COSMOS Chapter 8 Solution 118 | Michigan State University - Edubirdie

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

2024
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80

Pages:

1
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 118. Geometry and force notation: Note: θ = sin −1 βC = β D = π 2 π r = 30° = rad, so contact angles are: 2r 6 + π 6 = 2π , 3 βE = π (a) For all pulleys locked, slip impends at all contacts If WA impends downward, T1 = (16 lb ) e µs β E , T2 = T1e µs β D , WA = T2e µ s βC 0.20( 73π ) µ β +β +β so WA = (16 lb ) e s ( C D E ) = (16 lb ) e = 69.315 lb If WA impends upward all ratios are inverted, so WA = (16 lb ) e ( ) −0.20 73π = 3.6933 lb For equilibrium, 3.69 lb ≤ WA ≤ 69.3 lb W (b) If pulley D is free to rotate, T1 = T2 while the other ratios remain as in (a) 0.20( 53π ) µ β +β For WA impending down WA = (16 lb ) e s ( C E ) = (16 lb ) e WA = 45.594 lb For WA impending upward, WA = (16 lb ) e ( ) −0.2 53π = 5.6147 lb 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. 5.61 lb ≤ WA ≤ 45.6 lb W

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