COSMOS Chapter 4 Solution 152 | Answer Key - Edubirdie

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

2022
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1
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palangresq3v

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 152. Free-Body Diagram: From free-body diagram of beam ΣFx = 0: Bx = 0 so that B = By ΣFy = 0: A + B − (100 + 200 + 300 ) N = 0 A + B = 600 N or Therefore, if either A or B has a magnitude of the maximum of 360 N, the other support reaction will be < 360 N ( 600 N − 360 N = 240 N ) . (100 N )( d ) − ( 200 N )( 0.9 − d ) − ( 300 N )(1.8 − d ) ΣM A = 0: + B (1.8 − d ) = 0 d = or 720 − 1.8B 600 − B Since B ≤ 360 N, d = 720 − 1.8 ( 360 ) 600 − 360 ΣM B = 0: = 0.300 m or d ≥ 300 mm (100 N )(1.8) − A (1.8 − d ) + ( 200 N )( 0.9 ) = 0 d = or 1.8A − 360 A Since A ≤ 360 N, d = 1.8 ( 360 ) − 360 360 = 0.800 m or d ≤ 800 mm or 300 mm ≤ d ≤ 800 mm

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