Lesson Video: Pressure Produced by Fluids Physics • 9th Grade

In this video, we will learn how to use the formula 𝑃 = ρgh to calculate the pressure produced at different depths by different fluids that gravity acts on.


Video Transcript

In this video, we’re talking about the pressure produced by fluids. And it’s true! All fluids — and those include gases as well as liquids — produce pressure. In this video, we’ll learn what causes fluid pressure. And we’ll also learn how to calculate fluid pressure for different types of fluids at different depths.

Now here’s a bit of an interesting thought. Whenever a person steps outside and walks around, whether they realize it or not, they’re under the influence of fluid pressure. A fluid we can recall is any form of matter that can flow. This includes gases and liquids. This means that whenever we’re outside, we’re actually underneath many layers of gases in the atmosphere. Altogether the weight of all these layers of gases in the atmosphere exert a force. And this force spread over the area of our body as we walk around creates a pressure. We call this atmospheric pressure. And this is a pressure experienced by any object at sea level.

In general, in order to create pressure, we need to have two ingredients. The first is a force and the second is an area that that force is spread over. For example, if we have some surface with an area that we can call capital 𝐴 and then we exert a force, we’ll call it capital 𝐹 on that surface, then the average pressure — we’ll call it 𝑃 acting on the surface — is equal to the force divided by the area. So that’s what pressure is. It’s a force spread out over an area. And just like stacks of layers of air in the atmosphere all create a downward-acting force through their weight. And when that force is spread out over the area of our body, we experience a pressure. Just like that, pressure is a force spread out over an area. And just briefly so it doesn’t catch us by surprise when we see it in an example, let’s say a quick word about the units of pressure.

Since pressure is a force divided by an area, we know that those units will be the units of newtons, the SI unit of force, divided by meters squared, the unit of area. But we almost never see pressure quoted this way. We almost never see a pressure given as some number of newtons per square meter. Instead, the typical way to quote a pressure is in a unit called a pascal abbreviated Pa. Here is what a pascal of pressure means. Say we had some surface and the area of that surface was exactly one meter squared. And then say that we apply a force to that surface. We apply exactly one newton of force. Well, one newton of force spread out over one square meter of area is equal to one pascal of pressure. And just as a sidenote, one pascal of pressure is a very, very small amount.

For example, the atmospheric pressure we talked about experiencing anytime we step outside involves over 100000 pascals of pressure. But in any case, this is what one pascal is, a newton of force spread over a meter of area. We’ve said here that any fluid whether a gas or a liquid will produce pressure thanks to its weight. And when we consider the pressure produced by liquids specifically, this is something lots of us have experience with. Anyone who’s ever been inside the diving well of a swimming pool has probably experienced that as they go deeper into the wheel, the pressure that they feel increases. Often, the first place we feel this is in our ears, one of the more sensitive spots on our body to pressure. Practically then, we get this overall sense that as our depth in the water increases, the pressure grows along with it.

And if we think about the water in this tank in terms of layers just like we thought of air in the atmosphere that way, this makes sense. Say that we consider each one of these separate layers of water in the water tank. We know that the water in each layer will have some weight thanks to its mass and the force of gravity on it. And if we start out by just considering the water in this topmost layer here, we could say that the weight of that water could be modelled this way. It creates a downward-acting force below that layer. This means if we were underneath that one layer of water, then we would feel the weight from it by itself.

Well, strictly speaking, there is also the weight of the air in the atmosphere acting on us at this point. But let’s forget about that for the moment. So we’re going to say that at this position in the water tank, we’re only experiencing the downward force from this one layer of water. But then, say that we drop down a bit. At our new location, we’ll now under two layers of water. So now, the force of the top layer and the force of the second layer are both acting down on us. And then, if we continue to descend, more layers of water are now above us. And therefore, the additional weight of those layers is felt. And that weight, which is a force, when it’s spread over the area of our body is experienced as a pressure.

Now, we’ve said that pressure in general is a force spread over some area. It turns out though that the pressure created by a fluid — a gas or a liquid — can be expressed a different way. We can write that pressure as capital 𝑃 sub 𝑓 to signify that it applies specifically to fluids. This pressure is equal to the product of three different values. The first value is the density of the fluid. We use the Greek letter 𝜌 to symbolize density. Now density if we recall is equal to the ratio of the mass of an object divided by the volume that object takes up. So for example, air has a fairly low density not much mass per unit volume. While a solid material, like concrete or wood, will have a much higher density much more mass over some amount of volume. So fluid pressure has to do with density. And it also has to do with the acceleration due to gravity, lowercase 𝑔.

