COSMOS Chapter 3 Solution 11 | 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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170

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1
Author:

lankstoetp15

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 11. The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y ( d ) M D = 1152 N ⋅ m where (TAB max ) y Now sin θ = = TAB max sin θ = ( 2880 N ) sin θ 1.05 m (d  ∴ 1152 N ⋅ m = 2880 N    + 0.24 ) + (1.05 ) m 2 2 1.05 (d or or or + 0.24 ) + (1.05 ) 2 2   (d )   ( d + 0.24 )2 + (1.05)2 (d = 2.625d + 0.24 ) + (1.05 ) = 6.8906d 2 2 2 5.8906d 2 − 0.48d − 1.1601 = 0 Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. Since only the positive value applies here, d = 0.48639 m or d = 486 mm .

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