### Video Transcript

In this video, we’re going to learn about buoyancy. We’ll learn what buoyancy is and how to calculate the buoyancy of various objects. To get started, imagine that one day you’re out fishing on your boat on a lake. Just as you think you might be getting a nibble on a line, a wave approaches the boat from the opposite side and sends it in a motion. Caught off balance, you tip over the side of the boat and fall in.

Over the next few moments, you see some of the things you fell in with come to the surface, your fishing line, for example. But other items seem to have disappeared: your sunglasses, your phone, and the sandals you were wearing.

To understand why some of these items float to the surface and some end up on the bottom of the lake, we’ll want to know a bit about buoyancy. Buoyancy is the tendency of an object to experience an upward acting force when the object is in a fluid. It could be a liquid or a gas. For example, if we had a container full of water and put a table-tennis ball on the surface of that water, the ball would experience an upward acting force called the buoyant force, able to balance out the weight force of the ball.

On the other hand, if we took a nugget of gold and dropped it into the water, that gold would sink all the way to the bottom and rest on its base. Even though it doesn’t float, the gold also experiences a buoyant force. And there’s a principle called Archimedes principle that connects these two cases.

This principle says that the upward buoyant force on an object, whether it’s a floating object like a ping-pong ball or a submerged object like our nugget, that upward buoyant force is equal to the weight of fluid that the object displaces. We can expand this principle to consider specifically the two cases where, first, an object floats such as our ping-pong ball and second when an object sinks.

When an object floats in a fluid, it displaces its own weight in that fluid. If we were to consider a zoomed-in view of our floating table-tennis ball, we would see that the ball isn’t entirely on the surface of the water. There’s actually a very small fraction of the ball’s volume that falls below the equilibrium level of the water in the container.

If we could somehow take that volume of water that the ball replaces and put that amount on a scale, with the table-tennis ball itself on the other side of the scale, we would find that these two masses are equal. Their weight comes out the same.

The second corollary of Archimedes principle is that if an object doesn’t float but sinks, then when submerged, it displaces its volume in the weight of the fluid it’s in. So now if we put our gold nugget on one side of a scale and then on the other side put the amount of water which takes up the same volume as the nugget, when we released the scale, we would find an imbalance. Though the volumes are the same, that amount of gold weighs more than that amount of water.

Reflecting on Archimedes principle, we come to see that buoyancy is really about relative density. The reason that the table-tennis ball floats is because, on average, it’s less dense than the fluid it’s in, whereas the gold nugget sinks because, on average, it’s more dense than water.

We can symbolize density using the Greek letter 𝜌. And we recall that a density of an object is equal to the mass of that object divided by the volume it takes up. Here’s a question about buoyancy and density. Say that I have a tank of water. And into that tank I drop a board of solid wood. What happens to the board? Well, it flows, right?

Now say I take a bar of solid metal and I threw that into the tank. The metal will clearly sink. If that’s true, then how is it possible that boats made of metal are able to float? It took quite a long time for people to discover that something like this was possible. And it is possible. And the reason it’s possible has to do with average density.

If my metal hull boat is filled with air, then the average density of all the air contained inside along with the metal may actually be less dense than the fluid that it’s in, so that the boat floats. Keeping this idea of average object density in mind, let’s try a few examples involving buoyancy.

A chunk of iron with a mass in air of 390.0 grams is found to have an apparent mass of 350.5 grams when completely submerged in an unknown liquid. Assume that the iron has a density of 7.86 times 10 to the third kilograms per meter cubed. What mass of fluid does the iron replace? What is the volume of the iron? Calculate the fluid’s density.

In the statement, we’re told the mass of the iron in air, 390.0 grams. We will name that mass 𝑚. We’re told that when the iron is submerged completely in an unknown liquid, it has an apparent mass of 350.5 grams. We’ll name that 𝑚 sub 𝑎. Knowing that the density, 𝜌, of iron is 7.86 times 10 to the third kilograms per cubic meter, we want to solve for three things: first, the mass of fluid that the iron replaces. We’ll name that 𝑚 sub 𝑓. Second, we want to know the volume of the iron. We’ll name it 𝑉. And finally, we wanna solve for this unknown fluid’s density. We’ll name that 𝜌 sub 𝑓.

