0 .θ = ° 42 in./sB =v 2.8 sin , 16.260 10 β β= = ° /D B D B= +v v v Dv [42= ] /D Bv+  β  / 42 43.75 in./s cos D Bv β = = / 43.75 , 10 D B DB v DB ω = = 4.38 rad/sDB =ωωωω tan ,D Bv v β= 12.25 in./sD =v ( ) 90 .b θ = ° 42 in./sB =v 5.6 sin , 34.06 10 β 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
  • 61. 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. ( )c 180 .θ = ° 42 in./sBv = 2.8 sin , 16.26 10 β β= = ° Bar :BD /D B D B= +v v v Dv [42= ] /D Bv+  β  / 42 43.75 cos D Bv β = = / 43.75 10 D B DB v DB ω = = 4.38 rad/sDB =ωωωω tanD Bv v β= 12.25 in./sD =v
  • 62. 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 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= = + 2.8 sin , 22.9 , 10 2.8 A BD A r l r β β= = = ° − − 22.9θ = ° For 180 , sin sin ,θ β θ β= ° + = − sin sinBD A Al r rβ β= − 2.8 sin , 12.6 10 2.8 A BD A r l r β β= = = ° + + 192.6θ = ° ( ) For matching magnitudesb /D B Bv v= ( )( )2.8 20 , 5.6 rad/s 10 A A BD BD A A BD BD r l r l ω ω ω ω= = = = For 22.9 ,θ = ° 5.60 rad/sBD =ωωωω For 192.6 ,θ = ° 5.60 rad/sBD =ωωωω
  • 63. 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 59. Rod BE: 0Ev = / (192)(4) 768 mm/sB B E BEr ω= = =v Rod ABD: D Dv=v /D B D B= +v v v Dv 768= 360 ADω+ 60° Draw diagram for vector addition. (a) 768 360 sin30ADω= ° 4.2667 rad/sADω = 4.27 rad/sAD =ωωωω (b) 768 tan30Dv= ° 1330 mm/sDv = 1.330 m/sD =v (c) / 240A B ADω=v 60 1024 mm/s° = 60° / 768A B A B= + =v v v 1024+ 60 1557 mm/s° = 34.7° 1.557 m/sA =v 34.7°
  • 64. 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 60. Rod BE: 0E =v B Bv=v Rod ABD: 1.6 m/sD =v AD ADω=ωωωω /B D B D= +v v v Bv 1.6= 0.360 ADω+ 60° Draw diagram for vector addition. (a) 1.6 0.360 sin60 ADω = ° 5.13 rad/sAD =ωωωω (b) 1.6tan30 0.92376Bv = ° = 0.924 m/sB =v (c) / 0.240A B ADω=v 60 1.2317 m/s° = 60° / 0.92376A B B A= + =v v v 1.2317+ 60° [1.5396= ] [1.0667+ ] 1.873= 34.7° 1.873 m/sA =v 34.7°
  • 65. 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 61. 1000 rpmABω = ( )( )1000 2 104.72 rad/s 60 π = = ( )a 0 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r ( )( )/ 3 104.72 314.16 in./sB 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./sD D Bv v= = P D=v v 0P =v 314.16 8 B BD v l ω = = 39.3 rad/sBD =ωωωω ( )b 90 . . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r ( )( )/ 3 104.72 314.16 in./sB 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./sD B Dv v= = / , D B BD v l ω = 0BD =ωωωω 314.16 in./sP D= =v v 26.2 ft/sP =v
