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Hydrostatic Forces on Curved, Submerged Surfaces. x. Pressure is always acting perpendicular to the solid surface since there is no shear motion in static condition. P z. P. q. P x. Z. q. q. dA x =dAcos( q ). dA z =dAsin( q ). Projected Forces. x. h. dA z. Z. Integrated over

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## Hydrostatic Forces on Curved, Submerged Surfaces

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**Hydrostatic Forces on Curved, Submerged Surfaces**x Pressure is always acting perpendicular to the solid surface since there is no shear motion in static condition. Pz P q Px Z q q dAx=dAcos(q) dAz=dAsin(q)**Projected Forces**x h dAz Z Integrated over all elements**Buoyancy**Force acting down FD= rgV1 from Buoyancy = FU-FD =rg(V2-V1)=rgV V: volume occupied by the object Force acting up FU = rgV2 from**x**Horizontal Forces h Projected area Ax dAz Finding Fx is to determine the force acting on a plane submerged surface oriented perpendicular to the surface. Ax is the projection of the curved surface on the yz plane. Similar conclusion can be made to the force in the y direction Fy. Z Integrated over all elements h dAx Equivalent system: A plane surface perpendicular to the free surface**Examples**Determine the magnitude of the resultant force acting on the hemispherical surface. x 2 m R=0.5 m z equals minus**Line of Action**Horizontal direction: line of action goes through z’ zC z’ Projection in x-direction Vertical direction: the line of action is 3R/8 away from the center of the hemisphere z The resultant moment of both forces with respect to the center of the hemisphere should be zero: Fx(2.03125-2)-Fz(0.1875) =15386(0.03125)-2564(0.1875)=0 location of the centroid for a hemisphere is 3R/8=0.1875(m) away from the equator plane

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