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

2023
Views:

463

Pages:

1
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 19. For slipping, F = µ k N = 0.30 N FBD’s: (a) For cw rotation of drum, the friction force F is as shown. From FBD arm: ΣM A = 0: ( 6 in.)( 600 lb ) + ( 6 in.) F − (18 in.) N 600 lb + F − 3 F = =0 F =0 0.30 600 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 F is reversed ΣM A = 0: ( 6 in.)( 600 lb ) − ( 6 in.) F − (18 in.) N 600 lb − F − 3 =0 F =0 0.30 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.