  • 66. 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 62. ( )( )1000 2 1000 rpm 104.72 rad/s 60 ABω π = = = 60 , . (Rotation about )Crank AB Aθ = ° / 3 in.B A =r 30° ( )( )/ 3 104.72 314.16 in./sB B A ABω= = =v r 60° Rod BD. (Plane motion Translation with Rotation about .)B B= + Geometry. sin sinl rβ θ= 3 sin sin sin60 8 18.95 r l β θ β = = ° = ° /D B D B= +v v v [ Dv ] [314.16= ] /60 D Bv° +  ]β Draw velocity vector diagram. ( )180 30 90 78.95φ β= ° − ° − ° − = ° Law of sines. ( ) / sin sin30 sin 90 D BD B vv v φ β = = ° ° − sin 314.16 sin78.95 326 in./s cos cos18.95 B D v v φ β ° = = = ° P Dv v= 27.2 ft/sP =v / sin30 314.16sin30 166.08 in./s cos cos18.95 B D B v v β ° ° = = = ° / 166.08 8 D B BD v l ω = = 20.8 rad/sBD =ωωωω
  • 67. 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 63. Bar AB. (Rotation about A) ( ) ( ) ( )/ 4 0.25 1.00 m/sB AB B A= × = − × − = −v r k j iωωωω Bar ED. (Rotation about E) ( )/ 0.075 0.15 0.15 0.075D 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 0.2D 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 BDω ω ω− = − +i j i j Components: : 0.15 1.00, 6.6667 rad/sDE DEω ω= − = −i 6.67 rad/sDE =ωωωω : 0.075 0.2DE BDω ω− =j ( )( )0.075 6.6667 0.2 BDω − − = 2.50 rad/sBD =ωωωω
  • 68. 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 64. Bar AB. (Rotation about A.) ( ) 0.18B AB ABAB ω ω= =v 30° Bar DE. (Rotation about E.) ( ) 0.18D DE DEDE ω ω= =v Bar BGD. (Plane motion = Translation with B + Rotation about .B ) [/D B D B Dv= +v v v ] [ Bv= ]30° /D Bv+  ]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 ω ω ω= = = / / 1 0.09 2 G B GB BD D B ABv l vω ω= = = /G B G B= +v v v Draw vector diagram.
  • 69. 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. Law of cosines ( ) ( ) ( )( )2 22 0.18 0.09 2 0.18 0.09 cos60G AB AB AB ABv ω ω ω ω= + − ° 2 2 0.0243G ABv ω= ( )( )6.415 6.416 2.5AB Gvω = = 16.04 rad/sAB =ωωωω 16.04 rad/sBD =ωωωω 0.18D B ABv v ω= = 0.18 0.18 D AB DE AB DE v l ω ω ω= = = 16.04 rad/sDE =ωωωω
  • 70. 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 65. Bar AB. (Rotation about A.) 25 rad/sAB =ωωωω ( )( )/ 8 25 200 in./sB B A ABr ω= = =v Bar ED. (Rotation about E.) D Dv=v 8D DEv ω= Plate BDHF. (Translation with B + Rotation about .B ) [/D B D B Dv= +v v v ] [ Bv= ] /D Bv+  ]30° Draw velocity vector diagram. / 200 230.94 in./s cos30 cos30 B D B v v = = = ° ° / 230.94 14.4338 rad/s 16 B D BDHF BD v l ω = = = (a) 14.43 rad/sBDHF =ωωωω ( )( )/ 8 14.4338 115.47 in./sF B BDHFv BFω= = = / 115.47 m/sF B =v 30° [/ 200 in./sF B F B= + =v v v ] [115.47 in./s+ ]30° (b) [142.265 in./sF =v ] [100 in./s+ ] 173.9 in./sF =v 54.9°
  • 71. 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 66. Rod DE. (Rotation about E.) 35 rad/sDEω = ( )( )/ 8 35 280 in./sD D E DEr ω= = =v Rod AB. (Rotation about A.) B Bv=v 8B ABv ω= Plate BDHF. (Translation with D + Rotation about .D ) [/B D B D Bv= +v v v ] [ Dv= ] /B Dv+  ]30° Draw velocity vector diagram. / / 280 560 in./s sin30 sin30 560 35 rad/s 16 D D B D B BDHF DB v v v l ω = = = ° ° = = = (a) 35.0 rad/sBDHF =ωωωω Point of zero velocity lies above point D. / 280 8 in. 35 D C D BDHF v y ω = = = (b) 8 in. above point D.
