Force due to pressure on side of a container


by loto
Tags: container, force, pressure
loto
loto is offline
#1
Sep5-05, 03:42 PM
P: 17
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.
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Fermat
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#2
Sep5-05, 03:57 PM
HW Helper
P: 879
Quote Quote by loto
...
_
| |
| |_
|_ _|<--Force due to pressure needed for this side.

...
Is this a Plan view (a view from on top) or a Side view?
loto
loto is offline
#3
Sep5-05, 03:59 PM
P: 17
From the side, sorry.

Fermat
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#4
Sep5-05, 04:00 PM
HW Helper
P: 879

Force due to pressure on side of a container


Ok, hang on a min.
LeonhardEuler
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#5
Sep5-05, 04:00 PM
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P: 864
Quote Quote by 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.
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 [itex]P_0+\rho gh[/itex] where [itex]P_0[/itex] 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 [itex](P_0+\rho gh)ldh[/itex] 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
loto is offline
#6
Sep5-05, 04:07 PM
P: 17
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
Fermat is offline
#7
Sep5-05, 04:10 PM
HW Helper
P: 879
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.
Attached Files
File Type: pdf Force on end-plate - strips method.pdf (84.6 KB, 75 views)


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