Hydrostatic pressure and consequences of pascal's law

  • #26
Ken G
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But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.
We do, that's what pressure is. Many people think pressure is a force per unit area, but not always-- it only becomes a force when you strip away half the stuff, leaving the force from the other way unbalanced. But pressure usually isn't like that, it's only like that at a boundary and only if you look at only have the story. So you can think of pressure two different ways-- either say pressure is a force per unit area, and only exists at a boundary, so when you talk about it in the interior, you imply you have stripped away one side and created a boundary. That's what we mean when we say "a piston exerts a pressure", so it's best used for external pressure. Or, we can say that pressure is a momentum flux per unit area, coming from both sides, so exists in the interior of the fluid or solid, and is best thought of as an interior pressure. Your confusion is between these two rather different, yet quantitatively equal, meanings of pressure.
 
  • #27
Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?
 
  • #28
Ken G
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Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?
Excellent question. Consider a crowd of people hitting tennis balls against a wall, and bouncing back. Now consider a similar crowd on the other side of the wall, doing the same thing. There is no net force on the wall, by symmetry, but there is "pressure", and you can get that pressure by looking at the rate the tennis balls are "depositing momentum" into the wall from either side (not both, there is no net deposition from both as there is no net force). Now remove the wall-- what changes? Nothing! The balls look exactly the same on average. There is still no net force in the middle plane, just as there was not before, and there is still pressure there, just as there was before! You don't even need the balls to collide with each other, pressure has no direct requirement for collisions.

Now your question is, but wait a minute, when you have the wall, we are only counting the pressure from half the balls, and when we remove the wall, we are counting all the balls. Do we not make a factor of 2 error? No, we do not, because when there is a wall, the balls from one side are bouncing, so they deposit their momentum twice. When there is no wall, the walls do not bounce, but there are twice as many that participate in the counting. This is that "two different ways" to think about pressure, that give the same answer.
 
  • #29
Alright. So when we measure pressure without an object at that level, we must consider both forces, right?
 
  • #30
russ_watters
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Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?
It does experience both forces, but the situation hasn't changed any: now you have four forces instead of two, all of which are balanced and equal to 10N.

Instead of 10=10, you have 10=10=10=10.

In order for forces to be additive, they have to be applied at the same point and pointed in the same direction. Here, the forces are opposite each other.
 
  • #31
russ_watters
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Alright. So when we measure pressure without an object at that level, we must consider both forces, right?
There is no "both forces" with pressure in a fluid: pressure is just pressure. It is a scalar. At a macroscopic level, it is continuous: acting in all directions, not just two.
 
  • #32
Ken G
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I think what is troubling you is that you have been told that pressure is a force per unit area, but a force is a vector, so has a direction. If you put an infinitely thin plate inside a fluid, where is the force per unit area? Well, it is on both the top and bottom of that plate, but it cancels, so you wonder, what's the point?

The point is, now imagine the plate is not infinitely thin, it has an interior that is empty, it's thin flat metal box. Now you have a force per unit area on the top, pointing down, and a force per unit area on the bottom, pushing up. That's what pressure does-- it tries to crush empty boxes you put in there! Those two forces per unit area are both P, but you'd never add them to get 2P, because for the net force on the box, they add up to zero (no net force on the box), but on each surface of the box, there is a big force, and it better be a strong box or it will crush.

Also, if the pressure changes with height, due to gravity, then the pressure at the top of a box with a finite width is a little less then the pressure at the bottom. That results in a force that does not cancel out, but only because the box has a nonzero width-- the force is proportional to the width of the box. That is called the buoyancy force, and it is not a pressure, it is a force-- but it's there because the pressure changes, not only because there is pressure, and that force goes to zero as the width of the box goes to zero. Pascal's theorem is the recognition that there is a big difference between the pressure, and the change in pressure. If you put more water above the box we are talking about, the pressure will increase, and the box better be even stronger to avoid crushing, but the buoyancy force is not changed, because that deals not in the pressure but in the change in pressure, and Pascal's theorem says the changes in pressure won't be different if you just crank up the pressure by some external action you do.
 
  • #33
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Consider a point at a certain depth in an ideal fluid. We say that pressure at that point is due to the weight of the fluid above it. But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.
This is very much the same thing as the tension in a string. At any point in the string the tension is T, but if you imagine a planar cut at any point within the string, the tension on the left surface of the cut is T acting to the right, while the tension on the right surface of the cut T is acting to the left. Why does pressure act in a way so much similar to tension? Because they are both asoects if the second order stress tensor. The pressure is a compressive stress, and the tension in the string is a tensile stress. The stress tensor encompasses both of these.

Chet
 
  • #34
OK, got it. Just one last question. If I have a fluid in a container in outer space with an object in the fluid and I seal the container tight, is there any pressure acting on the object?
 
  • #35
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Yes. If there is liquid water and water vapor in the container, then the pressure on the object is equal to the equilibrium vapor pressure of water at the temperature in the container. If there is only liquid water present in the container, then the pressure on the object is higher than the equilibrium vapor pressure of water. The magnitude of the pressure then depends on how much the container is squeezing on the liquid, which, in turn, depends on how the container was filled.

Chet
 

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