Vector Mechanics For Engineers Dynamics- Ch 12 Solutions

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Chapter 12, Solution 1.

m = 20 kg, g = 3.75 m/s 2 W = mg = ( 20 )( 3.75 )

Vector Mechanics for 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.

W = 75 N W

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Chapter 12, Solution 2.

m = 2.000 kg W

At all latitudes, (a) φ = 0°,

(

)

g = 9.7807 1 + 0.0053 sin 2 φ = 9.7807 m/s 2 W = mg = ( 2.000 )( 9.7807 )

(

W = 19.56 N W

)

(b) φ = 45°, g = 9.7807 1 + 0.0053 sin 2 45° = 9.8066 m/s 2 W = mg = ( 2.000 )( 9.8066 )

(

W = 19.61 N W

)

(c) φ = 60°, g = 9.7807 1 + 0.0053 sin 2 60° = 9.8196 m/s2 W = mg = ( 2.000 )( 9.8196 )

Vector Mechanics for 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.

W = 19.64 N W

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Chapter 12, Solution 3.

Assume g = 32.2 ft/s 2 m=

W g

ΣF = ma : W − Fs =  a W 1 −  = Fs g 

or

Fs

W =

1−

a g

W a g

7

= 1−

2 32.2 W = 7.46 lb W

m=

W 7.4635 = = 0.232 lb ⋅ s 2 /ft g 32.2 ΣF = ma : Fs − W =

W a g

 a Fs = W 1 +  g  2   = 7.46 1 +  32.2  

Fs = 7.92 lb W

For the balance system B, ΣM 0 = 0: bFw − bFp = 0 Fw = Fp   a a But, Fw = Ww 1 +  and Fp = W p 1 +  g g   so that Ww = W p and mw =

Wp g

Vector Mechanics for 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.

mw = 0.232 lb ⋅ s 2 /ft W

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Chapter 12, Solution 4.

Periodic time:

τ = 12 h = 43200 s

Radius of Earth:

R = 3960 mi = 20.9088 × 106 ft

Radius of orbit:

r = 3960 + 12580 = 16540 mi = 87.33 × 106 ft

Velocity of satellite:

v=

2π r

τ

=

( 2π ) (87.33 × 106 ) 43200

= 12.7019 × 103 ft/s It is given that (a)

mv = 750 × 103 lb ⋅ s m=

mv 750 × 103 = = 59.046 lb ⋅ s 2 /ft 3 v 12.7019 × 10 m = 59.0 lb ⋅ s 2 /ft W

(b)

W = mg = ( 59.046 )( 32.2 ) = 1901 lb

W = 1901 lb W

Vector Mechanics for 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.

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Chapter 12, Solution 5.

+ ∑ Fy = ma y :

10 + 10 + 10 + 20 − 40 = ay =

ay =

40 ay 32.2

( 32.2 )(10 ) = 8.05 ft/s2 40

dv dy dv dv = =v dt dt dy dy

v dv = a y d y v

v

∫ 0 v dv = ∫ 0 a y d y v = 2a y y =

1 2 v = ay y 2

( 2 )(8.05)(1.5)

Vector Mechanics for 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.

v = 4.91 ft/s W

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Chapter 12, Solution 6.

Data: v0 = 108 km/h = 30 m/s, x f = 75 m (a)

Assume constant acceleration. a = v

dv dv = = constant dx dt

0 xf ∫ v0 v dv = ∫ 0 a dx

1 − v02 = a x f 2 a=−

v02 2x f

=−

(30) ( 2)( 75)

= − 6 m/s 2

0 tf ∫ v0 dv = ∫ 0 a dt

− v0 = a t f tf = − (b)

v0 − 30 = a −6

t f = 5.00 s W

+ ∑ Fy = 0: N − W = 0 N =W

∑ Fx = ma : µ=−

µ=−

− µ N = ma

ma ma a =− =− N W g

( − 6) 9.81

Vector Mechanics for 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.

µ = 0.612 W

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Chapter 12, Solution 7.

(a)

+ ∑ F = ma : a=− =−

Ff m

− F f + W sin α = ma +

Ff W sin α =− + g sin α m m

(

)

7500 N + 9.81 m/s 2 sin 4° = − 4.6728 m/s 2 1400 kg

a = 4.6728 m/s 2

4°

v0 = 88 km/h = 24.444 m/s From kinematics,

a=v

dv dx

xf 0 ∫ 0 a dx = ∫ v0 v dv

1 a x f = − v02 2

( 24.444 ) v02 =− 2a ( 2 )( − 4.6728) 2

xf = −

Vector Mechanics for 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.

x f = 63.9 m W

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Chapter 12, Solution 8.

(a) Coefficient of static friction. ΣFy = 0:

N −W = 0

N =W

v0 = 70 mi/h = 102.667 ft/s v 2 v02 − = at ( s − s0 ) 2 2 0 − (102.667 ) v 2 − v02 = = − 31.001 ft/s 2 2 ( s − s0 ) ( 2 )(170 ) 2

at =

For braking without skidding µ = µ s , so that µ s N = m | at | ΣFt = mat : − µ s N = mat

µs = −

mat a 31.001 = − t = W g 32.2

µ s = 0.963 W

(b) Stopping distance with skidding. Use µ = µk = ( 0.80 )( 0.963) = 0.770 ΣF = mat : µk N = −mat at = −

µk N m

= − µk g = − 24.801 ft/s 2

Since acceleration is constant,

( s − s0 ) =

0 − (102.667 ) v 2 − v02 = 2at ( 2 )( − 24.801)

2

s − s0 = 212 ft W

Vector Mechanics for 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.

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Chapter 12, Solution 9.

For the thrust phase,

ΣF = ma : Ft − W = ma =

W a g

F   2  a = g  t − 1 = ( 32.2 )  − 1 = 289.8 ft/s 2 W 0.2     At t = 1 s, v = at = ( 289.8 )(1) = 289.8 ft/s y =

1 2 1 2 at = ( 289.8 )(1) = 144.9 ft 2 2

For the free flight phase, t > 1 s. a = − g = − 32.2 ft/s v = v1 + a ( t − 1) = 289.8 + ( − 32.2 )( t − 1) At v = 0,

t −1 =

289.8 = 9.00 s, t = 10.00 s 32.2

v 2 − v12 = 2a ( y − y1 ) = −2 g ( y − y1 ) 0 − ( 289.8 ) v 2 − v12 =− = 1304.1 ft y − y1 = − 2g ( 2 )( 32.2 ) 2

(a)

ymax = h = 1304.1 + 144.9

(b) As already determined,

Vector Mechanics for 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.

h = 1449 ft W t = 10.00 s W

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Chapter 12, Solution 10.

Kinematics: Uniformly accelerated motion. ( x0 = 0, v0 = 0 ) x = x0 + v0t +

1 2 at , 2

a=

or

2 x ( 2 )(10 ) = = 1.25 m/s 2 2 2 t ( 4)

ΣFy = 0: N − P sin 50° − mg cos 20° = 0 N = P sin 50° + mg cos 20° ΣFx = ma : P cos 50° − mg sin 20° − µ N = ma or P cos50° − mg sin 20° − µ ( P sin 50° + mg cos 20° ) = ma P=

ma + mg ( sin 20° + µ cos 20° ) cos50° − µ sin 50°

For motion impending, set a = 0 and µ = µ s = 0.30. P=

( 40 )( 0 ) + ( 40 )( 9.81)( sin 20° + 0.30cos 20° ) cos50° − 0.30sin 50°

= 593 N

For motion with a = 1.25 m/s 2 , use µ = µk = 0.25. P=

( 40 )(1.25) + ( 40 )( 9.81)( sin 20° + 0.25cos 20° ) cos50° − 0.25sin 50°

P = 612 N W

Vector Mechanics for 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.

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Chapter 12, Solution 11.

Calculation of braking force/mass ( Fb / m ) from data for level pavement. v0 = 100 km/hr = 27.778 m/s v 2 v02 − = a ( x − x0 ) 2 2 a=

0 − ( 27.778 ) v 2 − v02 = 2 ( x − x0 ) ( 2 )( 60 )

2

= −6.43 m/s 2 ΣFx = ma : − Fbr = ma Fbr = −a = 6.43 m/s 2 m (a) Going up a 6° incline. (θ = 6° ) ΣF = ma : − Fbr − mg sin θ = ma a=−

Fbr − g sin θ m

= −6.43 − 9.81sin 6° = −7.455 m/s 2 0 − ( 27.778 ) v 2 − v02 = 2a ( 2 )( −7.455)

2

x − x0 =

x − x0 = 51.7 m W (b) Going down a 2% incline. ( tan θ = −0.02, θ = −1.145° ) ΣF = ma : − Fbr − mg sin θ = ma F a = − br − g sin θ m = − 6.43 − 9.81sin ( −1.145° ) = − 6.234 m/s 2 0 − ( 27.778 ) v 2 − v02 = 2a ( 2 )( −6.234 )

2

x − x0 =

Vector Mechanics for 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.

x − x0 = 61.9 m W

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Chapter 12, Solution 12.

Let the positive directions of x A and xB be down the incline. Constraint of the cable:

x A + 3xB = constant

a A + 3aB = 0

1 aB = − a A 3

or

For block A:

ΣF = ma : mA g sin 30° − T = mAa A

For block B:

ΣF = ma : mB g sin 30° − 3T = mB aB = − mB a A (2)

Eliminating T and solving for

(1)

aA , g

( 3mA g − mB g ) sin 30° =  3mA + 

mB   aA 3 

( 3mA − mB ) sin 30° = ( 30 − 8) sin 30° = 0.33673 aA = g 3m A + mB / 3 30 + 2.667 (a) a A = ( 0.33673)( 9.81) = 3.30 m/s 2 aB = −

1 ( 3.30 ) = −1.101 m/s2 3

a A = 3.30 m/s 2

30° W

a B = 1.101 m/s 2

30° W

(b) Using equation (1),  a  T = mA g  sin 30° − A  = (10 )( 9.81)( sin 30° − 0.33673) g   T = 16.02 N W

Vector Mechanics for 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.

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Chapter 12, Solution 13.

Let the positive directions of x A and xB be down the incline. Constraint of the cable:

x A + 3xB = constant 1 aB = − a A 3

a A + 3aB = 0

ΣFy = 0: N A − mA g cos 30° = 0

Block A:

ΣFx = ma : mA g sin 30° − µ N A − T = m Aa A Eliminate N A. mA g ( sin 30° − µ cos 30° ) − T = mAa A ΣFy = 0: N B − mB g cos 30° = 0

Block B:

ΣF = ma : mB g sin 30° + µ N B − 3T = mB aB = −

mB a A 3

Eliminate N B . mB g ( sin 30° + µ cos30° ) − 3T = −

mB a A 3

Eliminate T.

( 3mA g − mB g ) sin 30° − µ ( 3mA g + mB g ) cos 30° =  3mA + 

mB   aA 3 

Check the value of µ s required for static equilibrium. Set a A = 0 and solve for µ.

µ =

( 3mA − mB ) sin 30° ( 3mA + mB ) cos 30°

=

( 75 − 20 ) tan 30° = 0.334. ( 75 + 20 )

Since µ s = 0.25 < 0.334, sliding occurs. Calculate

aA for sliding. Use µ = µk = 0.20. g continued

Vector Mechanics for 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.

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( 3mA − mB ) sin 30° − µ ( 3mA + mB ) cos 30° aA = g 3mA + mB / 3 =

( 30 − 8) sin 30° − ( 0.20 )( 30 + 8) cos 30° 30 + 2.667

(a) a A = ( 0.13525 )( 9.81) = 1.327 m/s 2

1 aB = −   (1.327 ) = − 0.442 m/s 2 3

= 0.13525

a A = 1.327 m/s 2

30° W

a B = 0.442 m/s 2

30° W

(b) T = mA g ( sin 30° − µ cos 30° ) − mAa A = (10 )( 9.81)( sin 30° − 0.20cos 30° ) − (10 )(1.327 ) T = 18.79 N W

Vector Mechanics for 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.

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Chapter 12, Solution 14.

Data:

mA =

55000 lb = 1708.1 lb ⋅ s 2 / ft 32.2 ft/s 2

mB =

44000 lb = 1366.5 lb ⋅ s 2 / ft 32.2 ft/s 2

v0 = − 55 mi/h = − 80.667 ft/s (a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same acceleration.

∑ Fx = ∑ max : − Fb − Fb = mAax + mB ax ax =

Fb + Fb 7000 + 7000 = = 4.5534 m/s 2 mA + mB 1708.1 + 1366.5

ax = v

dv dx

xf 0 ∫ 0 ax dx = ∫ v0 v dv

ax x f =

v02 2

( −80.667 ) = − 751 ft v2 xf = − 0 = − 2ax ( 2 )( 4.5534 ) 2

715 ft to the left W (b) Use car A as free body. Fc = coupling force.

∑ Fx = ∑ max : Fc − Fb = mAax Fc = mAax − Fb = (1708.1)( 4.5534 ) + 7000 = 778 lb Fc = 778 lb tension W

Vector Mechanics for 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.

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Chapter 12, Solution 15.

mA =

55000 lb = 1708.1 lb ⋅ s 2 / ft 32.2 ft/s 2

mB =

44000 lb = 1366.5 lb ⋅ s 2 / ft 32.2 ft/s 2

Data:

v0 = − 55 mi/h = − 80.667 ft/s (a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same acceleration.

∑ Fx = ∑ max : − Fb − Fb = mAax + mB ax ax =

Fb 7000 = = 2.2767 m/s 2 mA + mB 1708.1 + 1366.5

ax = v

∫

xf 0

dv dx 0

ax dx = ∫ v v dv 0

ax x f =

v02 2

( −80.667 ) = 1429 ft v02 =− 2ax ( 2 )( 2.2767 ) 2

xf = −

1429 ft to the left W (b) Use car B as a free body. Fc = coupling force.

∑ Fx = ∑ max : − Fc = mB ax − Fc = (1366.5)( 2.2767 ) = 3110 lb Fc = 3110 lb. compression W

Vector Mechanics for 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.

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Chapter 12, Solution 16.

Constraint of cable:

2 x A + ( xB − x A ) = x A + xB = constant.

a A + aB = 0,

or

aB = −a A

Assume that block A moves down and block B moves up. Block B:

ΣFy = 0: N AB − WB cosθ = 0 ΣFx = ma : − T + µ N AB + WB sin θ =

WB aB g

Eliminate N AB and aB . −T + WB ( sin θ + µ cosθ ) = WB

Block A:

aB a = −WB A g g

ΣFy = 0: N A − N AB − WA cosθ = 0 N A = N AB + WA cosθ = (WB + WA ) cosθ ΣFx = m Aa A : − T + WA sin θ − FAB − FA = −WB ( sin θ + µ cosθ ) − WB

aA + WA sin θ − µWB cosθ g

− µ (WB + WA ) cosθ = WA

(WA − WB ) sinθ

WA aA g

aA g

− µ (WA + 3WB ) cosθ = (WA + WB )

aA g

Check the condition of impending motion.

µ = µs = 0.20,

a A = aB = 0,

θ = θs

(WA − WB ) sin θ s − 0.20 (WA + 3WB ) cosθ s

=0

continued

Vector Mechanics for 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.

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tan θ s =

0.20 (WA + 3WB ) ( 0.20 )(128 ) = = 0.40 64 WA − WB

θ s = 21.8° < θ = 25°. The blocks move. Calculate

aA using µ = µ k = 0.15 and θ = 25°. g

(WA − WB ) sin θ − µk (WA + 3WB ) cosθ aA = g WA + WB =

64sin 25° − ( 0.15 )(128 ) cos 25° 96

= 0.10048

a A = ( 0.10048 )( 32.2 ) = 3.24 ft/s 2

(a) aB = −3.24 ft/s 2 (b)

T = WB ( sin θ + µ cosθ ) + WB

a B = 3.24 ft/s 2

25° !

aA g

= 16 ( sin 25° + 0.15cos 25° ) + (16 )( 0.10048 ) T = 10.54 lb !

Vector Mechanics for 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.

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Chapter 12, Solution 17.

Constraint of cable:

2 x A + ( xB − x A ) = x A + xB = constant. a A + aB = 0,

or

aB = −a A

Assume that block A moves down and block B moves up. Block B:

ΣFy = 0: N AB − WB cosθ = 0 ΣFx = max : − T + µ N AB + WB sin θ =

WB aB g

Eliminate N AB and aB . −T + WB ( sin θ + µ cosθ ) = WB

Block A:

aB a = −WB A g g

ΣFy = 0: N A − N AB − WA cosθ + P sin θ = 0 N A = N AB + WA cosθ − P sin θ = (WB + WA ) cosθ − P sin θ

ΣFx = mAa A : − T + WA sin θ − FAB − FA + P cosθ = −WB ( sin θ + µ cosθ ) − WB

aA + WA sin θ − µWB cosθ g

− µ (WB + WA ) cosθ + µ P sin θ + P cosθ = WA

(WA − WB ) sin θ

WA aA g

aA g

− µ (WA + 3WB ) cosθ + P ( µ sin θ + cosθ ) = (WA + WB )

Check the condition of impending motion.

µ = µ s = 0.20, a A = aB = 0, θ = 25°

(WA

− WB ) sin θ − µ s (WA + 3WB ) cosθ + Ps ( µ s sin θ + cosθ ) = 0 continued

Vector Mechanics for 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.

aA g

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Ps =

=

µ s (WA + 3WB ) cosθ − (WA − WB ) sin θ µ s sin θ + cosθ

( 0.20 )(128) cos 25° − 64 sin 25° 0.20 sin 25° + cos 25°

= −3.88 lb < 10 lb

Blocks will move with P = 10 lb. Calculate

aA using µ = µ k = 0.15, θ = 25°, and P = 10 lb. g

(WA − WB ) sin θ − µk (WA + 3WB ) cosθ + P ( µk sinθ + cosθ ) aA = g WA + WB =

64 sin 25° − ( 0.15 )(128 ) cos 25° + (10 )( 0.15sin 25° + cos 25° ) 96

= 0.20149

a A = ( 0.20149 )( 32.2 ) = 6.49 ft/s 2

(a) aB = −6.49 ft/s 2 , (b)

T = WB ( sin θ + µ cosθ ) + WB

a B = 6.49 ft/s 2

25° !

aA g

= 16 ( sin 25° + 0.15cos 25° ) + (16 )( 0.20149 ) T = 12.16 lb. !

Vector Mechanics for 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.

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Chapter 12, Solution 18.

Assume a B > a A so that the boxes separate. Boxes are slipping.

µ = µk ΣFy = 0: N − mg cos15° = 0 N = mg cos15° ΣFx = ma : µ k N − mg sin15° = ma

µ k mg cos15° − mg sin15° = ma a = g ( µ k cos15° − sin15° ) ,

independent of m.

For box A, µ k = 0.30 a A = 9.81( 0.30cos15° − sin15° )

or a A = 0.304 m/s 2

15° W

For box B, µ k = 0.32 aB = 9.81( 0.32cos15° − sin15° )

Vector Mechanics for 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.

or a B = 0.493 m/s 2

15° W

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Chapter 12, Solution 19.

Let y be positive downward position for all blocks. Constraint of cable attached to mass A: y A + 3 yB = constant a A + 3aB = 0 Constraint of cable attached to mass C:

a A = −3aB

yC + yB = constant

aC + aB = 0

For each block

or or

aC = − aB

ΣF = ma :

Block A:

WA − TA = mAa A ,

or TA = WA − mAa A = WA − 3m AaB

Block C:

WC − TC = mC aC ,

or TC = WC − mC aC = WC − mC aB

Block B: WB − 3TA − TC = mB aB WB − 3 (WA − 3mAaB ) − (WC − mC aB ) = mB aB or (a) Accelerations.

