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

2022
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356

Pages:

2
Author:

crawniadxfo

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 57. Free-Body Diagram: First note T = tension in spring = ks s = elongation of spring where ( )θ − ( AB )θ = AB = 90° θ   90°  = 2l sin   − 2l sin   2  2    θ   1  = 2l sin   −     2   2    θ   1  ∴ T = 2kl sin   −     2   2  (1) (a) From free-body diagram of rod BC   θ  ΣM C = 0: T l cos    − P ( l sin θ ) = 0  2   Substituting T From Equation (1)   θ   1    θ  2kl sin   −   l cos    − P ( l sin θ ) = 0   2    2   2     θ   1   θ  θ   θ  2kl 2 sin   −   cos   − Pl  2sin   cos    = 0  2 2  2     2   2  Factoring out θ  2l cos   , leaves 2 continued COSMOS: Complete Online Solutions Manual Organization System   θ   1  θ  kl sin   −   − P sin   = 0  2   2   2  or 1  kl  θ  sin   =   2  kl − P  2   kl ∴ θ = 2sin −1    2 ( kl − P )  (b) P = kl 4  kl  2 kl − θ = 2sin −1  ( kl 4 )   kl  4   −1  4   = 2sin −1     = 2sin    3 2   2  3 kl   = 2sin −1 ( 0.94281) = 141.058° or θ = 141.1°

COSMOS Chapter 4 Solution 57

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