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

2021
Views:

357

Pages:

1
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 52. First note: JJJG BA = TBA BA TBA = ( 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: − 900 lb ⋅ ft = 300 = or 315 = Then, or (1) ( 2) ⇒ 52 + ( 0.5 − LBC ) TBA 52 + ( 0.5 − LBC ) For the k components: − 315 lb ⋅ ft = or TBA 2 2 52 + ( 0.5 − LBC ) 52 + ( 0.5 − LBC ) 2 52 + ( 0.5 − LBC ) 2 i 4 ft 4 j k 1.5 ft −12 ft ( 0.5 − LBC ) − 6 1.5 ( − 6 ) + 12 ( 0.5 − LBC )  lb ⋅ ft (1 + 4LBC ) TBA 4TBA TBA 2 (1)  4 ( 0.5 − LBC ) − 1.5 ( 4 )  lb ⋅ ft (1 + LBC ) (2) 300 1 + 4LBC = 315 4 (1 + LBC ) LBC = 59 ft 4 LBC = 14.75 ft .

COSMOS Chapter 3 Solution 52

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