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

2021
Views:

26

Pages:

1
Author:

hekeldigzjbz

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 53. FBD Block C: φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.8014° ΣFx = 0: RACx − RCFx = 0 ΣFy = 0: so RCFy − RACy − 175 lb = 0 RCFy RCFx − RACy RACx = 175 lb RACx cot ( 20° + φ ) − cot ( 32.2° ) > 0 φ < 12.2° < φ s = 21.8° so block C does not slip (or impend) FBD Block B: (a) φ B = tan −1 µ B = tan −1 ( 0.4 ) = 21.8014° RB 175 lb , = sin 41.8014° sin 46.3972° RB = 161.083 lb (b) φ B = tan −1 µ B = tan −1 ( 0.6 ) = 30.9638° RB 175 lb , = sin 50.9638° sin 37.2330° RB = 224.65 lb FBD Wedge: P RB = , sin 59.6028° sin 52.1986° (a) RB = 161.083 lb, (b) RB = 224.65 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. P = 1.09163 RB P = 175.8 lb P = 245 lb

COSMOS Chapter 8 Solution 53

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