# Hydrostatic Conceptual Problem

1. Apr 3, 2008

### brentoat

Hey guys I'm a year 12 student studying physics, and at the moment we are studying hydraulics. We learnt that pressure varies with depth, which led me to this question. If I have two plates almost infinitely close together, each with a cross sectional area of A, would the total force on each wall be the same when I place the small mass of water between the plates, as compared to if the plates were spaced a long distance apart. The thing that gets me is if this is true would it not be theoretically possible to have two plates closely together and fill it with a very small mass of water, which generates ridiculously large horizontal forces if the depth is great enough.

Regards,
Brent

2. Apr 3, 2008

### stewartcs

I'm not clear on what you are asking. Are you saying that you have two plates pressed together with a film of water between them?

CS

3. Apr 3, 2008

### brentoat

yeah thats pretty much what I mean. Is there some minimum spacing or dimension required to develop full hydrostatic pressure?

4. Apr 3, 2008

### brentoat

I just did a quick calc, say you have two glass plates each 1m x 1m, seperated by 1mm. The 1mm all filled up with water. This means the vertical load is about 9.81N. If these are cantilever walls could the horizontal reaction to hold them up really be 9.81kpa x 1 x 1/2 = 4.9kN? Iv read through the text and the net and the maths seems right but to me it just doesnt make physical sense?

Last edited: Apr 3, 2008
5. Apr 3, 2008

### stewartcs

I believe you've misunderstood what hydrostatic pressure is. Hydrostatic pressure is due to the weight of the fluid acting on some area. It is equal to $$\rho gh$$.

where,

rho is the density of the fluid
g is gravitational acceleration
h is the height of the fluid column

https://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

Hope that helps.

CS

Last edited by a moderator: May 3, 2017
6. Apr 3, 2008

### brentoat

yeah that all makes sense, but what about that calculation I showed. It seems ridiculous that if I have 2 1x1m cantilever panels holding 0.001m^3 of water that the horizontal reaction at the base of each panel is about 4.9kN, when the total weight of the water is around 9.8N. If you extrapolate this further you can pretty much have 4.9kN of lateral load with a near infentesimal volume of water if the panels continue to get closer to each other?

7. Apr 3, 2008

### stewartcs

The total load on the vertical wall would be determined by integration as in the example below. The moment at the botom of the wall will then be a function of the applied force and the height of the wall.

http://www.math.tamu.edu/~mbelk/StudyGuide9-6.pdf [Broken]

http://physics.uwstout.edu/StatStr/statics/Beams/beame43c.htm

CS

Last edited by a moderator: May 3, 2017
8. Apr 3, 2008

### brentoat

yeah but if its a linear distribution of load then the load on the wall will be 9.81kpa x 1 x 1/2 = 4.9kN if its a 1 x 1 metre panel, as integration is basically area under the curve. Does this not sound like a ridiculous lateral load, say if the volume of water between the panels in 1x10^-6 m^3??

9. Apr 3, 2008

### stewartcs

The force found in the integration is the total force acting on the wall.

It's not a linear distribution, the lateral load increases with the depth of the water. The maximum pressure is at the bottom.

CS

10. Apr 3, 2008

### brentoat

I know the maximum pressure is at the bottom, and zero at a depth of zero, which is why it is a linear load distribution on the wall. In my calculation the formula I used was derived from integration, but I did it from first principles again to reinforce. http://img98.imageshack.us/img98/2991/hydrodx5.jpg [Broken]
Using both ways the lateral load comes out to be 4.9kN, with a weight of water of only 9.81N. Extrapolating this as I did before, does this mean there can be a 4.9kN of load on the wall if the walls are placed closer again, with the total volume of water being say 1cm^3??

Last edited by a moderator: May 3, 2017
11. Apr 4, 2008

### stewartcs

Sorry, it is linearly distributed, but what I was trying to say was that the load on the wall would be varying (I guess is a better description) and increasing on the way down (but you already know that).

I can't view the link you gave at work so I'm not sure what you did. I'll take a look later. If the walls are open at the top to the atmosphere, then placing them closer together will just force the water up and increase the height of the fluid column. However, the total load on the wall is the same (unless the water spills over the top of the wall). I think that is what you are asking anyway. Of course I'm assuming the fluid is contained as in the case of a tank.

CS

Last edited by a moderator: May 3, 2017
12. Apr 4, 2008

### brentoat

yeah thats sort of what Im getting at, let me try to ask it another way. If I have 2 1x1m panels seperated by 1cm and filled up to the top with water, will the total horizontal force be the same as if I have 2 1x1m panels seperated by 1mm and filled to the top with water?

13. Apr 4, 2008

### stewartcs

If you have just two vertical panels, side by side, then the water will not stay between them. If you have four vertical panels that form a box (with a bottom of course) full of water then the load on the walls is a function of the wetted area of the walls, and depth and density of the fluid. The horizontal distance between the walls doesn't matter as far as the hydrostatic pressure of the fluid goes.

CS

14. Apr 4, 2008

### brentoat

yeah according to the maths I agree, just my intuition tells me its ridiculous that nearly 5kn can be generated on a 1x1m panel even if there is only say 50g of water between the two panels and their enclosed sides

15. Apr 4, 2008

### Q_Goest

Hi Brent,
Maybe thinking of it this way will help. The pressure at any depth is a function of the weight of the water above it. Let’s put a coordinate system on our 2 plates with Z being vertical, and with X and Y being the two horizontal directions. For some small plane, dx dy, the volume of the water above this thin column is dx dy Z. The weight then, is simply the volume times gravity, times the density (dx dy Z) g (rho). To find the pressure which this column is pushing down with, you take the weight and divide by the area. Now the dx and dy drop out, so you’re left with Z g rho which is how we come up with the pressure at any depth.

Now you can think of a large volume of water as having all these columns, dx dy, each exerting a pressure at some depth Z. From this, it becomes obvious that pressure is independent of volume, it depends only on the depth.

Pressure also has the interesting characteristic of pushing equally in all directions (at some point in the fluid). So the pressure of the water at any point pushes equally in all directions, and this pressure is independent of how much volume you have.

The conclusion is what you find so intuitively uninviting. A tiny drop of water can create a huge force tending to separate two plates if the drop is forced to spread out over a large area.

16. Apr 4, 2008

### brentoat

Thanks for that Goest, the maths has always made sense it was just my intuition telling me maybe there is a minimum dimension needed, depending on the viscosity of the fluid. However it makes sense that if the horizontal and vertical forces are different at any point then there will be a deviator stress causing shear within that infentesimal element, causing fluid motion, which obviously cant happen. Anyway thanks for your help