(b) Tensions.

aB W − 3WA − WC 60 − 60 − 20 = B = = − 0.076923 g WB + 9WA + WC 60 + 180 + 20

aB = ( − 0.076923)( 32.2 ) = − 2.477 ft/s 2

a B = 2.48 ft/s 2 W

a A = − ( 3)( − 2.477 ) = 7.431 ft/s 2

a A = 7.43 ft/s 2 W

aC = − ( − 2.477 ) = 2.477 ft/s 2

aC = 2.48 ft/s 2 W

20 ( 7.43) 32.2 20 TC = 20 − ( 2.477 ) 32.2 TA = 20 −

Vector Mechanics for 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.

TA = 15.38 lb W TC = 18.46 lb W

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Chapter 12, Solution 20.

Let y be positive downward for both blocks. Constraint of cable: y A + yB = constant a A + aB = 0

For blocks A and B,

aA =

aB = −a A

ΣF = ma :

Block A: WA − T =

Solving for a A ,

or

WA aA g

or

T = WA −

Block B: P + WB − T =

WB W aB = − B a A g g

P + WB − WA +

WA W a A = − B aA g g

WA aA g

WA − WB − P g WA + WB

(1)

2 v A2 − ( v A )0 = 2a A  y A − ( y A )0 

with

( v A )0 = 0

v A = 2a A  y A − ( y A )0  v A − ( v A )0 = a At t=

with

(2)

( v A )0 = 0

vA aA

(3) continued

Vector Mechanics for 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.

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(a) Acceleration of block A. System (1): WA = 100 lb,

( aA )1

By formula (1),

=

WB = 50 lb, P = 0

100 − 50 ( 32.2 ) 100 + 50

( a A )1 = 10.73 ft/s2

!

System (2): WA = 100 lb, WB = 0, P = 50 lb

( a A )2

By formula (1),

=

100 − 50 ( 32.2 ) 100

( a A )2

= 16.10 ft/s 2 !

( a A )3

= 0.749 ft/s 2 !

System (3): WA = 1100 lb, WB = 1050 lb, P = 0 By formula (1),

( a A )3

=

1100 − 1050 ( 32.2 ) 1100 + 1050

(b) v A at y A − ( y A )0 = 5 ft. Use formula (2). System (1):

( v A )1

=

( 2 )(10.73)( 5)

( v A )1

= 10.36 ft/s !

System (2):

( v A )2

=

( 2 )(16.10 )( 5)

( v A )2

= 12.69 ft/s !

System (3):

( v A )3

=

( 2 )( 0.749 )( 5)

( v A )3

= 2.74 ft/s !

(c) Time at v A = 10 ft/s. Use formula (3). System (1): t1 =

10 10.73

t1 = 0.932 s !

System (2): t2 =

10 16.10

t2 = 0.621 s !

System (3): t3 =

10 0.749

t3 = 13.35 s !

Vector Mechanics for 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.

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Chapter 12, Solution 21.

(a) Maximum acceleration. The cable secures the upper beam; only the lower beam can move.

For the upper beam, ΣFy = 0: N1 − W = 0 N1 = W = mg

For the lower beam, ΣFy = 0: N 2 − N1 − W = 0

or

N 2 = 2W

ΣFx = ma : 0.25 N1 + 0.30 N 2 = ( 0.25 + 0.60 )W = ma

a = 0.85

W = ( 0.85 )( 32.2 ) = 23.37 ft/s 2 m

For the upper beam,

a = 27.4 ft/s 2

!

ΣFx = ma : T − 0.25 N1 = ma

 3000  T = 0.25W + ma = ( 0.25 )( 3000 ) +   ( 23.37 ) = 2927 lb  32.2 

T = 2930 lb !

(b) Maximum deceleration of trailer. Case 1: Assume that only the top beam slips. As in Part (a) N1 = mg.

ΣF = ma : 0.25W = ma a = 0.25g = 8.05 ft/s 2 continued

Vector Mechanics for 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.

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Case 2: Assume that both beams slip. As before N 2 = 2W .

ΣF = ( 2m ) a :

( 0.30 )( 2W ) = ( 2m ) a

a = 0.30 g = 9.66 ft/s 2

The smaller deceleration value governs.

Vector Mechanics for 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 = 8.05 ft/s 2 !

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Chapter 12, Solution 22.

Since both blocks move together, they have a common acceleration. Use blocks A and B together as a free body. ΣF = Σ ma : P − mA g sin 30° − mB g sin 30° = ( mA + mB ) a a=

P 500 − g sin 30° = − 9.81 sin 30° mA + mB 50

= 5.095 m/s 2 Use block B as a free body. ΣF = mB a cos 30°:

F f = mB a cos30°

F f = (10 )( 5.095 ) cos 30° = 44.124 N ΣF = mB a sin 30°:

N B − mB g = mB a sin 30°

N B = mB ( g + a sin 30° ) = 10 ( 9.81 + 5.095 sin 30° ) = 123.575 N Minimum coefficient of static friction:

µ min =

Ff NB

=

44.124 123.575

Vector Mechanics for 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.

µ min = 0.357 W

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Chapter 12, Solution 23.

(a) Kinematics of the belt. vo = 0 a1 = 3.2 m/s 2

1. Acceleration phase with

v1 = vo + a1t1 = 0 + ( 3.2 )(1.5 ) = 4.8 m/s x1 = xo + vot1 + 2.

1 2 1 2 a1t1 = 0 + 0 + ( 3.2 )(1.5 ) = 3.6 m 2 2

Deceleration phase. v2 = 0 since the belt stops. v22 − v12 = 2a2 ( x2 − x1 ) 0 − ( 4.8) v22 − v12 = = −11.52 2 ( x2 − x1 ) 2 ( 4.6 − 3.6 ) 2

a2 =

t2 − t1 =

a 2 = 11.52 m/s 2

v2 − v1 0 − 4.8 = = 0.41667 s −11.52 a2

(b) Motion of the package. 1. Acceleration phase. Assume no slip.

( a p )1 = 3.2 m/s2

ΣFy = 0: N − W = 0 or N = W = mg

( )1

ΣFx = ma : F f = m a p The required friction force is F f .

The available friction force is µ s N = 0.35 W = 0.35 mg Ff m

( )1 < µms N

= ap

= µ s g = ( 0.35 )( 9.81) = 3.43 m/s 2

Vector Mechanics for 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.

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Since 3.2 m/s 2 < 3.43 m/s 2 , the package does not slip.

( v p )1 = v1 = 4.8 m/s

( x p )1 = 3.6 m.

and

2. Deceleration phase. Assume no slip.

( a p )2 = −11.52 m/s2

( )2

ΣFx = ma : − F f = m a p Ff m

µs N m

=

µ s mg m

( )2 = −11.52 m/s2

= ap

= µ s g = 3.43 m/s 2 < 11.52 m/s 2

Since the available friction force µ s N is less than the required friction force F f for no slip, the package does slip.

( a p )2 < 11.52 m/s2 ,

F f = µk N

( )2

( )2

ΣFx = m a p : − µk N = m a p

( a p )2 = − µmk N

= − µk g = − ( 0.25 )( 9.81) = −2.4525 m/s 2

( v p )2 = ( v p )1 + ( a p )2 ( t2 − t1 ) = 4.8 + ( −2.4525)( 0.41667 ) = 3.78 m/s2 ( x p )2 = ( x p )1 + ( v p )1 ( t2 − t1 )2 + 12 ( a p )2 ( t2 − t1 )2 = 3.6 + ( 4.8 )( 0.41667 ) +

1 ( −2.4525)( 0.41667 )2 = 5.387 m 2

Position of package relative to the belt

( x p )2 − x2 = 5.387 − 4.6 = 0.787

Vector Mechanics for 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.

x p/belt = 0.787 m

W

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Chapter 12, Solution 24.

Acceleration a1 : Impending slip.

F1 = µ s N1 = 0.30 N1

ΣFy = mAa y : N1 − WA = mAa1 sin 65° N1 = WA + mAa1 sin 65° = mA ( g + a1 sin 65° ) ΣFx = mAax : F1 = mAa1 cos 65° F1 = µ s N or mAa1 cos 65° = 0.30mA ( g + a1 sin 65° ) a1 =

0.30 g = (1.990 )( 9.81) = 19.53 m/s 2 cos 65° − 0.30sin 65°

a1 = 19.53 m/s 2

Deceleration a2 : Impending slip.

65° W

F2 = µ S N 2 = 0.30 N 2

ΣFy = ma y : N1 − WA = −m Aa2 sin 65° N1 = WA − mAa2 sin 65° ΣFx = max : F2 = mAa2 cos 65° F2 = µ S N 2 a2 =

or

mAa2 cos 65° = 0.30mA ( g − a2 cos 65° )

0.30 g = ( 0.432 )( 9.81) = 4.24 m/s 2 cos 65° + 0.30sin 65°

a2 = 4.24 m/s 2

Vector Mechanics for 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.

65° W

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Chapter 12, Solution 25.

Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P / T be the acceleration of the plywood relative to the truck. (a) Find the value of aT so that the relative motion of the plywood with respect to the truck is impending. aP = aT and F1 = µ s N1 = 0.40 N1 ΣFy = mP a y : N1 − WP cos 20° = −mP aT sin 20° N1 = mP ( g cos 20° − aT sin 20° ) ΣFx = max : F1 − WP sin 20° = mP aT cos 20° F1 = mP ( g sin 20° + aT cos 20° ) mP ( g sin 20° + aT cos 20° ) = 0.40 mP ( g cos 20° − aT sin 20° ) aT =

( 0.40cos 20° − sin 20° ) g cos 20° + 0.40sin 20°

= ( 0.03145 )( 9.81) = 0.309

aT = 0.309 m/s 2 (b) xP / T = ( xP / T )o + ( vP / T ) t + aP / T =

sW

1 1 aP / T t 2 = 0 + 0 + aP / T t 2 2 2

( 2 )(1) = 12.5 m/s2 , 2 xP / T = 2 t ( 0.4 )2

a P / T = 12.5 m/s 2

a P = aT + a P / T = ( aT → ) + (12.5 m/s 2

20° )

Fy = mP a y : N 2 − WP cos 20° = −mP aT sin 20° N 2 = mP ( g cos 20° − aT sin 20° ) continued

Vector Mechanics for 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.

20°

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ΣFx = Σmax : F2 − WP sin 20° = mP aT cos 20° − mP aP / T F2 = mP ( g sin 20° + aT cos 20° − aP / T ) For sliding with friction

F2 = µk N 2 = 0.30 N 2

mP ( g sin 20° + aT cos 20° − aP / T ) = 0.30mP ( g cos 20° + aT sin 20° ) aT =

( 0.30 cos 20° − sin 20° ) g

+ aP / T cos 20° + 0.30sin 20°

= ( −0.05767 )( 9.81) + ( 0.9594 )(12.5 ) = 11.43

aT = 11.43 m/s 2

Vector Mechanics for 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.

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Chapter 12, Solution 26.

At maximum speed a = 0.

F0 = kv02 = 0

k =

F0 v02

When the propellers are reversed, F0 is reversed.

ΣFx = ma : − F0 − kv 2 = ma − F0 − F0

v2 = ma v02 dx =

∫ x = −

mv02 F0

(

x dx 0

1 ln v02 + v 2 2

0 v0

(

F0 2 v + v2 2 0 mv0

vdv mv02vdv = a F0 v02 + v 2

(

= −

)

a−

)

)

mv02 0 vdv ∫ F0 v0 v02 + v 2

= −

mv02  2 mv02 ln v0 − ln 2v02  = ln 2  2 F0  2F0

( )

x = 0.347

Vector Mechanics for 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.

m0v02 W F0

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Chapter 12, Solution 27.

ΣF = ma : P − kv = ma dv P − kv =a= dt m m dv

m

m

t v ∫ 0 dt = ∫ 0 P − kv = − k ln ( P − kv ) 0 = − k ln ( P − kv ) − ln P 

t = −

v

m P − kv ln k P

or

P − kv = e −kt / m m t

x = ∫ 0 v dt =

= x =

(

Pt k

or t

− 0

ln v =

P − kv kt = − m m

(

P 1 − e− kt / m k

P  k − kt / m  − e  k m

)

(

t 0

Pt P − kt / m Pt P e + −1 = − 1 − e− kt / m k m k m Pt kv − , which is linear. k m

Vector Mechanics for 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.

)

) W

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Chapter 12, Solution 28.

Let y be the position coordinate of the projectile measured upward from the ground. The velocity and acceleration a taken to positive upward. D = kv 2. (a) Upward motion.

ΣFy = ma : − D − mg = ma

 D kv 2   a = −  g +  = −  g +  m m     dv kv 2  k  2 mg  v = −  g +  = −  v +  dy m  m k   v dv k = − dy mg m v + k 2

∫

0

v dv k =− mg v0 2 m v + k

∫

h 0

dy

0

1  2 mg  kh ln  v +  = − k v m 2  0

mg  1 1  kv 2 kh k ln = − ln  0 + 1 = − 2 v 2 + mg 2  mg m  0 k  ( 0.0024 )( 90 )2   m  kv02 4 + 1 = + 1 ln  ln  2k  mg ( 2 )( 0.0024 )  ( 4 )( 9.81)   = 335.36 m h = 335 m W

h =

continued

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ΣF = ma :

(b) Downward motion.

D − mg = ma D kv 2 −g = −g m m dv kv 2 k  mg  v = −g = −  − v2  dy m m k  a =

v dv k = − dy mg m − v2 k

∫

vf 0

v dv k =− mg m

∫

0

h

dy

vf

1  mg kh  ln  − v2  = 2  k m 0

 mg − v 2f 1  k − ln  2  mg  k 

  kh  = m  

 kv 2f  2 kh  = − ln 1 −  mg  m 

1− vf = ± vf =

kv 2f mg

= e− 2 kh/m

(

mg 1 − e−2 kh/m k

)

( 4 )( 9.81) 1 − e−( 2)( 0.0024)(335.36)/4  0.0024 

= 73.6 m/s

Vector Mechanics for 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.



v f = 73.6 m/s W

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Chapter 12, Solution 29. Choose the origin at point C, and let x be positive to the right. Then x is a position coordinate of the slider B, and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle L =

A2 + x 2

The elongation of the spring is e = L − A, and the magnitude of the force exerted by the spring is Fs = ke = k

A2 + x2 − A

)

x

cosθ =

By geometry,

(

A2 + x2

ΣFx = max : − Fs cosθ = ma −k

(

A2 + x2 − A

a = −

)

k x − m 

x

A + x2 2

= ma

  A 2 + x 2 

Ax

v 0 ∫ 0 v dv = ∫ x0 a dx 0  k 1 2 2 2  dx = −  x − A A + x  m2  x0 A 2 + x 2 

v 1 2 k 0 = − ∫x x − v m 0  2 0

Ax

1 2 k 1  v = −  0 − A 2 − x02 + A A 2 + x02  m 2 2  k v2 = 2A 2 + x02 − 2A A 2 + x02 m k  2 A + x02 − 2A A 2 + x02 + A 2  =  m 

(

)

(

)

answer: v =

Vector Mechanics for 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.

k m

(

)

A 2 + x02 − A W

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Chapter 12, Solution 30. Let yA, yB, and yC be the position coordinates of blocks A, B, and C respectively measured downward from the upper support. Then the corresponding velocities and accelerations are positive downward. Constraint of cable:

y A − yB + y A + 2 yB + yC = constant

Differentiating twice, a A − aB + a A + 2aB + aC = 0 2a A + aB + aC = 0

(1)

Draw free body diagrams of each of the blocks.

Block A.

ΣF = ma : m A g − 2T = m Aa A

aA = g −

(2)

ΣF = ma : mB g − T = mB aB

Block B.

aB = g −

Block C.

2T mA

T mB

(3)

T mC

(4)

ΣF = ma : mC g − T = mC aC aC = g −

Substitute (2), (3) and (4) into (1).  2T   T   T  2 g − =0  + g −  + g − mA   mB   mC    4 1 1 4g −  + +  mA mB mC 4 1 1 ( 4 )( 9.81) −  + +  T = 0  10 10 10 

T = 65.4 N

( 2 )( 65.4 ) = − 3.27 m/s2

Substitute into (2),

a A = 9.81 −

(a) Change in position.

y A − ( y A )0 = y A − ( y A )0 =

 T = 0 

10

1 a At 2 2

1 ( − 3.27 )( 0.5)2 = − 0.409 m 2 ∆y = 0.409 m !

(b) Tension in the cable.

Vector Mechanics for 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.

T = 65.4 N !

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Chapter 12, Solution 31.

WA = m A g = 98.1 N

WB = mB g = 49.05 N

Assume that block B slides downward relative to block A. Then the friction force F1 is directed as shown. Its magnitude is

F1 = µk N1 = 0.10 N1. ΣFy = 0: N1 − WB cos30° = 0, N1 = WB cos30° = 49.05cos30° = 42.48 N.

F1 = ( 0.10 )( 42.48 ) = 4.248 N. ΣFx = mB aB : WB sin 30° − F1 = mB aB

aB =

1 1 (WB sin 30° − F1 ) = ( 49.05sin 30° − 4.248) = 4.055 m/s2 5 mB

Assume that block A slides downward relative to the fixed plane. The friction force F2 is directed as shown. Its magnitude is F2 = µk N 2 = 0.20 N 2. Fy = 0: N 2 − N1 − WA cos30° = 0, N 2 = 42.48 + 98.1cos30° = 127.44 N. F2 = ( 0.20 )(127.44 ) = 25.49 N ΣFx = mAa A : WA sin 30° − F2 + F1 = mAa A

aA = =

1 (WA sin 30° − F2 + F1 ) mA 1 ( 98.1sin 30° − 25.49 + 4.248) = 2.781 m/s2 10

aB / A = aB − a A = 4.055 − 2.781 = 1.274 m/s 2 Since both aB / A and a A are positive, the directions of relative motion are as assumed above. (a) Acceleration of block A.

a A = 2.78 m/s 2

30° W

v B / A = 0.637 m/s

30° W

(b) Velocity of B relative to A at t = 0.5 s. vB / A = aB / At = (1.274 )( 0.5 )

Vector Mechanics for 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.

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Chapter 12, Solution 32.

Let the positive direction for position coordinates, velocities, and accelerations be to the right. Let the origin lie at the fixed anchor. Constraint of cable:

3 ( xC − x A ) + ( xC − xB ) + ( − xB ) = constant 4aC − 2aB − 3a A = 0

(1)

ΣFx = max : Block A:

3T = m Aa A

or

aA =

3T 3T = g mA 20

(2)

Block B:

2T = mB aB

or

aB =

2T 2T = g mB 10

(3)

Block C: P − 4T = mC aC

or

aC =

P − 4T P − 4T = g 20 20

(4)

Substituting (2), (3), and (4) into (1),

 P − 4T   2T   3T  4  − 2  −3   = 0  20   10   20  4 9  4P  16 + +  T = 20  20 10 20  T = ( 0.12121) P = ( 0.12121)( 50 ) = 6.0605 lb (a) From (2), a A = From (3), aB =

( 3)( 6.0605)( 32.2 ) 20

( 2 )( 6.0605)( 32.2 ) 10

= 29.3 ft/s 2

a A = 29.3 ft/s 2

W

= 39.0 ft/s 2

a B = 39.0 ft/s 2

W

aC = 41.5 m/s 2

W

50 − ( 4 )( 6.0605 )  ( 32.2 ) = 41.5 ft/s 2 From (4), aC =  20

(b) As determined above,

Vector Mechanics for 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.

T = 6.06 lb W

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Chapter 12, Solution 33.

mA = mC =

20 10 = 0.62112 lb ⋅ s 2 / ft, mB = = 0.31056 lb ⋅ s 2 / ft 32.2 32.2

Let the positive direction for position coordinates, velocities, and accelerations be to the right. Let the origin lie at the fixed anchor. Constraint of cable: 3 ( xC − x A ) + ( xC − xB ) + ( − xB ) = constant 4aC − 2aB − 3a A = 0

(1)

Block A: ΣFy = 0: N A − WA = 0 N A = WA , FA = µ k N A = µκ WA ΣFx = mAax : 3T − FA = mAa A aA = Block B: N B = WB ,

3T − µkWA 3T = − 0.20 g mA mA

(2)

FB = µkWB ΣFx = mB aB : 2T − FB = mB aB aB =

2T − µkWB 2T = − 0.20 g mB mB

(3)

Block C: NC = WC , FC = µ kWC ΣFx = mC a A : P − 4T − FC = mC aC Kinematics: xC = ( xC )0 + ( vC )0 + aC =

2  xC − ( xC )0  t

2

(4)

1 1 aC t 2 = 0 + aC t 2 2 2

=

( 2 )( 2.4 ) ( 0.4 )2

Vector Mechanics for 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.

