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

2023
Views:

419

Pages:

2
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 115. 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: 8 56 TBE + TBF − 10TCD cos φ cosθ = 0 3 9 i: 8 56 (840 N ) + ( 450 N ) = 10TCD cos φ cosθ 3 9 (1) 16 32 TBE − TBF − 10TCD cos φ sin θ = 0 3 9 k: 16 32 (840 N ) − ( 450 N ) = 10TCD cos φ sin θ 3 9 (a) (2)  Eq. ( 2)  gives:  Eq. (1)  16 (840) − 32 ( 450) 10TCD cos φ sin θ 3 9 = 8 10TCD cos φ cos θ (840) + 56 ( 450) 3 9 tan θ = 1 1.75 θ = 29.7° θ = 29.745° (b) Substituting into (1) gives: 8 56 (840 N ) + ( 450 N ) − 10TCD cos8° cos 29.745° = 0, 3 9 or TCD = 586.19 N or TCD = 586 N (c) ΣFx = 0: Ax − 2 4 (840 N ) + ( 450 N ) + ( 586.19 N ) cos8° sin 29.745° = 0 3 9 Ax = 72.0 N ΣFy = 0: Ay − 2 4 (840 N ) − ( 450 N ) − ( 586.19 N ) sin 8° = 0 3 9 Ay = 842 N ΣFz = 0: Az + 1 3 (840 N ) + ( 450 N ) − ( 586.19 N ) cos8° cos 29.745° = 0 3 9 Az = −126.0 N Therefore: A = ( 72.0 N ) i + ( 842 N ) j − (126.0 N ) k