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

2021
Views:

196

Pages:

2
Author:

palangresq3v

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 93. FBD Pipe 2: FBD Pipe 1: ΣFx' = 0: NG − 3 ( 220 lb ) = 0, 5 NG = 132 lb ΣFy′ = 0: ND − 4 ( 220 lb ) = 0, 5 N D = 176 lb ΣFx′ = 0: ΣFy′ = 0: ΣM B = 0: FBD BE: NF − NC − 3 ( 220 lb ) − 132 lb = 0, 5 4 ( 220 lb ) = 0, 5 N F = 264 lb NC = 176 lb ( 24 in.) E y − ( 32 in.) Ex + ( 5 in.) ( 264 lb ) = 0 4Ex − 3E y = 165 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. (1) COSMOS: Complete Online Solutions Manual Organization System FBD Frame with Pipes: ΣM A = 0: ( 48 in.) E y − (14 in.) Ex + ( 5 in.) ( 264 lb ) − ( 35 in. + 45 in.) (176 lb ) = 0 (2) solving with (1) above: E y = 355 ΣFx = 0: 2 lb, 3 Ax − 308 lb − E y = 356 lb 4 3 ( 264 lb ) + ⋅ 2 (176 lb ) = 0 5 5 A x = 308 lb ΣFy = 0: Ay + 355 ! ! 2 3 4 lb − ( 264 lb ) − ⋅ 2 (176 lb ) = 0 3 5 5 A y = 84.3 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. !