COSMOS Chapter 6 Solution 25 | Answer Key

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

2023
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284

Pages:

2
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 25. FBD Truss: ΣFx = 0: 180 lb − Ax = 0 A x = 180 lb ΣM A = (12 ft ) G y − ( 6 ft )( 480 lb ) − ( 2 ft )(180 lb ) − (8 ft)(120 lb) = 0, ΣFy = 0: G y = 350 lb Ay − 480 lb − 120 lb + 350 lb = 0 A y = 250 lb Joint FBDs: Joint A: ΣFx = 0: 1 FAC − 180 lb = 0, FAC = 180 2 lb 2 FAC = 255 lb T ! ΣFy = 0: Joint B: ΣFx = 0: ΣFy = 0: Solving: 1 (180 2 lb) + 250 lb − FAB = 0, 2 FAB = 430 lb C ! 4 2 FBD + FBC = 0 5 5 3 1 430 lb − FBD + FBC = 0 5 5 180 lb − FBC = 590 5 lb, FBC = 1319 lb T ! FBD = 1700 lb C ! Joint C: ΣFy = 0: 5 1 1 (590 5 lb) − (180 2 lb) = 0 FCD − 41 5 2 FCD = 154 41 lb, ΣFx = 0: FCD = 986 lb T ! 4 2 (154 41 lb) − (590 5 lb) 41 5 1 − (180 2 lb) = 0 12 FCE = 744 lb T ! FCE + By inspection of joint G: 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. FFG = 350 lb C and FEG = 0 ! COSMOS: Complete Online Solutions Manual Organization System Joint F: FDF F 350 = EF = ; 5 1 20 FDF = 1750 lb C ! FEF = 700 5 lb, FEF = 1565 lb T ! Joint E: ΣFy = 0: 5 1 (700 5 lb) = 0 FDE − 120 lb − 41 5 FDE = 164 41 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. FDE = 1050 lb T !