Lesson Video: Pascal’s Principle Physics

In this video, we will learn how to use Pascal’s principle to analyze the magnitude and direction of fluid pressure on an object.


Video Transcript

In this video, we’re talking about Pascal’s principle. This is a principle having to do with how pressure behaves in an incompressible fluid. Now, when we talk about an incompressible fluid, we know that a fluid is a substance that flows. Both gases as well as liquids are fluids.

But then, let’s consider what it means for a fluid to be incompressible. Say that we have a box, this one here. And we fill it with some gas. Inside this closed container, the gas will naturally expand to fill all of this volume. So then, we could say that the volume of this gas is equal to the interior volume of our box. But then, what if our box’s volume is adjustable? Say that it’s possible, for instance, for the top face of the box to be pushed downward.

With our gas on the inside, it is possible to push down on this upper face and decrease the box’s volume. If we do that, then this box in orange shows us the volume that was lost to the box’s interior. This is how much the box has been compressed. And therefore, it’s how much the gas inside the box has been compressed too. All this shows us that this gas is not an incompressible fluid, because we are able to compress it. We’re able to make some amount of it occupy a smaller volume.

But then, say that we do this. We take water. And we fill a large container most of the way to the top. And then, right at the surface of the water, attached to the sides of the container, we fasten a rubber membrane. This membrane, if we looked at it from a top-down view, would look like this. It’s a continuous sheet of rubber, except for two small holes. And these holes are designed to fit snugly around a person’s ankles. So the idea is this. If a person were to climb up to the top of this container of water and then put their feet right through these two small holes in the rubber membrane, the water that their feet were pressing down on would have nowhere to escape from this container. It would be sealed in.

So this person, if they could balance while standing up, would be able to have all their weight supported by the water that they’re standing on. And this is because, unlike the gas we saw earlier, this water is incompressible. Once it occupies a certain volume, even if we press on it to try to shrink that volume down, we’re not able to. So water is one example of an incompressible fluid. And as we said, it’s these fluids that Pascal’s principle specifically addresses.

Knowing that, let’s take a closer look at this tank of water. Even before the person climbs up and puts their feet through the holes on the rubber sheet, there’re some interesting things we can say about water pressure in this tank. First off, we know that the pressure is not the same everywhere for the water in this tank. That’s because pressure is created by the weight of the water on itself, which means that water towards the bottom of the tank, down here, is under more pressure than water towards the top. And that’s because the water towards the bottom has this much of a height of water pushing down on it whereas the water towards the top only has this depth of water pushing down on it.

We could say that the column of water pushing down on this upper point is much shorter than the column of water pushing down on this lower point. This means there’s much more water weight pushing down on the lower point, which means the pressure there is higher. And in fact, if we call the depth of a fluid below its surface level ℎ, then there’s even a mathematical relationship telling us what the pressure of that fluid will be in terms of ℎ, the acceleration due to gravity 𝑔, and 𝜌, the density of the fluid. In fact, if we multiply these three quantities together, then their product is equal to the pressure of the fluid at some depth, we’ve called it ℎ.

Going over to the left side of our tank, if we call this shorter depth ℎ one and the larger depth ℎ two. We can see that our equation for fluid pressure due to the weight of the fluid shows us that the difference in pressure between these two elevations in our fluid, we’ll call that pressure difference 𝛥𝑝. Is equal to the density of the fluid, we’ll call that 𝜌, multiplied by the acceleration due to gravity multiplied by the difference in height between ℎ one and ℎ two. All this shows us that, indeed, the pressure at various depths in a fluid isn’t the same. But rather, it varies with that depth.

And it’s worth pointing out that this equation we’ve used over here assumes that the density of our fluid is constant. In other words, that density doesn’t change as the depth of our fluid changes. This means we’re assuming that the fluid is incompressible. No matter how much of the fluid we stack up on top of itself, its density never increases. So if we were to model the pressure acting on the sides of this tank as depth increases, that pressure might look like this. Fairly light pressure towards the top of the container. But then, as we go down, pressure increases.

Another interesting fact about the pressure in this tank though is that it doesn’t only push out against the walls or the bottom of the container. If we were to pick a point at random within our water tank, say we pick this point right here. If we had to draw in the direction that pressure acts at that point, we might naturally draw an arrow down from that point. Since we know the weight of the water above that point is pressing down on it. And that’s true. Pressure does act that way at this or any other point. But interestingly, it also acts up in the opposite direction. This may seem strange since we’re saying that pressure, due to the weight of water above this point, is able to act up from that point. But it’s true.

And to see that it’s so, we could put a small, lightweight article, like a bit of sawdust, at that point in our tank. If the water in the tank was stationary and we put a few grains of sawdust at this point, then the dust would effectively stay in place. Meaning that the force pushing down on it, from the pressure due to the water column, is the same as the force pushing up on it. So at this point, or any other point in our tank, pressure acts down as well as up. And it also acts to the left and to the right and in every direction equally from this point we’ve picked. And recall that we pick this point at random. Any point that we would have chosen would show the same thing. Say, for example, that we had picked a point up here, closer to the surface.

At this second higher up point, the pressure would again act in all directions from that point. But it would just be less in magnitude than the pressure at the lower point we picked first. So, this is the condition of the pressure of the water in our tank before the person stands on top of it. Then, let’s say that they do climb up, put their feet through the holes in the rubber mat, and manage to stand upright on this water. What will happen? Well, we knew that the feet of this person will create a downward acting pressure at the points at which they contact the water. And it turns out that this added pressure, due to the weight of the person on the water, is transmitted all throughout the rest of the water in this tank. In other words, these arrows, indicating the forces created by the water pressure, all get a little bit longer.

