Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 12 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 12, Solution 1. m 20 kg, g 3 2 W mg ( 20 )( 3 ) Vector Mechanics for Engineers: Statics and Dynamics, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The...

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 1.

m = 20 kg, g = 3.75 m/s2 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 sin2 φ = 9.7807 m/s2 W = mg = ( 2.000 )( 9.7807 )

(b) φ = 45°,

(

)

g = 9.7807 1 + 0.0053 sin2 45° = 9.8066 m/s2 W = mg = ( 2.000 )( 9.8066 )

(c) φ = 60°,

W = 19.56 N W

(

W = 19.61 N W

)

g = 9.7807 1 + 0.0053 sin2 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/s2 m =

W g

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

or

W=

Fs a 1− g

W a g

7

= 1−

2 32.2 W = 7.46 lb W

m=

7.4635 W = = 0.232 lb ⋅ s 2 /ft 32.2 g Σ F = 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 = Wp  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.

m w = 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 × 10 6 ft

Radius of orbit:

r = 3960 + 12580 = 16540 mi = 87.33 × 10 6 ft

Velocity of satellite:

v=

2πr

τ

=

( 2π ) ( 87.33 × 106 ) 43200

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

mv = 750 × 10 3 lb ⋅ s m=

750 × 103 mv 2 = = 59.046 lb⋅ s /ft 3 v 12.7019 × 10 m = 59.0 lb ⋅ s2 /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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 5.

+ ∑ F y = ma y :

10 + 10 + 10 + 20 − 40 = ay =

ay =

( 32.2 )( 10) 40

40 ay 32.2

= 8.05 ft/s2

dv dy dv dv = =v dt dt dy dy

v dv = a y d y v v ∫ 0 v dv = ∫ 0 ay d y

v = 2 a yy =

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 0

dv dv = = constant dx dt

x

∫ v0 v dv = ∫ 0 f a dx 1 − v20 = a x f 2 a=−

v 20 2x f

=−

( 30) = − 6 m/s2 (2 )(75 )

t 0 ∫ v0 dv = ∫ 0f a dt

− v0 = a t f tf = − (b)

v0 − 30 = −6 a

t f = 5.00 s W

+ ∑ Fy = 0: N − W = 0 N=W

∑ Fx = ma: µ =−

µ=−

ma N

=−

− µ N = ma ma a =− 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 : Ff

a =− =−

m

− F f + W sin α = ma +

F W sinα = − f + g sin α m m

(

)

7500 N + 9.81 m/s2 sin 4° = − 4.6728 m/s2 1400 kg

a = 4.6728 m/s2

4°

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

a= v

dv dx

0 xf ∫ 0 a dx = ∫ v0 v dv

1 a xf = − v02 2 v2 (24.444 ) xf = − 0 = − 2a 2 ( )(− 4.6728 ) 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.

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 v 02 − = a t (s − s 0 ) 2 2 2

at =

0 − (102.667 ) v 2 − v 20 2 = = − 31.001 ft/s 2( s − s0 ) ( 2)( 170)

For braking without skidding µ = µs , so that µsN = m |a t | Σ Ft = mat : − µs N = mat

µs = −

31.001 mat a = − t = W g 32.2

µ s = 0.963 W

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

µkN m

= − µk g = −24.801 ft/s 2

Since acceleration is constant, 2

( s − s0 ) =

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

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  − 1 = 289.8 ft/s2 a = g  t − 1 = ( 32.2) 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 2g ( 2)( 32.2) 2

y − y1 = − (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 + v 0t +

1 2 at , 2

or

a =

2x ( 2 )( 10) = 1.25 m/s2 = 2 t (4 )2

ΣFy = 0: N − P sin 50° − mg cos20° = 0 N = P sin 50° + mg cos 20° Σ Fx = 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 andµ = µ s = 0.30. P =

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

= 593 N

For motion with a = 1.25 m/s 2, use µ = µ k = 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 v 02 − = a ( x − x0 ) 2 2 a=

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

2

= − 6.43 m/s2 ΣFx = ma : − Fbr = ma Fbr = −a = 6.43 m/s2 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/s2 2

0 − ( 27.778) v 2 − v02 = x − x0 = 2a (2 )( −7.455 )

x − x0 = 51.7 m W (b) Going down a 2% incline. ( tanθ = − 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/s2 2

x − x0 =

0 − ( 27.778) v 2 − v02 = 2a ( 2 )(− 6.234 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 xA and x B be down the incline. Constraint of the cable:

x A + 3 xB = constant

a A + 3a B = 0

or

1 aB = − aA 3

For block A:

ΣF = ma : mA g sin 30° − T = mA aA

For block B:

Σ F = ma : mB g sin 30° − 3T = m B a B = − m B a A (2)

Eliminating T and solving for

( 3mAg

(1)

aA , g

m   − mB g ) sin 30° =  3m A + B  a A 3  

aA ( 3 m A − mB )sin 30 ° (30 − 8) sin 30 ° = = = 0.33673 g 3m A + mB / 3 30 + 2.667

(a) a A = (0.33673) (9.81 ) = 3.30 m/s2 aB = −

1 (3.30 ) = −1.101 m/s 2 3

a A = 3.30 m/s2

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

aA + 3 aB = 0

ΣFy = 0: N A − mA g cos30° = 0

Block A:

ΣFx = ma : mA g sin 30 ° − µN A − T = mAa A Eliminate NA . mA g ( sin 30° − µ cos 30°) − T = mA aA ΣFy = 0: N B − mB g cos30° = 0

Block B:

Σ F = ma:

mB gsin 30° + µ NB − 3 T = mB aB = −

m Ba A 3

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

mB a A 3

Eliminate T.

