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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 Online Solutions 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 =
…ver más…

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 8.

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

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

(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 ) =

2 0 − (102.667 ) v 2 − v0 = 2at ( 2 )( − 24.801)

2

s − s0 = 212 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 9.

For the thrust phase,

ΣF = ma : Ft

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