COSMOS Chapter 6 Solution 106 | Answer Key - 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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2024
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1
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Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 106. \ FBD Frame: ΣM A = 0: (12 in.) Dx − ( 28 in.)( 411 lb ) = 0 D x = 959 lb FBD DF: Note that BF and CE are two-force members. ΣFx = 0: ΣM D = 0: Solving: 959 lb + 4 15 FCE − FBF = 0 5 17 (12 in.)  3   8  FCE  + ( 34.5 in.)  FBF  = 0 5   17  FBF = 357 lb, ΣM E = 01: ( 22.5 in.)  ( 357 lb ) − (12 in.) Dy = 0 D y = 315 lb , 8 17 so  (a) (b) 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. D = 1009 lb 18.18° FBF = 357 lb T

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