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Q & A: Pascal’s Law

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Q:
Could you give more information on pascal’s law please.A better understanding of how his theory works.
- Richard (age 35)
A:
Richard -

Pascal's Law (also called Pascal's Principle) says that "changes in pressure at any point in an enclosed fluid at rest are transmitted undiminished to all points in the fluid and act in all directions." ("Conceptual Physics," copyright 1993, HarperCollins College Publishers.) This sounds rather overwhelming at first, so let's break it down a bit.

When it says "enclosed fluid," that means that in order for Pascal's Law to be true, you have to be looking at a liquid in a closed container.

Pressure is basically a fancy word for how much something pushes on its container and on things in it. For example, air pressure is how hard air pushes on things. When you pump more and more air into your bike tire, you're increasing its pressure. If you increase its pressure too much, then it will be pushing out more than the plastic is capable of pushing in, and your tire will explode. Water pressure works the same way.

So lets say you have a long, closed tube of water with a piston at one end. (The piston is the piece of the tube at one end, except that it can slide back and forth through the tube... so it can be used to compress the water.) If you push on the piston, then you're changing the amount of space that the water has to take up, so it will push back on the piston even more. This means that you've changed it's pressure. By Pascal's Law, we know that not only did the pressure change right next to the piston, but it changed through the whole tube. So now, the water is pushing out on the entire tube more, and it's pushing out in every direction.

Let's say, instead, that we had a long tube of water with pistons at /both/ ends. Then, if we were to push on one piston, the pressure in the water would increase, like before, so the water would be pushing on the other piston, too. If there weren't anything holding the other piston down, then the water's pushing on it would cause it to slide through the tube away from the water. But because the pressure is transmitted "undiminished" through the water, it would actually move exactly as far as the first piston had moved.

Pressure is measured as the force felt by a certain amount of surface area. For example, you could have a pressure of 1 lb/cm^2. This means that the water is pushing on the container with a force of 1 pound on every square centimeter of the container. The reason that we care about this is because we can use it with Pascal's Law.

For example, let's say we have a container with a big open space in the middle, and two tubes sticking out of it. Like before, there's a piston at the end of each tube. The difference here, though, is that one of the tubes is a lot bigger than the other one. Let's say that the small piston has 1 square cm (1 cm^2) of surface area, and the big one has 100 square cm (100 cm^2) of surface area. (So the big one is 100 times as big as the small one is.) Then we push on the little one with a force of 1 lb. Since we were pushing on a 1 cm^2 with a force of 1 lb, we've increased the pressure in the entire container (by Pascal's Law) by 1 lb/cm^2. This means that the big piston will also be pushed on by a pressure of 1 lb/cm^2. But since the big piston has so much more area to be pushed on (100 cm^2), it will feel a force of 100 lb! (This is because 1lb / 1cm^2 = 100 lb / 100cm^2.)

This is really useful because it means that we can turn a small force into a big one without doing any extra work! This is the concept behind all kinds of hydraulic pumps. For example, if you've ever taken your car to the shop and had them put it up on a lift, this is probably how they did it!

This is just one example of how Pascal's Law works and why it's so important, but there's lots of others, too.

-Tamara

(republished on 08/02/06)

Follow-Up #1: why and when does Pascal's principle work?

Q:
Can you explain WHY Pascal's principle is true? If liquids/gases have so much space between molecules then why don't they compress unevenly and stronger closest to the source of the applied force?
- Jenna (age 24)
Sacramento
A:
That's a great question.

Let's look at some fluid in equilibrium. (If it's not in equilibrium, then Pascal's principle doesn't have to apply.) None of it is going anywhere. It isn't going to start accelerating off somewhere. So the net force on any little region is zero. The only external force acting on the fluid, except at its surfaces, is gravity. So if you look sideways, you better have the same pressure forces on each side of some little region in the fluid. At every horizontal plane, the pressure in the connected parts of the fluid has to be equal. What about up and down? The pressure pushing up on a region had better be just enough bigger than the pressure pushing down on it to cancel the downward force from gravity. So the pressure decreases as you go up, in a precise way that you can calculate.

A liquid like water is only very slightly compressible precisely because there isn't much space between the molecules. If you have a nearly incompressible fluid like liquid water, the pressure drop as you go up is just fixed by the gravity strength (g) and the fluid density (ρ). So if you know the pressure p0 at some point, you now know the pressure not only at every point at the same height (still p0) but also at every other height (h) from that, p0+ρgh. If you change p0 you change the pressure throughout the fluid by the same amount.

What about for something more compressible, like a gas, in which there are big spaces between molecules? The argument that the pressure at any height is constant still holds. In many cases you're not interested in heights big enough for ρgh to matter, since ρ is very small for a gas.So then it still works well enough for the whole gas.  If you are interested in greater heights (say a few kilometers) then ρgh does change and the pressure does vary a lot from the top to bottom.  In order to change p at some point, then either the density ρ must have changed or the temperature T, or both. In either case you can calculate that the pressure change at other heights, which depends on T, does not equal the pressure change at the starting point.

So you're right- in the case (gases) where the spaces between the molecules are big and variable, Pascal's principle doesn't apply quite as generally as it does for nearly incompressible liquids.

Mike W.

(published on 03/02/13)

Follow-up on this answer.