Force due to pressure on side of a container

[loto]

Hi gang,

I'm having an issue with one my homework problems. There is an L shaped tank, 3d in height, 1d in width. The verticle part of the L is 2d in height, and the horizontal part of the L is a 1d box. d = 8m
_
| |
| |_
|_ _|<--Force due to pressure needed for this side.

I need to find the pressure on the right wall of the horizontal part of the L. What I have managed to get so far is the equation:
F=(rho)(g)(width)(height^2)

Not very far, I know. I'm assuming I have to integrate this somehow, but I am unsure as to how to set it up. I'm not really looking for a solution, just a push in the right direction with regards to the integration, or in the correct direction if I am very, very wrong.

Thanks, in advance, for the help.

 

[Fermat]

...
_
| |
| |_
|_ _|<--Force due to pressure needed for this side.

...

Is this a Plan view (a view from on top) or a Side view?

 

[loto]

From the side, sorry.

 

[Fermat]

Ok, hang on a min.

 

[LeonhardEuler]

Hi gang,

I'm having an issue with one my homework problems. There is an L shaped tank, 3d in height, 1d in width. The verticle part of the L is 2d in height, and the horizontal part of the L is a 1d box. d = 8m
_
| |
| |_
|_ _|<--Force due to pressure needed for this side.

I need to find the pressure on the right wall of the horizontal part of the L. What I have managed to get so far is the equation:
F=(rho)(g)(width)(height^2)

Not very far, I know. I'm assuming I have to integrate this somehow, but I am unsure as to how to set it up. I'm not really looking for a solution, just a push in the right direction with regards to the integration, or in the correct direction if I am very, very wrong.

Thanks, in advance, for the help.

OK, I can explain this in general but I didn't completely follow all the dimentions you gave. From equilibrium considerations, the pressure at some depth, h is P_0+\rho gh where P_0 is the pressure at the surface. The force at this hieght is PA where A is the area. Along the surface of the wall, the pressure is changing with the depth. At a given depth, h, the force is (P_0+\rho gh)ldh where l is the length of the wall. Now it is a matter of integrating the infinitesimal forces at each height to get the total force.

 

[loto]

We are actually using gauge pressure, so it simplifies the formula you gave a bit. I did get the correct answer using your method, though, so thank you very much.

 

[Fermat]

Arrgh, too late!

Oh, well. What LeonhardEuler said.

I enclose a small piece of work explaining how to do these types of integrals for pressure over a submerged area.

It may be useful for yourself or anyone else reading these posts.