In this video, we’re talking about the buoyant force, which is the force that pushes up on an object whenever it’s in a fluid. We’re going to learn what causes this force. And we’ll also see how to calculate mathematically whether an object will float or sink in a given liquid.
To get started on this topic, say that we have a basin. And then we fill that basin with water to a particular level, that is, to a particular depth. And then imagine that we have a cube of wood and we toss that cube into the basin of water. Now, if our cube is light enough, we know what will happen when we do this. Instead of sinking to the bottom, our cube will float. But as we look closely, we notice that that’s not the only thing that goes on here.
Notice what’s happened to the level of water in the basin. It’s gone up from its original level. If we draw in a dashed line where the water level was at first, then we can see that by dropping this cube into the water, we’ve raised the average water level. More on that in a moment. But for now, let’s notice that something very interesting is happening here.
If we were to take a second identical wooden cube to our first cube and just drop that cube to the ground, nothing would stop it. We know the cube would fall to the ground. But here, in the case of our cube in the basin of water, the cube hasn’t fallen all the way to the bottom. Something is holding it up. That something is called the buoyant force. And we know it needs to be acting upward on our cube with a force equal in magnitude to the weight of this cube downward. That’s the requirement for our cube to float rather than sinking to the bottom of the basin.
If we think about what could be causing the buoyant force. As we look at our cube in the basin, we see the only thing the cube is in contact with is the water. It’s not touching the sides of the basin or the bottom or any other object. So somehow, the fluid in this basin must be exerting this buoyant upward force. If we were to pick a certain point within our basin of water. Say that point right there. We know that thanks to the weight of the water on itself, there’s pressure exerted at that point. If we were to use blue arrows to represent the pressure, we know that pressure acts downward at that point. But then it also acts to the side and upward and really in every direction equally from that location. This fact that pressure at a point within a fluid is equal in all directions is known as Pascal’s principle.
It’s this idea of Pascal’s principle that explains to us how there can be an upward-acting force on our floating cube. It’s because the pressure within a fluid doesn’t only act downward, but it also acts upward or to the sides or in any direction. So when it comes to the buoyant force in general, this upward-acting force on objects in fluids, we can say that this force is due to the pressure that’s exerted by the fluid that the object is in. That pressure acts on the object in the fluid and pushes up on it, hence the buoyant force.
Okay, so that’s what causes this upward-acting force. But then we wonder how do we know how big this force is. In other words, what determines just how much pressure is exerted on an object in a fluid? How do we know whether that object will float or sink? To answer that question, we’ll come back to what we noticed earlier, that the height of water in our basin changed when we put this wooden cube in it. When we dropped the cube in the basin, the water level went up by this much, which means that this cross section of water that we’re seeing here wasn’t there originally. This is water that’s been moved out of the way by the cube.
This water, which we could say was originally here in the basin before the cube was dropped in, was pushed out of the way when the cube landed. The fancy word for this is it was displaced. It’s this amount of water, the water that was displaced when we dropped our cube into the basin that contributes to the pressure exerted upward on the cube by the fluid. In particular, it’s the weight of this amount of water, this amount of displaced fluid, that equals the upward buoyant force on a cube. We could think of it this way.
Say that we have a measuring scale and say that we were able to put on one side of the scale all of the water that was displaced by dropping our wooden cube into the water basin. So all of this displaced water here that we’ve marked out in our sketch is collected on one side of the scale. And then on the other side of the scale, we put the object that we dropped into our fluid, in this case, the wooden block. Now, if the amount of fluid that our object displaced has the same weight as the object we dropped in the fluid, then our scale will balance out like we’re seeing here. And the object will float in the fluid. And then if we push the wood block down so it was completely underwater, say with our hand. In that case, the amount of fluid pushed out of the way would outweigh the object, because now we’re displacing slightly more fluid. So again, the block would float.
As a side note, when an object does float, that means it’s displacing its weight in fluid. So a partly submerged floating object like our block displaces less fluid than if the block was fully underwater while it floated. And then, of course, there’s a third possibility, which is that our object actually outweighs the amount of fluid that it displaces. In that instance, our scale would look something like this. And the result would be that this object in our fluid would sink. We can see that this third possibility is not happening in our case because this wooden block is actually floating in the water.
But the key point here is that the two things on either side of this scale, so to speak, the things that determine whether an object floats or sinks in a liquid are, on the one hand, the object itself and, on the other, the amount of fluid that that object pushes out of the way when it’s in the fluid. It’s the weight of that amount of pushed out of the way or displaced fluid that determines the buoyant force, the upward force acting on the object.
Now let’s consider a special case of a floating object. Say that we drop our wood cube into this water basin. And this time, the cube still floats. But the top of the cube is exactly level with the level of the water in the basin. In other words, the cube is completely submerged. It’s completely underwater, except for the very top surface. But it’s not sinking. It still floats. Now, if we were to once again consider the amount of water that was pushed out of the way or displaced by dropping our cube in the water and then put that amount on our scale along with the block. We would see what we saw in the very first instance that the weight of the water displaced by the cube is exactly equal to the weight of the cube.
