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

2024
Views:

472

Pages:

1
Author:

strofamay55

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 18. FBD’s: FBD Drum: (a) ΣM D = 0:  10   ft  F − 50 lb ⋅ ft = 0  12  F = 60 lb Impending slip: N = F µs = 60 lb = 150 lb 0.40 FBD arm: ΣM A = 0: ( 6 in.) C + ( 6 in.) F − (18 in.) N =0 C = − 60 lb + 3 (150 lb ) = 390 lb Ccw = 390 lb (b) Reversing the 50 lb ⋅ ft couple reverses the direction of F, but the magnitudes of F and N are not changed. Then, using the FBD arm: ΣM A = 0: ( 6 in.) C − ( 6 in.) F − (18 in.) N =0 C = 60 lb + 3 (150 lb ) = 510 lb Cccw = 510 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 18

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