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

2021
Views:

53

Pages:

2
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 36. FBD Truss: ΣFz = 0: Bz = 0 ΣM x = 0: − ( 0.47 m ) C y + ( 0.08 m )( 940 N ) = 0 C y = 160 Nj Joint FBDs: Joint D: where FAD = FAD FBD = FBD = FBD FCD = FCD ΣFy = 0: ΣFx = 0: ΣFz = 0: 4 FBD − 940 N = 0, 21 10 80 − FAD − FCD − 89 101 1 39 − FAD + FCD − 89 101 Solving: − 0.80 m i − 0.08 m k ( 0.8) 2 = FAD 2 + ( 0.08 ) m −10i − k 101 − 0.8 m i − 0.16 m j − 0.2 m k ( 0.8 m )2 + ( 0.16 m )2 + ( 0.2 m )2 − 20i + 4 j − 5k 21 − 0.8 m i + 0.39 m k ( 0.8 m ) 2 + ( 0.39 m ) FBD = 4935 N, 2 = FCD − 80i + 39k 89 FBD = 4.94 kN T ! 20 ( 4935 N ) = 0 21 5 ( 4935 N ) = 0 21 FAD = − 590 101 N, FAD = 5.93 kN C ! FCD = 1335 N, FCD = 1.335 kN T ! continued 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: Complete Online Solutions Manual Organization System Joint C: FBC = FBC 0.16 m j − 0.59 m k ( 0.16 m )2 + ( 0.59 m )2 = FBC 16 j − 59k 3737 16 FBC + 160 N = 0 3737 ΣFy = 0: FBC = −10 3737 N ΣFz = 0: − FAC − ( FBC = 611 N C ! ) 59 39 −10 3737 N − (1335 N ) = 0 89 3737 FAC = 5.00 N T ! Joint B: FAB = FAB ΣFy = 0: − 0.16 m j + 0.12 m k ( 0.16 m ) − 4 FAB + 5 2 + ( 0.12 m ) 2 ( − 4 j + 3k 5 = FAB ) 16 4 10 3737 N − ( 4935 N ) = 0 21 3737 FAB = 975 N C ! 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.