# Force due to pressure on side of a container

by loto
Tags: container, force, pressure
 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.
HW Helper
P: 876
 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?
 P: 17 From the side, sorry.
 HW Helper P: 876 Force due to pressure on side of a container Ok, hang on a min.
PF Gold
P: 864
 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 $P_0+\rho gh$ 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.
 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.
HW Helper
P: 876
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
 Force on end-plate - strips method.pdf (84.6 KB, 92 views)

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