COSMOS Chapter 3 Solution 96 | Michigan State University - Edubirdie

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

2020
Views:

132

Pages:

1
Author:

Sean Horton

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 96. First assume that the given force W and couples M1 and M 2 act at the origin. Now W = − Wj and M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. F =W (a) Have or F = − Wj = − ( 2.4 N ) j or F = − ( 2.40 N ) j W (b) Assume that the line of action of F passes through point P (x, 0, z). Then for equivalence M = rP/O × F rP/O = xi + zk where ∴ − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k i j k = x 0 z = (Wz ) i − (Wx ) k 0 −W 0 Equating the i and k coefficients, z = (b) For −M z cos 25° W and  M − M 2 sin 25°  x = − 1  W   W = 2.4 N, M1 = 0.068 N ⋅ m, M 2 = 0.065 N ⋅ m x= 0.068 − 0.065sin 25° = − 0.0168874 m − 2.4 or x = −16.89 mm W z = − 0.065cos 25° = − 0.024546 m 2.4 or z = − 24.5 mm W .

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