University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2024
Views:

269

Pages:

1
Author:

behend6abs

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 5. Tension in cord is equal to P from FBD of pulley. ( ) W = (10 kg ) 9.81 m/s 2 = 98.1 N ΣFy = 0: N − ( 98.1 N ) cos 20° + P sin 25° = 0 (1) ΣFx = 0: P cos 25° − ( 98.1 N ) sin 20° + F = 0 (2) For impending slip down the incline, F = µ s N = 0.3 N and solving (1) and (2), PD = 7.56 N For impending slip up the incline, F = − µ s N = − 0.3 N and solving (1) and (2), PU = 59.2 N so, for equilibrium 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. 7.56 N ≤ P ≤ 59.2 N

COSMOS Chapter 8 Solution 5

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