= 30 ft/s 2

(5)

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Substitute (2), (3) and (5) into (1).

( 4)(30) − ( 2) 

2T   3T  − 0.20 g  − ( 3)  − 0.20 g    0.62112  0.31056

 2T   3T  = ( 4)( 30) − ( 2)  − ( 0.2)( 32.2)  − ( 3)  − ( 0.2)( 32.2)   0.31056   0.62112  = 120 − 27.37 T + 32.2 = 0 T = 5.5608 lb P = 4T + FC + mC aC

From (4),

= ( 4 )( 5.5608 ) + ( 0.20 )( 20 ) + ( 0.62112 )( 30 ) = 44.877 lb

From (2),

aA =

From (3),

aB = (a)

( 3)( 5.5608) 0.62112

( 2 )( 5.5608) 0.31056

− ( 0.20 )( 32.2 ) = 20.42 ft/s 2 − ( 0.20 )( 32.2 ) = 29.37 ft/s 2

Acceleration vectors.

a A = 20.4 ft/s 2

W

a B = 29.4 ft/s 2

W

aC = 30 ft/s 2

W

Since a A , aB , and aC are to the right, the friction forces FA , FB , and FC are to the left as assumed. (b)

Tension in the cable.

(c)

Force P.

Vector Mechanics for 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.

T = 5.56 lb W P = 44.9 lb

W

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Chapter 12, Solution 34.

Let the positive direction of x and y be those shown in the sketch, and let the origin lie at the cable anchor. Constraint of cable: x A + yB / A = constant or a A + aB / A = 0, where the positive directions of a A and aB / A are respectively the x and the y directions. Then

aB / A = −a A

First note that a B = a A + a B / A = ( a A Block B:

(1) 20° ) + ( aB / A

20° )

ΣFx = mB ( aB ) x : mB g sin 20° − N AB = mB a A mB a A + N AB = mB g sin 20° 15 a A + N AB = 50.328

(2)

ΣFy = mB ( aB ) y : mB g cos 20° − T = mB aB / A mB aB / A + T = mB g cos 20° 15 aB / A + T = 138.276

Block A:

(3)

ΣFx = m A a A : m A g sin 20° + N AB + T = m A a A m Aa A − N AB + T = m A g sin 20° 25 a A − N AB + T = 83.880

(4)

Eliminate aB / A using Eq. (1), then add Eq. (4) to Eq. (2) and subtract Eq. (3). 55 a A = −4.068

or

a A = −0.0740 m/s 2 , a A = 0.0740 m/s 2

20° !

From Eq. (1), aB / A = 0.0740 m/s 2 From Eq. (3), T = 137.2 N

Vector Mechanics for 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.

T = 137.2 N !

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Chapter 12, Solution 35.

Motion of B relative to A. Particle B is constrained to move on a circular path with its center at point A. ( a B / A )t is the component of a B / A lying along the circle, say to the left in the diagram and ( a B / A )n is directed

toward point A. Initially, a B / A = 0, since the system starts from rest. (a) a B = a A + a B / A = ( a A

25° ) + ( aB / A

)

ΣFx = Σmax : 0 = mB aB / A − mB a A cos 25°

Crate B:

aB / A = a A cos 25° = 0.4cos 25° = 0.363

a B / A = 0.363 m/s 2

W

Fy = mB ( aB ) y : TAB − mB g = mB a A sin 25° TAB = mB ( g + a A sin 25° ) = 250 ( 9.81 + 0.4sin 25° ) = 2495 N (b) Trolley A:

ΣF = mAa A : TCD − (TAB + mA g ) sin 25° = mAa A

TCD = (TAB + mA g ) sin 25° + mAa A =  2495 + ( 20 )( 9.81)  sin 25° + ( 20 )( 0.4 )

TCD = 1145 N W

Vector Mechanics for 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.

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Chapter 12, Solution 36.

The ball moves at constant speed on a circle of radius

ρ = L sin θ a=

Acceleration (toward center of circle).

+ ΣFy = ma y : ΣFy = max :

v2

ρ

T cosθ − W = 0

T =

W cosθ

T sin θ = ma W mv 2 mv 2 sin θ = = cosθ L sin θ ρ

(1.5) = 0.38226 mv 2 v2 tan θ sin θ = = = WL gL ( 9.81)( 0.6 ) 2

(a)

θ = 34.21° (b)

T =

W mg = cosθ cosθ

θ = 34.2° W =

( 2 )( 9.81) cos 34.21°

Vector Mechanics for 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.

T = 23.7 N W

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Chapter 12, Solution 37.

Let ρ be the radius of the horizontal circle. The length of the wire is L =

ρ sin θ1

+

ρ sin θ 2

.

Solving for ρ ,

ρ =

L sin θ1 sin θ 2 sin θ1 + sin θ 2

ΣFy = 0: T cosθ1 + T cosθ 2 − mg = 0 T =

mg cosθ1 + cosθ 2

ΣFx = max : T sin θ1 + T sin θ 2 = man =

mv 2

ρ

mg ( sin θ1 + sin θ 2 ) mv 2 ( sin θ1 + sin θ 2 ) = cosθ1 + cosθ 2 L sin θ1 sin θ 2 v 2 = Lg

sin θ1 sin θ 2 sin 60° sin 30° = ( 2 )( 9.81) = 6.2193 m 2 /s 2 cosθ1 + cosθ 2 cos 60° + cos30°

v = 2.49 m/s W

Vector Mechanics for 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.

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Chapter 12, Solution 38.

θ 3 = θ1 − θ 2 = 50° − 25° = 25°

A1

sin θ 2

=

d sin θ 3

or

ρ = A1 sin θ1 = =

d sin θ 2 sin θ 3

A1 =

d sin θ 2 sin θ1 sin θ 3

( 4 )( sin 25° )( sin 50° ) sin 25°

= 3.0642 ft

ΣFy = 0: TAC cosθ 2 + TBC cosθ1 − W = 0

(1)

Wv 2 gρ

(2)

ΣFx = max : TAC sin θ 2 + TBC sin θ1 =

Case 1: TBC = 0.

TAC cosθ 2 − W = 0

or TAC =

TAC sin θ 2 = W tan θ 2 =

W cosθ 2

Wv 2 gρ

v 2 = g ρ tan θ 2 = ( 32.2 )( 3.0642 ) tan 25° = 46.01 ft 2 /s 2

v = 6.78 ft/s Case 2: TAC = 0. TBC cosθ1 − W = 0

or

TBC sin θ1 = W tan θ1 =

TBC =

W cosθ1

Wv 2 gρ

v 2 = g ρ tan θ1 = ( 32.2 )( 3.0642 ) tan 50° = 117.59 ft 2 /s 2 v = 10.84 ft/s

Vector Mechanics for 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.

6.78 ft/s ≤ v ≤ 10.84 ft/s W

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Chapter 12, Solution 39.

(a)

ΣFy = 0: T sin θ − W = 0 T =

(b)

W 16 = sin θ sin 60°

ΣFx = man : T cosθ = m

v2 =

=

ρT cosθ m

=

or T = 18.48 lb W

v2

ρ

ρW cosθ m sin θ

( 3)( 32.2 ) = 55.77 ft 2/s2 ρg = tan θ tan 60° v = 7.47 ft/s W

Vector Mechanics for 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.

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Chapter 12, Solution 40.

y =

r 2 dy , = r = 1 = tan θ 2 dx

or

θ = 45°

ΣFy = 0: N cosθ − mg = 0 N =

mg cosθ

ΣFx = max : N sin θ = man mg tan θ = m

v2 r

v 2 = gr tan θ (a) v 2 = ( 9.81)(1)(1.0000 ) = 9.81 m 2 /s 2 (b) N =

(1)( 9.81) mg = cos 45° cos 45°

Vector Mechanics for 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.

v = 3.13 m/s W

N = 13.87 N W

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Chapter 12, Solution 41.

Geometry

A OC 2 = A OB 2 + A BC 2 − 2A OBA OC cos 30° = ( 0.3) + ( 0.6 ) − ( 2 )( 0.3)( 0.6 ) cos 30° = 0.13823 m 2 2

2

A OC = 0.37179 m A OC

sin 30° sin β =

=

A OB

sin β

A OB sin 30° ( 0.3) sin 30° = = 0.40345 A OC 0.37179

β = 23.79° Acceleration components: at = 1.3 m/s 2 an =

v2

ρ

=

vC2

A BC

=

(1)2 0.6

= 1.667 m/s 2

Mass m = 1 kg ΣFt = mat : N cos β = (1)(1.3) = 1.3 N =

1.3 = 1.421 N cos 23.79°

ΣFn = man : T − N sin β = (1)(1.667 ) = 1.667 (a) T = 1.667 + 1.421 sin 23.79° (b) Force exerted by rod on collar is 1.421 N Force exerted by collar on rod:

Vector Mechanics for 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.

T = 2.24 N W

( 30° + β )

= 53.8°

1.421 N

53.8° W

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Chapter 12, Solution 42. W = mg = ( 0.5 )( 9.81) = 4.905 N

ρ = 150 mm = 0.150 m ΣFy = 0 : TDA cos 20° − TDE cos 30° − W = 0 0.93969 TDA − 0.86603 TDE = 4.905 TDA = 0.92160 TDE + 5.2198

(1a)

TDE = 1.08506 TDA − 5.6638

(1b)

ΣFx = man =

mv 2

ρ

: TDA sin 20° + TDE sin 30° =

0.5 2 v 0.150

v 2 = 0.10261 TDA + 0.15 TDE

(2)

Try TDA = 75 N. By Eq. (1b), TDE = 75.72 N > 75 N (unacceptable) Try TDE = 75 N. By Eq. (1a), TDA = 74.34 N < 75 N (acceptable) By Eq. (2), v 2 = ( 0.10261)( 74.34 ) + ( 0.15 )( 75 ) = 18.877 m 2 / s 2 v = 4.34 m Try TDA = 0. By Eq. (1b ) , TDE = −5.6638 Try TDE = 0. By Eq. (1a ) , TDA = 5.2198

( unacceptable ) ( acceptable )

By Eq. (2), v 2 = ( 0.10261)( 5.2198 ) + ( 0.15)( 0 ) = 0.5356 m 2 / s 2 v = 0.732 m/s For 0 ≤ TBA , TBC , TDA , TDE ≤ 75 N,

Vector Mechanics for 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.

0.732 m/s ≤ v ≤ 4.34 m/s W

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Chapter 12, Solution 43. W = mg = ( 5 )( 9.81) = 49.05 N

ρ = 0.9 m ΣFy = 0 : TCA cos 40° − TCB cos15° − W = 0 0.76604 TCA − 0.96593 TCB = W = 49.05 TCB = 0.79307 TCA − 50.780

(1a)

TCA = 1.26093 TCB + 64.030

(1b)

ΣFx = max : TCA sin 40° + TCB sin15° = man =

ρ

ρ

(TCA sin 40° + TCB sin15° ) m 0.9 = (TCA sin 40° + TCB sin15° ) 5 = 0.115702 TCA + 0.046587 TCB

v2 =

mv 2

(2)

Try TCB = 116 N. By (1b), TCA = 210.3 N (unacceptable) Try TCA = 116 N. By (1a), TCB = 41.216 N (acceptable) By (2),

v 2 = ( 0.115702 )(116 ) + ( 0.046587 )( 41.216 ) = 15.34 m 2 / s 2 v = 3.92 m/s

Try TCA = 0. By (1a ) , TCB = −50.78 N Try TCB = 0. By (1b ) , TCA = 64.03 N By (2),

( unacceptable ) ( acceptable )

v 2 = ( 0.115702 )( 64.030 ) + 0 = 7.408 m 2 / s 2

v = 2.72 m/s For 0 ≤ TCA , TCB ≤ 116 N,

Vector Mechanics for 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.

2.72 m/s ≤ v ≤ 3.92 m/s W

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Chapter 12, Solution 44.

a = 0

(a) Before wire AB is cut. ΣFx = 0 :

− TAB cos 50° + TCD cos 70° = 0

(1)

ΣFy = 0 :

TAB sin 50° + TCD sin 70° − W = 0

(2)

Solving (1) and (2) simultaneously, TCD = 0.742 W W

TAB = 0.395 W (b) Immediately after wire AB is cut. an =

v2

ρ

TAB = 0, v = 0

=0

+

Fn = man = 0 :

TCD − W cos 20° = 0 TCD = W cos 20° TCD = 0.940 W W

Vector Mechanics for 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.

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Chapter 12, Solution 45.

ΣFy = ma y : N − mg = − man = − N = mg −

mv 2

ρ

mv 2

ρ

ΣFx = max : Ft = mat Ft = µ s N

At onset of slipping,

 mvs2  mat = µ s  mg −  ρ  

 a  0.150   2 2 vs2 = ρ  g − t  = ( 0.300)  9.81 −  = 2.883 m /s  0.75  µs   vs = 1.6979 m/s Time at slipping.

ts =

vs 1.6979 = at 0.150

Vector Mechanics for 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.

ts = 11.32 s W

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Chapter 12, Solution 46.

Angle change over arc AB. θ =

8° π = 0.13963 rad 180°

Length of arc: s AB = ρθ = ( 4 ) (5280) ( 0.13963) = 2949 ft sBC = (0.5)(5280) = 2640 ft, s AC = 2949 + 2640 = 5589 ft or

4802 5402 − = at ( 5589 ) 2 2

or

vB2 5402 − = ( −5.475 )( 2949 ) 2 2

480 5589 ∫ 540 v dv = ∫ 0 at ds

at = −5.475 ft/s 2 vB 2949 ∫ 540 v dv = ∫ 0 at ds

vB2 = 259300 ft 2 /s 2

Mass of passenger: m = Just before point B.

vB = 509.2 ft/s

200 = 6.211 lb ⋅ s 2 /ft 32.2

v = 509.2 ft/s, ρ = (4)(5280) = 21120 ft

an =

v2

ρ

=

( 509.2 )2 21120

= 12.277 ft/s 2

ΣFy = N1 − W = −m ( an )1 : N1 = 200 − ( 6.211)(12.277 ) = 123.75 lb ΣFx = Ft = mat :

Just after point B.

( Ft )1 = ( 0.6221)( −5.474 ) = −34.01 lb

v = 509.2 ft/s, ρ = ∞,

ΣFy = 0 : N 2 − W = 0

an = 0

N 2 = W = 200 lb

ΣFx = mat : Ft = mat = ( 6.211)( −5.475 ) = −34.01 lb Ft does not change.

N increases by 76.25 lb magnitude of change of force = 76.3 lb W

Vector Mechanics for 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.

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Chapter 12, Solution 47.

(

)

s = 3 180t − t 2 m, at =

v=

dv = − 6 m/s 2 dt

ds = 3 (180 − 2t ) m/s, dt

ρ = (4)(5280) = 21120 ft

Length of arc AB. s AB = ρθ AB = ( 21120 )

8°π = 2949 ft 180°

vB s AB ∫ v A v dv = ∫ 0 at ds

vB2 v A2 − = 2at s AB or vB2 = v A2 + 2at s AB = 5402 + ( 2 )( −6 )( 2949 ) 2 2 vB = 506.2 ft/s

= 256212,

For passenger, W = 165 lb, m =

W 165 = = 5.124 lb ⋅ s 2 /ft 32.2 g

ΣF = man : W cosθ − N = N = W cosθ −

mv 2

ρ

mv 2

(1)

ρ

ΣF = mat : P − W sin θ = mat

P = W sin θ + mat (a) Just after point A, t = 0,

v = 180 m/s,

(2)

θ = 8°

2 5.124 )( 540 ) ( = 165cos8° −

= 92.65 lb

From Eq. (1),

N

From Eq. (2),

P = 165sin 8° + ( 5.124 )( −6 ) = −7.78 lb

F =

21120

N 2 + P 2 = 93.0 lb, tan β =

92.65 = 11.909, β = 85.2° 7.78

β − 8° = 77.2°

F = 93.0 lb

77.2° W

(b) At point B. θ = 0, v = 506.2 ft/s

( 5.124 )( 506.2 )2

From Eq. (1),

N = 165 −

From Eq. (2),

P = ( 5.124 )( −6 ) = −30.74 lb

F=

(102.83)

2

21120

+ ( 30.74 ) = 107.3 lb, tan β = 2

= 102.83 lb

102.83 = 3.345, β = 73.4° 30.74 F = 107.3 lb

Vector Mechanics for 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.

73.4° W

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Chapter 12, Solution 48.

(a) v = 160 km/h = 44.44 m/s Wheels do not touch the road. ΣFy = −man : − mg = −mv 2 / ρ

( 44.44 ) = 201.4 v2 = g 9.81 2

ρ =

ρ = 201 m W (b) v = 80 km/h = 22.22 m/s m = 70 kg for passenger ΣFy = −man : N − mg = −

mv 2

ρ

 v2  N = m  g −  ρ  

 22.22 2  = ( 70 )  9.81 −   201.4  

N = 515 N W

Vector Mechanics for 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.

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Chapter 12, Solution 49.

m = 0.2 kg, W = mg = (0.2)(9.81) = 1.962 N ΣFt = mat : W cosθ = mat

at =

W cosθ = g cosθ m

v s θ ∫ v0 v dv = ∫ 0 at ds = ∫ 0 at rdθ

1 2 1 2 v − v0 = 2 2

θ ∫ 0 g cosθ rdθ = gr sin θ

v 2 = v02 + 2 gr sin θ , where r = 0.6 m for 0 ≤ θ ≤ 90° v = vmax when the cord touches the peg or θ = 90°. 2 vmax = v02 + 2 gr

(1)

When the cord touches the peg, the radius of curvature of the path becomes ρ = 0.3 m. ΣFy = man : Tmax − W =

vmax 2 =

ρ m

(Tmax

mvmax 2

ρ

− W)

(2)

Eliminating vmax 2 from equations (1) and (2), v02 =

ρ (Tmax − W ) m

− 2 gr =

( 0.3)(10 − 1.962 ) 0.2

= 0.285 m 2 /s 2

Vector Mechanics for 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.

− ( 2 )( 9.81)( 6.0 )

v 0 = 0.534 m/s

W

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Chapter 12, Solution 50.

Mass of block B. Acceleration of block B.

mB =

0.5lb = 0.015528lb ⋅ s 2 /ft 32.2 ft/s 2

an =

v2

ρ

at = 0

=

(9 ft/s) 2 = 27 ft/s 2 3ft

since v = constant.

∑ F = ma :

Q − P − W sin θ = − man Q = P + W sin θ − man

But, Q ≥ 0.

P + W sin θ − man > 0

sin θ ≥

man P (0.015528)(27) 0.3 − = − 0.5 0.5 W W

sin θ ≥ 0.2385 13.8° ≤ θ ≤ 166.2°

Vector Mechanics for 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.

W

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Chapter 12, Solution 51.

At position A, the vertical component of apparent weight is shown as N A. ΣF = man : N A − W =

an =

W an g

NA − W 380 − 120 g = ( 32.2 ) = 69.77 ft/s2 120 W

v A2 = ρ an = ( 3600 )( 69.77 ) = 251.2 × 103 ft 2 /s 2

At position C, the vertical component of apparent weight is shown as NC . ΣF = man : NC + W =

an =

W an g

NC + W 80 + 120 g = ( 32.2 ) = 53.67 ft/s2 120 W

v C2 = ρ an = ( 3600 )( 53.67 ) = 193.2 × 103 ft 2 / s 2 Length of arc ABC:

s AC = πρ = π ( 3600 ) = 11310 ft Calculate at , using vC2 − v A2 = 2at s AC

at =

vC2 − v A2 193.2 × 103 − 251.2 × 103 = 2s AC ( 2 )(11310 )

= − 2.562 ft/s 2 At position B, s AB =

π 2

ρ =

π 2

( 3600 ) = 5655 ft

vB2 = v A2 + 2at s AB = 251.2 × 103 + ( 2 )( −2.562 )( 5655 ) = 222.2 × 103 ft 2 / s 2

Vector Mechanics for 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.

