COSMOS Chapter 6 Solution 45 | Answer Key - Edubirdie

University:
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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COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 45. FBD Truss: ΣFx = 0: Ax = 0 By load symmetry A y = L y = 840 N 0.4 2 = 3 15 3.5 7 IK = 3 = 4 24 2.3 23 IJ = = 3 30 slope: GI = FBD Section JKL: Resolve FIK and FGJ at L ΣM J = 0: ( 3 m )  7   24  FIK  − ( 0.4 m )  FIK  25 25     + ( 3 m )( 840 N − 120 N ) − (1 m )( 240 N ) = 0 FIK = 4210.53 N, ΣM I = 0: FIK = 4.21 kN C W    3.5   15  2 FGJ  −  FGJ  m   229  229   3  (4 m)  + ( 4 m )( 840 N − 120 N ) − ( 2 m )( 240 N ) = 0 FGJ = 3823.0 N, ΣFx = 0: FGJ = 3.82 kN T W 7 15 30 FIJ = 0 ( 4210.5 N ) − ( 3823.0 N ) − 25 229 1429 FIJ = 318 N T W 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.

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