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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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What do you mean by 'has its lowest generator horizontal'? –  copper.hat Jul 15 '12 at 4:44
    
Its lying horizontal on a plane. The term highest or lowest doesnt make much sense if its standing up. –  Soham Jul 15 '12 at 4:49
    
A cone is generally characterized by a height and a radius. You said 'vertical height'. Did you mean radius, diameter? What is a 'lowest generator'? Also, you might want to consider a more recent text, Loney's is from 1932... –  copper.hat Jul 15 '12 at 5:01
    
I am so sorry, it was a very bad typo. I meant vertical angle as $\pi/3$ and not vertical height. Here vertical angle means the apex angle. Can you suggest a book. –  Soham Jul 15 '12 at 5:06
    
I used Massey's "Mechanics of Fluids", but this was in the late '70s. I don't have enough point of comparison to know whether or no it is a 'good' book. It was focused more on the engineering issues rather than physics. –  copper.hat Jul 15 '12 at 5:23
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1 Answer 1

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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Thanks a lot for the links. True, when I saw their language I realized that they are confusing a vector with a scalar. When I posted the question, I was fully thinking that though P might be proportional to W, but dimensionally, the RHS will have the unit of Pa. I agree what you said, except for one area. Why should $\rho g h $ be multiplied by area of the base, and not area of the entire surface of the cone. –  Soham Jul 15 '12 at 9:20
    
@Soham: The idea is to calculate the force on the curved surface by subtracting the force on the base from the total force (i.e. the weight). To get the force on the base, you need to multiply the average pressure on the base (which is the pressure at half height) by the area of the base. Multiplying the pressure at half height of the base with the total surface area of the cone wouldn't lead to any meaningful quantity; the pressure on the curved surface is more complicated; that's why it's better not to calculate it directly but as the difference of two quantities that are easier to calculate –  joriki Jul 15 '12 at 10:16
    
Are you considering the cone to be vertically standing? –  Soham Jul 15 '12 at 16:32
    
@Soham: No, I'm considering as you described it and as it's depicted in the solutions I linked to. –  joriki Jul 15 '12 at 16:53
    
Sorry I mistook something else. One thing is clear :) I have to go through this chapter again after 10 odd years –  Soham Jul 16 '12 at 20:27
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