# Confused about a very basic concept: Pressure.

1. Jun 5, 2013

### Tomer

Hello everybody,

this is slightly embarrassing (for a physics student), but I realized in the last couple of days that I am somewhat confused with the concept of "pressure".

Pressure is defined in "school-level" as "force per area". So if I have a plate of 2m^2 on which a downwards force of 3N is exerted, the pressure on the plate is 3/2 N/m^2. Right?

First question: What happens if, in the scenario above, I exert the same force in the opposite direction (3N upwards) as well? Is my pressure 0 N/m^2 or 3 N/m^2 (double than before)?

The motivation for this question is really somewhat more complex: One says that the pressure on the surface of a star vanishes (P=0) and maximal in its center.

Now if I assume I have a massive ball, and I put aside all gas/radiation/whatever pressures and concentrate only on the gravitational pressure, i.e., the force/area exerted by gravitation-
Second question:
Why should the pressure in the center be maximal? All the forces cancel one another in the "exact center", so that there's practically 0 Force - why not therefore 0 pressure?
Furthermore, why is the pressure on the outer layer 0? The force there (on a certain gas element) is GMm/r^2, thus non-vanishing - why shouldn't there be any pressure? What if I'm simply holding a plate floating above the star - doesn't it have gravitation pressure on it?

So I've been working with pressure for years now, using it in thermodynamic equations as this quantity that tells me of the force transferred by gas molecules to normal surfaces, ideal gas equation and so one, but I realize, I don't really "dig" the definition of pressure.

Can someone enlighten me?

Thanks a lot.

2. Jun 5, 2013

### Staff: Mentor

Makes sense to me. Assuming that the load is evenly distributed over the plate.

I think what you're describing is this: A plate with a force of 3 N inward on both sides, giving a net force of zero. Correct?

Realize that this is really the same scenario as above, unless you meant for the plate to be floating in air or something. The plate was resting on a table, say, and you pushed down with a force of 3 N. Of course the table is pushing up with an equal force. The pressure is still 3/2 N/m^2.

A couple of points:

(1) Just because the net force is zero, doesn't mean there's no pressure. Imagine an elephant stepping on your toe, squishing it against the ground. The net force is zero--your toe isn't accelerating, is it? Yet I'm sure you'll agree that the pressure is far greater than zero.

(2) Imagine a tall stack of thin books. Thousands of them stacked up. The pressure on the bottom book is due to the weight of all the books piled on top of it. As you go higher in the stack, the pressure is reduced. At the highest point, the pressure goes to zero. (Imagine a very thin book.) That's why the pressure increases as you get closer to the center of the star--you have the weight of the entire star to support.

3. Jun 5, 2013

### Tomer

Thanks so much for the fast reply!

Well, in the first scenario I actually did mean a "floating plate" in this sense. I did not want to write it, but I wanted to imagine the situation of a force acting on only one side of the plate, as opposed to the second scenario.

What happens if the plate just "floats"? Is there in this case no pressure? Why do we say that pressure is force/area then?
And if there is pressure (3/2 N/m^2) - how come the existence of a force from the opposite direction doesn't affect (cancel?) it?

I agree of course with your example, but what does it mean? What is the definition of pressure in this scenario? Obviously, not the total force (0!) per area of my toe? The net force is zero, and yet my toe being squashed. How can this be described if we're dealing an ideal 2D plate?

Again, I understand the analogy. But here we have the Earth pulling all the books downwards. Going back to the star, how do the facts that:
1. The force outside of the star is maximal, i.e. GMm/r^2, and
2. The force in the center is 0,
agree with the fact that the pressure outside is 0, and the force in the center is maximal?
Again - if I hold a plate in my hands (forgetting about the "normal force" for a second) - does the gravitational field of Earth not exert a pressure on this plate?

I hope you'll (or others) have the patience to get me out of this disturbing thought-loop, I realize what I'm saying is contra-intuitive, and yet I don't understand how these fact add up from a theoretical/definition point of view.

Thanks!

4. Jun 5, 2013

### Staff: Mentor

Well, if you only push one side of the plate, it will go flying. But of course there would still be pressure on it.

Sure there is pressure. You're pushing on the surface.

Pressure is force on a surface. Pushing a different surface doesn't directly affect the pressure on first surface.

You need to describe the force exerted on the surface. Not the net force on the object (toe or plate) as a whole.

You are describing the gravitational field strength, not the pressure. The gravitational force on a given mass will be greatest on the surface and zero at the center. But that's different than the total pressure pushing down on any layer.

Well, depends what you mean. At what point in the plate? Imagine the plate is an inch thick. At the bottom surface, you must exert an average upward pressure to support the entire weight of the plate. But what about in the middle of the plate? That "surface" only has to exert an upward force to support half the weight of the plate. See how the pressure varies with depth? What pressure would you expect on the top surface of the plate?

Note: You can't forget about normal force. You are holding the plate. If you just let it go, then there won't be any pressure as it would be in free fall. (Ignoring complications such as air resistance.)

5. Jun 5, 2013

### Tomer

Thanks!
Ok. Let's remember this statement.

So the two sides of an ideal 2D plates are considered to be 2 different surfaces? If this is the case, it definitely helps me understand why the two forces from both ends don't add up/cancel the pressures.

Why? Why is the pressure a layer "feels" due to gravitation not the gravitational force divided by the area?

I understand what you mean. and yet, what you're describing is how the different layers of the plate react to the weight of the layers above them, which are attracted to the star due to its gravitational field. But what I think about is pressure the gravitational force itself exerts on a certain layer, without considering the other layers pressing on it. In other words, I'm imagining a flat 2D plate. Wouldn't the gravitational force exert a pressure on it, just because of the fact there's a force acting on it?

