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Buoyant force example problems edited

Explore the concept of buoyant force! Learn how the weight difference of an object in and out of water helps calculate its volume. Discover how the density of an object determines the percentage submerged in water.
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Created by Sal Khan.

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Video transcript

- [Voiceover] Let's say that when, I have some object and when it's outside of water, its weight is, so, weight outside of water is, I don't know, 10 newtons. And let's say then I submerge it in water, I put it on a weighing machine, in water, it's weight, so let's call it weight in water, is, I don't know, it's two newtons. So what must be going on here? Well, the water must be exerting some type of upward force, to counteract at least eight newtons of the person's original, of, not the person. The object's original weight. This would be a very small person. Most people's weights, in newtons, are in the hundreds. But anyway. And that difference is the buoyant force. Right, so, a way to think about it is once you put the object in the water, so let's say it's a, you know, it could be a cube, doesn't have to be a cube, could be anything. And it's in the water, so let me... draw the water. This is the water, up here. We know that we have a downward, there's a downward weight. Right. That is 10 newtons. But we know that once it's in the water, the net weight is two newtons, so there must be some force acting upwards on the object, of eight newtons. Right? And that's the buoyant force, that we learned about in the previous video, and the video about Archimedes' principle. This is the... the buoyant force. So the buoyant force is equal to 10 minus two, is equal to eight, right, that's how much the water is pushing up. And what is that also equal to? That equals the volume, the weight of the water displaced. So, eight newtons is equal to weight of water displaced, and what is the weight of the water displaced? That's the volume of the water displaced, times the density of water, times gravity. All right. And so what is the volume of water displaced? Well, it's just the volume of water, divide eight newtons by the density of water, is 1,000 kilograms per meter cubed. And this is eight newtons, so a newton is what, kilogram... kilogram... meter per second squared. And then what's gravity? It's 9.8 meter per second squared. And if we look at all the units, they actually do turn out with just, you end up having just meters cubed, but let's do the math. Equals 8.2 times 10 to the minus four, meters. Cubic meters. So just knowing the difference in the weight of an object, I can figure out, the difference when I put it in water, I can figure out the volume. And so this could be a fun game to do next time your friends come over, is weigh yourself outside of water, then get some type of spring or waterproof weighing machine, put it at the bottom of your pool, stand on it, and figure out what your weight is. Assuming that you're dense enough to go all the way into the water. But you could figure out somehow, your weight in water. And then you would know your volume. There's other ways. You could just figure out how much the surface of the water increases, and take that water away. But anyway, this was interesting. Just knowing how much the buoyant force of the water was, or how much lighter we are, when the object goes into the water, we can figure out the volume of the object. And this might seem like a very small volume, but just keep in mind, in a meter cubed, in a meter cubed, you have, let's see, you have 27 square feet. Right? You have 27 square feet. So if we multiply that number times 27... it equals .02... 0.02 square feet, roughly, and .02 square feet, how many, in a square foot, there's actually let's see, 12, to the third power, times 12, times 12, is equal to 1728, times 0.02. So this is actually 34 square inches. So the object isn't as small as you may have thought it to be. It's actually, you know, maybe a little bit bigger than three inches by three inches by three inches. So it's a reasonably sized object. Anyway. Let's do another problem. Let's say I have some balsa wood, and I know that the density of balsa wood is 130 kilograms per meter cubed. Meter cubed. That's the density of balsa wood. And I have some big cube of balsa wood, and what I want to know is, if I put that, so let me draw the water. That's the water. And I have some big cube of balsa wood, which I'll do in brown. So I have a big cube of balsa wood, and the water should go on top of it, just so you see that it's submerged in the water. All right. I want to know, what percentage of the cube goes below the surface of the water. Interesting question. So how do we do that? Well, for the object to be at rest, for this big cube to be at rest, there must be a zero net forces on this object, right? So in that situation, the buoyant force must completely equal the weight, or you know, the force of gravity. Well, what's the force of gravity going to be? Well, the force of gravity is just the weight of the object, and that's the volume of, let's just say, the wood. Volume of the balsa wood, times the density of the balsa wood, times the density of balsa wood, times gravity. And what's the buoyant force? The buoyant force is equal to the volume of the displaced water, right. Volume of displaced water. But that's also, well, the volume of displaced water, and it's the volume of the cube that's been submerged. The part of the cube that's submerged, that's volume that's also equal to the amount of volume of water displaced, right? So we could say that's the volume of the block submerged, which is the same thing, remember, as the volume of the water displaced, times the density of water, times gravity, remember this is density of water. So remember, the buoyant force is just equal to the weight of the water displaced, and that's just the volume of the water displaced, times the density of water, times gravity. Of course, the volume of the water displaced is the exact same thing as the volume of the block that's actually submerged. And since the block is stationary, it's not accelerating upwards or downwards, we know that these two quantities must equal each other. So V, volume of the wood, the entire volume, not just the amount that's submerged, times the density of the wood, times gravity, must equal the volume of the wood submerged, which is equal to the volume of the water displaced, times the density of water, times gravity, but we have the acceleration of gravity, we have that on both sides, so we can cross it out. Let me switch colors, to ease the monotony. And then, let's see. What happens if we... divide both sides by the volume of the balsa wood? You get, let's put this over here. Let's divide both... I'm just rearranging this equation, I think you'll figure it out. We can get, divide both sides by that, you get the volume submerged, divided by the volume of the balsa wood, right, I just divided both sides by VB, and switched sides, is equal to, and now lets, it's equal to the density of the balsa wood, divided by the density of water. That make sense? I just did a couple of quick algebraic operations. But hopefully, that, you know, get rid of the g, and that should make sense to you. And now we're ready to solve our problem. 'Cause my original question is what percentage of the object is submerged? And so that's exactly this number, right? If we say this is the volume submerged, over the total volume, this is the percent submerged. And so that equals the density of balsa wood, which is 130 kilograms per meter cubed, divided by the density of water, which is 1,000 kilograms per meter cubed. So 130 divided by 1,000, is 0.13. So VS over VB is equal to 0.13, which is the same thing as 13%. So exactly 13% of this object will be, of this balsa wood block, will be submerged in the water. That's pretty neat to me. And it actually didn't have to be a block. It could have been shaped like a horse. I'll see you in the next video.