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Effective forces at B: man =

mat =

W vB2 120 222.2 × 103 = = 230 lb g ρ 32.2 3600

W 120 at = ( −2.562 ) = −9.5 lb 32.2 g

ΣF = mat :

P − W = mat or P = W + mat = 120 − 9.5 = 110.5 lb ΣF = man : N B = man = 230 lb Force exerted by seat:

F =

NB2 + P2 = tan β =

2302 + 110.52 = 255 lb 110.5 230

β = 25.7° F = 255 lb

Vector Mechanics for 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.

25.7° W

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Chapter 12, Solution 52.

The road reaction consists of normal component N and friction component F. The resultant R makes angle φs with the normal. Case 1: v = vmax ΣFy = 0: R cos (θ + φs ) − mg = 0 R =

mg cos (θ + φs )

ΣFx = man : R sin (θ + φs ) = man man = mg tan (θ + φs ) vmax 2 = g tan (θ + φs ) r vmax =

gr tan (θ + φs )

Case 2: v = vmin ΣFy = 0: R cos (θ − φs ) − mg = 0

R =

mg cos (θ − φs )

ΣFx = man : R sin (θ − φs ) = man man = mg tan (θ − φs ) vmin 2 = g tan (θ − φs ) r vmin =

gr tan (θ − φs ) gr tan (θ − φs ) ≤ v ≤

Vector Mechanics for 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.

gr tan (θ + φs ) W

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Chapter 12, Solution 53.

Weight. Acceleration.

W = mg a=

v2

ρ

F + W sin θ = ma cosθ

∑ Fx = max :

F =

mv 2

ρ

cosθ − mg sin θ

(1)

N − W cosθ = ma sin θ

∑ Fy = ma y :

N=

mv 2

ρ

sin θ + mg cosθ

(2)

(a) Banking angle. Rated speed v = 180 km/h = 50 m/s. F = 0 at rated speed. 0= tan θ =

mv 2

ρ

cosθ − mg sin θ

v2 (50) 2 = = 1.2742 ρ g (200) (9.81)

θ = 51.875° (b) Slipping outward. F = µN

θ = 51.9° W

v = 320 km/h = 88.889 m/s

µ=

F v 2 cosθ − ρ g sin θ = N v 2 sin θ + ρ g cosθ

µ=

(88.889) 2 cos51.875° − (200) (9.81)sin 51.875° (88.889)2 sin 51.875° + (200) (9.81) cos 51.875°

= 0.44899

µ = 0.449 W continued

Vector Mechanics for 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.

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(c) Minimum speed.

F = −µ N

−µ=

v 2 cosθ − ρ g sin θ v 2 sin θ + ρ g cosθ

v2 =

ρ g (sin θ − µ cosθ ) cosθ + µ sin θ

=

(200) (9.81) (sin 51.875° − 0.44899 cos 51.875°) cos 51.875° + 0.44899sin 51.875°

= 1029.87 m 2 /s 2 v = 32.09 m/s

Vector Mechanics for 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.

v = 115.5 km/h W

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Chapter 12, Solution 54.

Rated speed: vR = 75 mi/h = 110 ft/s, 125 mi/h = 183.33 ft/s From Sample Problem 12.6,

(110 ) = 2674 ft vR2 = 32.2 tan 8° g tan θ 2

vR2 = g ρ tan θ

or

ρ =

Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

mv 2

ρ

cos (θ + φ )

(a) φ = 0.  v2  cos (θ + φ ) − sin (θ + φ )  Fs = W   gρ   (183.33) 2  cos8° − sin 8° =W  ( 32.2)( 2674)  Fs = 0.247 W !

= 0.247 W

(b) For Fs = 0, v2 cos (θ + φ ) − sin (θ + φ ) = 0 gρ v2 (183.33) = 0.39035 tan (θ + φ ) = = g ρ ( 32.2)( 2674) 2

θ + φ = 21.3° φ = 21.3° − 8°

Vector Mechanics for 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.

φ = 13.3° !

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Chapter 12, Solution 55.

Rated speed: vR = 75 mi/h = 110 ft/s, 125 mi/h = 183.33 ft/s From Sample Problem 12.6,

(110 ) = 2674 ft vR2 ρ = = 32.2 tan 8° g tan θ 2

vR2

= g ρ tan θ

or

Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

Solving for Fs ,

mv 2

ρ

cos (θ + φ )

 v2  cos (θ + φ ) − sin (θ + φ )  Fs = W   gρ 

v2 (183.33) = 0.39035 and F = 0.12 W so that = s gρ ( 32.2 )( 2674 ) 2

Now

0.12 W = W 0.39035 cos (θ + φ ) − sin (θ + φ ) 

Let u = sin (θ + φ ) . Then, cos (θ + φ ) = 1 − u 2 . 0.12 = 0.39035 1 − u 2 − u

0.39035 1 − u 2 = 0.12 + u

or

(

)

Squaring both sides, 0.15237 1 − u 2 = 0.0144 + 0.24u + u 2 or

1.15237u 2 + 0.24u + 0.13797 = 0

The positive root of the quadratic equation is u = 0.2572. Then, θ + φ = sin −1 u = 14.90°

φ = 14.90° − θ = 14.90° − 8°

Vector Mechanics for 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.

φ = 6.90° !

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Chapter 12, Solution 56.

If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω Normal acceleration. an =

v2

ρ

=

ρ 2ω 2 = (r sin θ ) ω 2 ρ ∑ Fy = ma y :

N cosθ − mg = 0 mg N = cosθ ΣFx = max : N sin θ = ma mg sin θ = (m r sin θ )ω 2 cos θ

Either

sin θ = 0

or

cos θ =

g rω 2

θ = 0° or 180° or

cosθ =

9.81 = 0.3488 (0.5) (7.5)2

θ = 0°, 180°, and 69.6° W

Vector Mechanics for 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.

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Chapter 12, Solution 57.

If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω From Prob. 12.56

r = 500 mm = 0.500 m, ω = 7.5 rad/s, m = 250g = 0.250 kg.

Normal acceleration:

an =

v2

ρ

=

ρ 2ω 2 = (r sin θ )ω 2 ρ

ΣF =

ma:

F − mg sin θ = − m (r sin θ ) ω 2 cos θ F = m ( g − rω 2 cos θ )sin θ ΣF =

ma :

N − mg cos θ = m(r sin θ )ω 2 sin θ N = m( g cos θ ) + r ω 2 sin 2 θ )

(a) θ = 75°.

F = (0.25) 9.81 − (0.500 cos 75°)(7.5)2  sin 75° = 0.61112 N

N = (0.25) 9.81cos 75° + (0.500sin 2 75°)(7.5)2  = 7.1950 N

µ s N = (0.25)(7.1950) = 1.7987 N Since F < µ s N , the collar does not slide. F = 0.611 N continued

Vector Mechanics for 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.

75° W

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F = (0.25) 9.81 − (0.500cos 40°) (7.5)2  sin 40°

(b) θ = 40°.

= − 1.8858 N

N = (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5)2  = 4.7839 N

µ s N = (0.25) (4.7839) = 1.1960 N Since F > µ s N , the collar slides. Since the collar is sliding, F = µk N . ∑ Fn = +

ma :

N − mg cosθ = man sin θ N = mg cos θ + m (r sin θ )ω 2 sin θ = m  g cos θ + (r sin 2 θ )ω 2  = (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5) 2  = 4.7839 N

F = µk N = (0.20) (4.7839) = 0.957 N F = 0.957 N

Vector Mechanics for 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.

40° W

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Chapter 12, Solution 58.

Draw the free body diagrams of the block B when the arm is at θ = 150°. v = at = 0,

g = 32.2 ft/s 2

ΣFt = mat : − W sin 30° + N = 0

N = W sin 30° ΣFn = man : W cos 30° − F = m

F = W cos 30° −

Form the ratio

v2

ρ

=

Wv 2 ρg

Wv 2 ρg

F , and set it equal to µ s for impending slip. N

cos 30° − ( 4.2 ) /(1)(32.2) cos 30° − v 2 / ρ g F = = µs = sin 30° sin 30° N 2

µ s = 0.636 W

Vector Mechanics for 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.

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Chapter 12, Solution 59.

Let β be the slope angle of the dish. tan β = At r = 6ft, tan β = 1

or

dy 1 = r dr 6

β = 45°

Draw free body sketches of the sphere. ΣFy = 0: N cos β − µs N sin β − W = 0

N =

W cos β − µs sin β

ΣFn = man : N sin β + µs N cos β =

W ( sin β + µs N cos β ) cos β − µs sin β v2 = ρg

=

mv 2

ρ

=

Wv 2 ρg

Wv 2 ρg

sin β + µs cos β sin 45° + 0.5cos 45° = ( 6)( 32.2) = 579.6 ft 2 /s 2 cos β − µs sin β cos 45° − 0.5sin 45° v = 24.1 ft/s W

Vector Mechanics for 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.

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Chapter 12, Solution 60.

D 5 ft = 5 in. = 2 12

Uniformly accelerated motion on a circular path. ρ = W = 60 × 10−6 oz =

60 × 10−6 lb 16

g = 32.2 ft/s 2 m=

W = 116.46 × 10−9 lb s 2 / ft g

(a) For uniformly accelerated motion, v = v0 + at t = 0 + (12 )( 4 )

v = 48 ft/s W (b)

(

)

ΣFt = mat : Ft = 116.46 × 10−9 (12 ) = 1.3975 × 10−6 lb.

ΣFn = man : Fn = man =

mv 2

ρ

(116.46 × 10 ) ( 48) = −9

2

5 /12

= 644.0 × 10−6 lb

Magnitude of force: F =

Ft 2 + Fn 2 = 10−6

( 644.0 )2 + (1.3975)2 F = 644 × 10 −6 lb W

Vector Mechanics for 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.

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Chapter 12, Solution 61.

Uniformly accelerated motion on a circular path. ρ = 8 ft

v = v0 + at t = 0 + ( 0.75 )(12 ) = 9 ft/s Ft = mat =

W a 0.75 at : Ft = W t = W = 0.0233 W g g 32.2

Fn = man =

W (9) Wv 2 = 0.3144 W : Fn = gρ ( 32.2 )(8) 2

F =

Ft 2 + Fn 2 = 0.315 W

This is the friction force available to cause the trunk to slide. The normal force N is calculated from equilibrium of forces in the vertical direction. ΣFy = 0: N − W = 0 Since sliding is impending,

µs =

F = 0.315 W

Vector Mechanics for 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.

N =W

µ s = 0.315 W

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Chapter 12, Solution 62.

For constant speed, at = 0 an =

vB2

ρ

with vB = 0.7 m/s, ρ = 0.2 m

ΣFx = max: F = man cosθ ΣFy = ma y : N − W = −man sin θ

Ratio

N = mg − man sin θ

F man cosθ cosθ cosθ = = = g g ρ N mg − man sin θ − sin θ − sin θ an vB 2

With

g ρ ( 9.81)( 0.2 ) F cosθ = = 4.0041, the ratio becomes = 2 2 N 4.0041 − sin θ vB ( 0.7 )

For no impending slide, µs ≥

F cos θ = N 4.0041 − sin θ

To find the value of θ for which the ratio is maximum set the derivative with respect to θ equal to zero. ± cosθ 1 − 4.0041 sin θ d   = 0   = ± dθ  4.0041 − sin θ  ( 4.0041 − sinθ )2 sin θ =

θ = 14.446°,

F N

=

1 = 0.24974 4.0041 cos14.446° 4.0041 − 0.24974

θ = 180° − 14.446° = 165.554°,

(a) Minimum value of µs for no slip. (b) Corresponding values of θ .

Vector Mechanics for 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.

= 0.258

F = 0.258 N

( µs )min

= 0.258 W

θ = 14.5° and 165.5° W

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Chapter 12, Solution 63.

For constant speed, at = 0 an =

vB 2

ρ

with vB = 0.7 m/s, ρ = 0.2 m

ΣFx = max : F = man cosθ ΣFy = ma y : N − W = −man sin θ Ratio

N = mg − man sin θ

F man cosθ cosθ cosθ = = = g g ρ N mg − man sin θ − sin θ − sin θ an vB 2

Let u =

gρ F cosθ = so that N u − sin θ vB 2

Determine the value of θ at which F/N is maximum. cos 2 θ − ( u − sin θ )( sin θ ) d  cosθ  1 − u sin θ = = = 0   2 dθ  u − sin θ  ( u − sin θ ) ( u − sin θ )2 The corresponding ratio

F . N

F ± 1 − u −2 = = N u − u −1

± u −1 1−u

(a) For impending sliding to the left:

−2

= ±

sin θ = ± tan θ cosθ

F = tan θ = µ s = 0.35 N

θ = arctan ( 0.35) = 19.29°, u −1 =

vB 2 = sin θ , gρ

vB 2 = ( 9.81)( 0.2 ) sin19.29° = 0.648 m 2 /s 2 vB = 0.805 m/s W

Vector Mechanics for 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.

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For impending motion to the right:

F = − tan θ = µ s = 0.35 N

θ = arctan ( −0.35 ) = 160.71° u −1 =

v2 2 = sin θ , gρ

vB 2 = ( 9.81)( 0.2 ) sin160.71° = 0.648 m 2 /s 2 = 0.805 m/s W

(b)

For impending sliding to the left, θ = 19.3° W For impending sliding to the right, θ = 160.7° W

Vector Mechanics for 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.

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Chapter 12, Solution 64. Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = l at the right edge of the plate. At the screen, x =

l 2

+ L

Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion, x = v0t ,

so that

t1 =

l

and

v0

t2 =

l 2v0

L . v0

+

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between plates ( 0 ≤ t ≤ t1 ) , the vertical force on the electron is Fy = eV / d . After it passes the plates ( t1 ≤ t ≤ t2 ) , it is zero. ΣFy = ma y : a y =

For 0 ≤ t ≤ t1,

( )0 + a y t

vy = vy

( )1 vy

For t1 ≤ t ≤ t2 ,

=

eVt1 md

and

m

=

eV md

eVt md

1 2 eVt 2 a yt = 0 + 0 + 2 2md

( )0

t +

y1 =

eVt12 2md

y = y0 + v y

At t = t1,

= 0+

Fy

ay = 0

( )1 ( t − t1 )

y = y1 + v y

( )1 ( t2 − t1 )

y2 = δ = y1 + v y

At t = t2

δ =

eVt12 eVt1 eVt 1 + ( t2 − t1 ) = 1  t2 − t1  md md  2md 2 

=

eV l  l L 1 l + −   mdv0  2v0 v0 2 v0 

Vector Mechanics for 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.

or δ =

eV lL ! mdv02

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Chapter 12, Solution 65.

Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = l at the right edge of the plate. At the screen,

x =

l

+ L 2 Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion,

x = v0t ,

so that t1 =

l

t2 =

and

v0

l 2v0

+

L . v0

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between the plates ( 0 ≤ t ≤ t1 ) , the vertical force on the

electron is Fy = eV / d . After it passes the plates ( t1 ≤ t ≤ t2 ) , it is zero. For 0 ≤ t ≤ t1,

ΣFy = ma y : a y =

( )0 + a y t

vy = vy

= 0+

At t = t1, But y <

So that

l v0

, y =

m

=

eV md

eVt md

( )0 t + 12 a yt 2

y = y0 + v y

Fy

= 0+0+

eVt 2 2md

eV l 2 2mdv02

d − 0.075d = 0.425d 2

eV l 2 < 0.425d 2mdv02 d2 1 eV eV > = 1.176 2 2 2 0.425 2mv0 l mv0

Vector Mechanics for 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.

d eV
> 1.085 l mv02

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Chapter 12, Solution 66.

θ = 30°, θ& = 2 rad/s, θ&& = 0 r = 0.6 m, W = mg Block B: Only force is weight Fr = W cos 30°,

(

Fθ = −W sin 30°

)

(a) Fθ = maθ = m rθ&& + 2r&θ& :

2r&θ& = r& = −

Fθ mg sin 30° − rθ&& = − − rθ&& = − g sin 30° − rθ&& m m

( 9.81) sin 30° + ( 0.6 )( 0 ) = −1.226 m/s g sin 30° + rθ&& = − & 2θ ( 2 )( 2 ) v B / rod = 1.226 m/s

(

)

r − rθ& 2 : (b) Fr = mar = m &&

&& r = rθ& +

Fr mg cos 30° = rθ& 2 + = rθ& 2 + g cos 30° m m 2

= ( 0.6 )( 2 ) + ( 9.81) cos 30° = 10.90 m/s 2

a B / rod = 10.90 m/s 2

Vector Mechanics for 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.

60°


60°


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Chapter 12, Solution 67.

θ = 45°, vr = r = 0,

(a)

θ = 10 rad/s2

r = 0.8 m,

vθ = rθ = 0,

W = mg

(

ΣFθ = maθ : N − W cos 45° = m rθ + 2rθ

(

N = mg cos 45° + m rθ + 2rθ

)

)

= (0.5)(9.81) cos 45° + 0.5 ( 0.8 )(10 ) + 0 

= 7.468 (b)

N = 7.47 N

(

ΣFr = mar : mg sin 45° = m r − rθ 2

r =

45° W

)

mg sin 45° + rθ 2 = g sin 45° + rθ 2 m

= ( 9.81) sin 45° + 0 = 6.937 m/s 2

a B / rod = 6.94 m/s 2

Vector Mechanics for 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.

45° W

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Chapter 12, Solution 68.

Use radial and transverse components of acceleration. r = 3t 2 − t 3 ft

θ = 2t 2 rad

r = 6t − 3t 2 ft/s

θ = 4t rad/s

r = 6 − 6t ft/s 2

θ = 4 rad/s 2

ar = r − rθ 2 = 6 − 6t − (3t 2 − t 3 )(16t 2 )

= 16t 5 − 48t 4 − 6t + 6 ft/s 2 aθ = rθ + 2rθ = (3t 2 − t 3 )(4) + (2)(6t − 3t 2 )(4t )

= 60t 2 − 28t 3 ft/s 2

Mass: m =

W 4 = = 0.12422 lb ⋅ s 2 /ft 32.2 g

(a) t = 0.

ar = 6 ft/sec2 aθ = 0

Apply Newton’s second law. Fr = mar = (0.12422)(6)

Fr = 0.745 lb W

Fθ = maθ = (0.12422)(0)

Fθ = 0 W

(b) t = 1 s. ar = − 32 ft/s 2 aθ = 32 ft/s 2

Apply Newton’s second law. Fr = mar = (0.12422)(−32)

Fr = − 3.98 lb W

Fθ = maθ = (0.12422)(32)

Fθ =

Vector Mechanics for 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.

3.98 lb W

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Chapter 12, Solution 69.

Use radial and transverse components of acceleration. r = 6(1 + cos 2π t ) ft

θ = 2π t rad

r = −12π sin 2π t ft/s

θ = 2π rad/s

r = − 24π 2 cos 2π t ft/s 2

θ =0

ar = r − rθ 2 = − 24π 2 cos 2π t − (6 + 6 cos 2π t )(2π )2

= − 24π 2 − 48π 2 cos 2π t ft/s 2 aθ = rθ + 2rθ = 0 + (2)(−12π sin 2π t )(2π )

= − 48π 2 sin 2π t

Mass: m =

W 1 = = 0.031056 lb ⋅ s 2/ft 32.2 g

(a) t = 0. ar = − 710.61 ft/s 2 aθ = 0

Apply Newton’s second law. Fr = mar = (0.031056)(710.61) Fθ = maθ = 0

Fr = − 22.1 lb W Fθ = 0 W

(b) t = 0.75 s. ar = − 236.87 ft/s2 aθ = 473.74 ft/s 2

Apply Newton’s second law. Fr = mar = (0.031056)(−236.87)

Fr = − 7.36 lb W

Fθ = maθ = (0.031056)(473.74)

Fθ = 14.71 lb W

Vector Mechanics for 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.

