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

2023
Views:

63

Pages:

2
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 121. Free-Body Diagram: Express tension in terms of rectangular components: uuuur DG = − ( 4.8 in.) j − ( 9 in.) k uuuur − 4.7 j − 9k DG 8 15 TDG = TDG = TDG = − TDG j − TDG k 2 2 DG 17 17 ( − 4.8) + ( − 9 ) Equilibrium: ΣM F = 0: ( 6.4 in.) i + ( − 2.4 in.) j × TDG + ( 5.2 in.) j × E + ( 7.6 in.) j + ( 9.6 in.) k  × ( − 55 lb ) i = 0 or 15   8   15   TDG i − ( 6.4 in.)   TDG k + ( 6.4 in.)   TDG j  17   17   17  ( 2.4 in.)  − ( 5.2 in.) Ex k + ( 5.2 in.) Ez i + ( 7.6 in.) (55 lb)k − ( 9.6 in.) (55 lb)j = 0 Setting the coefficients of the unit vectors equal to zero: (a) j: 15 TDG ( 6.4 in.) − ( 55 lb )( 9.6 in.) = 0 17 TDG = 93.5 lb continued COSMOS: Complete Online Solutions Manual Organization System (b) k: 8  (93.500 lb)  ( 6.4 in.) = 0 17   ( 55 lb )( 7.6 in.) − Ex ( 5.2 in.) −  E x = 26.231 lb i:  15  Ez ( 5.2 in.) +  (93.500 lb)  ( 2.4 in.) = 0 17  E z = − 38.077 lb ΣFx = 0: − 55 lb + 26.231 lb + Fx = 0 Fx = 28.796 lb ΣFy = 0: Fy − 8 ( 93.500 lb ) = 0 17 Fy = 44.000 lb ΣFz = 0: − 38.077 lb + Fz − 15 ( 93.500 lb ) = 0 17 Fz = 120.577 lb Therefore: E = ( 26.2 lb ) i − ( 38.1 lb ) k F = ( 28.8 lb ) i + ( 44.0 lb ) j + (120.6 lb ) k