One way to understand why this acceleration should be here in this formula for pressure is to reconsider our diving well example. In that case, we saw that it was the weight of these layers of water which created pressure on the diver. And the reason those layers have any weight at all is because of the attractional gravitational force on them. The Earth is pulling downward on these layers. And it does so through the gravitational field created by the Earth. So gravity’s acceleration is a vital part of calculating fluid pressure.

The last piece of this equation is the height ℎ of our fluid. And when we talk about the height of a fluid, we have to be a little bit careful. Let’s say that looking at our diving well, we want to calculate the pressure created by the water at this location. We might think that the height of that location will be measured from the bottom of the diving well. After all, if we had a cup of water and we’re asked to measure the height of the water, we would probably measure it that way. But when we’re calculating pressure, things are different.

Recall that the pressure created on our diver was thanks to the water that was above the diver rather than below them. It’s only the layers of water above a certain point that can contribute pressure to that point. And this means if we would have measured the height of this point we’ve worked out, that height will be measured actually from the surface of the water. That’s how tall the stack of fluid is that’s responsible for creating pressure at that point.

Although this idea of measuring height starting at the top rather than at the bottom of a liquid may at first seem strange, we can at least see that measuring the height this way makes physical sense. Given our equation for fluid pressure, we know that the deeper an object is below a fluid surface, the more pressure it will experience. If we have a small height, a small distance below the surface, then our pressure will be relatively smaller. But as we drop deeper and deeper down into a fluid, the pressure will grow. In fact, we could draw in arrows representing the pressure exerted on the wall of the diving well as the depth of the well increases.

At the top of the diving well, the pressure in the wall is very small. But that pressure grows as our depth increases until at the bottom of the well the pressure on the wall is fairly significant. This is one reason why if you ever see a concrete dam used to hold a lot of water in place, the bottom of the dam will be much thicker than the top part. It’s because the pressure of the fluid acting on the dam lower down is much greater and therefore requires more reinforcement. Knowing all this about fluid pressure, let’s get some practice now through a couple of examples.

Which of the following formulas correctly shows the relationship between the pressure exerted by a fluid, the height ℎ of the fluid above the point at which pressure is measured, the density 𝜌 of the fluid, and the gravitational acceleration rate 𝑔? A) 𝑃 is equal to 𝑔 times ℎ divided by 𝜌. B) 𝑃 is equal to 𝜌 times 𝑔 divided by ℎ. C) 𝑃 is equal to 𝑔 divided by 𝜌 times ℎ. D) 𝑃 is equal to 𝜌 times 𝑔 times ℎ. E) 𝑃 is equal to ℎ divided by 𝜌 times 𝑔.

So in this exercise, we’re trying to find the correct mathematical relationship between four quantities: pressure, density, height, and gravity. We’re told that one of our five answer options shows this correct relationship. To figure out which one it is, there are two ways we can go about it. The first way — and certainly the quickest way — is to recall off the top of our head the equation for the pressure produced by a fluid. But let’s say that we’re not able to recall that equation. Well, there is still a way we can solve for the correct answer. The way we can do it is by looking at the units on either side of each one of these equations.

Here’s what we mean by that. We can see that each one of these five equations has four terms in it. Each one has pressure, it has height, ℎ, it has density, 𝜌, and it has acceleration due to gravity, 𝑔. Now we can take each one of these terms and we can look at the units that that term involves. For example, the units of pressure are pascals, abbreviated Pa. And a pascal we can recall is equal to a newton of force spread over a square meter of area. And we can further recall that because a Newton is equal to a kilogram meter per second squared, then that means the units of pressure are kilograms meters per second squared per meter squared, which simplifies in the most basic of base units to a kilogram per meter second squared.

The base unit of height on the other hand is meters. The base unit of density is kilograms per cubic meter. And we can recall that the units of the acceleration due to gravity are meters per second squared. Now here’s the reason we’ve gone through all this. If we look at the left-hand side of each one of our five candidate equations, we see they all involve the pressure 𝑃 by itself. We’ve seen from our analysis that the units of pressure are kilograms per meter second squared. This means that these are the units on the left-hand side of all of our candidate equations.

And for an equation to be true, that is for it to be accurate, that means that the units on the right-hand side must match these. They must also be kilograms per meter second squared. If they’re not, then that means the two sides of the equation can’t be equal. And that means that that particular equation doesn’t correctly represent the relationship between these four values. So here is what we’ll do. We’ll go through these equations one by one, starting with equation A. And we’ll analyse the units on the right-hand side. If those units match up to the units of pressure, then we’ll mark that equation as a possibly correct solution.

