COSMOS Chapter 3 Solution 51 - Edubirdie

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

2020
Views:

149

Pages:

1
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 51. First note: TBA JJJG BA = TBA BA = ( 70 lb ) = ( 70 lb ) ( 4 ) i + 1.5 − ( LBC + 1)  j + ( − 6 ) k 2 ( 4 )2 + 1.5 − ( LBC + 1)  + ( − 6 ) 2 4 i + ( 0.5 − LBC ) j − 6 k 52 + ( 0.5 − LBC ) 2 rA = ( 4 ft ) i + (1.5 ft ) j − (12 ft ) k Have M O = rA × TBA = For the i components: − 763 lb ⋅ ft = 70 52 + ( 0.5 − LBC ) 2 70 lb 52 + ( 0.5 − LBC ) 10.9 52 + ( 0.5 − LBC ) = 3 + 12 LBC or (10.9 )2 52 + ( 0.5 − LBC )2  = 9 + 72LBC or 25.19L2BC + 190.81LBC − 6198.8225 = 0 Taking the positive root . j k 1.5 ft −12 ft ( 0.5 − LBC ) − 6 1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft or Then 2 i 4 ft 4 2 LBC = −190.81 ± + 144 L2BC (190.81)2 − 4 ( 25.19 )( − 6198.8225) 2 ( 25.19 ) LBC = 12.35 ft

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