COSMOS Chapter 3 Solution 111 - Edubirdie

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

2019
Views:

236

Pages:

2
Author:

elongeaborblid

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 111. Equivalent force-couple at A due to belts on pulley A ΣF : −120 N − 160 N = RA Have ∴ R A = 280 N ΣM A : − 40 N ( 0.02 m ) = M A Have ∴ M A = 0.8 N ⋅ m Equivalent force-couple at B due to belts on pulley B ΣF : Have ( 210 N + 150 N ) ∴ R B = 360 N 25° = R B 25° ΣM B : − 60 N ( 0.015 m ) = M B Have ∴ M B = 0.9 N ⋅ m Equivalent force-couple at F ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j) Have = ( 326.27 N ) i − (127.857 N ) j R RF R2 + Fy ( 326.27 )2 (127.857 )2 350.43 N  RFy  −1  −127.857   = tan   = −21.399°  326.27   RFx  θ = tan −1  or R F = R = 350 N . 21.4° COSMOS: Complete Online Solutions Manual Organization System Have ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m − ( 360 N ) cos 25°  ( 0.010 m ) + ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m M F = − ( 3.5056 N ⋅ m ) k To determine where a single resultant force will intersect line FE, M F = dRy ∴ d = MF −3.5056 N ⋅ m = = 0.027418 m = 27.418 mm Ry −127.857 N or d = 27.4 mm .

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