COSMOS Chapter 3 Solution 28 | Michigan State University - Edubirdie

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

2024
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116

Pages:

1
Author:

imophulebf13

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 28. Have where M O = TAC d d = Perpendicular distance from O to rope AC with M O = rA/O × TAC and rA/O = ( 30 ft ) j + ( 3 ft ) k TAC = − ( 62 lb ) cos10° i − ( 62 lb ) sin10° j = − ( 61.058 lb ) i − (10.766 lb ) j Then MO i j k 0 30 3 lb ⋅ ft = − 61.058 −10.766 0 = ( 32.298 lb ⋅ ft ) i − (183.174 lb ⋅ ft ) j + (1831.74 lb ⋅ ft ) k and MO = ( 32.298 lb ⋅ ft )2 + ( −183.174 lb ⋅ ft )2 + (1831.74 lb ⋅ ft )2 = 1841.16 lb ⋅ ft ∴ 1841.16 lb ⋅ ft = ( 62 lb ) d or d = 29.7 ft W .

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