COSMOS Chapter 6 Solution 158 | Michigan State University

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

2023
Views:

246

Pages:

2
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 158. FBD Truss: ΣM F = 0: (10 m ) G y − ( 7.5 m )( 80 kN ) − ( 8 m )( 30 kN ) = 0 G y = 84 kN ΣFx = 0: − Fx + 30 kN = 0 Fx = 30 kN ΣFy = 0: Fy + 84 kN − 80 kN = 0 Fy = 4 kN FEG = 0 ! By inspection of joint G: FCG = 84 kN C ! 84 kN F F = CE = AC = 10.5 kN 8 61 29 Joint FBDs: FCE = 82.0 kN T ! FAC = 56.5 kN C ! ΣFy = 0: 2 6 FAE + (82.0 kN ) − 80 kN = 0 5 61 FAE = 19.01312 ΣFx = 0: − FDE − 1 (19.013 kN ) + 5 FDE = 43.99 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. FAE = 19.01 kN T ! 5 (82.0 kN ) = 0 61 FDE = 44.0 kN T ! COSMOS: Complete Online Solutions Manual Organization System ΣFx = 0: 1 FDF − 30 kN = 0 5 FDF = 67.082 kN ΣFy = 0: 2 ( 67.082 kN ) − FBF − 4 kN = 0 5 FBF = 56.00 kN ΣFx = 0: 30 kN + ΣFy = 0: 56 kN − Solving: ΣFy = 0: FDF = 67.1 kN T ! 5 FBD − 61 6 FBD − 61 FBF = 56.0 kN C ! 5 FAB = 0 29 3 FAB = 0 29 FBD = 42.956 kN FBD = 43.0 kN T ! FAB = 61.929 kN FAB = 61.9 kN C ! 6 2 ( 42.956 N ) + ( FAD − 67.082 N ) = 0 61 5 FAD = 30.157 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. FAD = 30.2 N T !