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

2024
Views:

104

Pages:

2
Author:

Sean Horton

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 129. Free-Body Diagram: Express the tension in terms of its rectangular components: uuuur DG = − ( 4.8 in.) j − ( 9 in.) k TDC = TDG ΣM E = 0: uuuur DG = TDG DG − 4.8 j − 9 k ( − 4.8) 2 + ( − 9) 2 =− 8 15 TDG j − TDG k 17 17 M E + ( 6.4 in.) i + ( − 7.6 in.) j × TDG + ( 2.4 in.) j + ( 9.6 in.) k  × ( − 44 lb ) i = 0 15   8   15   TDG i − ( 6.4 in.)   TDGk + ( 6.4 in.)   TDG j  17   17   17   or ( M Ex i + M Ey j + M Ez k ) + ( 7.6 in.)  + ( 2.4 in.)( 44 lb ) k − ( 9.6 in.)( 44 lb ) j = 0 Setting the coefficient of the unit vector j equal to zero: (a) j:  15   TDG  ( 6.4 in.) − ( 44 lb )( 9.6 in.) = 0 17   TDG = 74.800 lb or TDG = 74.8 lb continued COSMOS: Complete Online Solutions Manual Organization System (b) ΣFx = 0: E x − 44 lb = 0, or E = 44.000 lb ΣFy = 0: Ey − 8 ( 74.8 lb ) = 0, or E y = 35.200 lb 17 ΣFz = 0: Ez − 15 ( 74.8 lb ) = 0, or Ez = 66.000 lb 17 or E = ( 44.0 lb ) i + ( 35.2 lb ) j + ( 66.0 lb ) k Using the moment equation again and setting the coefficients of the unit vectors i and k equal to zero: i:  15  M Ex + ( 7.6 in.)   ( 74.800 lb ) = 0  17  M Ex = −501.60 lb ⋅ in. k: 8  M Ez + ( 44 lb )( 2.4 in.) −  ( 74.8 lb )  ( 6.4 in.) = 0 17  M Ex = 119.680 lb ⋅ in. or M E = − ( 502 lb ⋅ in.) i + (119.7 lb ⋅ in.) k