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Chapter 12, Solution 70.

Kinematics:

dr = r = 1.5 ft/s, r = 0 dt

r t ∫ 0 dr =∫ 0 r dt

or

θ = 0.8t rad/s,

r = 1.5t ft

θ = 0.8 rad/s 2

ar = r − rθ 2 = 0 − (1.5t )( 0.8t ) = −0.96t 3 ft/s 2 2

aθ = rθ + 2rθ = (1.5t )( 0.8 ) + ( 2 )(1.5 )( 0.8t ) = 3.6t ft/s 2

Kinetics: Sketch the free body diagrams for the collar. ΣFr = mar : − T = mar ΣFθ = maθ : Q = maθ Set T = Q to obtain the required time. −mar = maθ

or

− ar = aθ

t2 =

3.6 = 3.75 s 2 0.96

Using the calculated expressions 0.96t 3 = 3.6t,

t = 1.936 s W

Vector Mechanics for 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.

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Chapter 12, Solution 71.

θ& = 10t rad/s, θ&& = 10 rad/s2 m = 0.5/32.2 = 0.015528 lb " s 2 /ft

Before cable breaks: Fr = −T and && r = 0.

(

Fr = mar : − T = m && r − rθ& 2

mrθ& 2 = mr&& + T

or

θ& 2 =

)

0+4 mr&& + T = = 171.733 rad 2 /s 2 mr ( 0.015528)(1.5)

θ& = 13.105 rad/s Immediately after the cable breaks: Fr = 0, r& = 0 (a) Acceleration of B relative to the rod.

(

)

m && r − rθ& 2 = 0

or

&& r = rθ& 2 = (1.5 )(13.105 ) = 257.6 ft/s 2 2

a B / rod = 258 ft/s 2 radially outward !

(b) Transverse component of the force.

(

Fθ = maθ : Fθ = m rθ&& + 2r&θ&

( 0.015528) (1.5)(10 ) + ( 2 )( 0 )(12 ) = 0.233

Vector Mechanics for 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.

) Fθ = 0.233 lb !

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Chapter 12, Solution 72.

θ = 15 rad/s, m = 230 g = 0.230 kg, θ = 0, Fθ = 9 N, r = −12 m/s2

Due to the spring, Fr = −kr, k = 60 N/m

(

ΣFr = Fr = mar : − kr = m r − rθ 2

)

( k − mθ ) r = −mr 2

(a) Radial coordinate. r =−

( 0.230 )( −12 ) mr =− 2 2 k − mθ 60 − ( 0.230 )(15 )

= 0.33455 m

r = 335 mm

(

ΣFθ = maθ : Fθ = m rθ + 2rθ 2rθ =

r =

)

Fθ − rθ m

Fθ − mrθ 9−0 = = 1.304 m/s 2mθ ( 2 )( 0.230 )(15)

(b) Radial component of velocity. vr = r

Vector Mechanics for 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.

vr = 1.304 m/s W

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Chapter 12, Solution 73.

At point A.

an =

v2

ρ

tan θ = −

=

(3.8)2 = 115.52 m/s 2 0.125

125 175 + 125

θ = − 22.62°

Draw the free body diagram of the collar.

+ ΣF = ma :

at =

Fs cos 22.6° = mat

Fs cos 22.6° 70cos 22.62° = = 43.077 m/s 2 m 1.5

Acceleration vector. a = an2 + at2 = 115.522 + 43.0772 = 123.29 m/s 2

tan φ =

115.52 43.077

φ = 69.55°

φ − 22.62° = 46.93° ar = a cos 46.93°

= 123.29cos 46.93° = 84.2 m/s 2 in negative r-direction ar = − 84.2 m/s 2
aθ = a sin 46.93°

= 123.29sin 46.93° aθ = 90.1 m/s 2


Vector Mechanics for 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.

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Chapter 12, Solution 74.

Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Constraint of cable:

r + yB = constant, r + vB = 0,

r + aB = 0

or

r = −aB (1)

For block A,

+ ΣFx = mAa A : T cosθ =

WA W a A or T = A a A secθ (2) g g

For block B,

ΣFy =

WB W aB : WB − T = B aB (3) g g

Adding Eq. (1) to Eq. (2) to eliminate T, WB =

WA W a A secθ + B aB (4) g g

Radial and transverse components of a A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

r − rθ 2 = ar = a A ⋅ er = −a A cosθ

(5)

Noting that initially θ = 0, using Eq. (1) to eliminate r , and changing signs gives

aB = a A cosθ

(6)

Substituting Eq. (6) into Eq. (4) and solving for a A , aA =

WB g ( 50)( 32.2) = = 17.991 ft/s 2 WA sec θ + WB cos θ 40sec30° + 50 cos 30°

From Eq. (6), aB = 17.991cos 30° = 15.581 ft/s 2 (a) From Eq. (2), T = ( 40/32.2 )(17.991) sec30° = 25.81 T = 25.8 lb W (b) Acceleration of block A. (c) Acceleration of block B.

Vector Mechanics for 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 A = 17.99 ft/s 2

W

a B = 15.58 ft/s 2 W

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Chapter 12, Solution 75.

Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Radial and transverse components of v A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

r = vr = v A ⋅ er = −v A cos 30°

= −6cos 30° = −5.19615 ft/s rθ = vθ = v A ⋅ eθ = −v A sin 30° = 6sin 30° = 3 ft/s

θ = Constraint of cable:

vθ 3 = = 1.25 rad/s r 2.4

r + yB = constant, r + vB = 0,

r + aB = 0

For block A, + ΣFx = mAa A : T cosθ = For block B,

or

r = −aB (1)

WA W a A or T = A a A secθ (2) g g

ΣFy =

WB W aB : WB − T = B aB (3) g g

Adding Eq. (1) to Eq. (2) to eliminate T, WB =

WA W a A secθ + B aB (4) g g

Radial and transverse components of a A. Use a method similar to that used for the components of velocity.

r − rθ 2 = ar = a A ⋅ er = −a A cosθ

(5)

Using Eq. (1) to eliminate r and changing signs gives

aB = a A cosθ − rθ 2

Vector Mechanics for 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)

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Substituting Eq. (6) into Eq. (4) and solving for a A ,

aA =

(

WB g + rθ 2

)

WA secθ + WB cosθ

=

( 50 ) 32.2 + ( 2.4 )(1.25)2  40sec30° + 50cos 30°

= 20.086 ft/s 2

From Eq. (6), aB = 20.086 cos 30° − ( 2.4 )(1.25 ) = 13.645 ft/s 2 2

(a)

From Eq. (2), T = ( 40/32.2 )( 20.086 ) sec30° = 28.8

(b)

Acceleration of block A.

(c)

Acceleration of block B.

Vector Mechanics for 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.

T = 28.8 lb W a A = 20.1 ft/s 2

W

a B = 13.65 ft/s 2 W

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Chapter 12, Solution 76.

Since the particle moves under a central force, h = constant.

Using Eq. (12.27),

h = r 2θ = h0 = r0v0

rv r v cos 2θ v = 0 cos 2θ or θ = 0 20 = 0 0 2 r0 r r0 Radial component of velocity. vr = r =

= r0

dr  d  r0 θ =  dθ dθ  cos 2θ

sin 2θ

( cos 2θ )

3/ 2

 sin 2θ θ θ = r0 3/ 2  cos 2 θ ( )

v0 cos 2θ r

vr = v0

sin 2θ W cos 2θ

Transverse component of velocity. vθ =

h r0v0 = cos 2θ r r0

Vector Mechanics for 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.

vθ = v0 cos 2θ W

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Chapter 12, Solution 77.

Since the particle moves under a central force, h = constant. Using Eq. (12.27),

h = r 2θ = h0 = r0v0

rv r v cos 2θ v = 0 cos 2θ or θ = 0 2n = 0 0 2 r0 r r0

Differentiating the expression for r with respect to time, r =

 sin 2θ sin 2θ v0 sin 2θ dr  d  r0 θ = θ = r0 cos 2θ = v0  θ = r0 3/ 2 3/ 2 dθ dθ  cos 2θ  cos 2θ ( cos 2θ ) ( cos 2θ ) r0

Differentiating again,  r =

(a) vr = r = v0

dr  d  sin 2θ   2cos 2 2θ + sin 2 2θ  v0 2 2cos 2 2θ + sin 2 2θ θ = θ =  v0 θ = v0 dθ dθ  r0 cos 2θ  cos 2θ ( cos 2θ )3 / 2 sin 2θ vr = 0 sin 2θ r0 cos 2θ v=

vθ = rθ =

( vr )2 + ( vθ )2

=

v0r cos 2θ r0

v0r sin 2 2θ + cos 2 2θ r0

v 2 2cos 2 2θ + sin 2 2θ ar =  r − rθ 2 = 0 − r0 cos 2θ =

v=

v0r W r0

r0 v0 2 cos 2 2θ cos 2θ r0 2

v0 2 cos 2 2θ + sin 2 2θ v0 v 2r = = 02 r0 r0 cos 2θ r0 cos 2θ Fr = mar =

mv02r : r02

Fr =

Since the particle moves under a central force, aθ = 0 continued

Vector Mechanics for 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.

mv02r W r02

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Magnitude of acceleration: a =

ar 2 + aθ 2 =

Tangential component of acceleration.

Normal component of acceleration.

r  But cos 2θ =  0  r (b) But an =

v2

ρ

2

or

Hence, an =

ρ =

at =

at =

v0 2r r0 2

dv d v r v v 2r =  0  = 0 r = 0 2 sin 2θ dt dt  r0  r0 r0

a 2 − at 2 =

v0 2r v0 2r cos 2θ 2 1 sin 2 − θ = r0 2 r0 2

v0 2 r

v2 v 2r 2 r = 02 ⋅ 2 an r0 v0

Vector Mechanics for 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.

ρ =

r3 W r02

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Chapter 12, Solution 78.

Since the particle moves under a central force, h = constant. Using Eq. (12.27),

h = r 2θ = h0 = r0v0 rv rv v0 or θ = 0 20 = 2 0 0 2 = r r0 cos 2 θ r0 cos θ

Radial component of velocity.

vr = r =

Transverse component of velocity. Speed.

v=

vr 2 + vθ 2 = r0θ =

d ( r0 cosθ ) = − ( r0 sinθ )θ dt

vθ = rθ = ( r0 cosθ )θ r0v0 r0 cos 2 θ

Vector Mechanics for 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.

v=

v0 W cos 2 θ

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Chapter 12, Solution 79.

Since the particle moves under a central force, h = constant Using Eq. (12.27),

h = r 2θ = h0 = r0v0

Radial component of velocity.

vr = r =

Transverse component of velocity. Speed.

v=

vr 2 + vθ 2 = r0θ =

θ =

r0v0 rv v0 = 2 0 02 = 2 r r0 cos 2 θ r0 cos θ

d ( ro cosθ ) = − ( r0 sin θ )θ dt

vθ = rθ = ( r0 cosθ )θ

r0v0 v0 = 2 cos 2 θ r0 cos θ

Tangential component of acceleration. at =

( −2 )( − sin θ )θ = 2v0 sin θ ⋅ v0 = 2v02 sin θ dv = v0 dt r0 cos 2 θ r0 cos5 θ cos3 θ cos3 θ

Tangential component of force. Ft = mat : Ft = (a) θ = 0, (b) θ = 45°,

2mv0 2 sin θ r0 cos5 θ

Ft = 0 Ft =

Ft = 0 W 2mv0 sin 45° cos5 45°

Vector Mechanics for 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.

Ft =

8mv0 2 W r0

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Chapter 12, Solution 80.

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

or

v=

GM r

Let τ be the periodic time to complete one orbit. vτ = 2π r

Solving for τ ,

But,

Then,

M =

τ =

τ =

or

τ

GM = 2π r r

2π r 3/ 2 GM

4 3 π R ρ, 3 3π  r    Gρ  R 

hence

GM = 2

π 3

G ρ R3 / 2

3/ 2

Using r = 3R as given leads to

τ = 33/ 2

3π π =9 Gρ Gρ

Vector Mechanics for 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.

τ = 9 (π / G ρ )

1/ 2

W

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Chapter 12, Solution 81.

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

v=

or

GM r

Let τ be the period time to complete one orbit. vτ = 2π r

But Then

GM τ 2 r = 4π 2 3

v 2τ 2 =

or

GM τ 2 = 4π 2r 2 r 1/ 3

 GM τ 2  r =  2    4π 

or

Data: τ = 23.934 h = 86.1624 × 103 s

(a) In SI units: g = 9.81 m/s 2 , R = 6.37 × 106 m

(

GM = gR 2 = ( 9.81) 6.37 × 106

(

)

2

)(

 398.06 × 1012 86.1624 × 103  r = 4π 2 

= 398.06 × 1012 m3/s 2

)

2

1/ 3

   

= 42.145 × 106 m

altitude h = r − R = 35.775 × 106

h = 35800 km W

In US units: g = 32.2 ft/s 2 , R = 3960 mi = 20.909 × 106 ft

(

GM = gR 2 = ( 32.2 ) 20.909 × 106

(

)(

 14.077 × 1015 86.1624 × 103 r =  4π 2 

)

2

)

2

= 14.077 × 1015 ft 3/s 2 1/ 3

   

= 138.334 × 106 ft

altitude h = r − R = 117.425 × 106 ft

Vector Mechanics for 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.

h = 22200 mi W

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(b) In SI units:

v=

GM = r

398.06 × 1012 = 3.07 × 103 m/s 6 42.145 × 10

v=

GM = r

14.077 × 1015 = 10.09 × 103 ft/s 138.334 × 106

v = 3.07 km/s W

In US units:

Vector Mechanics for 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.

v = 10.09 × 103 ft/s W

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Chapter 12, Solution 82.

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

or

v=

GM r

Let τ be the periodic time to complete one orbit.

vτ = 2π r

from which GM = But

GM = gR 2 ,

τ

or

GM = 2π r r

4π 2r 3

τ2 hence, g =

4π 2r 3 R 2τ 2

(1) 1/ 3

 gR 2τ 2  r =  2    4π 

Solving for r,

W

Data: r = 417,000 mi = 2.202 × 109 ft R = 44, 400 mi = 234.4 × 106 ft

τ = 3.551 days = 85.224 h = 306.8 × 103 s Using (1), g =

(

4π 2 2.202 × 109

)

3

(306.8 × 10 ) ( 234.4 × 10 ) 3

2

6

2

= 81.5 ft/s 2

Vector Mechanics for 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.

g = 81.5 ft/s 2 W

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Chapter 12, Solution 83.

Let M be the mass of the sun and m the mass of Venus. For the circular orbit of Venus,

GMm mv 2 = man = 2 r r

GM = rv 2

where r is radius of the orbit.

r = 67.2 × 106 mi = 354.8 × 109 ft

Data:

v = 78.3 × 103 mi/h = 114.84 × 103 ft/s

(

)(

GM = 354.8 × 109 114.84 × 103 M =

(a) Mass of sun.

)

2

= 4.6792 × 1021 ft 3/s 2

GM 4.6792 × 1021 ft 3/s 2 = G 34.4 × 10−9 ft 4 /lb ⋅ s 4 M = 136.0 × 1027 lb ⋅ s 2 /ft W

R = 432 × 103 mi = 22.81 × 109 ft

(b) At the surface of the sun,

GMm = mg R2 g =

GM 4.6792 × 1021 = 2 R2 22.81 × 109

(

)

Vector Mechanics for 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.

g = 899 ft/s 2 W

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Chapter 12, Solution 84.

For gravitational force and a circular orbit, Fr = But

v=

2π r

τ

,

2π r

hence,

τ

=

GMm mv 2 = r r2

GM r

or

v=

or

GM r

GM τ 2 4π 2

r3 =

1/ 3

 GM τ 2  r =  2    4π 

Solving for r, 1/ 3

 4π 2  GM v = = GM  2  r  GM τ  2

(1)

1/ 3

 4π 2 ( GM )2   =   τ2  

1/ 3

 2π GM  v=   τ 

(2)

For earth: R = 6.37 × 106 m, g = 9.81 m/s 2

(

GM = gR 2 = ( 9.81) 6.37 × 106

(

)

2

= 398.06 × 1012 m3/s 2

)

For Jupiter: GM = ( 319 ) 398.06 × 1012 = 126.98 × 1015

(a) For Ganymede: τ = 7.15 days = 171.6 h = 617.76 × 103 s

By Eq. (1),

(

)(

 126.98 × 1015 617.76 × 103  r= 4π 2 

)

2

1/ 3

   

= 1.071 × 109 m

r = 1.071 × 106 km W

(b) For Callisto: τ = 16.69 days = 400.56 h = 1.4420 × 106 s

By Eq. (2),

(

 2π 126.98 × 1012 v=  1.4420 × 106 

) 

1/ 3

 

= 8.209 × 103 m/s

Vector Mechanics for 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.

v = 8.21 km/s W

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Chapter 12, Solution 85.

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

v=

or

GM r

Let τ be the periodic time to complete one orbit. vτ = 2π r

or

GM = 2π r r

τ

1/ 3

 GM τ 2  r =  2    4π 

Solving for r,

For earth, R = 6370 km = 6.37 × 106 m, g = 9.81 m/s 2

(

GM = gR 2 = ( 9.81) 6.37 × 106

)

2

= 398.06 × 1012 m3/s 2

Data: τ = 120 min = 7200 s

(

)

1/ 3

 398.06 × 1012 ( 7200 )2   r=   4π 2   (a) altitude

h = r − R = 1.685 × 106 m

(b) cosθ =

R 6.37 × 106 = = 0.7908 r 8.055 × 106

= 8.055 × 106 m

h = 1685 km W

θ = 37.74° t AB =

( 75.48)( 7200 ) = 1509.6 s 2θ τ = 360° 360 t AB = 25.2 min W

Vector Mechanics for 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.

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Chapter 12, Solution 86.

GMm = man r2

an =

(

GM r2

)(

GM = 34.4 × 10−9 ft 4 /lb ⋅ s 4 5.03 × 1021 lb ⋅ s 2 /ft

)

= 173.032 × 1012 ft 3/s 2 an = g

(a) At the surface of the moon,

r = R = 1080 mi = 5.7024 × 106 ft g=

(

)(

34.4 × 10−9 ft 4 /lb ⋅ s 4 5.03 × 1021 lb ⋅ s 2 /ft GM = 2 R2 5.7024 × 106 ft

(

)

) g = 5.32 ft/s 2 W

(b) Orbit of space vehicle. r = 1080 + 200 = 1280 mi = 6.7584 × 106 ft an = v=

τ =

τ =

v 2 GM = 2 r r GM r

2π r 2π r 3/2 = v GM

(

2π 6.7584 × 106

)

3/2

12

173.032 × 10

= 8.3923 × 103 s

Vector Mechanics for 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.

τ = 2.33 h W

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Chapter 12, Solution 87.

GMm mv 2 = man = 2 r r 2π r

v= r3

τ

2

v2 = 4π 2r 2

τ

τ

2

=

GM r GM r

GM = constant 4π 2

=

r = 183.3 × 103 mi = 967.8 × 106 ft

Tethys:

τ = 1.888 days = 163.12 × 103 s

( (

) )

967.8 × 106 GM = 4π 2 163.12 × 103

3 2

= 34.068 × 1015 ft 3/s 2

τ = 4.52 days = 390.53 × 103 s

Rhea: r3 =

GM 2 τ 4π 2

(

)(

= 34.068 × 1015 390.53 × 103

)

2

= 5.1958 × 1027 ft 3

(a) r = 1.732 × 109 ft

r = 328 × 103 mi W

(b) Mass of Saturn. M =

=

4π 2  GM    G  4π 2 

(

4π 2 34.068 × 1015 34.4 × 10

−9

) = 39.1 × 10

24

M = 39.1 × 1024 lb ⋅ s 2 /ft W

Vector Mechanics for 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.

