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

2019
Views:

468

Pages:

1
Author:

Sean Horton

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 164. FBD Frame: Q = 65 N P = 15 N ΣM D = 0: ( 0.25 m )( P + Q ) − (.15 m )( P + Q ) − ( 0.08 m ) Ex Ex = 1.2 ( P + Q ) = 100 N ΣFx = 0: (b) Dx − Ex = 0 = Dx − 100 N (a) ΣFy = 0: E x = 100.0 N =0 W Dx = 100 N D x = 100.0 N W E y − Dy − 2P − 2Q = 0 E y = D y + 2 ( P + Q ) = Dy + 160 N FBD member BF: ΣM C = 0: ( 0.15 m )( 65 N ) − ( 0.1 m ) Dy − ( 0.04 m )(100 N ) − ( 0.25 m )(15 N ) = 0 Dy = 20 N From above D y = 20.0 N W E y = 20 N + 160 N = 180 N (b) E y = 180.0 N W ΣFx = 0: (a) − C x + 100 N = 0 (a) ΣFy = 0: C x = 100.0 N W − 65 N + C y − 20 N − 15 N = 0 (a) 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. C y = 100.0 N W

COSMOS Chapter 6 Solution 164

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