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Published byBeatrix Lindsey Modified about 1 year ago

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

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Projected Forces x Z Integrated over all elements h dA z

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Buoyancy Force acting down F D = gV 1 from Force acting up F U = gV 2 from Buoyancy = F U -F D = g(V 2 -V 1 )= gV V: volume occupied by the object

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Horizontal Forces x Z Integrated over all elements h dA z h dA x Equivalent system: A plane surface perpendicular to the free surface Projected area A x Finding F x is to determine the force acting on a plane submerged surface oriented perpendicular to the surface. A x is the projection of the curved surface on the yz plane. Similar conclusion can be made to the force in the y direction F y.

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Examples Determine the magnitude of the resultant force acting on the hemispherical surface. 2 mR=0.5 m z x minusequals

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Line of Action Horizontal direction: line of action goes through z’ z zCzC Vertical direction: the line of action is 3R/8 away from the center of the hemisphere Projection in x-direction location of the centroid for a hemisphere is 3R/8=0.1875(m) away from the equator plane The resultant moment of both forces with respect to the center of the hemisphere should be zero: F x (2.03125-2)-F z (0.1875) =15386(0.03125)-2564(0.1875)=0 z’

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