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

2019
Views:

171

Pages:

1
Author:

lankstoetp10

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 126. FBD AB: ΣM A = 0: ( 6 in.) FBC cos 30° − 660 lb ⋅ in. = 0, FBC = 127.017 lb T FBD Piston: 180 lb C cos φ ΣFy = 0: 180 lb − FCE cos φ = 0, FCE = ΣFy = 0: 180 lb cos φ + (127.017 lb ) sin 30° cos φ − FCD cos φ = 0, FBD Joint C: FCD cos φ = 243.51 lb ΣFx = 0: (1) (127.017 lb ) cos30° − 180 lb sin φ − FCD sin φ = 0 cos φ FCD sin φ = 110.000 lb − (180 lb ) tan φ divide (2) by (1) (2) tan φ = 0.45173 − 0.73919 tan φ tan φ = 0.25974, 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. φ = 14.56°

COSMOS Chapter 6 Solution 126

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