Fisica
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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