Now here’s something interesting because our cube is completely submerged underwater. That means that the amount of water it displaces is equal to the volume of the cube itself. After all, all of the cube is underwater, just barely. This means that really, the displaced water on our scale would take up exactly as much space as the wooden cube. They would be the same size. So let’s think about this. If the water takes up exactly as much space as the wooden block and based on the fact that our scale is balanced, they must have the same weight. Then that means the displaced water and the wooden block have the same volume and they have the same mass. Now it’s at this point that we can recall the definition for the density symbolized using the Greek letter 𝜌 of an object.
The density of a material is equal to the mass of that material divided by the space it takes up, its volume. But we’ve just said that both the mass and the volume of the wooden block and the displaced water are equal when the wooden block is completely submerged under the water but still floating. And if they have the same mass and volume, then they must also have the same density. So when the top of our wooden block is exactly on level with the top of the water in the basin, we can say that the density of the block — we’ll call it 𝜌 sub b — is equal to the density of the water — we’ll call it 𝜌 sub w.
But then, what if these two densities weren’t equal to one another? Well, there are two possibilities for that. In the one case, the density of the block could be greater than the density of the water. In that case, when we tossed the block in the water, the buoyant force created by the water wouldn’t be enough to hold up the block. So the block would sink to the bottom of the basin. The other possibility is that the density of the water is greater than the density of the block. In that scenario, what we saw earlier takes place. The block not only floats, but part of it floats above the level of the water. So in answer to our question of how do we know whether an object will sink or float when we drop it in a certain fluid, we see that the answer comes down to the density of the object compared to the density of that fluid.
What we’re finding then is that it’s the differences in density between the object and the fluid the object is in that determines whether the object floats or whether it sinks. And not only this, but we see we can now extend our definition of the buoyant force. This force, we know, is due to the pressure exerted by a fluid on an object. And the buoyant force is equal to the weight of the fluid the object displaces or pushes out of the way. So going over here to our mass of water that’s displaced by the object, the weight of that mass of water we can call it capital 𝑊 is equal to the mass of the water times the acceleration due to gravity.
Whenever we use the weight of a displaced fluid to calculate the buoyant force, we need to be careful. That’s because of this term here, the mass in this equation. It’s important to realize that this mass, in the case of a buoyant force calculation, is not equal to the overall mass of our object, but rather it’s equal to the mass of the fluid it displaces. So if we call the mass of fluid that’s displaced by an object 𝑚 sub f, we can say that 𝑚 sub f multiplied by 𝑔 is equal to the buoyant force — we’ll call it 𝐹 sub B — created by that displaced fluid. The important point to remember is not to confuse the mass of the fluid displaced with the mass of the object that’s displacing it. That all may sound a bit confusing, but let’s look at an example which should help clarify these ideas.
A cube-shaped object with sides that are 130 centimeters long has a density of 950 kilograms per cubic meter. The cube is placed in a body of water. The water has a density of 1000 kilograms per cubic meter. What is the volume of the object in cubic meters? What is the mass of the object? Answer to the nearest kilogram.
Alright, so in this scenario, we have this cube-shaped object. And then if we mark out the height and width and depth of the cube, we know that because it’s a cube, these three quantities are all equal to one another. And we’re told in the problem statement that they’re equal to 130 centimeters. Our first question asks, what is the volume of this object in cubic meters? That volume, which we can call 𝑉, is equal to the total amount of space that this cube takes up. In other words, it’s equal to the height of the cube multiplied by its width multiplied by its depth. And we know that each one of those values is 130 centimeters.
So if we took that amount 130 centimeters and then cubed it, that would be equal to the volume of the cube but in units of cubic centimeters rather than cubic meters like our question asks. So before we multiply out the right-hand side of this expression, let’s convert 130 centimeters into whatever that distance is in meters. To do that, let’s recall that 100 centimeters is equal to one meter. Which means that to do this conversion, we’ll take the decimal place in our current value of 130 centimeters, and we’ll move it two spots to the left.
By doing that, we convert to a unit of meters and our value for that distance is 1.30. When we go ahead and cube this value, we find a result of 2.197 cubic meters. That’s how much space this cubic object takes up. Knowing this, we can move on to our second question, what is the mass of the object?
In the problem statement, we’re told about this object’s density, the value of that. And we can recall that density — symbolized using the Greek letter 𝜌 — is equal to the mass of an object divided by how much space it takes up, its volume. It’s not density we want to solve for for this object, but rather the mass. So to do that, let’s multiply both sides of the equation by the volume 𝑉. When we do, that term cancels out on the right-hand side and we find that the mass of an object is equal to its density multiplied by its volume.
Now, if we call the mass of our cube 𝑚 sub c, then we know that that’s equal to the density of the cube. We’ll call it 𝜌 sub c multiplied by the cube’s volume 𝑉. 𝜌 sub c is equal to 950 kilograms per cubic meter and 𝑉 the cube’s volume we just solved for earlier. So the cube’s mass is equal to the product of these two values. And before we multiply them, notice what happens to the units. We have kilograms per cubic meter multiplied by cubic meters, so that unit of cubic meters cancels out and we’re left with units of kilograms. That’s perfect because we’re calculating a mass after all. And to the nearest kilogram, this number comes out to be 2087. That’s the total mass of our cube-shaped object. We can now write that result off to the side. And then, we’ll move on to the last two questions in our scenario.