Starting out with equation A, the right-hand side here has 𝑔 times ℎ divided by 𝜌. Looking over at the units of each one of these terms, we see that the units of 𝑔 are meters per second squared, the units of ℎ are meters, and the units of density are kilograms per cubic meter. In the numerator of this fraction, these two factors of meters combine. And then if we multiply this fraction by meters cubed divided by meters cubed, then that term meters cubed cancels out in our denominator. And we have meters to the fifth divided by second squared divided by kilograms or simply meters to the fifth divided by second squared kilograms. When we compare these units to the units we have for pressure, we see that there’s not a match, which means that option A is not a reasonable candidate for the correct mathematical relationship.

Moving on to the right-hand side of option B. Here, we have units of kilograms per cubic meter for 𝜌 multiplied by meters per second squared for 𝑔 divided by meters for ℎ. This numerator simplifies to kilograms per meter squared second squared. And then if we divide both numerator and denominator by meters, we have a final result of kilograms per cubic meter second squared. This is close to the units for pressure, but it’s not quite right. Therefore, option B isn’t our answer either.

Moving to the right-hand side of option C. Here we have meters per second squared, the units of 𝑔, divided by kilograms per cubic meter, the units of 𝜌, multiplied by meters, the units of height. A meter divided by meters cubed is equal to one over meters squared. And then, if we multiply top and bottom of this fraction by the inverse of the denominator, in other words multiply it by meter squared per kilogram, then the meter squared per kilogram, the denominator, cancels out with kilograms per meter squared. We’re left with a final result of meters cubed per kilogram second squared, also not equal to the units of pressure. So we cross option C off our list.

Looking at the right side of equation D. Here we have units of kilograms per cubic meter multiplied by meters per second squared multiplied by meters. Combining all these factors of meters, we get a result of kilograms per meter second squared. And we see that this is a match for the units of pressure. So let’s put a star by option D. And we’ll keep that in mind as a possible answer choice.

Looking finally at the right-hand side of option E. Here we have units of meters divided by kilograms per cubic meter multiplied by meters per second squared. This denominator simplifies to kilograms per meter squared second squared. And if we multiply both top and bottom by meters squared second squared, this term cancels out in our denominator, leaving us with meters cubed second squared per kilogram. We see this is not a match for the units of pressure. And therefore, option E is off the table. And this means that option D is our answer. The pressure exerted by a fluid is equal to the density of that fluid multiplied by 𝑔 multiplied by ℎ.

Let’s look now at a second example involving fluid pressure.

What is the pressure exerted by water at a depth of 2.5 meters? Use a value of 1000 kilograms per cubic meter for the density of water.

So in this example, say that we have a column of water. And we’re interested in the pressure exerted by that water at a depth of 2.5 meters below the surface. So let’s say that’s a point here in our water column. The pressure at this point we’ve marked is created by the weight of all the water that’s above that point in our water column. And by the way, it doesn’t make a difference how wide the column is. However wide or narrow it is, the pressure will be the same as long as we have this certain depth, 2.5 meters.

To answer this question of what is the pressure exerted by the water at that point, we can recall that the pressure created by a fluid is equal to the density of that fluid multiplied by its height below the surface of the fluid times 𝑔, the acceleration due to gravity. Recalling that 𝑔 is 9.8 meters per second squared, when it comes to the density of our fluid, we’re given that in our problem statement, 1000 kilograms per cubic meter. And we’re also given the height, ℎ, 2.5 meters. And this means we can get right to calculating this pressure. It’s equal to the density of the water multiplied by the height below the surface of the water times the acceleration due to gravity.

Now before we multiply these numbers together, notice the units involved. That all the units are in base unit form. We see that in the numerator of our units, we have these two factors of 𝑚, the distance in meters. While in the denominator, we have meters cubed. This means if we were to multiply all the units involved together, we would get an overall result of kilograms per meter second squared. This is equivalent to a newton per meter squared. And we can recall that a newton per meter squared is equal to the unit pascal, which is a pressure unit. This means that the units we’ll end up with after we do our calculation are pascals. When we multiply these three numbers together, we find a result of 24500 pascals. That’s the pressure exerted by the water at this depth.

Let’s summarise now what we’ve learned in this lesson.

We started out talking about pressure. And we saw that pressure is equal to a force spread out over an area. Written as an equation, we say that 𝑃, the pressure, is equal to 𝐹, the force, divided by 𝐴, the area. We saw further that pressure is measured in units of pascals, where one pascal is equal to a newton per meter squared. We learned in this lesson that fluids — liquids and gases — exert pressure due to their weight. And this pressure is experienced by any object under the fluid. We saw that the pressure created by fluids — sometimes called 𝑃 sub 𝐹 — is equal to the density of the fluid multiplied by the acceleration due to gravity multiplied by the height of that fluid. And lastly, we saw that the height in this equation is measured not from the bottom of a fluid but rather from the top down. And this is because pressure comes from the weight of a fluid layer acting down on some point.

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