To start in on our solution, let’s consider the difference between the apparent mass and the true mass of the iron. Say that we have our chunk of iron and we suspend it by an effectively weightless chord from a scale. The scale reads out a mass. And we’ve called that mass in air 𝑚. Then imagine we take the same setup but we submerged the chunk of iron in our unknown fluid. When we do that, the scale records the value we’ve named 𝑚 sub 𝑎, the apparent mass of the iron.

By a corollary of Archimedes principle, the mass of the fluid displaced by the iron must be equal to the difference between the true mass — we can call it the mass in air of the iron — and its apparent mass. Written as an equation, we can see that 𝑚 sub 𝑓 is equal to the mass of the iron in air minus the apparent mass or the measured mass when it’s in the fluid. That’s 390.0 grams minus 350.5 grams, or 39.5 grams. That’s the mass of fluid displaced by the iron.

Next, we wanna solve for the volume 𝑉 of this chunk of iron. We recall that the density of an object, 𝜌, is equal to its mass divided by its volume. Rearranging that expression for our case, we can write that the volume of the iron is equal to the mass of the sample divided by its density. We’ll use the mass of the sample measured in air. We can call that the true mass of the iron.

Substituting in our values for that mass of the iron in units of kilograms and the iron’s density, when we calculate this fraction, we find it’s 49.6 centimeters cubed. That’s the volume taken up by this chunk of iron.

Finally, we want to solve for the density of the unknown fluid that the iron is submerged in. To figure this out, we can again rely on our expression for density. The density of the fluid is equal to the fluid’s mass divided by its volume, which is the same as the volume of the submerged iron. Plugging in for these two values, when we calculate this fraction, we find it’s 0.796 grams per centimeter cubed. That’s the density of the fluid the iron is submerged in.

Now let’s look at an example involving the relative densities of a fluid and a floating object. A woman floats in freshwater at density 1.00 times 10 to the third kilograms per cubic meter. 4.00 percent of her volume is above the water’s surface. What is the density of the woman? What percent of the woman’s volume would be above water if she floated in seawater of density 1.03 times 10 to the third kilograms per meter cubed?

In part one, we want to solve for the density of the woman. We can call that 𝜌 sub 𝑤. And in part two, we want to solve for the percent of the woman’s volume that would be above water if she was in seawater. We’ll call that percent capital 𝑃. We’re told the density of freshwater as well as saltwater. And we can name those densities 𝜌 sub 𝑓 and 𝜌 sub 𝑠, respectively.

In the initial example, where the woman is in freshwater, we’re told that 4.00 percent of her volume is above that water when she floats. This means that, compared to the density of freshwater, the density of the woman, 𝜌 sub 𝑤, is four percent less. Written as an equation, we can say that 𝜌 sub 𝑤 equals 𝜌 sub 𝑓 minus four percent of 𝜌 sub 𝑓, 0.04 times that density. This is equal to 0.96 times 𝜌 sub 𝑓 or 0.96 times 1.00 times 10 to the third kilograms per cubic meter. This product equals 960 kilograms per cubic meter. That’s the density of the woman.

In part two, we imagine that the woman moves from freshwater to saltwater. And now we want to know what percentage of her volume is above that level of the water. We’ll figure this out using the density of saltwater, 𝜌 sub 𝑠, and the density of the woman from part one, 𝜌 sub 𝑤.

Here’s one way to think of this. Imagine that 𝜌 sub 𝑤 and 𝜌 sub 𝑠 were the same, so that this fraction is equal to one. In that case, 100 percent of the woman would be submerged. And zero percent would be above water. So to solve for 𝑃, the percent above water, we’ll want to subtract this ratio from 100 percent. And then more than that, we’ll want to turn this ratio from a decimal into a percentage value. And we’ll do that by multiplying it by 100.

Now we have an expression, which when we calculate it, we’ll solve for the percentage of the woman’s volume that is above the level of the saltwater. When we plug in the densities for the woman and for saltwater and calculate this expression, we find it’s equal, to three significant figures, to 6.80 percent. That’s the percentage of the woman’s volume that would be above water level if she was floating in seawater.

Let’s summarize what we’ve learned so far about buoyancy. We’ve seen that buoyancy describes the upward force that acts on objects in fluid. Archimedes principle tells us that a floating object displaces its weight and a submerged object displaces its volume in the fluid it’s floating or submerged in. And we’ve seen that buoyancy comes down to relative average densities between the object that may float or sink and the fluid that it may float or sink in.