COSMOS Chapter 3 Solution 116 | Answer Key

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

2021
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1
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COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 116. Let ( R, M D ) be the equivalent force-couple system at O. Now ΣF : R = ΣF = (1.8 lb )( − sin 40°i − cos 40°k ) + (11 lb )( − sin12° j − cos12°k ) + (18 lb )( − sin15° j − cos15°k ) or R = − (1.157 lb ) i − ( 6.95 lb ) j − ( 29.5 lb ) k Note that each belt force may be replaced by a force-couple that is equivalent to the same force plus the moment of the force about the shaft (x axis) of the sander. Then ... ΣM O : M O = ΣM O i j k = (1.8 lb ) 0 0.75 in. 2.2 in. − sin 40° 0 − cos 40° − ( 2.5 in.)(11 lb )  i − ( 9 in.) i × (11 lb )( − sin12° j − cos12°k ) + ( 2.5 in.)(18 lb )  i − ( 9 in.) i × (18 lb )( − sin15° j − cos15°k ) = (1.8 )( −0.75cos 40°i − 2.2sin 40° j + 0.75sin 40°k ) − 27.5i + ( 99 )( sin12°k − cos12° j) + 45i + (162 )( sin15°k − cos15° j)  ( lb ⋅ in.) = ( −1.03416 − 27.5 + 45 ) i + ( −2.5454 − 96.837 − 156.480 ) j + ( 0.86776 + 20.583 + 41.929 ) k  ( lb ⋅ in.) or M O = (16.47 lb ⋅ in.) i − ( 256 lb ⋅ in.) j + ( 63.4 lb ⋅ in.) k .

COSMOS Chapter 3 Solution 116

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