COSMOS Chapter 4 Solution 23 | Answer Key - Edubirdie

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

2019
Views:

220

Pages:

1
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 23. Free-Body Diagram: From free-body diagram for (a): − B ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0 ΣΜ A = 0: Β = −187.50 Ν ΣFx = 0: or B = 187.5 N −187.5 N − 50 N + Ax = 0 Ax = 237.50 N ΣFy = 0: Ay − 100 N = 0 Ay = 100.000 N A= and: Ax2 + Ay2 = θ = tan −1 Ay Ax ( 237.5)2 + (100 )2 = tan −1 = 257.69 N 100 = 22.834° 237.5 ∴ A = 258 N 22.8° From For (b) ΣΜ A = 0: − B cos 45° ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0 Β = 265.17 Ν or B = 265.17 N 45° − ( 265.17 N ) cos 45° − 50 N + Ax = 0 ΣFx = 0: Ax = 237.50 N Ay + ( 265.17 ) sin 45° − 100 N = 0 ΣFy = 0: Ay = −87.504 N and: A= Ax2 + Ay2 = θ = tan −1 Ay Ax (237.50)2 + (−87.504)2 = 253.11 N = tan −1 87.504 = 20.226° 237.50 ∴ A = 253 N 20.2°

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