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

2022
Views:

198

Pages:

2
Author:

Campbell Wagner

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 66. The pressure p at an arbitrary depth ( y sin θ ) is p = γ ( y sin θ ) so that the hydrostatic force dP exerted on an infinitesimal area dA is dP = (γ y sin θ ) dA The magnitude P of the resultant force acting on the plane area is then P = ∫ dP = ∫ γ y sin θ dA = γ sin θ ∫ ydA = γ sin θ ( yA) Now p = γ y sin θ ∴ P = pA Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then requires ΣM x: − yP P = −∫ ydP Now 2 ∫ ydP = ∫ y (γ y sin θ ) dA = γ sin θ ∫ y dA = ( γ sin θ ) I x Then or yP P = (γ sin θ ) I x yP = (γ sin θ ) I x γ sin θ ( yA ) or yP = ΣM y : xP P = ∫ xdP Now Ix Ay ∫ xdP = ∫ x (γ y sin θ ) dA = γ sin θ ∫ xydA = (γ sin θ ) I xy ( Equation 9.12 ) continued COSMOS: Complete Online Solutions Manual Organization System Then xP P = (γ sin θ ) I xy or xP = (γ sin θ ) I xy γ sin θ ( yA ) or xP = Now I x = I x′ + Ay 2 From above I x = ( Ay ) yP By definition I x′ = k x2′ A Substituting ( Ay ) yP I xy Ay = k x2′ A + Ay 2 yP − y = Rearranging yields k x2′ y Although k x′ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on the depth. ∴ ( yP − y ) = f ( depth )