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COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 1.

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

W = 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.

COSMOS: Complete OnlineSolutions Manual Organization System

Chapter 12, Solution 2.

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 )

m = 2.000 kg

(

)

W = 19.56 N

(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

(c) φ = 60°, g = 9.7807 1 + 0.0053 sin 2 60° = 9.8196 m/s2
W = mg= ( 2.000 )( 9.8196 ) W = 19.64 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 3.

Assume g = 32.2 ft/s 2 m= W g W a g

ΣF =ma : W − Fs =  a W 1 −  = Fs g  or W = Fs 1− = 1− 7

a g

2 32.2 W = 7.46 lb

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

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

mw = 0.232lb ⋅ s 2 /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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 4.

Periodic time: Radius of Earth: Radius of orbit: Velocity of satellite:

τ= 12 h = 43200 s
R = 3960 mi = 20.9088 × 106 ft r = 3960 + 12580 = 16540 mi = 87.33 × 106 ft
v= 2π r =

( 2π ) (87.33 × 106 )
43200

τ

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

W = 1901 lb

Vector Mechanics for Engineers: Statics andDynamics, 8/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.

+ ∑ Fy = ma y :

10 + 10 + 10 + 20 − 40 = ay =

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

40 ay 32.2

ay =

dv dy dv dv = =v dt dt dy dy

vdv = a y d y

∫ 0 v dv = ∫ 0 a y d y
v = 2a y y =

v

v

1 2 v = ay y 2

( 2 )(8.05)(1.5)

v = 4.91 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.

COSMOS: Complete Online Solutions Manual Organization SystemChapter 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 2 − v0 = a x f 2 a=−
2 v0 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

+ ∑ Fy = 0: N − W = 0
N =W

∑ Fx = ma :
µ=−

− µ N = ma

ma ma a=− =− N W g

µ=−

( − 6)
9.81

µ = 0.612

Vector Mechanics for Engineers: Statics and Dynamics, 8/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 7.

(a)

+ ∑ F = ma :
a=− =− Ff m +...
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