The first of those two questions says this: How many cubic meters of water have a mass equal to that of the object?
At this point, we can recall from our problem statement that our cubic object is placed in a body of water and we’re given the density of that water. Now, when we talk about this question of how many cubic meters of water have a mass equal to the mass of the object, what we’re saying is what is the volume of water, whose mass will be equal to the mass of our cubic object. Though we don’t yet know what that volume is, we do know a condition on that volume. The mass of that volume of water — we’ll call that mass 𝑚 sub w — must be equal to the mass of our cubic object, 𝑚 sub c.
That’s what it means when we say that this amount of water, however much it is, has a mass equal to the mass of our object. And at this point, we can again recall the definition of density. That density is equal to an object’s mass divided by its volume. And as we saw earlier, this means that the density of an object times its volume is equal to its mass. This means that we can rewrite our equation here for the mass of the water and the mass of the cubic object. We can replace these terms with the densities and volumes of the water and cubic object, respectively.
Here’s what that would look like. We can replace the mass of the water with the density of the water — we’ll call it 𝜌 sub w — times the volume of the water, 𝑉 sub w. And then, we’ll replace the mass of the cubic object 𝑚 sub c with the density of the cubic object 𝜌 sub c multiplied by its volume 𝑉. And now that we have this second equation, we see that it’s 𝑉 sub w the volume of water that we want to solve for. That’s how many cubic meters of water have a mass equal to that of the object. To go about doing that, we’ll rearrange this equation by dividing both sides by the density of water 𝜌 sub w. That cancels that term on the left-hand side.
So the volume of water we’re after is equal to the density of our cubic object divided by the density of water multiplied by the volume of our cubic object. And this whole equation was made possible by the fact that we’re setting the mass of our water equal to the mass of our object. And we’re asking the question just how much water in volume is that? Well, there’s good news. The density of our cubic object is given to us in the problem statement. And so is the density of water. It’s 1000 kilograms per cubic meter. We’re not given the volume of our cubic object. But remember we solved for that earlier. So it is a known quantity.
When we plug in for all these values, here’s what we get. Notice that in our numerator, the units of cubic meters cancel out. And along with that, the units of kilograms cancel from numerator and denominator. When all the dust settles, we’ll be left with units of cubic meters, units of volume. This is a good sign to us that we’re on the right track with our calculation. And when we compute this value, we find it’s 2.087 cubic meters. That’s the volume of water that would have a mass equal to the mass of the cubic object. Now, let’s move on to the last question in this exercise.
How far below the water’s surface must the base of the object be in order to displace a massive water equal to that of the object? Answer to the nearest centimeter.
Okay, so now, we’re finally facing the fact that this cube is not sitting off by itself, but rather is placed in a body of water. So we can say that this line represents the water level, and we want to know how far below that surface must the base of our cubic object be so that the object displaces a massive water equal to the mass of the object. Now, before we go any further, let’s give a name to this distance. Let’s call this distance 𝑑. And here’s what we’re saying with this distance, with our cube submerged at distance 𝑑 under the surface of the water, some fraction of that cube’s volume is pushing water out of the way. It’s displacing that water. And the idea is that that mass of water displaced is equal to the total mass of the cube. Let’s clear some space on screen to get into this a bit more.
What we have is a cube with this total volume, but only part of that volume is submerged underwater. It’s only that part of the volume, which is underwater, which displaces or pushes water out of the way. Because the cube is floating, that means that this amount of water displaced is equal to the total mass of the cube. That’s what it means for the weight force of the object pulling it down to be balanced out by the buoyant force, pushing it up. For these two forces to be equal, here’s what needs to be true. The total mass of the cube 𝑚 sub c needs to be equal to the mass of the displaced water. We’ll call it 𝑚 sub dw.
We know the total cube mass from earlier. And using our modified density equation, we can say that the mass of the displaced water from the cube is equal to the volume of the cube underwater multiplied by the density of water. And the volume of the cube underwater is equal to 𝑑, what we want to solve for, times the width of the cube times its depth.
As a next step then, we can rearrange this equation to solve for lowercase 𝑑. We’ll do it by dividing both sides by the width of the cube times its depth times the density of water. Doing so cancels these terms out on the right-hand side. When we plug in for all these values on the left, using a distance of 1.30 meters for both 𝑊 and 𝑑, and then calculate this term. We find a result of 1.23 meters or 123 centimeters. That’s how far below the surface the bottom edge of this cube is.
Let’s summarize now what we’ve learned in this lesson. First off, we learned that buoyant force is caused by pressure within a fluid. Buoyant force is equal to the weight of fluid displaced by an object. And lastly, we saw that the relative densities of an object and fluid determine whether the object sinks or floats in that fluid.