COSMOS Chapter 3 Solution 88 | Michigan State University - Edubirdie

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

2019
Views:

59

Pages:

1
Author:

Arielle Armstrong

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 88. (a) Let R be the single equivalent force. Then R = (120 N ) k ΣM B : R = 120 N − a (120 N ) = − ( 0.165 m )( 90 N ) cos15° + ( 0.201 m )( 90 N ) sin15° a = 0.080516 m ∴ The line of action is y = 201 mm − 80.516 mm = 19.984 mm 2 or y = 19.98 mm (b) ΣM B : Then − ( 0.201 − 0.040 ) m  (120 N ) = − ( 0.165 m )( 90 N ) cosθ + ( 0.201 m )( 90 N ) sin θ or cosθ − 1.21818sin θ = 1.30101 or cos 2 θ = (1.30101 + 1.21818sin θ ) or 1 − sin 2 θ = 1.69263 + 3.1697sin θ + 1.48396sin 2 θ or 2.48396sin 2 θ + 3.1697 sin θ + 0.69263 = 0 sin θ = −3.1697 ± 2 ( 3.1697 )2 − 4 ( 2.48396 )( 0.69263) 2 ( 2.48396 ) or θ = −16.26° . and θ = −85.0°

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