# Hydrostatic Pressure on Curved Circular Surface

1. Nov 5, 2014

### Tom Hardy

1. The problem statement, all variables and given/known data
Picture:

3. The attempt at a solution

Thanks

Edit:

Where it says net force I meant to say net force per unit width.

Edit2: My integration is wrong, it should read Density*gravitational constant*R^2(-1+cos(sin^-1(h/R)))

Last edited: Nov 5, 2014
2. Nov 5, 2014

### Staff: Mentor

Hi Tom,

The pressure at a given depth is isotropic, meaning that it acts equally in all directions (i.e., perpendicular to any surface of arbitrary orientation). You correctly determined the force per unit width between Θ and Θ+dΘ as $\rho g z Rd\theta$, where z is the local depth. This pressure force is acting perpendicular to the surface of the gate at depth z. So, you need to resolve this local force into components in the horizontal and vertical directions. You then integrate the horizontal and vertical components over the wetted surface of the gate.

Chet

3. Nov 8, 2014

### Tom Hardy

Hi Chet,

Thank you for replying, I appreciate it a lot. Does this solution look better to you? (the first F/width is in the x direction and the one after in the y direction).

4. Nov 9, 2014

### Staff: Mentor

Hi Tom,

This solution really isn't correct. But I think I can help you to arrive at the correct solution by asking you a sequence of leading questions.

At a given angle Θ, what is the depth of the location on the gate below the surface (in terms of R, h, and Θ)?

After you answer this, more questions to follow.

Chet

5. Nov 9, 2014

### Tom Hardy

Thanks for sticking with this, I thought h was the local depth (variable?), maybe I misunderstood the diagram. If we call h is a constant (representing the distance between the crest level and the surface of the water) then the local depth would just be $$z = h - Rsin \Theta$$
So the force per unit width acting perpendicular to the gate at $d \Theta$ would be $\rho gzRd \Theta = \rho gR(h-R \sin \Theta) d \Theta$ is that correct? If so, should I resolve that into it's horizontal and vertical components and then integrate from there?

6. Nov 9, 2014

### Staff: Mentor

Tom! You're the man!!! Excellent!!!

Chet

7. Nov 9, 2014

### Tom Hardy

Awesome, thanks so much for your help, I really appreciate it.

8. Mar 29, 2015

### Tanya Sharma

Hello Chet

Should the horizontal force per unit width on the curved surface be $\frac{1}{2}ρgh^2$ ?

Last edited: Mar 29, 2015
9. Mar 29, 2015

### Staff: Mentor

Hi Tanya.

Yes. That's what Tom's integral comes out to be. And, of course, the answer could have been obtained much more easily in another way (which is the way, I'm sure, that you thought of). But I thought it would be useful for Tom to continue evaluating it by the integration. Would you like to reveal for Tom the much simpler way you figured it?

Chet

10. Mar 29, 2015

### Tanya Sharma

Please have a look at the attached figure . The cylinder is lying on a frictionless surface with liquid of density 2ρ on left and liquid of density 3ρ on right . The cylinder is in equilibrium and our task is to find 'h' .

I am having problem applying this result to the situation in the figure.I know how to solve this problem wihout much maths,but would like to solve it with the above mentioned integration approach .

Let h = x +R

Using the result in post 8 , doing the horizontal force balance on the cylinder

$\frac{1}{2}(2ρ)gx^2 + \frac{1}{2}(2ρ)gR^2 = \frac{1}{2}(3ρ)gR^2$

This gives , $x = \frac{R}{\sqrt{2}}$ and $h = R + \frac{R}{\sqrt{2}}$ .

But value of h should come out to be $\sqrt{\frac{3}{2}}R$ .

Could you help me in figuring out the mistake ?

#### Attached Files:

• ###### cylinder.PNG
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11. Mar 29, 2015

### Staff: Mentor

Your error is on the left hand side of the equation. It should read: $\frac{1}{2}(2ρ)g(R+x)^2=\frac{1}{2}(2ρ)gh^2$

Chet

12. Mar 29, 2015

### Tanya Sharma

I understand what you are saying , but I think i haven't explained myself properly .

I will go a bit slowly in order to find the mistake . Isn't the horizontal force exerted by fluid on the left side from bottom till height R = $\frac{1}{2}(2ρ)gR^2$ ?

Last edited: Mar 29, 2015
13. Mar 29, 2015

### Staff: Mentor

No it isn't. What is the pressure at height R?

Chet

14. Mar 29, 2015

### Tanya Sharma

Thanks .

I have found the error in the maths involved in the problem . Now I would like to understand it a bit conceptually .

If the liquid on the left side were up to height R only , then the horizontal force would have been $\frac{1}{2}(2ρ)gR^2$ . Right ?

But this changes (i.e force due to liquid from bottom to height R ) from $\frac{1}{2}(2ρ)gR^2$ to $\frac{1}{2}(2ρ)g(R^2+2xR)$ , if the liquid on the left has height h = x +R , x ≥ 0 .

Could you explain this qualitatively.

Last edited: Mar 29, 2015
15. Mar 29, 2015

### Staff: Mentor

So the real question is, "what does $\frac{1}{2}(2ρ)g(2xR)$ represent physically?"

What if we wrote this in a slightly different way: $[(2ρ)gx]R$? What does $[(2ρ)gx]$ represent physically?

Chet

16. Mar 29, 2015

### Tanya Sharma

No . This isn't what I intended to ask . Sorry for not being clear with my questions .

Nevertheless , I have found the error in my reasoning .

Thank you very much . I am glad I discussed my doubts with you .