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# Hydrostatic Forces on Curved, Submerged Surfaces

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

## Hydrostatic Forces on Curved, Submerged Surfaces

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1. 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)

2. Projected Forces x h dAz Z Integrated over all elements

3. 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

4. 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

5. Examples Determine the magnitude of the resultant force acting on the hemispherical surface. x 2 m R=0.5 m z equals minus

6. 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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