University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2024
Views:

152

Pages:

1
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 68 FBD Section ABDC: Since only DE can provide the leftward force necessary for equilibrium, it must be in tension, and CF must be slack, FCF = 0 ΣFx = 0: ΣM D = 0: 20 kN − 4 FDE = 0, 5 FDE = 25.0 kN C ! ( 3.2 m ) FCE − ( 2.4 m )( 20 kN ) = 0 FCE = 15.00 kN T ! ΣFy = 0: FDF − 15.00 kN − 3 ( 25 kN ) = 0 5 FDE = 30.0 kN C ! 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.

COSMOS Chapter 6 Solution 68

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