COSMOS Chapter 3 Solution 106 | Michigan State University - Edubirdie

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

2024
Views:

321

Pages:

1
Author:

Caroline Mcfarland

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 106. (a) ΣM I : 0 = ( 0.32 m )( 80 N ) + ( 0.1 m )( 40 N ) − ( 0.1 m )( 60 N ) − ( 0.36 m )( 90 N ) cos α − ( 0.16 m )( 90 N ) sin α or 4sin α + 9cos α = 6.5556 ( 9cosα )2 = ( 6.5556 − 4sin α ) ( 2 ) 81 1 − sin 2 α = 42.976 − 52.445sin α + 16sin 2 α 97sin 2 α − 52.445sin α − 38.024 = 0 Solving by the quadratic formula gives for the positive root sin α = 0.95230 α = 72.233° or α = 72.2° Note: The second root (α = − 24.3° ) is rejected since 0 < α < 90°. (b) ΣF : R = ( 80 N ) i − ( 40 N ) j − ( 60 N ) j + ( 90 N )( − sin 72.233°i − cos 72.233° j) = − ( 5.7075 N ) i − (127.463 N ) j R= ( − 5.7075 N )2 + ( −127.463 N )2 = 127.6 N tan θ = −127.463 − 5.7075 θ = 87.4° or . R = 127.6 N 87.4°

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