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

2021
Views:

321

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 21. Joint FBDs: FDF F 500 lb = EF = 916 916 480 Joint F: FDF = FEF = 954.17 lb FDF = 954 lb T ! or FEF = 954 lb C ! Joint D: FBD FDE 954.17 lb = = 884 240 916 FBD = 920.84 lb FDE = 250.00 lb FBD = 921 lb T ! or FDE = 250 lb C ! By symmetry of joint A vs. joint F FAB = 954 lb T ! FAC = 954 lb C ! Joint B: ΣFx = 0: 920.84 lb − 884 4 ( 954.17 lb ) + FBE = 0 916 5 FBE = 0 ! ΣFy = 0: FBC − 240 ( 954.17 lb ) = 0 916 FBC = 250.00 lb or 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. FBC = 250 lb C ! COSMOS: Complete Online Solutions Manual Organization System Joint C: ΣFx = 0: 884 ( 954.17 lb ) − FCE = 0 916 FCE = 920.84 lb ΣFy = 0: FCH − 240 (954.17 lb) − 250.00 lb = 0 916 FCH = 500.00 lb ΣFx = 0: FCE = 921 lb C ! or 920.84 lb − or FCH = 500 lb C ! 884 4 ( 954.17 lb ) − FHE = 0 916 5 FHE = 0 ! Joint E: ΣFy = 0: FEJ − 240 (954.17 lb) − 250.00 lb = 0 916 FEJ = 500.00 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. or FEJ = 500 lb C !

COSMOS Chapter 6 Solution 21

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