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

2024
Views:

412

Pages:

1
Author:

Campbell Wagner

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 40. Free-Body Diagram: Equilibrium for rod: (a) ΣM E = 0: ( 6 lb ) cos 60° ( dOE ) − (T cosθ ) ( dOE ) = 0 T = 3 lb cosθ (1) Thus T is minimum when cos θ is maximum: (b) (c) With θ = 0°, (1) gives: T = 3 lb ΣFx = 0: θ = 0° T = 3.00 lb 3 lb − ( 6 lb ) cos 60° − N A sin 45° − N D sin 45° = 0 N A = ND ΣFy = 0: (2) − ( 6 lb ) sin 60° + N A cos 45° + N D cos 45° = 0 N A + N D = 7.3485 lb (3) Solving (2) and (3) gives: N A = N D = 3.6742 lb Therefore: N A = 3.67 lb 45° N D = 3.67 lb 45°