  • 72. 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 67. (15 rad/s)AB = − kωωωω BD BDω= kωωωω DE DEω=ω k ( )/ 0.2mB A = −r j ( ) ( )/ 0.6 m 0.25 mD B = − −r i j ( )/ 0.2 mD E = −r i ( ) ( ) ( )/ 0 15 0.25 3 m/sB A AB B A= + × = + − × − = −v v ω r k j i ( ) ( )/ / 6 0.2 0.25 0.6D 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 BDω ω= + = − + −v v v i i j (1) ( ) ( )/ /0 0.2 0.2D 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 0BDω− + =i 12 rad/sBDω = : 0.6 0.2BD DEω ω− = −j 36 rad/sDEω = ( )a Angular velocity of rod BD. 12 rad/sBD =ωωωω (b) Velocity of the midpoint M of rod BD. ( ) ( )/ / 1 0.3 m 0.125 m 2 M B D B= = − −r r i j ( )/ / 3 12 0.3 0.125M 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°
  • 73. 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 68. Bar AB. ( ) ( )/ 0.300 m + 0.125 m ,B A =r i j ( )3 rad/sAB = kωωωω 0A =v / /0B 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 mD B = −r j BD BDω= kωωωω ( )0.375 + 0.9 + 0.325D B D/B BDω= + = − × −v v v i j k j 0.375 + 0.9 + 0.325 BDω= − 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 DEω= + × − + =v k i j 0.375 0.9 0.325 0.15 0.2 0BD DE DEω ω ω− + + − − =i j i j i Components: j: 0.9 0.15 0DEω− = 6 rad/sDEω = i: 0.375 0.325 0.2 0BD DEω ω− + − = ( )( )0.325 0.375 0.2 6BDω = + 4.85 rad/sBDω = 4.85 rad/sBD =ωωωω 6.00 rad/sDE =ωωωω
  • 74. 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 69. 80 km/h 22.222 m/sA = =v 0C =v 560 mm 280 mm 0.28 m 2 d d r= = = = 22.222 79.364 rad/s 0.28 Av r ω = = = / / /B A D A E Av v v rω= = = ( )( )0.28 79.364 22.222 m/s= = [/ 22.222 m/sB A B A= + =v v v ] [22.222 m/s+ ] 44.4 m/sB =v [/ 22.222 m/sD A D A= + =v v v ] [22.222 m/s+ ]30° 42.9 m/sD =v 15.0° [/ 22.222 m/sE A E A= + =v v v ] [22.222 m/s+ ] 31.4 m/sE =v 45.0°
  • 75. 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 70. (a) 0β = .Wheel AD 0, 45 in./sC D= =v v 45 11.25 rad/s 4 D AD v CD ω = = = ( ) ( ) 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 Av+  ]ϕ 16.88 in./sB =v / 0B Av = 0ABω = (b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v 2.5 tan , 32.005 4 DA DC γ γ= = = ° 4.7170 in. cos DC CA 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 , 18.663 12.5 φ φ= = ° Plane motion = Translation with A + Rotation about A.
