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

2019
Views:

438

Pages:

1
Author:

salcignofi11

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 156. Free-Body Diagram: (a) From free-body diagram of rod AB ΣM C = 0: P ( l cosθ ) + P ( l sin θ ) − M = 0 or sin θ + cosθ = (b) For M Pl M = 150 lb ⋅ in., P = 20 lb, and l = 6 in. sin θ + cosθ = 150 lb ⋅ in. 5 = = 1.25 20 lb 6 in. ( )( ) 4 sin 2 θ + cos 2 θ = 1 Using identity ( sin θ + 1 − sin 2 θ (1 − sin θ ) 2 1 2 ) 1 2 = 1.25 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0 Using quadratic formula sin θ = = or − ( −2.5 ) ± − 2 ( 2) 0.5625) 2.5 ± 1.75 4 sin θ = 0.95572 ∴ θ = 72.886° and and sin θ = 0.29428 θ = 17.1144° or θ = 17.11° and θ = 72.9°