COSMOS Chapter 6 Solution 160 | Answer Key - Edubirdie

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

2019
Views:

450

Pages:

1
Author:

Dixie Summers

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 160. FBD Section: p = 8.4 ( 2.7 m ) = 1.89 m Note: BG 12.0 ΣM A = 0: ( 2.7 m ) FFG − ( 3.6 m )(8 kN ) − ( 7.8 m )(8 kN ) − (12 m )( 4 kN ) = 0 FFG = 51.56 kN ΣM G = 0: FFG = 51.6 kN C W (1.89 m )  12  FAB  − ( 4.2 m )( 8 kN ) − ( 8.4 m )( 4 kN ) = 0  12.3  FAB = 36.44 kN ΣM D = 0: FAB = 36.4 kN T W ( 4.2 m )(8 kN ) + (8.4 m )(8 kN ) − ( 8.4 m )  3  FAG  = 0 5   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. FAG = 20.0 kN T W

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