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Chapter 12, Solution 88.

R = 6370 km = 6.37 × 106 m rA = 6370 + 960 = 7330 km = 7.33 × 106 m rB = 6370 + 8300 = 14670 km = 14.67 × 106 m

mrAv A = mrB ( vB )θ

( vB )θ =

(

)(

)

7.33 × 106 10.4 × 103 rAv A = = 5.196 × 103 m/s 6 rB 14.67 × 10 x A = 0,

Parabola AB.

yA = 0

xB = rB = 14.67 × 106 m yB = rA = 7.33 × 106 m For a parabolic trajectory, y = kx 2

Differentiating with respect to x,

At point B,

from which

k =

yB xB 2

dy 2 yB x = 2kx = dx xB 2

(

)

( 2) 7.33 × 106 dy 2 yB = tan φ = = = 1.0000 dx xB 14.67 × 106 φ = 45° vB =

( vB )θ

sin 45°

=

5.196 × 103 = 7.349 × 103 m/s sin 45° v B = 7.35 × 103 m/s

Vector Mechanics for 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.

45° W

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Chapter 12, Solution 89.

( GM )earth = gR 2 = ( 32.2 ) ( 20.909 × 106 )

For earth, R = 3960 mi = 20.909 × 106 ft

(

)(

2

= 14.077 × 1015 ft 3/s 2

)

For sun, ( GM )sun = 332.8 × 103 14.077 × 1015 = 4.6849 × 1021 ft 3/s 2 For circular orbit of earth, rE = 92.96 × 106 mi = 490.8 × 109 ft

( GM )sun

vE = For transfer orbit AB,

rE

=

4.6849 × 1021 = 97.70 × 103 ft/s 490.8 × 109

rB = rM = 141.5 × 106 mi = 747.12 × 109 ft

rA = rE ,

v A = vE + ( ∆v ) A = 97.70 × 103 + (1.83)( 5280 ) = 107.36 ft/s

mrAv A = mrBvB vB =

(

)(

)

490.8 × 109 107.36 × 103 rAv A = = 70.527 ft/s rB 747.12 × 103

For circular orbit of Mars, rM = 141.5 × 106 mi = 747.12 × 109 ft vM =

( GM )sun rM

=

4.6849 × 1021 = 79.187 × 103 ft/s 747.12 × 109

Speed increase at B.

( ∆v ) B = v M

− vB = 79.187 × 103 − 70.527 × 103 = 8.660 × 103 ft/s

( ∆v )B = 1.640 mi/s

Vector Mechanics for 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.

W

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Chapter 12, Solution 90.

Circular orbits: v =

GM r rA = 1400 mi = 7.392 × 106 ft

( vA )1 =

( 34.4 × 10 )( 5.03 × 10 ) −9

21

7.392 × 10

6

= 4.8382 × 103 ft/s

rB = 1300 mi = 6.864 × 106 ft

( vB ) 2 =

( 34.4 × 10 )( 5.03 × 10 ) −9

21

6.864 × 10

6

= 5.0208 × 103 ft/s

(a) Transfer orbit AB.

( v A )2

= ( v A )1 + ( ∆v ) A = 4.8382 × 103 − 86 = 4.7522 × 103 ft/s

mrA ( v A )2 = mrB ( vB )1

( vB )1 =

rA ( v A )2 rB

( 7.392 × 10 )( 4.7522 × 10 ) = 5.1178 × 10 = 6

3

3

6.864 × 103

ft/s

( vB )1 = 5.12 × 103 ft/s W (b) Speed change at B.

( ∆vB ) = ( vB )2 − ( vB )1 = 5.0208 × 103 ft/s − 5.1178 × 103 ft/s = −97.0 ft/s Speed reduction at B.

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∆vB = 97.0 ft/s W

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Chapter 12, Solution 91.

R = 6370 km = 6.37 × 106 m, rA = rD = 6370 + 610 = 6980 km = 6.98 × 106 m

rB = 6370 + 290 = 6660 km = 6.66 × 106 m, rC = 6900 km = 6.90 × 106 m

( vA )circ ( vB )circ

=

gR 2 = rA

=

gR 2 = rB

( 9.81) ( 6.37 × 106 ) 6.98 × 10

2

= 7.55173 × 103 m/s

6

( 9.81) ( 6.37 × 106 )

2

6.66 × 106

= 7.73102 × 103 m/s

For path BC. vB = ( vB )circ + ( ∆v ) B = 7.73102 × 103 + 85 = 7.81602 × 103 m/s

mrBvB = mrC ( vC )1

( vC )1 =

(

)(

)

6.66 × 106 7.81602 × 103 rBvB = = 7.54416 × 103 m/s 6 rC 6.90 × 10

For path CD.

( vC )2 = ( vC )1 + ( ∆v )C

= 7.54416 × 103 + 79 = 7.62316 × 103 m/s

mrC ( vC )2 = mrD vD vD =

( ∆v ) D

rC ( vC )2 rD

( 6.90 × 10 )( 7.62316 × 10 ) = 7.53579 × 10 6

=

3

3

6.98 × 106

m/s

= ( v A )circ − vD = 7.55173 × 103 − 7.53579 × 103 = 15.94 m/s

( ∆v )D

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= 15.94 m/s W

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Chapter 12, Solution 92.

Masses: mA = mB =

2.6 = 0.08075 lb ⋅ s 2 /ft 32.2

Let y be the position coordinate of B, positive upward with origin at O. Constraint of the cord: r − y = constant (a) Kinematics:

( aB ) y

=  y =  r

and

 y =  r

or

( a A )r

=  r − rθ 2

Collar B:

ΣFy = mB aB : T − WB = mB  y = mB  r

Collar A:

ΣFr = m A ( a A )r : − T = mA  r − rθ 2

(

(1)

)

(2)

Adding (1) and (2) to eliminate T, −WB = ( mA + mB )  r + m Arθ 2 a A / rod =  r =

( 0.08075)( 0.6 )(10 )2 − ( 2.6 ) m Arθ 2 − WB = mA + mB 0.08075 + 0.08075 a A / rod = 13.90 ft/s2

From (1),

W

T = mB (  r + g ) = ( 0.08075 )(13.90 + 32.2 ) T = 3.72 lb W

(b) Conservation of angular momentum of collar A: ( H 0 )2 = ( H 0 )1

mAr1(vθ )1 = mAr2 (vθ )2 (vθ )2 =

r1(vθ )1 r12θ (0.6) 2 (10) = = = 4.00 r2 r2 0.9

( vθ )2

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= 4.00 ft/s W

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Chapter 12, Solution 93.

Masses: mA = mB =

2.6 = 0.08075 lb ⋅ s 2 /ft 32.2

(a) Conservation of angular momentum of collar A: ( H 0 )2 = ( H 0 )1

mAr1(vθ )1 = mAr2 (vθ )2 (vθ )2 =

r1(vθ )1 r12θ1 (0.6)2 (12) = = = 3.6 r2 r2 1.2

( vθ )2 θ2 =

( vθ )2 rA

=

= 3.60 ft/s W

3.6 = 3.00 rad/s 1.2

(b) Let y be the position coordinate of B, positive upward with origin at O. Constraint of the cord: r − y = constant

 y =  r

or

Kinematics:

( aB ) y

=  y =  r

and

( a A )r

=  r − rθ 2

Collar B:

ΣFy = mB aB : T − WB = mB  y = mB  r

Collar A:

ΣFr = mA ( a A )r : − T = mA  r − rθ 2

(

(1)

)

(2)

Adding (1) and (2) to eliminate T, −WB = ( mA + mB )  r + m Arθ 2 a A / rod =  r=

2 mArθ 2 − WB ( 0.08075 )(1.2 )( 3.00 ) − ( 2.6 ) = = −10.70 ft/s 2 0.08075 + 0.08075 mA + mB

T = mB (  r + g ) = ( 0.08075)( −10.70 + 32.2 )

T = 1.736 lb W

a A / rod = 10.70 ft/s 2 radially inward. W

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Chapter 12, Solution 94.

Since friction and mass of the rod are neglected, the resultant force acting on the collar is the spring force, which is a central force. The angular momentum of the collar about the shaft is constant. mrvθ = mr0v0 = mr02θ0 (a)

vθ =

r02θ0 (0.150)2 (12) = r 0.600

vθ = 0.450 m/s W

Calculate the spring force. Elongation: Force:

e = 0.600 − 0.750 = − 0.150 m Fr = − ke = − (5)(−0.150) = 0.750 N

Apply Newton’s second law. Fr = mar (b)

ar =

Fr 0.750 = m 0.300

ar = 2.50 m/s 2 W

Fθ = 0 aθ = (c)

Fθ m

aθ = 0

W

Acceleration of B relative to the rod. v2 ar =  r − rθ 2 =  r− θ r aB/rod =  r = ar +

vθ2 (0.450)2 = 2.50 + r 0.600 aB/rod = 2.84 m/s 2 W

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Chapter 12, Solution 95.

Data from Prob. 12.94.

θ0 = 12 rad/s, l0 = 750 mm = 0.750 m, r0 = 150 mm = 0.150 m, r = 600 mm = 0.600 m.

m = 300 g = 0.300 kg,

k = 5 N/m,

Since friction and mass of the rod are neglected, the resultant force acting on the collar is the spring force, which is a central force. The angular momentum of the collar about the shaft is constant. mrvθ = mr 2θ = mr02θ0

θ = (a)

r02θ0 r2

Radial component vr of the collar. r − rθ 2 Kinematics: ar =  Newton’s second law: Fr = − k (r − l0 ) = mar k  r − rθ 2 = − (r − l0 ) m

Eliminating ar,

 r=

dr dr dr  dr  1 d (r) 2 = =   r = dt dr dt  dr  2 dr

1 d k (r)2 = rθ 2 − (r − l0 ) 2 dr m =

r04θ02 k − (r − l0 ) m r3

Separate variables and integrate with respect to r, noting that r = 0 when r = r0.  r 2  r 2θ 2 k d   = 0 0 dr − (r − l0 )dr r m  2

r 2 2

r

= r02θ02 0

∫

r

dr k − 3 r0 r m

∫

r r0

(r − l0 ) dr continued

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r

r

1 2 k 1  1   r = r04θ02  − 2  −  (r − l0 ) 2  2  2r  r m  2 r 0

0

 1 1  k r 2 = r02θ02  2 − 2  −  (r − l0 )2 − (r0 − l0 )2  r  m  r0

1  5   1 (0.600 − 0.750) 2 − (0.150 − 0.750)2  = (0.150)4 (12)2  − − 2 2 0.600  0.3   0.150 = 3.0375 + 5.625 = 8.6625 m 2/s 2 r = 2.9432 m/s (b)

vr = 2.94 m/s W

Value of θ. F aθ = rθ + 2rθ = θ = 0 m

θ = −

2rθ r

r 2θ (0.150) 2 (12) where θ = 0 2 0 = = 0.750 rad/s (0.600) 2 r

θ = −

(2)(2.9432)(0.750) 0.600

Vector Mechanics for 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.

θ = − 7.36 rad/s 2 W

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Chapter 12, Solution 96.

u =

1 2 − cosθ = , r r0

d 2u 2 F +u = = 2 r dθ mh 2u 2 0

Solving for F,

F =

du sin θ = , dθ r0

d 2u cosθ = 2 r0 dθ

by Eq. (12.37 )

2mh 2u 2 2mh 2 = r0 r0r 2

Since m, h, and r0 are constants, F is proportional to

1 , or inversely proportional to r 2. r2

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Chapter 12, Solution 97.

u =

1 6cosθ − 5 = , r r0

d 2u 5 F +u =− = 2 r dθ mh 2u 2 0

Solving for F,

F =−

du 6sin θ =− , dθ r0

d 2u 6cosθ = dθ r0

by Eq. (12.37 ) .

5mh 2u 2 5mh 2 =− r0 r0r 2

1 , or inversely proportional to r 2. The minus sign r2 indicates that the force is repulsive, as shown in Fig. P12.97.

Since m, h, and r0 are constants, F is proportional to

Vector Mechanics for 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.

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Chapter 12, Solution 98.

From Fig. P12.98, But

y =

x2 4r0

x = r sin θ ,

y = r0 − r cosθ

or r0 − r cosθ =

(

r 2 1 − cos 2 θ r 2 sin 2 θ = 4r0 4r0  1 − cos 2 θ  4r0 

)

 2  r + ( cosθ ) r − r0 = 0 

Solving the quadratic equation for r, r =

=

 2  2r0  − cosθ ± cos 2 θ − 4  1 − cos θ  4r0 1 − cos 2 θ   

   ( −r0 )    

2r (1 − cosθ ) 2r0 − cosθ ± 1) = 0 2 ( 1 − cos θ 1 − cos 2 θ

since r > 0. Simplifying gives

r =

2r0 1 + cosθ

1 1 + cosθ = =u r 2r0

or

du sin θ =− dθ 2r0

and

(1)

d 2u cosθ =− 2r0 dθ 2

d 2u 1 F +u = = 2 2r0 dθ mh 2u 2

Solving for F,

F =

mh 2u 2 mh 2 = 2r0 2r0r 2

Since m, h, and r0 are constants, F is proportional to

1 , or inversely proportional to r 2. r2

1 GM = 2 (1 + ε cosθ ) = u r h Comparing with (1) shows that ε = 1 and

By Eq. (12.39′ ) ,

GM 1 = 2 2 r0 h

Vector Mechanics for 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.

h=

2GMr0 W

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Chapter 12, Solution 99.

u=

1 1 −bθ = e r r0

du b = − e−bθ dθ r0 d 2u b 2 −bθ = e r0 dθ 2 d 2u b 2 + 1 −bθ F + u = e = 2 r0 dθ mh 2u 2 F=

=

(b 2 + 1)mh 2u 2 −bθ e r0 (b 2 + 1)mh 2u 2 (b 2 + 1)mh 2 = r r3

Since b, m, and h are constants, F is proportional to

1 , or inversely proportional to r 3. r3

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Chapter 12, Solution 100.

For a parabolic trajectory, the velocity v0 at the perigee is equal to the escape velocity. v0 = vesc =

2GM r0 M =

Solving for M, Data:

r0v02 2G

r0 = 350 × 103 km = 350 × 106 m v0 = 26.9 km/s = 26.9 × 103 m/s G = 66.73 × 10−12 m3/kg ⋅ s 2 M =

(350 × 106 )(26.9 × 103 )2 (2)(66.73 × 10−12 )

Vector Mechanics for 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.

M = 1.898 × 1027 kg W

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Chapter 12, Solution 101.

(

)(

)

GM = 34.4 × 10−9 ft 4 /lb ⋅s 4 334 × 1021 lb ⋅ s2 /ft = 11.490 × 1015 ft 3/s 2 R = 3761 mi = 19.858 × 106 ft r0 = 3761 + 175 = 3936 mi = 20.782 × 106 ft (a) Velocity of probe as it approaches A.

v0 =

2GM = r0

( 2 ) (11.490 × 1015 ) 20.782 × 10

6

= 33.252 × 103 ft/s or v0 = 6.30 mi/s W

vcir =

GM = r0

11.490 × 1015 = 23.513 × 103 ft/s 20.782 × 106

(b) Decrease velocity at A. ∆v = vcirc − v0 = −9.739 × 103 ft/s = −1.845 mi/s or ∆v = 1.845 mi/s W

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Chapter 12, Solution 102.

For earth, R = 6.37 × 106 m

g = 9.81 m/s 2

(

GM = gR 2 = ( 9.81) 6.37 × 106

(

)

2

= 398.06 × 1012 m3/s 2

)

For Io, GM = ( 0.01496 ) 398.06 × 1012 = 5.955 × 1012 m3/s 2 r0 = 2.820 × 106 m,

v0 = 15 × 103 m/s

h = r0v0 = 42.3 × 109 m 2 /s 1 GM = 2 (1 + ε cosθ ) r h

(

1 GM = 2 (1 + ε ) r0 h

and

)

2

42.3 × 109 h2 1+ε = = = 106.5 r0GM 2.820 × 106 5.955 × 1012

(

)(

)

ε = 105.5 W

Vector Mechanics for 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.

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Chapter 12, Solution 103.

R = 6.37 × 106 m

Earth:

g = 9.81 m/s 2

GM e = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3/s 2

Jupiter: GM = (318)(398.06 × 1012 ) = 126.583 × 1015 m3/s 2 Orbit of satellite. a = 23.6 × 109 m 1 GM = 2 (1 + ε ) r0 h r0 =

h2 GM (1 + ε )

2a = r0 + r1 =

1 GM = 2 (1 − ε ) r1 h r1 =

h2 GM (1 − ε )

2h 2 GM (1 − ε 2 )

h 2 = (1 − ε 2 )GMa = (1 − 0.452 )(126.583 × 1015 )(23.6 × 109 ) = 2.3824 × 1027 m 4 /s 2 h = 48.810 × 1012 m/s r0 =

2.3824 × 1027 = 12.980 × 109 m (126.583 × 1015 )(1 + 0.45)

r1 =

2.3824 × 1027 = 34.220 × 109 m (126.583 × 1015 )(1 − 0.45)

vmax =

h 48.810 × 1012 = = 3.760 × 103 9 r0 12.980 ×10

vmin =

h 48.810 × 1012 = = 1.426 × 103 r1 34.220 ×109

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vmax = 3760 m/s ! vmin = 1426 m/s !

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Chapter 12, Solution 104.

Using Eq. (12.39 ) , But

1 GM = 2 + C cosθ A rA h

and

1 GM = 2 + C cosθ B . rB h

θ B = θ A + 180°, so that cosθ A = − cosθ B .

Adding,

1 1 1 1 2GM + = + = rA rB ro r1 h2

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Chapter 12, Solution 105. For earth, R = 3960 mi = 20.909 × 106 ft

(

)

GM = gR 2 = ( 32.2 ) 20.909 × 106 = 14.077 × 1015 ft 3/s 2

rA = 3960 + 11870 = 15830 mi = 83.582 × 106 ft vcirc =

GM = rA

vesc =

2GM = rA

14.077 × 1015 = 12.978 × 103 ft/s 83.582 × 106 2 vcirc = 18.353 × 103 ft/s

(a) Increase in speed at A. ∆v = vesc − vcirc = 5.375 × 103 ft/s

Elliptical orbit with Using Eq. (12.37 ) , But

rB = 3960 + 3960 = 7920 mi = 41.818 × 106 ft. 1 GM = 2 + C cosθ A rA h

and

1 GM = 2 + C cosθ B . rB h

θ B = θ A + 180°, so that cosθ A = − cosθ B

Adding,

h=

∆v = 5.38 × 103 ft/s !

1 1 r + rB 2GM + = A = rA rB rArB h2

2GMrArB = rA + rB

( 2 ) (14.077 × 1015 )(83.582 × 106 )( 41.818 × 106 ) 125.4 × 106

= 885.848 × 109 ft 2 /s vA =

h 885.848 × 109 = = 10.599 × 103 ft/s rA 83.582 × 106

(b) Decrease in speed. ∆v = vcirc − v A = 2.379 × 103 ft/s (c)

∆v = 2380 ft/s !

1 1 r − rB − = A = C cosθ B − C cosθ A = 2C rB rA rArB

C =

By Eq. (12.40),

rA − rB 41.764 × 106 = = 5.974 × 10−9 ft −1 2rArB ( 2 ) 83.582 × 106 41.818 × 106

(

(

)(

)(

5.974 × 10−9 885.848 × 109 Ch2 = ε = GM 14.077 × 1015

)

)

2

ε = 0.333 !

Vector Mechanics for 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.

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Chapter 12, Solution 106.

