# Pressure of fluids

1. Apr 6, 2009

### Niles

Hi all.

There's a couple of things I have thought about, and I hope you can clarify this for me.

1) In Torricelli's experiment with the tube and mercury, the vacuum-part in the tube has zero pressure (as vacuum always does). Does this mean that it is never possible to "suck" the mercury out of the tube (not unless there is negative pressure)?

2) Please take a look here: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]. At some point, the author explains why different-shaped containers with water have the same pressure at a height h. The author says:

"The pressure P is the same on the bottom of each vessel. Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things:
a. Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel.
b. A fluid can not support its self without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth.
c. The pressure at given level is transmitted equally throughout the fluid to be the same value at that level."

I cannot see how part B and part C explain this fact. Why do these 2 points even come into consideration? Isn't part A enough?

3) This is regarding #2. Is it only the pressure at the bottom of the vessels that are equal, or the pressure at any random height h also?

Last edited by a moderator: May 4, 2017
2. Apr 6, 2009

### arildno

In order to understand what b says, consider the following scenario:

Replace the fluid in the container with a (n idealized) rigid body of the same shape.

Since the constituent molecules of the rigid body is NOT free to jump about, hitting the container wall all the time, it follows that the walls exert NO pressure at all upon the rigid body!
(We assume that the bottom is the only place that can exert an upwards normal force upon the body)

Thus, the stress distribution, or if you like, pressure distribution will be totally different in the two cases of a contained fluid and a "contained" rigid body of the same shape.

Do you agree to that?

3. Apr 6, 2009

### Niles

I agree 100%. But I cannot see the link between B (and C for that matter also) and the fact that differently shaped vessels will have the same pressure at the bottom.

4. Apr 6, 2009

### arildno

Well, to take c first:

If you look at the interior of the fluid domain, the only other force (than pressure) acting upon a fluid element will be gravity.

Thus, in the horizontal directions (i.e, at any particular height level), the pressure gradient must be zero, i.e, constant pressure at any particular height.

That is at least an argument for the truth of c, if not for the moment an argument for the relevance of c.

Agreed?

5. Apr 6, 2009

### Niles

Yes, I agree.

6. Apr 6, 2009

### arildno

Very fine!

1. We both agree that the air pressure at the top of each container is the same! (Right?)

2. Now, consider a needle-thin column of water, whose radius is tiny enough to fit comfortably into BOTH containers, stretching downwards some height h.

3. This column of water is in EQUILIBRIUM, in both containers. They will weigh the same, and since only the bottom of the fluid column will be able to counteract the force of gravity, the pressure there must be the same, IRRESPECTIVE the size or shape of the surrounding container!

Note that the really crucial argument here is neither a,b, or c, but the fact that pressure always acts in the NORMAL direction of a surface!

Agreed thus far?

7. Apr 6, 2009

### Niles

I have an appointment at the doctor in 15 minutes, so I will be back in 45-60 minutes to read your reply.

8. Apr 6, 2009

### arildno

Okay, I'll just say that a,b, and c are pseudo-explanations; b, for example, answers the question "given that the pressure is equal at both bottoms (levels), how must the induced pressure from the walls be?".

It is rather tangential to the explanation of why the pressure will be equal at the same depth, to put it mildly..

9. Apr 6, 2009

### Niles

This argument I do not fully agree with, but it is mostly because I don't think I understand it properly. This is how I have understood it:

We have 2 arbitrary shaped containers of height H, and we also have 2 identical needle-thin water columns of height h. We place 1 needle-thin water column in each container, and since they are so thin, both containers are still in equilibrium.

From here I am a little lost.

10. Apr 6, 2009

### arildno

No, first off:

We do not "place" a column of water there in the first place, they are there.

Because neither of them moves, they must both experience an equilibrium of forces acting upon them.

Agreed?

11. Apr 6, 2009

### Niles

Yes, 100%.

12. Apr 6, 2009

### arildno

So, under the crucial assumption that the pressure on top of both columns is the same (i.e, air pressure), it follows that the pressure at the bottom of each column must also be the same, as long as the columns are of equal height (depth into to the fluid domain).

Agreed?

13. Apr 6, 2009

### Niles

Yes, this I agree on, since they are in equilibrium.

And since the height of our column can be arbitrary, this goes for any height h, so the answer to me question #3 is "any height", correct?

Last edited: Apr 6, 2009
14. Apr 6, 2009

### arildno

Precisely!

Now, it follows from the previous established fact of constant pressure at any particular height level that this constant value will "spread out" at that level until it reaches the container wall WHEREVER that might be.

Agreed?

15. Apr 6, 2009

### Niles

Yes, 100%.

16. Apr 6, 2009

### arildno

Okay!

So, note that this effectively proves the second part of b: "Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth."

It is not an "explanation" of the previous argument, rather, at best, one should think of it as an elucidation of how pressure will distribute itself along the geometry of the container.

Understood?

17. Apr 6, 2009

### Niles

Yes, I understand this now. This is interesting.

18. Apr 6, 2009

### Niles

And thank you for answering my questions.

19. Apr 6, 2009

### arildno

You are welcome!