COSMOS Chapter 4 Solution 72 | Michigan State University - Edubirdie

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

2024
Views:

188

Pages:

2
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 72. Free-Body Diagram: We note from the free-body diagram that the ladder is a three-force body. Point D in the free-body diagram is the intersection between the lines of action of the three forces. It then follows that: BD = 9.2 m cosθ = ( 4.6 m ) cosθ =, but also that 2 1.75  BD = 1.75 tan θ + tan θ   m  Therefore BD = 9.2 m cosθ = ( 4.6 m ) cosθ 2 This implies: 4.6 cosθ = 1.75 tan θ + 1.75 tan θ 92sin θ = 35(1 + tan 2 θ ) = 35 sec2 θ 92sin θ cos 2 θ = 35 continued COSMOS: Complete Online Solutions Manual Organization System Solving numerically for the smallest possible root: θ = 31.722° Then sin 31.722° = 1.75 9.2 − a a = 5.8717 m or a = 5.87 m (b) From the force triangle FW = W cos 31.722° FW = 611.24 N or FW = 611 N 58.3°

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