For earth, R = 6370 km = 6.370 × 106 m

(

GM = gR 2 = ( 9.81) 6.37 × 106

)

2

= 398.06 × 1012 m3/s 2

rA = 6370 + 10,000 = 16370 km = 16.370 × 106 m rB = 6370 + 140, 000 = 146,370 = 146.37 × 106 m For a circular orbit with r0 = rA ,

GM = r0

vcirc =

398.06 × 1012 = 4.931 × 103 m/s 6 16.370 × 10

Elliptic orbit. 1 GM = 2 + C cosθ A rA h

Using Eq. (12.39), But

and

1 GM = 2 + C cosθ B . rB h

θ B = θ A + 180°, so that cosθ B = − cosθ A

Adding,

1 1 r + rB 2GM + = A = rA rB rArB h2 h=

2GMrArB = rA + rB

( 2 ) ( 398.06 × 1012 )(16.370 × 106 )(146.37 × 106 ) 162.74 × 106

= 108.27 × 109 m 2 /s vA =

h 108.27 × 109 = = 6.614 × 103 m/s rA 16.370 × 106

(a) Increase in speed at A. ∆v A = v A − vcirc = 1.683 × 103 m/s

∆v A = 1.683 × 103 m/s W

(b) Speed of observatory at B. vB =

h 108.27 × 109 = rB 146.37 × 106

Vector Mechanics for 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.

vB = 7.40 × 103 m/s W

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Chapter 12, Solution 107.

For earth, R = 6.37 × 106 m

(

GM = gR 2 = ( 9.81) 6.37 × 106

(

)(

)

2

= 398.06 × 1012 m3/s 2

)

For sun: GM = 332.8 × 103 398.06 × 1012 = 132.474 × 1018 m3/s 2

rA = 325 × 109 m, rB = 148 × 109 m

For elliptic orbit AB, Using Eq. (12.39),

1 GM = 2 + C cosθ A rA h

But θ B = θ A + 180°, Adding,

so that

and

1 GM = + C cosθ B . rB hAB 2

cosθ A = − cosθ B .

1 1 r + rB 2GM + = A = rA rB rArB hAB 2 hAB =

2GMrArB = rA + rB

( 2 ) (132.474 × 1018 )( 325 × 109 )(148 × 109 ) 473 × 109

= 5.1907 × 1015 m 2 /s (a)

( vA )1 =

hAB 5.1907 × 1015 = = 15.971 × 103 rA 325 × 109

For transfer orbit AB′,

( vA )1 = 15.97 × 103 m/s

rB′ = 137.6 × 109 m continued

Vector Mechanics for 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.

W

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2GMrArB′ = rA + rB′

hAB′ =

( 2 ) (132.474 × 1018 )( 325 × 109 )(137.6 × 109 ) 462.6 × 109

= 5.0609 × 1015 m 2 /s

( v A )2

=

hAB′ 5.0609 × 1015 = = 15.572 × 103 m/s 9 rA 325 × 10

(b) Decrease of speed at A.

( ∆vA ) = ( vA )1 − ( vA )2

hAB′ 5.0609 × 1015 = = 36.780 × 103 m/s rB′ 137.6 × 109

( vB )1 =

rA′ = 264.7 × 109 m

For elliptic orbit B′A′,

hB′A′ =

∆v A = 399 m/s W

= 399 m/s

2GMrB′rA′ = rB′ + rA′

( 2 ) (132.474 × 1018 )(137.6 × 109 )( 264.7 × 109 ) 402.3 × 109

= 4.8977 × 1015 m 2 /s

( vB ′ ) 2

=

hB′A′ 4.8977 × 1015 = = 35.594 × 103 m/s rB′ 137.6 × 109

Decrease in speed at B′:

( vB′ )1 − ( vB′ )2

= 1.186 × 103

Vector Mechanics for 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.

∆vB = 1.186 × 103 m/s W

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Chapter 12, Solution 108.

For Earth,

(

GM earth = gR 2 = ( 9.81) 6.37 × 106

)

2

= 398.06 × 1012 m3/s 2

For Venus, GM = 0.82 GM earth = 326.41 × 1012 m3/s 2 For a parabolic trajectory with rA = 15 × 106 m

( v A )1 = vesc =

2GM = rA

( 2 ) ( 326.41 × 1012 ) 15 × 106

= 6.5971 × 103 m/s

rB = 300 × 106 m

First transfer orbit AB. At point A, where θ = 180°

1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB

(1)

At point B, where θ = 0° 1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB

Adding,

(2)

1 1 r + rA 2GM + = B = 2 rA rB rArB hAB

Solving for hAB, hAB =

2GMrArB = rB + rA

( 2 ) ( 326.41 × 1012 )(15 × 106 )( 300 × 106 ) 315 × 106

= 96.571 × 109 m 2 /s continued

Vector Mechanics for 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.

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( v A )2 =

hAB 96.571 × 109 = = 6.4381 × 103 m/s rA 15 × 106

( vB )1 =

hAB 96.571 × 109 = = 321.90 m/s rB 300 × 106

rC = 9000 km = 9 × 106 m

Second transfer orbit BC. At point B, where θ = 0.

1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC

At point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC

Adding,

1 1 r + rC 2GM + = B = 2 rB rC rB rC hBC

hBC =

2GMrB rC = rB + rC

( 2 ) ( 326.41 × 1012 )( 300 × 106 )( 9 × 106 ) 309 × 106

= 75.5265 × 109 m 2 /s

( vB )2 =

hBC 75.5265 × 109 = = 251.76 m/s rB 300 × 106

( vC )1 =

hBC 75.5265 × 109 = = 8.3918 × 103 m/s 6 rC 9 × 10

Final circular orbit. rC = 9 × 106 m

( vC )2 =

GM = rC

326.41 × 1012 = 6.0223 × 103 m/s 9 × 106

Vector Mechanics for 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.

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Speed reductions. (a) At A.

( v A )1 − ( vA )2 = 6.5971 × 103 − 6.4381 × 103 ∆v A = 159.0 m/s !

(b) At B.

( vB )1 − ( vB )2 = 321.90 − 251.76 ∆vB = 70.1 m/s !

(c) At C.

( vC )1 − ( vC )2 = 8.3918 × 103 − 6.0223 × 103 ∆vC = 2.37 × 103 m/s !

Vector Mechanics for 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.

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Chapter 12, Solution 109.

Data from Problem 12.108: rC = 9000 km,

M = 0.82 M earth

(

)

2

For Earth,

GM earth = gR 2 = ( 9.81) 6.37 × 106

For Venus,

GM = 0.82 GM earth = 326.409 × 1012 m3/s 2

Transfer orbit AB:

= 398.059 × 1012 m3/s 2

v A = 6500 m/s, rA = 15 × 106 m

(

)

hAB = rAv A = 15 × 106 ( 6500 ) = 97.5 × 109 m 2 /s At point A, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB At point B, where θ = 0° 1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB Adding,

1 1 2GM + = 2 rA rB hAB

( (

)

( 2 ) 326.409 × 1012 1 2GM 1 1 = 2 − = − = 2.0057 × 10−9 m −1 2 6 9 rB rA hAB 15 × 10 97.5 × 10 (a) Radial coordinate rB.

( vB )1 =

)

rB = 498.56 × 106 m

hAB 97.5 × 109 = = 195.56 m/s rB 498.56 × 106

Vector Mechanics for 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.

rB = 499 × 103 km W

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Second transfer orbit BC.

rC = 9 × 106 m

At point B, where θ = 0 1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC At point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC Adding,

1 1 r + rC 2GM + = B = 2 rB rC rB rC hBC hBC =

2GMrB rC = rB + rC

( 2 ) ( 326.409 × 1012 )( 498.56 × 106 )( 9 × 106 ) 507.56 × 106

= 75.968 × 109 m 2 /s

( vB ) 2 =

hBC 75.968 × 109 = = 152.37 m/s rB 498.56 × 106

( vC )1 =

hBC 75.968 × 109 = = 8.4409 × 103 m/s rC 9 × 106

Circular orbit with

( vC )2 =

rC = 9 × 106 m. GM = rC

326.409 × 1012 = 6.0223 × 103 m/s 9 × 106

(b) Speed reductions at B and C. At B

( vB )1 − ( vB )2 = 195.56 – 152.37 ∆vB = 43.2 m/s W

At C

( vC )1 − ( vC )2 = 8.4409 × 103 − 6.0223 × 103 ∆vC = 2.42 × 103 m/s W

Vector Mechanics for 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.

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Chapter 12, Solution 110.

For earth, R = 3960 mi = 20.909 × 106 ft

(

GM = gR 2 = ( 32.2 ) 20.909 × 106

(

)

)

2

= 14.077 × 1015 ft 3/s 2

For Mars, GM = ( 0.1074 ) 14.077 × 1015 = 1.51188 × 1015 ft 3/s 2 rA = 5625 mi = 29.7 × 106 ft, For the parabolic approach trajectory at A,

( 2 ) (1.51188 × 1015 )

2GM = rA

( vA )1 =

29.7 × 106

First elliptic transfer orbit AB. 1 GM Using Eq. (12.39), = 2 + C cosθ A rA hAB But

θ B = θ A + 180°,

Adding,

so that

rB = 112500 mi = 594 × 106 ft

= 10.0901 × 103 ft/s

1 GM = 2 + C cosθ B . rB hAB

and

cosθ A = − cosθ B .

1 1 r + rB 2GM + = A = 2 rA rB rArB hAB

( 2 ) (1.51188 × 1015 )( 29.7 × 106 )( 594 × 106 )

2GMrArB = rA + rB

hAB =

623.7 × 106 hAB = 292.45 × 109 m 2 /s

a=

(

b=

τ AB

rArB =

( 29.7 × 10 )( 594 × 10 ) = 132.82 × 10

2π ab h 1 π ab = τ = 2 h

Periodic time for full ellipse:

For half ellipse AB,

)

1 1 ( rA + rB ) = 29.7 × 106 + 594 × 106 = 311.85 × 106 ft 2 2 6

6

6

ft

τ =

τ AB =

(

)(

π 311.85 × 106 132.82 × 106 9

292.45 × 10

) = 444.95 × 10

Vector Mechanics for 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.

3

s

τ AB = 123.6 h W

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Chapter 12, Solution 111.

From Keplers third law, 2

 TH   aH    =   TE   aE 

3

Solve for aH. T  aH = aE  H   TE 

aH =

Also,

2/3

 76 years  = rE    1 year 

2/3

= 17.942 rE

1 ( rmin + rmax ) 2

Solve for rmax.

rmax = 2aH − rmin = ( 2 )(17.942 rE ) −

1 rE 2

= 35.3844 rE or

(

rmax = ( 35.3844 ) 92.6 × 106 mi

) rmax = 3.28 × 109 mi W

Vector Mechanics for 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.

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Chapter 12, Solution 112.

For earth, R = 3960 mi = 20.909 × 106 ft ,

g = 32.2 ft/s 2

(

GM = gR 2 = ( 32.2 ) 20.909 × 106

)

2

= 14.077 × 1015 ft 3/s

For circular orbit of satellite,

r0 = 3960 + 310 = 4270 mi = 22.546 × 106 ft/s v0 =

GM 14.077 × 1015 = = 24.988 × 103 ft/s r0 22.546 × 106

(

)

6 2π r0 ( 2π ) 22.546 × 10 τ0 = = = 5.6692 × 103 s 3 v0 24.988 × 10

For elliptic orbit of spacecraft it is given that 3 2

τ = τ 0 = 8.5038 × 103 s a= Using Eq. (12.39), But

1 GM = 2 + C cosθ A rA h

θ B = θ A + 180°,

Adding,

1 ( rA + rB ) , 2

so that

τ =

(

rArB

1 GM = 2 + C cosθ B . rB h

cosθ A = − cosθ B .

1 1 r + rB 2a 2GM + = A = 2 = rA rB rArB b h2

Periodic time:

or

h=

GMb 2 a

2π ab 2π ab a 2π a 3 / 2 = = h GM GMb 2

)(

14.077 × 1015 8.5038 × 103 GM τ 2 a = = 4π 2 4π 2 3

and

b=

)

2

= 25.786 × 1021 ft 3

Vector Mechanics for 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.

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a = 29.543 × 106 ft,

rA = r0 = 22.546 × 106 ft

rB = 2a − rA = 36.541 × 106 ft, h=

2π ab

τ

=

(

b = rArB = 28.703 × 106 ft

)(

2π 29.543 × 106 28.703 × 106

vA =

3

8.5038 × 10

) = 626.54 × 10

9

ft 2 /s

h 626.54 × 109 = = 27.789 × 103 ft/s rA 22.546 × 106

(a) Increase in speed at A. ∆v A = v A − v0 = 27.789 × 103 − 24.988 × 103

∆v A = 2.80 × 103 ft/s W

(b) Periodic time for elliptic orbit. As calculated above τ = 8.5038 × 103 s

Vector Mechanics for 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.

τ = 141.7 min W

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Chapter 12, Solution 113.

For earth’s orbit about the sun, v0 =

GM 2π RE 2π RE 3/ 2 , τ0 = = RE v0 GM

GM =

or

2π RE 3/ 2

τ0

(1)

For the comet Hyakutake, 1 GM = 2 = (1 + ε ) , r0 h

1 GM = 2 (1 − ε ) , r1 h

1 r ( r0 + r1 ) = 0 , 2 1−ε

a=

h=

b=

r0r1 =

r1 =

1+ε r0 1−ε

1+ε r0 1−ε

GMr0 (1 + ε )

2π r0 2 (1 + ε ) 2π ab = h (1 − ε )3/ 2 GMr0 (1 + ε ) 1/ 2

τ =

2π r03/ 2

=

GM (1 − ε )

3/ 2

 r  = 0   RE 

3/ 2

1

(1 − ε )3/ 2

= ( 0.230 )

3/ 2

(

=

2π r03/ 2τ 0

2π RE 3 (1 − ε )

3/ 2

τ0

1

(1 − 0.999887 )

3/ 2

τ 0 = 91.8 × 103 τ 0

)

Since τ 0 = 1 yr, τ = 91.8 × 103 (1.000 )

Vector Mechanics for 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.

τ = 91.8 × 103 yr W

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Chapter 12, Solution 114.

For earth, R = 3960 mi = 20.909 × 106 ft, g = 32.2 ft/s 2

(

GM = gR 2 = ( 32.2 ) 20.909 × 106

)

2

= 14.077 × 1015 ft 3/s 2

At point A, rA = 4560 mi = 24.077 × 106 ft v A = 6.48 mi/s = 34.214 × 103 ft/s from which

h = rAv A = 823.78 × 109 ft 2 /s

For trajectory BAC,

1 GM = 2 (1 + ε cosθ ) with ε = 1 r h

At point A, θ = 0 while at B and C , θ = ± 90° 1 1 GM = = 2 rB rC h

or

rB = rC =

h2 = 2rA GM

As the spacecraft travels from B to C, the area swept out is a parabolic area A. A=

(

2 8 8 ( rB + rC ) rA = rA2 = 24.077 × 106 3 3 3 dA 1 = h dt 2 tBC

or

(

A=

)

2

= 15.458 × 1015 ft 2

1 1 h dt = ht ∫ 2 2

)

15 2 A ( 2 ) 15.458 × 10 = = = 3.753 × 103 s h 823.78 × 109

Vector Mechanics for 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.

τ BC = 1.043 h W

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Chapter 12, Solution 115.

g = 32.2 ft/s 2 ,

For earth,

R = 3960 mi = 20.9088 × 106 ft

(

GM = gR 2 = ( 32.2 ) 20.9088 × 106

)

2

= 14.077 × 1015 ft 3/s 2

r0 = 3960 + 182 = 4142 mi = 21.8698 × 106 ft

For the orbit,

1 GM = 2 (1 − ε ) r1 h

1 GM = 2 (1 + ε ) r0 h r1 = r0 a=

1+ ε 1.0356 − 21.8698 × 106 = 23.4844 × 106 ft 1−ε 0.9644

(

)

1 ( r0 + r1 ) = 22.6770 × 106 ft 2

b = r0r1 = 22.6627 × 106 ft

h=

(1 + ε ) GMr0

=

(1.0356 ) (14.077 × 1015 )( 21.8698 × 106 )

= 564.64 × 109 ft 2 /s

(

)(

6 6 2π ab 2π 22.6770 × 10 22.6627 × 10 τ = = h 564.64 × 109

)

= 5.7188 × 103 s

Vector Mechanics for 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.

τ = 95.3 min !

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Chapter 12, Solution 116.

For the circular orbit, r0 = rA = nR GM = r0

v0 =

GM nR

The crash trajectory is elliptic.

β 2GM

v A = β v0 =

nR

h = rAv A = nRv A =

β 2n GMR

GM 1 = 2 2 β nR h 1 GM 1 + ε cosθ = 2 (1 + ε cosθ ) = r β 2nR h At point A, θ = 180° 1 1 1−ε = = 2 rA nR β nR

β2 = 1 − ε

or

ε = 1− β2

or

At impact point B, θ = π − φ 1 1 = rB R 1 1 + ε cos (π − φ ) 1 − ε cos φ = = R β 2nR β 2nR

ε cos φ = 1 − nβ 2

or

cos φ =

1 − nβ 2

ε

=

1 − nβ 2 1 − β2

(

)(

)

φ = cos −1  1 − nβ 2 / 1 − β 2  W 

Vector Mechanics for 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.

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Chapter 12, Solution 117

For earth, R = 3960 mi = 20.909 × 106 ft

(

GM = gR 2 = ( 32.2 ) 20.909 × 106 For the moon,

(

)

2

= 14.077 × 1015 ft 3/s 2

)

GM = ( 0.01230 ) 14.077 × 1015 = 173.149 × 1012 ft 3/s 2

For elliptic orbit AB, rA = 1110 mi = 5.861 × 106 ft, rB = 2240 mi = 11.827 × 106 ft Using Eq. (12.39), But

1 GM = 2 + C cosθ A rA h

θ B = θ A + 180°,

Adding,

so that

and

1 GM = 2 + C cosθ B . rB h

cosθ B = − cosθ A

1 1 r + rB 2GM + = A = rA rB rArB h2 h=

2GMrArB = rA + rB

( 2 ) (173.149 × 1012 )( 5.861 × 106 )(11.827 × 106 ) 17.688 × 106

( vB )1 = For crash trajectory BC, At B, θ = 180°, r = rB , At C, θ = 70°, r = rC ,

= 36.839 × 109 ft 2 /s

hAB 36.839 × 109 = = 3.1148 × 103 ft/s rA 11.827 × 106

1 GM = 2 (1 + ε cosθ ) r h 1 GM = 2 (1 − ε ) rB hBC 1 GM = 2 (1 + ε cos 70° ) rC hBC

Vector Mechanics for 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.

(1)

(2)

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Dividing Eq. (2) by Eq. (1), rB 1 + ε cos 70° = rC 1−ε

ε = From Eq. (1), hBC =

GM (1 − ε ) rB =

( vB ) 2

=

ε =

or

rB / rC − 1 ( rB / rC ) + cos 70°

2240 /1080 − 1 = 0.44455 2240 /1080 + cos 70°

(173.149 × 10 ) ( 0.55545) (11.827 × 10 ) = 33.726 × 10 12

6

9

ft 2 /s

hBC 33.726 × 109 = = 2.8516 × 103 ft/s 6 rB 11.827 × 10

∆vB = ( vB )2 − ( vB )1 = −263.2 ft/s

Vector Mechanics for 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.

∆vB = 263 ft/s W

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Chapter 12, Solution 118.

For earth,

R = 3960 mi = 20.909 × 106 ft

(

GM = gR 2 = ( 32.2 ) 20.909 × 106 For the trajectory,

)

2

= 14.077 × 1015 ft 3/s 2

rC = 3960 + 930 = 4890 mi = 25.819 × 106 ft rA = rB = R = 20.909 × 106 ft,

rC 25.819 × 106 = = 1.2348 rA 20.909 × 106

Range A to B: s AB = 3700 mi = 19.536 × 106 ft

φ=

s AB 19.536 × 106 = = 0.9343 rad = 53.534° R 20.909 × 106

1 GM = 2 (1 + ε cosθ ) r h φ 1 GM θ = 180° − = 153.233°, = 2 (1 + ε cos153.233°) 2 rA h

For an elliptic trajectory, At A,

(1)

1 GM = 2 (1 − ε ) rC h Dividing Eq. (1) by Eq. (2), rC 1 + ε cos153.233° = = 1.2348 rA 1−ε At C,

θ = 180° ,

ε = From Eq. (2),

h=

(2)

1.2348 − 1 = 0.6866 1.2348 + cos153.233°

GM (1 − ε ) rC h=

(14.077 × 10 ) ( 0.3134) ( 25.819 × 10 ) = 337.500 × 10

(a) Velocity at C.