( 3mAg − mB g ) sin 30° − µ ( 3m Ag + mB g ) cos30° =

mB    3m A + a A 3  

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

µ =

( 3m A − m B ) sin 30° ( 3mA + mB ) cos30°

=

(75 − 20) tan 30° = 0.334. ( 75 + 20)

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

aA for sliding. Use µ = µ k = 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 3m A + m B / 3 =

(a)

(30 − 8 )sin 30° − (0.20 )(30 + 8 )cos30° 30 + 2.667

= 0.13525

aA = (0.13525 )(9.81 ) = 1.327 m/s 2

a A = 1.327 m/s2

30° W

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

a B = 0.442 m/s2

30° W

(b) T = m Ag ( sin 30° − µ cos 30° ) − m Aa 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 ⋅ s2 / ft 2 32.2 ft/s

mB =

44000 lb = 1366.5 lb ⋅ s2 / ft 32.2 ft/s2

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 = mA ax + mB ax ax =

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

ax = v

dv dx

0 xf ∫ 0 a x dx = ∫ v0 v dv

ax x f =

v02 2

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

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

∑ Fx = ∑ max : Fc − Fb = mA ax Fc = mA ax − Fb = (1708.1)( 4.5534 ) + 7000 = 778 lb F c = 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 15.

Data:

mA = mB =

55000 lb 32.2 ft/s 2 44000 lb 32.2 ft/s

2

= 1708.1 lb ⋅ s2 / ft = 1366.5 lb ⋅ s2 / ft

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 = mA ax + mB ax ax =

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

ax = v

dv dx

0 xf ∫ 0 a x dx = ∫ v0 v dv

ax x f =

v02 2

v2 ( −80.667 ) = xf = − 0 = − 1429 ft 2a x ( 2)( 2.2767) 2

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

or

aB = −aA

Assume that block A moves down and block B moves up. Σ Fy = 0:

Block B:

N AB − WB cos θ = 0

ΣFx = ma: − T + µ N AB + WB sin θ =

WB aB g

Eliminate N AB and a B. −T + WB (sin θ + µ cos θ ) = WB

aB a = −WB A g g

ΣF y = 0: N A − N AB − W A cos θ = 0

Block A:

N A = N AB + W A cos θ = ( WB + WA ) cos θ ΣFx = m Aa A : − T + WA sin θ − FAB − FA = −WB (sinθ + µ cos θ ) − WB

aA + WA sin θ − µWB cos θ g

−µ ( W B + W A) cosθ = W A

(WA

WA aA g

aA g

− WB ) sinθ − µ (WA + 3WB ) cos θ = (W A + WB )

aA g

Check the condition of impending motion.

µ = µs = 0.20,

aA = aB = 0,

θ = θs

(W A − W B )sinθ s − 0.20 (W A + 3W B )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 θs =

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

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

aA using µ = µk = 0.15 and θ = 25 °. g aA (W A − W B ) sin θ − µk (W A + 3W B ) cos θ = g WA + WB =

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

= 0.10048

a A = (0.10048 )(32.2 ) = 3.24 ft/s2

(a) aB = − 3.24 ft/s 2 (b)

T = WB ( sin θ + µ cos θ ) + WB

a B = 3.24 ft/s 2

25° !

aA g

= 16 ( sin 25° + 0.15 cos 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 17.

Constraint of cable: 2 x A + ( xB − x A) = xA + xB = constant. a A + a B = 0,

or

a B = −a A

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

ΣFy = 0: N AB − WB cosθ = 0 ΣF x = ma x : −T + µ N AB + W B sinθ =

WB aB g

Eliminate N AB and aB . −T + WB (sin θ + µcos θ ) = 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 + W A) c osθ − P sin θ

Σ Fx = mA aA : − T + WA sin θ − FAB − FA + Pcos θ = −WB ( sinθ + µ cos θ ) − WB

aA + W A sin θ − µW B cos θ g

−µ (WB + W A )cos θ + µP sin θ + P cos θ = W A

(W A − W B )sin θ

WA a g A

aA g

− µ (W A + 3W B )cos θ + P ( µsin θ + cos θ ) = (W A + W B )

Check the condition of impending motion.

µ = µ s = 0.20, a A = aB = 0, θ = 25°

(WA − WB )sin θ

− µs (WA + 3WB )cos θ + Ps ( µs sin θ + cos θ ) = 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 =

=

µs ( W A + 3WB ) cosθ − (W A − W B ) sin θ µs sin θ + cos θ

(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 = µk = 0.15, θ = 25° , and...