COSMOS Chapter 4 Solution 45 | Michigan State University - Edubirdie

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

2023
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338

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2
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Adowj7b

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 45. Note: Weight of block is W = (10 kg)(9.81 m/s2) = 98.1 N (a) Free-Body Diagram: ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 98.1 N = 0 Ay = 98.1 N or A = 98.1 N Therefore: ΣM A = 0: M A − ( 98.1 N )( 0.45 m ) = 0 M A = 44.145 N ⋅ m or M A = 44.1 N ⋅ m (b) Free-Body Diagram: ΣFx = 0: Ax − 98.1 N = 0 Ax = 98.1 N ΣFy = 0: Ay − 98.1 N = 0 Ay = 98.1 N continued COSMOS: Complete Online Solutions Manual Organization System Thus: A= Ax2 + Ay2 = ( 98.1)2 + ( 98.1)2 = 138.734 N or A = 138.7 N ΣM A = 0: 45° M A + ( 98.1 N )( 0.45 m + 0.1 m ) = 0 M A = 44.145 N ⋅ m or M A = 44.1 N ⋅ m (c) Free-Body Diagram: ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 98.1 N − 98.1 N = 0 Ay = 196.2 N ΣM A = 0: or A = 196.2 N M A − ( 98.1 N )( 0.45 m − 0.1 m ) − ( 98.1 N )( 0.45 m + 0.1 m ) = 0 M A = 88.290 N ⋅ m or M A = 88.3 N ⋅ m

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