Complete COSMOS: Online Solutions Manual Organization System
Chapter 8, Solution 19.
FBD's: For slipping, F = k N = 0.30 N (a) For cw rotation of drum, the friction force F is as shown. From FBD arm:
M A = 0:
( 6 in.)( 600 lb ) + ( 6 in.) F - (18 in.) N
600 lb + F - 3 F =0 0.30
=0
F =
600 lb 9
Moment about D = (10 in.) F = 666.67 lb in.
M cw = 55.6 lb ft
(b) For ccw rotation of drum, the friction force is F reversed
M A = 0:
( 6 in.)( 600 lb ) - ( 6 in.) F - (18 in.) N
600 lb - F - 3 F =0 0.30
=0
F =
600 lb 11
10 600 Moment about D = ft lb = 45.45 lb ft 12 11
M ccw = 45.5 lb 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.