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A Cone whose vertical angle is $\pi/3$ has its lowest generator horizontal and filled with a liquid. Prove that the pressure on the curved surface is $\frac{W\sqrt{19}}{2}$, where $W$ is the weight of the liquid.

From my view: I would also like to find out the center of pressure, for my conceptual clarity.

SUBNOTE: I am using SL Loney "The elements of Statics and Dynamics" and it doesnt have hydrostatics portion. Can you suggest me some good book, which handles hydrostatics etc ?

Soham

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    @copper.hat I also have an engineering background, and dealt with things from a more engineering perspective.2012-07-15

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You can find a full solution here and here. Both books commit the same abuse of language as you do by calling the total force on a surface a "pressure". The total force on the curved surface is calculated as the (vector) difference between the total force exerted by the liquid on the cone, which is the weight of the liquid, and the total force on the plane surface, which can be calculated by multiplying the average pressure $\rho gh/2$ by the area $\pi r^2$, where $\rho$ is the density of the liquid, $h$ is the height of the highest point above the horizontal generator and $r$ is the radius of the base. The rest is trigonometry.

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    Sorry I mistook something else. One thing is clear :) I have to go through this chapter again after 10 odd years2012-07-16