COSMOS Chapter 6 Solution 143 - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2021
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1
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COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 143. FBD Joint B: α = tan − 1 1 = 5.7106° 10 ΣFx′ = 0: FBC cos ( 30° + 5.7106° ) − ( 54 lb ) cos 30° = 0 FBC = 57.595 lb T FCD = 57.595 lb By symmetry: FBD Joint C: ΣFx′′ = 0: 2 ( 57.595 lb ) sin ( 5.7106° ) − P = 0 P = 11.4618 lb so 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 = 11.46 lb 30.0° W

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