COSMOS Chapter 3 Solution 70 | Michigan State University - Edubirdie

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

2020
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261

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1
Author:

slypery5hy

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 70. (a) Note when a = 8 in., rC/F is perpendicular to the inclined 10 lb forces. Have M = ΣFd ( = −( ) ) a + ( ) − ( )  a 2+ ( ) a = 8 in., For M = − (10 lb )(18 in. + 24.627 in.) = − 426.27 lb ⋅ in. or M = 426 lb ⋅ in. (b) Have M = 480 lb ⋅ in. Also M = Σ ( M + Fd ) ( ) = Moment of couple due to horizontal forces at A and D + Moment of force-couple systems at C and F about C. Then − 480 lb ⋅ in. = −10 lb  a + 8 in. + 2 (1 in.)  +  M C + M F + FX ( a + 8 in.) + Fy ( 2a )  Where M C = − (10 lb )(1 in.) = −10 lb ⋅ in. M F = M C = −10 lb ⋅ in. Fx = −10 lb 2 Fy = −10 lb 2 ∴ − 480 lb ⋅ in. = −10 lb ( a + 10 in.) − 10 lb ⋅ in. − 10 lb ⋅ in. − 10 lb 10 lb ( a + 8 in.) − ( 2a ) 2 2 303.43 = 31.213 a or a = 9.72in. .

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