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

2020
Views:

375

Pages:

2
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 67. (a) Free-Body Diagram: (α = 90° ) The bracket is a three-force body and A is the intersection of the lines of action of the three forces.  6 θ = tan −1   = 26.565°  12  From the force triangle: A = (75 lb) cot θ = (75 lb) cot 26.565° = 150.000 lb C = 75 lb 75 lb = = 167.705 lb sin θ sin 26.565° or A = 150.0 lb or C = 167.7 lb 63.4° continued COSMOS: Complete Online Solutions Manual Organization System (b) Free-Body Diagram: (α = 45° ) Let E be the intersection of the lines of action of the three forces acting on the bracket. Triangle ABE is isosceles and therefore AE = AB = 16 in. From triangle CEF  CF  −1  6   = tan   = 12.0948°  EF   28  θ = tan −1  From force triangle: β = 180° − 135° − θ = 180° − 135° − 12.0948° = 32.905° Using the law of sines: A C 75 lb = = sin 32.905° sin135° sin12.0948° Solving for A and C: A = 194.452 lb C = 253.10 lb or A = 194.5 lb or C = 253 lb 77.9°

COSMOS Chapter 4 Solution 67

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