Force due to pressure on side of a container

In summary: Thanks for your time.In summary, the pressure on the right wall of the horizontal part of the L is 8m. The pressure is due to the force needed for this side.
  • #1
loto
17
0
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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  • #2
loto said:
...
_
| |
| |_
|_ _|<--Force due to pressure needed for this side.

...
Is this a Plan view (a view from on top) or a Side view?
 
  • #3
From the side, sorry.
 
  • #4
Ok, hang on a min.
 
  • #5
loto said:
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.
 
  • #6
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.
 
  • #7
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.
 

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1. What is force due to pressure?

Force due to pressure refers to the amount of force per unit area that is exerted perpendicular to the surface of an object or container. It is caused by the collisions of particles, such as molecules or atoms, against the surface.

2. How is force due to pressure calculated?

Force due to pressure can be calculated by multiplying the pressure (in pascals) by the surface area (in square meters) that is experiencing the pressure. The formula is F = P x A, where F is force, P is pressure, and A is area.

3. What causes pressure on the side of a container?

Pressure on the side of a container is caused by the constant collisions of particles inside the container against its walls. As the number of collisions increases, so does the pressure on the container's surface.

4. How does force due to pressure affect the stability of a container?

Force due to pressure can affect the stability of a container by exerting an outward force on its walls. This can cause the container to expand or even burst if the pressure becomes too great. To ensure stability, containers are designed to withstand a certain amount of pressure.

5. Can force due to pressure be controlled?

Force due to pressure can be controlled by adjusting the surface area or the pressure itself. For example, a larger surface area can distribute the force over a larger area, reducing the overall pressure. Similarly, decreasing the pressure will also decrease the force exerted on the container's walls.

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