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

2024
Views:

459

Pages:

1
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 120. Geometry and force notation: θ = sin −1 5 in. π = 30° = rad, so contact angles are: 10 in. 6 βC = π − π 6 = 5π π π 2π = , βD = + , 6 2 6 3 βE = π 2 (a) All pulleys locked with impending slip at all. If WA impends upward, T1 = WAe µs βC , T2 = T1e µs β D , WA = (16 lb ) e (16 lb ) = T2e µs β E , − µ s ( βC + β D + β E ) so = (16 lb ) e ( ) −0.20 56 + 46 + 63 π WA = 4.5538 lb If WA impends downward all ratios are inverted so WA = (16 lb ) e +0.20( 2π ) = 56.217 lb 4.55 lb ≤ WA ≤ 56.2 lb For equilibrium, (b) Pulley D is free to rotate so T1 = T2 , other ratios are the same If WA impends upward, WA = (16 lb ) e − µ s ( βC + β E ) = (16 lb ) e ( ) −0.20 43π WA = 6.9229 lb If WA impends downward, ratios are inverted, WA = (16 lb ) e ( ) +0.20 43π WA = 36.979 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. 6.92 lb ≤ WA ≤ 37.0 lb