# Hydrostatic pressure and consequences of pascal's law

1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

UltrafastPED
Gold Member
For many purposes the hydrostatic pressure is equal everywhere.

But as the scale of the containment increases it must be realized that the weight of the water column increases the pressure ... and more so the further down you go.

This is very significant in the oceans; insignificant in a barrel of water.

You have found an example of physical laws with limitations to their applicability; this happens all the time.

Chestermiller
Mentor
1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

Number (1) is stated incorrectly. It should read "pressure within an incompressible fluid acts equally in all directions at a given point in the fluid." This is Pascal's law observation.

Chet

If it acts equally in all directions at a given point, then why isn't the net pressure at all points in a fluid zero?

It's at a given point not at all points.

1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

I think 1) is a reasonable restatement of Pascal's Law.

For 2) you are essentially summing up all the pressures from the weight of the fluid. So as you work your way down the effect is cumulative.

I don't see a contradiction between the two, rather 2) follows as a consequence of 1).

Alright, consider one point at a certain depth in a fluid. If the pressure along all directions is equal, then the net pressure there is zero. So why do fluids have pressure at all?

Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

Chestermiller
Mentor
If it acts equally in all directions at a given point, then why isn't the net pressure at all points in a fluid zero?
If you take a tiny cube of fluid at a given location in the fluid as a free body and the fluid is static, then the pressure on each of the 6 sides of the cube is p. The cube is in static equilibrium, and the net force on the cube is zero, but that doesn't mean that the force on each of its sides is equal to zero.

Chet

True, but the NET pressure at every point is zero. Then why does fluid pressure exist?

Chestermiller
Mentor
True, but the NET pressure at every point is zero. Then why does fluid pressure exist?
If you have a solid cube (e.g., not submerged in liquid), and you apply equal normal forces on all six of its sides so that the NET force on the cube is zero, are you asking why force exists?

Chet

russ_watters
Mentor
Alright, consider one point at a certain depth in a fluid. If the pressure along all directions is equal, then the net pressure there is zero. So why do fluids have pressure at all?
Pressure is a scalar so it doesnt sum to zero, but even if it did, summing to zero doesn't mean the force is zero, it just means there is no motion. If you put your hand in a vise and crank it down, the forces sum to zero too - but it still hurts!

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russ_watters
Mentor
Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.
In all directions, not at all points.

Do you feel more pressure sitting on top of a pile of bricks or under it.

russ_watters
Mentor
True, but the NET pressure at every point is zero. Then why does fluid pressure exist?
There is no such thing as "net pressure". Pressure is a scalar, not a vector.

UltrafastPED
Gold Member
There is no such thing as "net pressure". Pressure is a scalar, not a vector.

I always thought that pressure is defined as force per unit area; thus pressure can have a direction.

See http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

This is usually most important at the boundaries - for example, inside the bubble the pressure is pointing outwards at the surface, and the air pressure is pointing inwards.

For ball under water the external pressure is all pointing inwards - until the ball (or submarine) is crushed.

Chestermiller
Mentor
Here is another example to illustrate item (1). Suppose you have a dam. At any given depth (location) right next to the dam, the water pressure is not only acting up and down. It is also pressing sideways (horizontally) on the face of the dam. This is all that item (1) is saying.

Another example is a submarine hull. The water pressure doesn't only press down on the top of the hull and up on the bottom of the hull. It also presses sideways on the sides of the hull.

The same happens to you when you go down to the bottom of a swimming pool at the deep end. If you are standing on the bottom at the deep end (say, held down by weights), the water presses horizontally on your sides as well as on the top of your head, even though your sides are nearly vertical.

Chet

Chestermiller
Mentor
I always thought that pressure is defined as force per unit area; thus pressure can have a direction.

See http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

This is usually most important at the boundaries - for example, inside the bubble the pressure is pointing outwards at the surface, and the air pressure is pointing inwards.

For ball under water the external pressure is all pointing inwards - until the ball (or submarine) is crushed.
Actually, if we are getting technical, pressure is the isotropic part of the (second order) stress tensor. Under hydrostatic conditions, the stress tensor is equal to p times the identity tensor. To get the force per unit area acting on a surface, one simply dots (contracts) the stress tensor with a unit normal vector to the surface. This yields the pressure times the unit normal. So the pressure always acts normal to surfaces, as UltrafastPED has indicated.

Chet

Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

Gravity does not apply pressure equally it's stronger the nearer you are to the body that's attracting. Pascal's law allows pressure to be to be equally transmitted when gravity is ignored or pressure is applied in it's absence.

Chestermiller
Mentor
Gravity does not apply pressure equally it's stronger the nearer you are to the body that's attracting. Pascal's law allows pressure to be to be equally transmitted when gravity is ignored or pressure is applied in it's absence.
Variation of gravitiational attraction with distance between the attracting bodies is typically not a significant contributor to hydrostatic pressure variations in practice (on the scale of a swimming pool or a glass of water, or even the depth of the ocean). The main contributor is the weight of the overlying fluid. Irrespective of this, I stand by what I said. The pressure at any given location is pushing equally in all directions. This applies whether the pressure is the result of a hydrostatic column of liquid, or whether the liquid is in a cylinder being squeezed by a piston. In fact, it applies not only to incompressible liquids but also to compressible gases.

Chet

Variation of gravitiational attraction with distance between the attracting bodies is typically not a significant contributor to hydrostatic pressure variations in practice (on the scale of a swimming pool or a glass of water, or even the depth of the ocean). The main contributor is the weight of the overlying fluid. Irrespective of this, I stand by what I said. The pressure at any given location is pushing equally in all directions. This applies whether the pressure is the result of a hydrostatic column of liquid, or whether the liquid is in a cylinder being squeezed by a piston. In fact, it applies not only to incompressible liquids but also to compressible gases.