  • 76. 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. [/B A B A Bv= +v v v ] [ Av= ] [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 A vv v δ γ φ = = ° + ( ) ( )53.066 sin 39.332sin sin 90 sin108.663 A B v v δ φ ° = = ° + ° 35.5 in./s= 35.5 in./sB =v ( )/ sin (53.066)sin32.005 sin 90 sin108.663 A B A v v γ φ ° = = ° + ° 29.686 in./s= / 29.686 2.37 rad/s 12.5 B A AB v AB ω = = = 2.37 rad/sAB =ωωωω
  • 77. 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 71. 0 180 km/h = 50 m/s=v ( )( )180 2 180 rpm = 18.85 rad/s 60 π ω = = Top View 0v zω= 0 50 2.65 m 18.85 v z ω = = = Instantaneous axis is parallel to the y axis and passes through the point 0x = 2.65 mz =
  • 78. 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 72. 1.5 1.0 1 rad/s 1.5 3 E D ED v v l ω − − = = = 1 3 1.0 3 mD CE v l ω = = = (a) 1.5 2 3 0.5 mACl = + − = lies 0.500 m to the right ofC A (b) ( ) 1 0.5 0.1667 m/s 3 A ACv l ω   = = =    A 0.1667 m/s=v
  • 79. 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 73. Contact points: 1 between gears A and B. 2 between gears B and C. Arm ABC: 4 rad/sABCω = ( )( )15 4 60 in./sAv = = ( )( )15 4 60 in./sCv = = Gear B: 8 rad/sBω = ( )( )1 10 8 80 in./sv = = ( )( )2 5 8 40 in./sv = = Gear A: 1 80 60 5 5 A A v v ω − − = = 4 rad/sAω = 60 15 in. 4 A A A v ω = = =l
  • 80. 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. Gear C: 2 60 40 10 10 C C v v ω − − = = 2 rad/sCω = 60 30 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. 4.00 rad/sA =ωωωω 2.00 rad/sC =ωωωω
  • 81. 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 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 Cv r ω= / 6 6 rad/s 1 B B C v r ω = = = 6.00 rad/s=ωωωω (b) Velocity of point A. /A A Cv 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.=
  • 82. 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 75. 12 1.5 35 rad/s 0.3 O A AO v v l ω − − = = = (a) 31.5 42.86 10 m 35 A CA v l 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 2 0.3 0.34286 0.45558 mCDl = + = ( )( )0.45558 35 15.95 m/sD CDv l ω= = = 0.3 tan , 41.2 0.34286 OD CO l l θ θ= = = ° 15.95 m/sD =v 41.2°
  • 83. 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 76. 12 1.5 45 rad/s 0.3 O A AO v v l ω − + = = = (a) 31.5 33.33 10 m 45 A AC v l 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 2 0.3 0.26667 0.4014 mCEl = + = ( )( )0.4014 45 18.06 m/sE CEv l ω= = = 0.3 tan , 48.4 0.26667 OE OC l l θ θ= = = ° 18.06 m/sE =v 48.4°
  • 84. 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 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= + 1.6667 rad/sω = − /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./sE Av v− = − =
  • 85. 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 78. 10 in./sD =v , 8 in./sB =v 10 8 4 rad/s 4.5 D Bv v BD ω + + = = = 10 2.5 in. 4 Dv CD ω = = = 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.
  • 86. 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 79. Rod AD. ( )( )/ 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 sin30 0.180 mB Cr = ° = / 0.768 4.2667 0.180 B AD B C v r ω = = = 4.27 rad/sAD =ωωωω " (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 = ° = 0.600sin30 0.300 mCEl = ° = 0.20785 tan 0.300 β = 34.7β = ° ( ) ( )2 2 0.20785 0.300 0.36497 mCAl = + = ( )( )0.36497 4.2667 1.557 m/sA CA ADv l ω= = = 1.557 m/sA =v 34.7° "
