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

2020
Views:

228

Pages:

2
Author:

crawniadxfo

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 92. FBD Pipe 2: ΣFx′ = 0: NG − 3 ( 220 lb ) = 0, 5 NG = 132 lb ΣFy′ = 0: ND − 4 ( 220 lb ) = 0, 5 N D = 176 lb FBD Pipe 1: θ = 90° − 2 tan −1 ΣFx′ = 0: 3 = 16.2602° 4 N F cos (16.2602° ) − 132 lb − 3 ( 220 lb ) = 0 5 N F = 275.00 lb ΣFy′ = 0: N C + ( 275 lb ) cos (16.2602° ) − 4 ( 220 lb ) = 0 5 NC = 99.00 lb FBD Frame & Pipes: ΣM A = 0:   4 5 ( 48 in.) E y − 24 in. + 24 in. + (10 in.)  (220 lb) = 0   E y = 256 ΣFy = 0: ΣFx = 0: 2 lb, 3 Ay − 2 ( 220 lb ) + 256 Ay = 183  1 lb 3 Ax − E x = 0, 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. E y = 257 lb A y = 183.3 lb 2 lb = 0 3 Ax = E x COSMOS: Complete Online Solutions Manual Organization System FBD BE: ΣM B = 0: ( 24 in.)  256  2   15  lb  − (18 in.) E x +  in.  ( 275 lb ) = 0 3   4  E x = 399.5 lb, E x = 400 lb From above, A x = 400 lb 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.

COSMOS Chapter 6 Solution 92

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