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I came across some problems related to finding the center of pressure. Please guide me how to go about them. Posting them all so that I can find out the intricacies involved when there are so many different geometrical shapes involved.

  1. A quadrant of the ellipse $$x^2+4y^2 = 4$$ is just immersed vertically in a homogeneous liquid with the major axis in the surface. Find the center of pressure.

  2. Find the depth of the center of pressure of a triangular lamina with vertex in the surface of the liquid and the other two vertices at depths $b$ and $c$ from the surface.

  3. A circular area of radius $a$ is immersed with its plane vertical and its center at a depth $c$. Find the position of its center of pressure.

Thanks in advance!!

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As @udiboy says, take the shape of the portion of the body that is under water, assume that shape is made of water, and then find the shape's center of gravity. – Mike Dunlavey Aug 6 '13 at 12:22

From what I understand, the center of pressure is the point where the net force due to fluid pressure is acting on an immersed body. Now we know that the net force of pressure is equivalent to the Buoyant force felt by the object. So the center of pressure will be the point of application of the buoyant force.

The Buoyant force can also be described as the force exerted on a body of that same fluid having exactly the same geometry as the volume of body submerged(Because force of fluid pressure only depends on the geometry).

In effect, the body immersed can be replaced by a body of that fluid occupying the same volume, and having the same geometry.

This is used to prove that the buoyant force is equal to the weight of the fluid displaced(The weight of that replaced body of fluid).

Similarly, it can be observed that the point of application(Center of pressure) of this Buoyant force will only depend on the geometry, and so will be the same in both cases.

In the second case, when we consider the body of fluid, the point of application will simply be the center of mass of that fluid(Because the center of mass of this fluid is at rest, and there is no rotation about it too).

For a body of uniform density the center of mass is the same as the geometrical center.

So if you want to find the center of pressure of a body of any mass distribution, find the center of mass of a uniform body, having the same geometry as the volume submerged. For example, the center of pressure for the triangular lamina will be its centroid. The process is the same for all other cases.

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Thanks I'll try that out....If you could post the solution to one of the problems above that would be useful. Thanks. – Sheetal Sarin Aug 5 '13 at 7:20
@SheetalSarin, I did mention that for the triangular lamina it will be the centroid of that triangle. Others are a little complicated. You'll have to do the math for them. – udiboy1209 Aug 5 '13 at 7:57

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