15

vC =

6

h 337.5 × 109 = = 13.07 × 103 ft/s rC 25.819 × 106

(b) Eccentricity of trajectory.

Vector Mechanics for 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.

9

ft 2 /s

vC = 2.48 mi/s W

ε = 0.687 W

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Chapter 12, Solution 119.

Radius of Earth

R = 3960 mi = 20.9088 × 106 ft

At point A,

rA = R + altitude = 4160 mi = 21.9648 × 106 ft

For Earth,

GM = gR 2 = ( 32.2 ) 20.9088 × 106

(

)

2

= 14.0771 × 1015 ft 3/s 2

For circular orbit at point A,

( vA )1 =

GM 14.0771 × 1015 = = 25.3159 × 103 ft/s 6 rA 21.9648 × 10

After speed reduction at A,

( vA )2 = ( 0.94 ) ( 25.3159 × 103 ) = 23.7969 × 103 ft/s For trajectory AB,

(

)(

hAB = rA ( v A )2 = 21.9648 × 106 23.7969 × 103

)

= 522.6946 × 109 ft 2 /s At point A, θ = 0. 1 GM GM = 2 + C cos θ = 2 + C rA hAB hAB Solving for C,

C=

1 GM 1 14.0771 × 1015 − 2 = − 6 rA hAB 21.9648 × 10 522.6946 × 109

(

)

2

= − 5.997512 × 10−9 ft −1

continued

Vector Mechanics for 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.

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At point B, θ = 50° 1 GM = 2 + C cos 50° rB hAB =

(

14.0771 × 1015 522.6946 × 109

)

2

+ ( −5.997512 cos 50° )

= 47.6698 × 10−9 ft −1 rB = 20.9777 × 106 ft

Altitude at B.

rB − R = 68.9 × 103 ft

altitude = 13.05 mi W

Vector Mechanics for 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.

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Chapter 12, Solution 120.

Using Eq. (12.39),

1 GM = 2 + C cosθ A rA h

But θ B = θ A + 180°, Adding,

so that

1 GM = 2 + C cosθ B . rB h

and

cosθ A = − cosθ B .

1 1 2GM + = rA rB h2

At points A and B the radial direction is normal to the path. an =

But

Fn =

v2

ρ

=

h2 r2ρ

GMm mh 2 ma = = n r2 r2ρ

1

ρ

=

GM 1 1 1 =  +  2 2  rA rB  h

Vector Mechanics for 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.

1

ρ

=

1 1 1  + W 2  r0 r1 

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Chapter 12, Solution 121.

1 GM = 2 (1 + ε cosθ ) r h 1 GM = 2 = (1 + ε ) rA h

At A,

θ =0

At B,

θ = 180°

or

1 GM = 2 = (1 − ε ) rB h

rA =

h2 GM (1 + ε )

or

rB =

h2 GM (1 − ε )

rB 1 + ε r = = 1 rA 1 − ε r0

ε =

(a)

r1 − r0 W r1 + r0

RE = 93 × 106 mi r0 = 0.230 RE = 21.39 × 106 mi (b)

r1 =

1+ ε 1 + 0.999887 21.39 × 106 r0 = 1−ε 1 − 0.999887

(

)

Vector Mechanics for 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.

r1 = 379 × 109 mi W

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Chapter 12, Solution 122.

For an ellipse 2a = rA + rB and b =

Using Eq. (12.39), But

1 GM = 2 + C cosθ A rA h

θ B = θ A + 180°,

Adding,

rArB

so that

1 GM = 2 + C cosθ B . rB h

and

cosθ A = − cosθ B .

1 1 r + rB 2a 2GM + = A = 2 = rA rB rArB b h2 h=b

GM a

By Eq. (12.45),

τ = τ2 =

2π ab 2π ab a 2π a3/ 2 = = h b GM GM 4π 2a3 GM

For orbits 1 and 2 about the same large mass,

τ12 =

4π 2a13 GM

and

τ 22 =

4π 2a23 GM 2

Forming the ratio,

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3

 τ1   a1    =  W  τ2   a2 

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Chapter 12, Solution 123.

1 GM = 2 (1 + ε cosθ ) r h

By Eq. (12.39′ ) ,

At A,

θ = 0°

1 GM = 2 = (1 + ε ) rA h

or

rA =

h2 GM (1 + ε )

At B,

θ = 180°

1 GM = 2 = (1 − ε ) rB h

or

rB =

h2 GM (1 − ε )

Adding,

rA + rB =

But for an ellipse,

h2 1  2h 2  1 = + = GM  1 + ε 1 − ε  GM 1 − ε 2

(

)

rA + rB = 2a 2a =

2h 2

(

GM 1 − ε 2

)

Vector Mechanics for 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.

h=

(

GMa 1 − ε 2

)W

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Chapter 12, Solution 124.

(a)

Front wheel drive with N f = 0.65W . ΣF = ma : µ s N F =

a=

W a g

( 32.2 )( 0.80 )( 0.65W ) = 16.74 ft/s2 g µs N F = W W

For constant acceleration, 2 vmax = 2ax = ( 2 )(16.74 )(1320 ) = 44.204 × 103 ft 2 /s 2

vmax = 210 ft/s

vmax = 143.2 mi/h W

(b) Rear wheel drive with N R = 0.42 W.

ΣF = ma : µ s N R =

a=

W a g

( 32.2 )( 0.80 )( 0.42W ) = 10.82 ft/s2 g µs N R = W W

For constant acceleration, 2 vmax = 2ax = ( 2 )(10.82 )(1320 ) = 28.563 × 103 ft 2 / s 2

vmax = 169.0 ft/s

Vector Mechanics for 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.

vmax = 115.2 mi/h W

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Chapter 12, Solution 125.

v0 = 90 km/h = 25 m/s

ΣFx = ma : − 60 × 103 − 16 × 103 = ( 7900 + 6800 ) a a=−

For constant acceleration,

76 × 103 = −5.170 m/s 2 14700

v 2 v02 − = a ( x − x0 ) 2 2

0 − ( 25 ) v 2 − v02 (a) x − x0 = = 2a ( 2 )( −5.170 ) 2

x − x0 = 60.4 m W

ΣFx = ma : − FH − 16 × 103 = ( 6800 )( −5.170 ) (b) FH = ( 6800 )( 5.170 ) − 16 × 103 = 19.16 × 103 N

Vector Mechanics for 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.

FH = 19.16 kN W

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Chapter 12, Solution 126.

a B = a A + a B / A , where a B / A is directed along the inclined contact surface. Block B:

ΣFx = Σmax : T − WB sin 25° = mB a A cos 25° + mB aB / A

30  30  aB / A = 50 − 30sin 25° cos 25°  a A +  32.2  32.2  0.84439 a A + 0.93168 aB / A = 37.321

(1)

ΣFy = ΣmB a y : N AB − WB cos 25° = −mB a A sin 25°  30  sin 25°  a A + N AB = 30cos 25° or 0.39374 a A + N AB = 27.189 (2)   32.2  Block A:

ΣFx = max : T − T cos 25° + N AB sin 25° = mAa A 55 a A − ( sin 25° ) N AB = 50 (1 − cos 25° ) 32.2 1.70807 a A − 0.42262 N AB = 4.6846

(3)

Using (2) and (3) to eliminate N AB and solve for a A , (a) Acceleration of block A.

a A = 8.63 ft/s2

W

Substituting for a A into (1) and solving for aB / A , (b) Acceleration of B relative to A.

Vector Mechanics for 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 B / A = 32.2 ft/s 2

25° W

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Chapter 12, Solution 127.

m =

W 180 = = 5.59 lb ⋅ s 2 /ft g 32.2

ρ = 40 ft ΣFn = man : T − W cosθ = man =

T =

mv 2

ρ

mv 2

ρ

+ W cosθ

(a) At top of swing. θ = 30°, v = 0 T = 0 + 180cos30° = 155.9

T = 155.9 lb W

(b) At bottom of swing. θ = 0°, v = 18.6 ft/s T =

( 5.59 )(18.6 )2 40

+ 180 = 228

Vector Mechanics for 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.

T = 228 lb W

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Chapter 12, Solution 128.

The road reaction consists of normal component N and friction component F. The resultant R makes angle φs with the normal. tan φs = µs = 0.65

or

φs = 33.02°

v0 = 110 km/h = 30.56 m/s

(a) θ = 10°. ΣFy = 0 : R cos (θ + φs ) − mg = 0 R =

mg cos (θ + φs )

ΣFx = man : R sin (θ + φs ) = man man = mg tan (θ + φs ) vmax 2 = g tan (θ + φs ) r vmax = rg tan (θ + φs ) =

( 50)( 9.81) tan (10° + 33.02°)

= 21.39 m/s

Speed reduction: ∆v = v0 − vmax = 30.56 − 21.39 = 9.16 m/s ∆v = 33.0 km/h W

(b) θ = −5°. The equation derived in part (a) applies also for negative value of θ . vmax =

rg tan (θ + φs ) =

( 50)(9.81) tan ( −5° + 33.02°) = 16.16 m/s

Speed reduction: ∆v = v0 − vmax = 30.56 − 16.16 = 14.40 m/s ∆v = 51.8 km/h W

Vector Mechanics for 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.

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Chapter 12, Solution 129.

α = 40°,

r = 24 in. = 2 ft, µ s = 0.35

F = ± µ s N , where the upper sign applies for downward impending motion and the lower sign for upward impending motion.

+ΣFy = 0 : N sin α + F cos α − W = 0 N ( sin α ± µ S cosα ) = W W sin α ± µ S cosα

N =

ΣFx = man : N cos α − F sin α = N ( cos α ∓ µ s sin α ) =

v 2 = gr

N ( cos α ∓ µ s sin α ) W

W v2 g r

W v2 g r

= gr

cos α ∓ µ s sin α sin α ± µ s cos α

For impending motion downward, v 2 = ( 32.2 )( 2.0 )

cos 40° − 0.35sin 40° = 38.25 ft 2 /s 2 sin 40° + 0.35cos 40° v = 6.18 ft/s

For impending motion upward, v 2 = ( 32.2 )( 2.0 )

cos 40° + 0.35sin 40° = 170.34 ft 2 /s 2 sin 40° − 0.35cos 40°

v = 13.05 ft/s Range of speeds for which the collar will not slide: 6.18 ft/s ≤ v ≤ 13.05 ft/s W

Vector Mechanics for 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.

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Chapter 12, Solution 130.

(a) At t = 0,

(

)

r = ( 0.5 + 0.3sin π t ) m

θ = 2π t 2 − 2t rad

r = 0.3π cos π t m/s

θ = 4π ( t − 1) rad/s

r = −0.3π 2 sin π t

θ = 4π rad/s 2

r = 0.5 m

θ = 0

r = 0.94248 m/s

θ = −12.5664 rad/s

r = 0

θ = 12.5664 rad/s 2 ar = r − rθ 2 = 0 − ( 0.5 )( −12.5664 ) = −78.957 m/s 2 2

aθ = rθ + 2rθ = ( 0.5 )(12.5664 ) + ( 2 )( 0.94248 )( −12.5664 ) = −17.4040 m/s

Fr = mB ar : Fr = ( 0.4 )( −78.957 ) = −31.6

Fr = −31.6 N W

Fθ = mB aθ : Fθ = ( 0.4 )( −17.4040 ) = −6.96

Fθ = −6.96 N W

(b) At t = 0.8 s, r = 0.67634 m

θ = −6.0319 rad

r = −0.76248 m/s

θ = −2.5133 rad/s

r = −1.74040 m/s 2

θ = 12.5664 rad/s 2

ar = r − rθ 2 = −1.74040 − ( 0.67634 )( −2.5133) = −6.0126 m/s 2 2

aθ = rθ + 2rθ = ( 0.67634 )(12.5664 ) + ( 2 )( −0.76248 )( −2.5133 ) = 12.332 m/s 2

Fr = mB ar : Fr = ( 0.4 )( −6.0126 ) = −2.41 Fθ = mB aθ : Fθ = ( 0.4 )(12.332 ) = 4.93

Vector Mechanics for 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.

Fr = −2.41 N W Fθ = 4.93 N W

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Chapter 12, Solution 131.

θ = 10 rad/s, θ = 0, b = 9 in. = 0.75 ft, W = 4 oz = 0.25 lb b b sin θ , r = Kinematics: r = θ cosθ cos 2 θ r =

=

(

b  cos 2 θ b sin θ  θ + 2 cos θ

(

b 1 + sin 2 θ

)

cos θ 3

ar = r − rθ 2 =

) ( cosθ ) − (sinθ )( 2cosθ )( − sinθ ) θ

2

cos θ 4

with θ = 0

(

b 1 + sin 2 θ cos3 θ

)θ

2

−

b 2b sin 2 θ 2 θ2 = θ cosθ cos3 θ

= 2b tan 2 θ secθ θ 2 b sin θ 2 θ = 2b tan θ secθ θ 2 cos 2 θ (a) Radial and transverse components of effective forces. W 2Wb Fr = ar : Fr = tan 2 θ secθ θ 2 g g aθ = rθ + 2rθ = 0 + 2

=

( 2 )( 0.25)(10 )2 tan 2 θ secθ

Fr = 1.165tan 2 θ secθ lb W

32.2

Fθ =

W 2Wb aθ : Fθ = tan θ secθ θ g g

( 2 )( 0.25)(10 )2 tanθ secθ

Fθ = 1.165 tan θ secθ lb W 32.2 (b) Forces P and Q exerted on the pin by the arm OA and the wall of the slot DE, respectively. =

ΣFy = P cosθ = Fr sin θ − Fθ cosθ

P = 1.165 tan θ sec3 θ lb

θ W

ΣFr = Q cosθ = Fr

Q = 1.165 tan 2 θ sec2 θ lb

Vector Mechanics for 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.

W

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Chapter 12, Solution 132.

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

v=

or

GM r

Let τ be the periodic time to complete one orbit. vτ = 2π r

Solving for M,

M =

or

τ

GM = 2π r r

4π 2r 3 Gτ 2

Data: r = 384.5 × 106 m

τ = 27.32 days = 655.68 h = 2.3604 × 106 s G = 66.73 × 10−12 m3/kg ⋅ s 2

M=

(

4π 2 384.5 × 106

)

3

( 66.73 × 10 )( 2.3604 × 10 ) −12

6

2

= 6.04 × 1024

M = 6.04 × 1024 kg W

Vector Mechanics for 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.

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Chapter 12, Solution 133.

For earth:

Re = 6370 km = 6.37 × 106 m, g e = 9.81 m/s 2

For the given orbit:

re = 6370 + 4500 = 10870 km = 10.870 × 106 m

( GM )e (a) F =

( GM )e m r2

(

= g e Re 2 = ( 9.81) 6.37 × 106

)

2

= 398.06 × 1012 m3/s 2

( 398.06 × 10 ) (540) = 1819 N = (10.870 × 10 ) 12

6

F = 1.819 kN W

2

For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

v=

or

GM r

Let τ be the periodic time to complete one orbit. vτ = 2π r

GM = 2π r r

τ

or

or

τ2 =

4π 2r 3 GM

Since earth orbit and moon orbit have the same periodic time,

τ2 = 1/ 3

M  (b) rm =  m   Me 

re = ( 0.01230 )

1/ 3

4π 2re3 4π 2rm3 = GM e GM m

(10.870 × 10 ) = 2.509 × 10 6

6

m rB = 2510 km W

GM e = ge Re 2 G = 2

(c)

gm

g e Re 2 g R 2 = m m Me Mm

and

GM m = g m Rm 2

with Rm = 1740 km = 1.740 × 106 m 2

 6.37 × 106   M  R  =  m  e  g e = ( 0.01230 )  6  ( 9.81)  M e  Rm   1.740 × 10 

g m = 1.617 m/s 2 W

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Chapter 12, Solution 134.

Let r and θ be polar coordinates with the origin lying at the shaft. Constraint of rod: θ B = θ A + π radians; θ&B = θ&A = θ&; θ&&B = θ&&A = θ&&. (a) Components of acceleration Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them. Owing to frictionless sliding of B along the rod, ( FB )r = 0. Radial component of acceleration of B. Fr = mB ( aB )r :

( aB ) r

= 0


Transverse components of acceleration.

( a A )θ

= rAθ&& + 2r&Aθ& = raθ&&

( aB )θ

= rBθ&& + 2r&Bθ&

(1)

Since the rod is massless, it must be in equilibrium. Draw its free body diagram, applying Newton’s 3rd Law. ΣM 0 = 0: rA ( FA )θ + rB ( FB )θ = rAmA ( a A )θ + rB mB ( aB )θ = 0

(

)

rAmArAθ&& + rB mB rBθ&& + 2r&Bθ& = 0

θ&& = At t = 0,

r&B = 0

−2rB r&Bθ& mArA2 + mB rB 2

so that θ&& = 0.

( aB )θ

From Eq. (1),

=0


(b) Acceleration of B relative to the rod. At t = 0,

( v A )θ

( vA )θ 96 = 8 ft/s = 96 in./s, θ& = = = 9.6 rad/s 10 rA && rB − rBθ& 2 = ( aB )r = 0 2

&& rB = rBθ& 2 = ( 8 )( 9.6 ) = 737.28 in./s 2

&& rB = 61.4 ft/s 2
Continued

Vector Mechanics for 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.

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(c) Speed of A. Substituting

(

)

d mr 2θ& for rFθ in each term of the moment equation dt

gives

(

)

(

)

d d mArA2θ& + mB rB 2θ& = 0 dt dt Integrating with respect to time,

(

mArA2θ& + mB rB 2θ& = mArA2θ&

) + ( m r θ& ) B B

0

2

0

Applying to the final state with ball B moved to the stop at C,

 WA 2 WB 2  & W W 2 rA + rC θ f =  A rA2 + B ( rB )0  θ&0  g g g g    

θ& f =

( vA ) f

2

WArA2 + WB ( rB )0 & (1)(10 ) + ( 2 )(8) 9.6 = 3.5765 rad/s θ0 = ( ) 2 2 WArA + WB rC (1)(10 )2 + ( 2 )(16 )2 2

= rAθ& f = (10 )( 3.5765 ) = 35.765 in./s

Vector Mechanics for 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.

2

( vA ) f

= 2.98 ft/s


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Chapter 12, Solution 135.

(

GM = gR 2 = ( 9.81) 6.37 × 106

)

2

= 398.06 × 1012 m3/s 2

rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m For the circular orbit through point A, GM = rA

vcirc =

398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106

For the descent trajectory, v A = vcirc + ∆v = 7.5773 × 103 − 152 = 7.4253 × 103 m/s

(

)(

)

h = rAv A = 6.933 × 106 7.4253 × 103 = 51.4795 × 109 m 2 /s 1 GM = 2 (1 + ε cosθ ) r h At point A,

θ = 180°,

r = rA 1 GM = 2 (1 − ε ) rA h

(

)

2

51.4795 × 109 h2 1−ε = = = 0.96028 GM rA 398.06 × 1012 6.933 × 106

(

)(

)

ε = 0.03972 1 GM = 2 (1 + ε cosθ B ) rB h

(

)

2

51.4795 × 109 h2 1 + ε cosθ B = = = 1.02567 GM rB 398.06 × 1012 6.491 × 106

(

cosθ B =

θ B = 49.7°

1.02567 − 1

ε

)(

)

= 0.6463

AOB = 180° − θ B = 130.3°

Vector Mechanics for 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.

AOB = 130.3° W