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

2021
Views:

64

Pages:

1
Author:

pravitate8aj9

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 94. Free-Body Diagram: Note that the rod is a three-force body. In the free-body diagram, E is the intersection of the lines of action of the three forces. (a) Using triangle DBC which is isosceles DB = a and using triangle BDE ED = DB tan 2θ = a tan 2θ From triangle GED ED = a tan 2θ = ( L − a) , tan θ and therefore L−a , or tan θ a (tan θ tan 2θ + 1) = L (1) From triangle BCD: a= L a 1.25 L 2 cos θ = 1.6 cos θ (2) Using (2) in (1): 1.6 cos θ = 1 + tan θ tan 2θ (3) ( ) 2 1 − cos 2 θ sin θ sin 2θ sin θ 2sin θ cosθ = = , (3) gives Noting that tan θ tan 2θ = cosθ cos 2θ cosθ 2cos 2 θ − 1 2cos 2 θ − 1 1.6 cosθ = 1 + ( 2 1 − cos 2 θ 2 ) , or 2 cos θ − 1 3 3.2 cos θ − 1.6 cos θ − 1 = 0 or θ = 23.5° Solving numerically, θ = 23.515° (b) Substituting into (2) for L = 8 in., a= 5 ( 8 in.) = 5.4528° 8 cos 23.515° or a = 5.45 in.