COSMOS Chapter 6 Solution 113 | Michigan State University - Edubirdie

University:
Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2022
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1
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 113. Note that, if we assume P is applied to BF, each individual member FBD looks like: Fshort = 2Flong = so 2 Fload 3 (by moment equations about S and L). Labeling each interaction force with the letter corresponding to the joint of application, we have: F = 2(P + E) 3 = 2B C D = 3 2 B A C = = 3 2 E = so 2(P + E) = 2 B = 6C = 18E 3 P + E = 27 E P 26 E= so D = 2E = A = 2C = 3E F = 2 P P +  3 26  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. A= F= P 13 3P 13 9P 13

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