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Published byBeatrix Lindsey Modified over 2 years ago

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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)

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Projected Forces x h dAz Z Integrated over all elements

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**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

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**Horizontal Forces x Z 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

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

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**Line of Action zC z’ Projection in x-direction**

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( )-Fz(0.1875) =15386( )-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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Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 4.2 - 1 3.2 Slope of a Line.

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 4.2 - 1 3.2 Slope of a Line.

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