  • 87. 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 80. 15 rad/sABω = ( ) ( )( )0.200 15 3 m/sB ABv AB ω= = = B Bv=v D Dv=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 12 rad/s 0.25 B BD v BC ω = = = 12.00 rad/sBD =ωωωω (b) Locate point M, the midpoint of rod BD. Draw CM. ( ) ( )2 2 0.6 0.25 0.65 mBD = + = 0.25 tan 22.62 90 67.38 0.6 β β β= = ° ° − = ° ( ) 1 0.325 m 2 CM DM MB BD= = = = ( ) ( )( )0.325 12 3.9 m/sMv CM ω= = = 3.90 m/sM =v 67.4°
  • 88. 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 81. Bar DC. (rotation about D) ( )( )( ) 18 10C CDv CDω= = 180 in./s= 180 in./sC =v 30° Bar AB. (rotation about A) B Bv=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= = 180 18 rad/s 10 C BC v IC ω = = = ( ) ( )( )10 3 18 311.77 in./sB BCv IB ω= = = (a) 311.77 31.177 rad/s 10 B AB v AB ω = = = 31.2 rad/sAB =ωωωω (b) 18.00 rad/sBC =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/sM =v 30°
  • 89. 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 82. Bar AB. (rotation about A) B Bv=v 60° Bar CD. (rotation about D) 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 9.36 rad/s 10 M BC v IM ω = = = (a) ( ) ( )( )10 3 9.36 162.12 in./sB BCv IB ω= = = 162.12 16.21 rad/s 10 B AB v AB ω = = = 16.21 rad/sAB =ωωωω (b) 9.36 rad/sBC =ωωωω (c) ( ) ( )( )10 9.36 93.6 in./sC BCv IC ω= = = 93.6 9.36 rad/s 10 C CD v DC ω = = = 9.36 rad/sCD =ωωωω
  • 90. 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 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 Av and BC perpendicular to Bv . (a) 800 3.0792 rad/s 300cos30 B ABD v BC ω = = = ° 3.08 rad/sABD =ωωωω ( ) ( )2 2 600cos30 300sin30 540.83 mmCDl = ° + ° = 300sin30 tan 16.10 90 73.9 600cos30 γ γ γ ° = = ° ° − = ° ° (b) ( )( ) 3 540.83 3.0792 1.665 10 mm/sD CD ABDv l ω= = = × 1.665 m/sD =v 73.9°
  • 91. 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 84. 40 ,θ = ° 0.6 m/sB =v , A Av=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) 0.6 0.46672 rad/s 1.28557 B ABD v BC ω = = = 0.467 rad/sABD =ωωωω / / /D C B C D B= +r r r [1.28557 m= ] [2 m+ ]40° 2.9930 m= 30.79° 30.79β = ° (b) ( )( )/ 2.9930 0.46672D D C ABDv r ω= = 1.397 m/s= 1.397D =v β 90 59.2β° − = ° 1.397 m/sD =v 59.2°
  • 92. 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 85. 2 2 320 240 400 mmDE = + = 240 tan 0.75 320 β = = 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 mm tan AC β = = ( )720 320 100 300 mmBC AC AB= − = − + = Since triangles FCB and BDK are similar, 300 100 b DK BC BK = = ( )( )300 300 900 mm. 100 b = = ( )a Distance b. 0.900 mb = 2 2 300 400 500 mm 0.5 mCD = + = = 6 1.2 rad/s 5 D BDF v CD ω = = = ( )( )0.900 1.2 10.8 m/sF BDFv bω= = = ( )b Velocity of point F. 10.80 m/sF =v
  • 93. 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 86. Locate the instantaneous center I of rotation of bar ABD as the intersection of line AI perpendicular to Av and line BI perpendicular to Bv Triangle IAB is equilateral. 300 mmIA IBl l= = (a) 900 mm/s 3 rad/s 300 mm A ABD IA v l ω = = = 3.00 rad/sABD =ωωωω (b) By the law of cosines, ( )600cos30 mmIDl = ° ( )( )600cos30 3 1559 mm/sD ID ABDv l ω= = ° = 1.559 m/sD =v
