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

2024
Views:

488

Pages:

2
Author:

Caroline Mcfarland

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 128. Free-Body Diagram: Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net moment about this point is zero. (a) ΣFx = 0: 3 lb + ( 3 lb ) cos 30° − T = 0 T = 5.5981 lb or T = 5.60 lb (b) ΣFy = 0: ΣFz = 0: Dy = 0 Dz − ( 3 lb ) sin 30° = 0 D = (1.500 lb ) k ΣM D = 0: M D + ( 0.72 in.) j − (1.2 in.) k  × ( −T ) i + ( 0.88 in.) j − ( 3 in.) k  × ( 3 lb )(1 + cos 30° ) i − ( 3 lb ) sin 30°k  = 0 i or M Dx i + M Dy j + M Dz k + 0 −T ( ) j ( 0.72 in.) 0 k i 0 ( −1.2 in.) + 0 1 + cos 30° j k ( 0.88 in.) ( − 3 in.) ( 3 lb ) = 0 − sin 30° 0 continued COSMOS: Complete Online Solutions Manual Organization System Setting the coefficients of the unit vectors equal to zero: i: M Dx − ( 3 lb ) sin 30° ( 0.88 in.) = 0 M Dx = 1.3200 lb ⋅ in. j: M Dy + ( 5.5981 lb )(1.2 in.) − ( 3 lb )( 3 in.)(1 + cos 30° ) = 0 M Dy = 10.0765 lb ⋅ in. k: M Dz + ( 5.5981 lb )( 0.72 in.) − ( 3 lb )( 0.88 in.)(1 + cos 30° ) = 0 M Dz = 0.89568 lb ⋅ in. or M D = (1.320 lb ⋅ in.) i + (10.08 lb ⋅ in.) j + ( 0.896 lb ⋅ in.) k