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

2022
Views:

379

Pages:

2
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 114. Free-Body Diagram First express tensions in terms of rectangular components: uuur − 8i − 8j + 4k BE 2 2 1 TBE = TBE = TBE = − TBE i − TBE j + TBE k 2 2 2 BE 3 3 3 ( − 8) + ( − 8) + ( 4 ) TBF = TBF uuur BF = TBF BF 8i − 8j + 14k (8) 2 2 + ( − 8 ) + (14 ) 2 = 4 4 7 TBF i − TBF j + TBF k 9 9 9 TCD = TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) ΣM A = 0: or rAB × TBE + rAB × TBF + rAC × TCD = 0 8j × TBE T ( − 2i − 2 j + k ) + 8j × BF ( 4i − 4 j + 7k ) 3 9 + 10 j × TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) = 0 continued COSMOS: Complete Online Solutions Manual Organization System Equating the coefficients of the unit vectors to zero: (a) i: 8 56 TBE + TBF − 10TCD cos φ cosθ = 0 3 9 (1) k: 16 32 TBE − TBF − 10TCD cos φ sin θ = 0 3 9 (2) − 2 ×  Eq. (1)  + Eq. ( 2 ) gives:   56  32  − 2   −  TBF − 10 ( 600 ) cos10° ( − 2 cos 30° + sin 30° ) = 0   9  9 TBF = 455.00 N Using this in Eq. (1), 8 56 TBE + ( 455.00 ) − 10 ( 600 ) cos10° cos 30° = 0 3 9 TBE = 857.29 N Therefore: TBE = 857 N TBF = 455 N (b) ΣFx = 0: ΣFy = 0: Ax − 2 4 (857.29 N ) + ( 455.00 N ) + ( 600 N ) cos10° sin 30° = 0 3 9 Ax = 73.9 N 2 4 Ay − ( 857.29 N ) − ( 455.00 N ) + ( 600 N ) sin10° = 0 3 9 Ay = 878 N ΣFz = 0: Az + 1 7 (857.29 N ) + ( 455.00 N ) − ( 600 N ) cos10° cos 30° = 0 3 9 Az = −127.9 N Therefore: A = ( 73.9 N ) i + ( 878 N ) j − (127.9 N ) k