  • 94. 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 87. A Av=v 45 ,° 7.5 ft/sB =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 Av and BC perpendicular to .Bv Let 24 in. 2 ftl AB= = = Law of sines for triangle ABC. 2.8284 ft sin75 sin60 sin 45 b a l = = = ° ° ° 2.4495 ft, 2.73205 fta b= = 7.5 2.7452 rad/s 2.73205 Bv b ω = = = (a) ( )( )2.4495 2.7452 6.724 ft/sAv aω= = = 6.72 ft/sA =v 45.0° (b) 2.75 rad/s=ωωωω (c) Let M be the midpoint of AB. Law of cosines for triangle CMB. ( ) ( ) ( )( )( ) 2 2 2 2 2 2 cos60 2 2 2.73205 1 2 2.73205 1 cos60 2.3942 ft l l m b b m   = + − °   = + − ° = Law of sines. 2 sin sin60 1sin60 , sin , 21.2 2.3942l m β β β ° ° = = = ° ( )( )2.3942 2.7452 6.573 ft/s,Mv mω= = = 6.57 ft/sM =v 21.2°
  • 95. 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 88. Bar DE. ( )( )24 8 192 in./sEDDv eω= = = 192 in./sD =v Bar AB. 8ABB ABv aω ω= = 8B ABω=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. 24 48 in. sin120 sin30 sin30 sin30 b d l = = = = ° ° ° ° 41.569 in., 24 in.b d= = (a) 192 8 rad/s 24 D BD v d ω = = = 8.00 rad/sBD =ωωωω (b) ( )( )41.569 8 332.55 in./sB BDv bω= = = 332.55 41.6 rad/s 8 B AB v a ω = = = 41.6 rad/sAB =ωωωω (c) Law of cosines for triangle CMD. ( ) ( )( )( ) 2 2 2 22 2 cos120 2 2 24 12 2 24 12 cos120 31.749 in. l l m d d m   = + − °   = + − ° = Law of sines. ( ) 2 12 sin120sin sin120 , sin , 19.1 31.749l m β β β °° = = = ° Velocity of M. ( )( )31.749 8 253.99 in./sM BDv mω= = = 21.2 ft/sM =v 19.1°
  • 96. 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 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 to Bv and DC perpendicular to .Dv Law of sines for triangle . sin150 sin15 sin15 CD BC BD BCD = = ° ° ° 180sin150 180 mm 347.73 mm sin15 BC BD CD ° = = = = ° 200 0.57515 rad/s 347.73 D DBE v CD ω = = = ( ) ( )( )180 0.57515 103.528 mm/sB DBEv BC ω= = = 103.528 0.57515 rad/s 180 B ABF v AB ω = = = ( ) ( )( )300 0.57515 172.546 mm/sF ABFv AF ω= = = Law of cosines for triangle DCE. ( ) ( ) ( ) ( )( )2 2 2 2 cos15CE CD DE CD DE= + − ° ( ) ( )( )( )2 2 2 347.73 300 2 347.73 300 cos15 , 96.889 mmCE CE= + − ° =
  • 97. 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. sin15 300 sin15EH DE= ° = ° 300sin15 cos 36.7 96.889 EH CE β 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°
  • 98. 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 90. 3 rad/sDE =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 Bv and DC perpendicular to .Dv ( )120 mm, cos30 120cos30BD DK BD= = ° = ° 120 cos30 cos , 49.495 , 180 30 100.505 160 DK ED β β φ β ° = = = ° = ° − ° − = ° Law of sines for triangle BCD. sin30 sin sin CD BC BD φ β = = ° ( ) ( ) sin30 120sin30 78.911 mm sin sin sin 120sin 155.177 mm sin sin BD CD BD BC β β φ φ β β ° ° = = = = = =
  • 99. 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. Law of cosines for triangle ABC. ( ) ( ) ( ) ( )( )2 2 2 2 cos150AC BC AB AB BC= + − ° ( ) ( )( )( )2 2 2 155.177 200 2 155.177 200 cos150 , 343.27 mmAC AC= + − ° = Law of sines. sin sin150 200 sin150 sin , 16.9 343.27AB AC γ γ γ ° ° = = = ° (a) 480 6.0828 rad/s 78.911 D ABD v CD ω = = = 6.08 rad/sABD =ωωωω (b) ( ) ( )( )343.27 6.0828 2088 mm/sA ABDv AC ω= = = 2.09 m/sA =v 73.1°
