COSMOS Chapter 6 Solution 62 | Michigan State University - 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
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 62. FBD Section above a-a: ΣM G = 0:  27  FIK  − ( 2.7 m )( 40 kN ) − ( 5.4 m )( 40 kN ) = 0  793  ( 5.9 m )  FIK = 57.275 kN, ΣFy = 0: FIK = 57.3 kN C W 27 ( 57.275 kN − FGJ ) = 0 793 FGJ = 57.275 kN T FBD Section ACIG: 27 ( 57.275 kN − 57.275 kN ) 793 ΣFy = 0: + ΣFx = 0: 18 ( FHK − FHJ ) = 0, 949 3 ( 40 kN ) − 2 FHJ = FHK 25 FHK = 0 949 FHK = 53.884 kN 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. FHK = 53.9 kN C W

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