Chet
What causes the variation in the pressure of the overlying fluid.
Why is there more pressure at the bottom of a column of water than the top.I agree that at any given location or point the pressure is equall but if you measured the pressure at a piont at the bottom of a column of water and then at another point at the top won't the two locations be pushing with different amounts of force when compared.

Chestermiller
Mentor
What causes the variation in the pressure of the overlying fluid.
Why is there more pressure at the bottom of a column of water than the top.
Suppose you have a cylindrical tank full of water. The pressure at the top surface of the tank is just atmospheric pressure. The pressure at the bottom of the tank is atmospheric pressure plus the weight of the water in the tank divided by the area of the base A. The weight of the water in the tank is equal to the mass of water in the tank times g. The mass of water in the tank is equal to the density of water ρ times the volume of water in the tank. The volume of water in the tank is equal to the area A of the base times the depth of the water h. So,

Pbottom=Patmospheric+ (ρAh)g/A =Patmospheric+ ρgh
I agree that at any given location or point the pressure is equall but if you measured the pressure at a piont at the bottom of a column of water and then at another point at the top won't the two locations be pushing with different amounts of force when compared.

Sure. But that's not what item (1) is saying. Item (1) is saying that, at any point near the top of the tank, the pressure is the same in all directions and, at any point near the bottom of the tank, the pressure is the same in all directions.

Chet

Ken G
Gold Member
But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.
The answer to this is that gravity does not apply pressure, it is a force with a very different nature than pressure. Gravity is what is known as a "body force", which means, if you think in terms of time rate of change of momentum, gravity acts to remove momentum from every gram of material at a fixed rate. Since it in some sense "acts on the body of the system", since it removes momentum per gram, it is a body force. What this means is, you should think of it as source or sink of momentum that does not need to move any momentum through any boundaries, it just plain deposits or removes it (depending on the sign you are imagining) at every point in the body, like magic if you will.

Pressure is quite different, because it is a surface force. The key difference is that when you think about how pressure adds or removes momentum from something, it always has to pass that momentum through the surface of that something, it cannot just make momentum appear like magic the way gravity does.

In the problem you are thinking about, like water in a pool, all the surfaces we need to think about are horizontal surfaces, and the only direction we need to worry about is vertical. This simplifies the situation, we don't need to worry about the fact that pressure is isotropic, we can just look at what it is doing with momentum in the up and down directions, and that suffices to see what is going on in a pool.

So imagine any horizontal surface inside the water in a pool, and ask, what kind of momentum is passing through that surface? The answer is, pressure is causing upward momentum to pass upward through that surface, and an equal amount of downward momentum to pass downward (that's the "isotropic" aspect). Let's call this equal rate of momentum crossing the surface in either direction P1A/2 (the notation is motivated by the fact that P is pressure and A is area, but it doesn't matter, it's just a number I'm talking about). It doesn't matter how it is doing that, this is just what pressure does, but if you want to know how, realize it is the action of all the little water molecules that are moving up and down across that surface all the time-- an upward molecule is obviously going to transport upward momentum upward across the surface, and a downward molecule is going to transport downward momentum downward across the surface. Now your question is, if an equal amount of upward momentum is moving upward as downward momentum is moving downward, why does anyone care?

Good question, and it has a good answer. Next imagine a second surface that is a bit deeper down below that first one. Again the same amount of upward momentum is crossing upward as downward momentum that is crossing downward, but the numerical value of this rate of crossing of momentum is different from before, now call it P2A/2, and you will not be surprised to hear that P1 < P2, since I mentioned that P is really pressure here.

Now for the reason you care about these P: think about an imaginary box that is bounded above by the first surface, and below by the second surface, and think about the rate that momentum is entering this box. There are 4 terms you have to add up, two from the top surface and two from the bottom. The top surface has P1A/2 rate of downward momentum coming down into it, so that's a net accumulation of downward momentum. It has the same rate of upward momentum leaving it upward, but here's the reason they don't cancel-- upward momentum has the opposite sign from downward momentum! So having upward momentum leave has the same effect as having downward momentum enter. Thus the two terms add up to a total of P1A, not to zero. In short, there is downward momentum entering the imaginary box through its upper surface, which is a downward force on the box, due to the P at its upper boundary.

Play the same game at the lower boundary, and you will find that P2A is the rate that upward momentum is entering from below, but that's the same as -P2A rate of downward momentum, since momentum is a vector. So the top and bottom momentum fluxes are the things that might cancel out, not the momentum passing through a given surface into our imaginary box. But if there's gravity, that body force that is adding downward momentum to the mass in the interior of our box, then we can only get force balance if pressure is removing downward momentum (or adding upward momentum, it's the same thing), and that's what the surface momentum fluxes that pressure controls is doing. That's why P1A < P2A, and that's why pressure matters, even though it is locally isotropic. It matters because it is different in different places, and Pascal's law says that once you have set up those differences, if you exert additional pressure (like adding weight to the top, for example), it can't affect the differences in pressure, because those were already what they needed to be to move a net flux of momentum around as needed to balance gravity.

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Thanks for all the help! Hydrostatics was a little tough for me to grasp.

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New doubt!

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.

russ_watters
Mentor
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.
All forces come in pairs, so you'd be counting the same force twice. For example, if you squeeze your thumb and finger together with 10N force, that requires one finger pushing one way at 10N and the other pushing in the opposite direction with the same force.