  • 100. 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 91. AB = 20 in. Instantaneous centers: at I for BC. at J for BD. Geometry ( ) 12 11 8.25 in. 16 IC   = =    ( ) 12 21 15.75 in. 16 JD   = =    ( ) 11 20 13.75 in. 16 AI   = =   20 in. 13.75 in. = 6.25 in.BI AB AI= − = − 6.25 in.BJ BI= = Member BC. 33 in./sC =v 33 4 rad/s 8.25 C BC v IC ω = = = ( ) ( )( )6.25 in. 4 rad/s 25 in./sB BCv BI ω= = = Member BD. 25 in./s 4 rad/s 6.25 in. B BD v BJ ω = = = (a) ( ) ( )( )= = 15.75 in. 4 rad/sD BDJD ωv 63.0 in./sD =v Member AB. (b) 25 in./s 20 in. B AB v AB ω = = 1.250 rad/sAB =ωωωω
  • 101. 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 92. 6 in./sA =v , B Bv=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 Av and BC perpendicular to Bv . Triangle ACB. Law of sines. sin 30 sin 30 sin120 AC CB AB = = ° ° ° 10sin 30 5.7735 in. sin120 AC CB ° = = = ° (a) 6 1.03923 rad/s 5.7735 A AB v AC ω = = = 1.039 rad/sAB =ωωωω ( ) ( )( )5.7735 1.03923 6 in./sB ABCB 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 to Bv and DI perpendicular to Dv . Triangle BID. Law of sines. sin120 sin 30 sin 30 BI DI BD = = ° ° ° 10 sin120 17.3205 in. sin 30 BI ° = = ° , 10 in.DI =
  • 102. 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. 6 0.34641 rad/s 17.3205 B BD v BI ω = = = 0.346 rad/sBD =ωωωω (b) ( ) ( )( )10 0.34641 3.46 in./sD BDv DI ω= = = 3.46 in./sD =v
  • 103. 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 93. Method 1 Assume Dv 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 2 0.27 0.36 0.45 mAD = + = ( ) ( )2 2 0.18 0.135 0.225 mBD = + = 0.27 0.18 sin 0.6, sin 0.8 0.45 0.225 θ 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 cos c θ β = 0.225sin cos d φ β = 0.36 0.27tan 0.135 0.18tana bβ β= − = + .Kinematics 0.4 m/s, 1 m/sA Bv v= = ,A D AD v v a c ω = = B D BD v v b d ω = = A B D cv dv v a b = = 0.45sin cos 0.4 0.6 cos 0.225sin 1 A B cv a b b b dv θ β β φ = = ⋅ ⋅ = ( )0.36 0.27tan 0.6 0.135 0.18tanβ 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 m cos cos c d β β = = = =
  • 104. 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. ( )a 0.4 2.4891 rad/s 0.1607 A AD v a ω = = = 2.49 rad/sAD =ωωωω ( )b 1 3.733 rad/s 0.2679 B BD v b ω = = = 3.73 rad/sBD =ωωωω (c) ( )( )0.3356 2.4891 0.835 m/s,D ADv cω= = = 0.835 m/sD =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/sA A B A= =v v 0.27 tan , 36.87 0.36 θ θ= = ° 0.36 0.45 m cos AD θ = = / 0.45D A ADω=v θ 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 ( ) 0.18 0.3 m sin cos 0.135 0.375 m CD BC CD θ θ = = = + = / 1.4 3.73 rad/s 0.375 B A BD v CB ω = = = ( ) ( )( )/ 0.3 3.73 1.120 m/sD A BDv CD ω= = = ( )a / 1.120 0.45 D A AD v AD ω = = 2.49 rad/sAD =ωωωω ( )b 3.73 rad/sBD =ωωωω ( )c [/ 0.4 m/sD A D A= + =v v v  [1.120+ ]θ 0.835 m/s= 53.6° 0.835 m/sD =v 53.6°
  • 105. 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 94. ( )( )0.2 0.2 4.5 0.9 m/sB ABω= = =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 b ω = = 0.9 D BD a a b ω= =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 bc ω = = ( ) ( )0.9 0.2 0.2 0.7E DE a d v d bc ω − = − = = (1) 0.25 0.15 0.25 , , 0.25tan , 0.15tan cos cos 0.15 a a c b d c β β β β = = = = = Substituting into (1) ( ) 0.25 0.2 0.15tan 0.9 0.7 or 1.2 0.9tan 0.7tan 0.15 0.25tan β β β β −  = − =    1.2 tan 0.75 36.87 cos 0.8 1.6 β β β= = = ° = 0.25 0.3125 m, 0.1875 m, 0.1875 m, 0.1125 m 0.8 a b c d= = = = = ( )a 0.9 4.8 rad/s 0.1875 BDω = = 4.80 rad/sBD =ωωωω ( )b ( )( ) ( )( ) 0.9 0.3125 8 rad/s 0.1875 0.1875 DEω = = 8.00 rad/sDE =ωωωω ( )c ( )( )0.3125 4.8 1.5 m/sDv = = 1.5 m/sD =v 53.1°
