If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Physics library

### Course: Physics library>Unit 9

Lesson 2: Buoyant Force and Archimedes' Principle

# Buoyant force example problems

A couple of problems involving Archimedes' principle and buoyant forces. Created by Sal Khan.

## Video transcript

Let's say that I have some object, and when it's outside of water, its weight is 10 newtons. When I submerge it in water-- I put it on a weighing machine in water-- its weight is 2 newtons. What must be going on here? The water must be exerting some type of upward force to counteract at least 8 newtons of the object's original weight. That difference is the buoyant force. So the way to think about is that once you put the object in the water-- it could be a cube, or it could be anything. We know that we have a downward weight that is 10 newtons, but we know that once it's in the water, the net weight is 2 newtons, so there must be some force acting upwards on the object of 8 newtons. That's the buoyant force that we learned about in the previous video, in the video about Archimedes' principle. This is the buoyant force. So the buoyant force is equal to 10 minus 2 is equal to 8. That's how much the water's pushing up. And what does that also equal to? That equals the weight of the water displaced, so 8 newtons is equal to weight of water displaced. What is the weight of the water displaced? That's the volume of the water displaced times the density of water times gravity. What is the volume of water displaced? It's just the volume of water, divide 8 newtons by the density of water, which is 1,000 kilograms per meter cubed. A newton is 1 kilogram meter per second squared. Then, what's gravity? It's 9.8 meters per second squared. If we look at all the units, they actually do turn out with you just ending up having just meters cubed, but let's do the math. We get 8 divided by 1,000 divided by 9.8 is equal to 8.2 times 10 to the negative 4. V equals 8.2 times 10 to the minus 4 cubic meters. Just knowing the difference in the weight of an object-- the difference when I put it in water-- I can figure out the volume. This could be a fun game to do next time your friends come over. 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. 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. 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. This might seem like a very small volume, but just keep in mind in a meter cubed, you have 27 square feet. If we multiply that number times 27, it equals 0.02 square feet roughly. In 0.02 square feet, how many-- in a square foot, there's actually-- 12 to the third power times 12 times 12 is equal to 1,728 times 0.02. So this is actually 34 square inches. The object isn't as small as you may have thought it to be. It's actually maybe a little bit bigger than 3 inches by 3 inches by 3 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. I have some big cube of balsa wood, and what I want to know is if I put that-- let me draw the water. 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 that you see it's submerged in the water. I want to know what percentage of the cube goes below the surface of the water? Interesting question. So how do we do that? For the object to be at rest, for this big cube to be at rest, there must be zero net forces on this object. In that situation, the buoyant force must completely equal the weight or the force of gravity. What's the force of gravity going to be? The force of gravity is just the weight of the object, and that's the volume of the balsa wood times the density of the balsa wood times gravity. What's the buoyant force? The buoyant force is equal to the volume of the displaced water, but that's also the volume of the 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. 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. Since the block is stationary, it's not accelerating upwards or downwards, we know that these two quantities must equal each other. So V, the 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. 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. What happens if we divide both sides by the volume of the balsa wood? I'm just rearranging this equation. I think you'll figure it out. We divide both sides by that, and you get the volume submerged divided by the volume of the balsa wood-- I just divided both sides by VB and switched sides-- is equal to the density of the balsa wood divided by the density of water. Does that make sense? I just did a couple of quick algebraic operations, but hopefully that got rid of the g, and that should make sense to you. Now we're ready to solve our problem. My original question is what percentage of the object is submerged? That's exactly this number. If we say this is the volume submerged over the total volume, this is the percent submerged. 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. Vs over VB is equal to 0.13, which is the same thing as 13%. So, exactly 13% percent of this balsa wood block will be submerged in the water. That's pretty neat to me. 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.