COSMOS Chapter 4 Solution 71 | Michigan State University - Edubirdie

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

2024
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161

Pages:

2
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 71. Free-Body Diagram: We note from the free-body diagram that the ladder is a three-force body. Point C in the free-body diagram is the intersection between the lines of action of the three forces. It then follows that: sin θ = 1.75 m ( 9.2 − 1.8) m θ = 13.6793° Also:  9.2  AG =  − 1.8  m = 2.8 m  2   9.2  BD =  m  cosθ = ( 4.6 m ) cosθ  2  CD = CG + GD = = AG sin θ 2.8 sin θ + BG sin θ + 9.2 2 sin θ continued COSMOS: Complete Online Solutions Manual Organization System Then 2.8 + 4.6sin13.6793° CD tan φ = = sin13.6793° BD 4.6cos13.6793° φ = 70.928° Now using the law of sines on the force triangle: FW B W = = sin(90° − φ ) sin θ sin [φ + (90° − θ )] FW B W = = cos φ sin θ sin(φ − θ ) (a) From the law of sines and noting that W = ( 53 kg ) (9.81 m/s 2 ) = 519.93 N FW 519.93 N = cos 70.928° cos(70.928° − 13.6793°) FW = 314.03 N or FW = 314 N 76.3° or B = 227 N 70.9° (b) In the same way B 519.93 N = sin13.6793° cos(70.928° − 13.6793°) B = 227.28 N

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