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

2021
Views:

360

Pages:

2
Author:

Kundulu8mbj

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 143. FBD block: ΣFn = 0: NC − ( 200 lb ) cos 30° = 0; N = 100 3 lb ΣFt = 0: TC − ( 200 lb ) sin 30° ∓ FC = 0 TC = 100 lb ± FC FBD Drum: (1) where the upper signs apply when FC acts (a) For impending motion of block , FC , and ( ) FC = µ s NC = 0.35 100 3 lb = 35 3 lb ( ) TC = 100 − 35 3 lb So, from Equation (1): TC = WAe µk β But belt slips on drum, so −0.25( WA =  100 − 35 3 lb  e   ( ) 2π 3 ) WA = 23.3 lb W and FC = µ s NC = 35 3 lb (b) For impending motion of block , FC From Equation (1): Belt still slips, so ( ) TC = 100 + 35 3 lb −0.25( WA = TC e− µk β =  100 + 35 3 lb  e   ( ) 2π 3 ) WA = 95.1 lb W continued 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 PROBLEM 8.143 CONTINUED (c) For steady motion of block , FC , and FC = µk NC = 25 3 lb ( ) T = 100 + 25 3 lb. Then, from Equation (1): Also, belt is not slipping on drum, so −0.35( WA = TC e− µ s β =  100 + 25 3 lb  e   ( ) 2π 3 ) WA = 68.8 lb W 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 Chapter 8 Solution 143

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