This idea that pressure contributed at one point in an incompressible fluid is transmitted all throughout the fluid is known as Pascal’s principle. We can write out this principle as follows. A pressure change at any point in a confined, incompressible fluid, and that’s what we have here in the case of the water in our tank, is transmitted throughout the fluid. So the same change occurs everywhere. That is, the pressure increase created by our person standing on top of the water was transmitted all throughout this fluid so that all pressures everywhere increased by that same amount.

When we think about a mathematical equation for pressure, we know that, in general, pressure 𝑃 is equal to a force 𝐹 spread over an area 𝐴. And partly, in honor of the development of this principle, the S.I. base unit of pressure is named after Blaise Pascal. That unit is the Pascal. Now, let’s say that when this person stood on top of the water, because of their weight, they added a pressure of 7500 Pascals. If we apply Pascal’s principle to the water in this tank, it tells us that all the pressures at every other point also increased by this same amount, 7500 Pascals. That’s because this pressure change is transmitted throughout the fluid. So the same change occurs everywhere. Every pressure at every point in the tank goes up by this same amount.

It’s possible to use Pascal’s principle to exert very strong forces. This principle, recall, has to do with pressures. But pressures are related to forces and areas. To see how Pascal’s principle can help us understand how to exert a strong force, let’s change the shape of our water container. Let’s say that our water is now in a sealed container that looks like this. These green lines that we’ve drawn on top are pistons that can move up or down within the container walls. We can see though that the pistons aren’t the same size. The one on the left is clearly smaller than the one on the right.

Now, let’s call the area of the piston on the left 𝐴. And we’ll say that the area of the piston on the right is 100 times that, 100𝐴. So, in other words, this piston on the right is 100 times bigger in area than the piston on the left. And let’s say further that a person with a mass of 75 kilograms stands on the piston to the left. Because this person has a mass that’s in a gravitational field, we know that they will create a downward acting weight force. And this weight 𝑊 will be equal to their mass multiplied by the acceleration due to gravity 𝑔.

Now, this weight 𝑊 is a force. It’s measured in units of newtons. And we can see that this weight force is being exerted over an area, an area we’ve called 𝐴. Therefore, by our equation for pressure, we know that this force spread over this area adds pressure to our liquid. That pressure, and we can call it 𝑃, is equal to the mass of the person times 𝑔 divided by the area 𝐴.

And at this point, let’s recall that Pascal’s principle tells us that because this pressure is being added to an enclosed incompressible fluid, then this amount of pressure is added to all of the points within that fluid. And say that we focus specifically on the pressure that’s added to the piston on the right, the much larger one. The result of this added pressure there is to push up on the piston. And Pascal’s principle tells us that the pressure pushing up on the piston is the same as the pressure we added by standing on the smaller left piston.

But then, let’s consider the right side of this equation. These values, the weight force created by the person and the area of the smaller piston, are specific to the left side of our system. On the right-hand side, we know that we have a much larger piston, with an area 100 times as big. And we have some force spread out all across that piston acting up. If we call that total upward force acting on the rightmost piston 𝐹 sub up, then we know that the added pressure to our system is equal to that force divided by the area of the second piston, 100𝐴. So here we have the force and the area of the piston on the right, compared to the force and the area of the piston on the left. And these ratios of force to area are both equal to the same pressure.

And now, notice what happens if we solve for 𝐹 sub up. We can do that by multiplying both sides by the area of the rightmost piston, 100𝐴. That causes 100𝐴 to cancel from numerator and denominator on the right. And it also causes the area 𝐴 to cancel from numerator and denominator on the left. What we find is that the force acting up on the rightmost piston is 100 times the force acting down due to the person’s weight force on the left piston. What we’ve created then is essentially a 100𝑥 force multiplier. And we’ve done it by making the area of the larger piston 100 times the area of the small one.

Here’s how we can write this in a general way. Because of Pascal’s principle, where the pressure added to a system is added equally all throughout that system, so long as it’s an enclosed incompressible fluid. This means that the added force over area at one location, which we know is the additional pressure at that location, equals the additional force over area at another location, where the force and area may be different from the first. By making these areas 𝐴 one and 𝐴 two very different, we’ve seen that we can create powerful force multipliers, where we use a relatively small force, in this case, the weight of a person, to create a relatively large one. And this is possible because pressure is transmitted undiminished in an enclosed incompressible fluid.

Let’s take a moment now to summarize what we’ve learned about Pascal’s principle. Starting off, we saw that gases are compressible fluids while liquids are mostly incompressible. We learned further that pressure varies in an incompressible fluid according to the fluid depth as well as its density. At a given depth below its surface level ℎ, the pressure of a fluid is equal to that depth times the acceleration due to gravity multiplied by fluid density 𝜌.

We saw that, at any given point within a fluid, pressure is the same in all directions. And this led us into learning Pascal’s principle, which tells us that a pressure change at any point in a confined, incompressible fluid is transmitted throughout the fluid. So the same change occurs everywhere. And lastly, when we combine the equation, that pressure is equal to force divided by area, with Pascal’s principle. We found that this let us take the force divided by an area of some part of our incompressible fluid and equate that to a different force over another area of the same fluid. And by setting up the areas 𝐴 one and 𝐴 two, we saw it was possible to create force multipliers using this method.

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