COSMOS Chapter 9 Solution 60 - Edubirdie

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

2019
Views:

376

Pages:

1
Author:

bratrochanter

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 60. From Sec. 9.2: R = γ ∫ ydA M AA′ = γ ∫ y 2dA Let yP = Distance of center of pressure from AA′. We must have RyP = M AA ' : where yP = 2 M AA′ γ ∫ y dA I AA′ = = R yA γ ∫ ydA I AA′ = ( I AA′ )1 + ( I AA′ )2 2 2   π π  4r   π 2  4r   3 1 =  ( 2r )( r )  +   r 4 − r 2  + r r +      2  3π   2  3π   3    8  = And 2 4 π 8 9 π 4 r + − + + + 3 2 3 8π  8 9π 5π  4  4  r = 2 + r 8    4r  π 2   r   YA = ΣyA =  ( 2r × r )  +  r +  r   3π  2   2   π 2  5 π  = 1 + +  r 3 =  +  r 3 2 3  3 2 Then 5π  4  2 + r 8  = 1.2242r yP =  5 π  3  + r 3 2 or yP = 1.224r

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