  • 106. 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 95. 5 rad/sABCω = , 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 26.565 18 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 5 rad/s 20.125 D DE JD v l ω = = = (a) 5.00 rad/sDE =ωωωω 9 cos 27 in.JE JDl l φ= + = ( )( )27 5E JE DEv l ω= = (b) 135 in./sE =v
  • 107. 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 96. 12 in./sA =v B Bv=v Point C is the instantaneous center of bar AB. 12 20cos30 B AB v AC ω = = ° 0.69282 rad/s= 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 0.4 rad/s 20cos30 D DE v DI ω = = = ° 0.400 rad/sDE =ωωωω ( )b ( ) ( )( )20sin30 0.4 4 in./sE DEv EI ω= = ° = 0.333 ft/sE =v 30°
  • 108. 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 97. Let points A, B, and C move to , , andA B C′ ′ ′ as shown. Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack. : lower rackspace 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 gearbody centrode A
  • 109. 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 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 Bv=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 cosCy l β= ( )22 2 2 300 mmC 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 cos2 2 sin cos sin sin 2 2 C C x AC l l y AC l l β β β β β β β = = = + = = = ( ) 2 2 22 2 2 150 150 2 2 C C C C l l x y x y     − + = = − + =        The body centrode is a semi circle of 150 mm radius centered midway between A and B.
  • 110. 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 99. 80 km/h 22.222 m/sAv = = 0C =v 1 560 mm, 0.280 mm 0.28 m 2 d r d= = = = Point C is the instantaneous center. 22.222 79.364 rad/s 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 ( ) 1 30 15 2 γ = ° = ° ( )( )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°
  • 111. 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 100. 900 rpm 30 rad/sOAω π= = ( ) ( )( )10 30 300 mm/sA OAOA 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 ( )sin30 10sin30 sin , 1.79 160 OA AB β β ° ° = = = ° ( ) ( )cos30 cos 10cos30 160cosOB OA AB β β= ° + = ° + 168.582 mm= 168.582 10 184.662 mm cos30 cos30 OB AC OA= − = − = ° ° ( )tan30 97.377 mmBC OB= ° = A B AB v v AC BC ω = = ( )( )97.377 300 497 mm/s, 184.662 B A BC v v AC π  = = =    497 mm/sB =v
  • 112. 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 101. Bar AB. 4 rad/sABω = ( )( )0.25 4B AB ABv l ω= = 1 m/sB =v Bar DE. 75 tan 26.565 150 φ φ= = ° 0.15 = 0.167705 m cos EDl φ = 0.167705D ED ED EDv l ω ω= = 0.167705D EDω=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 m tan IBl φ = = 0.4 0.44721 m cos IDl φ = = 1 2.5 rad/s 0.4 B BD IB v l ω = = = 2.50 rad/sBD =ωωωω 1.11803 m/sD ID BDv l ω= = 6.67 rad/sD ED ED v l ω = = 6.67 rad/sED =ωωωω
  • 113. 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 102. (a) 0β = .Wheel AD 0, 45 in./sC D= =v v 45 11.25 rad/s 4 D AD v CD ω = = = ( ) ( ) 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 0ABω = (b) 90β = ° .Wheel AD 0, 11.25 rad/sC ADω= =v 2.5 tan , 32.005 4 DA DC γ γ= = = ° 4.7170 in. cos DC CA 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 , 18.663 12.5 φ φ= = °