COSMOS Chapter 6 Solution 44 | 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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2023
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COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 44. FBD Truss: ΣFx = 0: Ax = 0 By load symmetry: A y = L y = 840 N ΣM E = 0: 2 24 (1.75 m ) FBD + ( 2 m )( 240 N ) 3 25 − ( 4 m )( 840 N − 120 N ) = 0 FBD = FBD Section ABC: ΣM B = 0: FBD = 2.14 kN C W  1.75  m  FCE − ( 2 m )( 840 N − 120 N ) = 0   3  FCE = ΣM A = 0: 15000 N, 7 17280 N, 7 FCE = 2.47 kN T W ( 4 m )  7  FBE  − ( 2 m )( 240 N ) = 0  25  FBE = 3000 N, 7 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. FBE = 429 N C W

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