How does this agree with your first claim? You wrote that if the plate would just be flying and accelerating due to a one-sided force (me pushing it, gravitation or whatever) - there would be pressure. Now you wrote that there's no pressure in free fall.

Thank you, waiting for your response :)

6. Jun 5, 2013

7. Jun 5, 2013

### SteamKing

Staff Emeritus
For a star, the condition P = 0 at the surface is a mathematical statement that the star is surrounded by the vacuum of space. P is a maximum at the center due to the fact that the entire mass of the star is attracted gravitationally to the center of the star.

8. Jun 5, 2013

### Staff: Mentor

Right. Pressure is always pressure DIFFERENCE. Saying there is only pressure on one side is equivalent to saying the other side is exposed to a vacuum.

9. Jun 5, 2013

### Staff: Mentor

How can I forget?

Again, pressure is defined on a surface. The net force on the object may be zero, but the pressure on each side is not.

If I bury you in a pile of dirt, the pressure you feel is the contact force of stuff above you pushing down on you, not your weight.

Pushing a plate is not the same as free fall. In free fall, the only force acting is gravity.

(Note that pressure is a contact force per unit area, not a gravitational force. It's often due to there being gravity, of course, but is not the same as gravity.)

10. Jun 5, 2013

### Staff: Mentor

Good point. For simplicity, in my examples I ignored things like atmospheric pressure acting on the plate.

11. Jun 5, 2013

### Tomer

I get stepped on by elephants, buried under dirt... :)
Intuitively, of course I understand. From a definition point-of-view, I have a plate, a force F is exerted on it due to gravitation, and thus the pressure should be F/A, only it's not. Perhaps the reason for the vanishing pressure is the fact that both sides of the 2D plate feel exactly the same "gravitational" pressure, and as our friend russ_watters suggested, the pressure difference in this case is 0?

Ok, that's something I didn't know. I didn't know pressure is defined only through contact force. (whatever contact is? Is it not merely an electromagnetic force?).

12. Jun 5, 2013

### Staff: Mentor

If the plate is laying on a table, then the average pressure on the bottom surface due to its own weight would equal F/A. There's nothing pushing down on the top surface, so why would there be a pressure there? (As before, I am ignoring atmospheric pressure.)

No. In the example I discuss above, there is pressure on the bottom of the plate but not the top. If you have another scenario in mind, describe it.

13. Jun 5, 2013

### Tomer

The scenario is the same one as described before: a 2D plate is free-falling due to a gravitational force. A force F is thus exerted on
the plate. And yet there's no pressure.

You then mentioned that pressure is only defined for "contact" forces (something I'm not seeing written in Wiki), which then of course means that the gravitational force directly cannot exert any pressure. But I was never aware of this distinction.

14. Jun 5, 2013

### Staff: Mentor

That's right, because the plate is in free fall and thus accelerating. Time for another example with books.

Put two books, one atop the other, on a table. Each book weighs W. So, the contact force between the bottom book and the table is 2W, and between the two books is W.

Take those same two books and drop them. Now the books are in free fall. What is the force between them? (When they were on the table, they exerted a force equal to W on each other. What about when they are falling?)

Yet another example. You are on an elevator, standing on a scale that reads the force you exert on the elevator floor. When the elevator is stationary (or moving with constant speed) the scale reads a force equal to your weight. When the elevator accelerates upward, the scale reads higher than your weight; when accelerating downward, it reads lower. And if all the cables (and brakes) were removed, so that you and the elevator were in free fall, what would the scale read then?

15. Jun 5, 2013

### Tomer

:)
I understand of course where you're going at and I agree that the pressure would be intuitively zero in these free-falling scenarios. What I still don't understand is, how does this agree with the definition " Pressure is the amount of force acting per unit area" (Wiki. Not that Wiki must be the best source of knowledge, but in such basic cases it's usually accurate enough). When the 2D plate is free-falling, the gravitational force is exerted on it. If I stick to the naive definition, the pressure would be the force divided by the area of the plate.

If the resolution to this discrepancy is "but gravitation isn't exerting any contact force": ok. But I would love to know what the definition for a contact force is. Does radiation exert contact force? Is the electromagnetic force a "contact force"? What is the difference?

If the resolution is another one, I'd really love it if you tell me what it is. I think we had enough examples with elephants and books :) My intuition tells me I'm wrong, the definitions don't.

16. Jun 5, 2013

### Staff: Mentor

The gravitational force is not being applied to the surface, so it can't be a pressure.

17. Jun 5, 2013

### Tomer

What is it applied to, if the plate is being accelerated by it? :)

18. Jun 5, 2013

### Dmobb Jr.

I think that the problem here is a bad definition. Clearly when something is freefalling there is a force on it or it would not accelerate. There is no pressure despite this. It seems like in every example here the force is a normal force.

So I think the definition should be Normal Force/Area, not Force/Area

19. Jun 5, 2013

### Staff: Mentor

Every individual particle inside the plate.

20. Jun 5, 2013

### Staff: Mentor

Implicit in that definition is that the force must be applied to a surface.

Before worrying about the accelerating plate, let's be sure you understand how pressure varies in a non-accelerating plate.

Imagine you have a (thick) plate resting on a table. Total weight W, area A. Describe to me the pressure at various points throughout the plate. How would you calculate that pressure?

(Too bad you didn't like the books and elephants. If you understood those, you'd understand this.)