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

2019
Views:

185

Pages:

1
Author:

MoGhoadoAnotozj7

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 98. FBD Frame: ( 0.625 m ) F − ( 0.75 m )( 4 kN ) − (1.25 m )( 3 kN ) = 0 ΣM A = 0: Fx = 10.8 kN ΣFx = 0: Ax − 10.8 kN = 0, ΣFy = 0: Ay − 4 kN − 3 kN = 0, FBD BF: (I) A x = 10.80 kN W A y = 7.00 kN W FBD ABD: (II) I: ΣM C = 0: ( 0.375 m )(10.8 kN ) − ( 0.25 m ) Bx = 0, II: ΣM D = 0: ( 0.25 m )(10.8 kN + 16.2 kN ) + ( 0.5 m ) By − (1.00 m )( 7.0 kN ) = 0, Bx = 16.2 kN, By = 0.5 kN, ΣFx = 0: −10.8 kN − 16.20 kN + Dx = 0, Dx = 27 kN, ΣFy = 0: 7.0 kN − 0.5 kN − Dy = 0, Dy = 6.5 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. B x = 16.20 kN B y = 500 N D x = 27.0 kN W W W D y = 6.50 kN W