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

2023
Views:

433

Pages:

1
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 113. FBD Drum: (a) With M E = 125 lb ⋅ ft ΣM E = 0: ( 7 in.) (TA − TC ) − (125 lb ⋅ ft ) = 0 TA − TC = 214.29 lb Belt slip: TA = TC e µk β = TC e ( ) 0.30 76π = 3.0028 TC so 2.0028 TC = 214.9 lb, TC = 106.995 lb TA = 321.28 lb FBD Lever: ΣM B = 0: P= (15 in.) P + ( 2 in.) TC − ( 7.5 in.) TA = 0 (1) ( 7.5 in.)( 321.28 lb ) − ( 2 in.)(106.995 lb ) 17 in. P = 129.2 lb (b) With M E = 125 lb ⋅ ft , the drum analysis will be reversed, and will yield TA = 106.995 lb, TC = 321.28 lb Eqn. (1) will remain the same, so P= ( 7.5 in.)(106.995 lb ) − ( 2 in.)( 321.28 lb ) 17 in. P = 9.41 lb 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 113

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