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Archimedes principle and buoyant force

Introduction to Archimedes' principle and buoyant force. Created by Sal Khan.
Video transcript
Let's say we have a cup of water. Let me draw the cup. This is one side of the cup, this is the bottom of the cup, and this is the other side of the cup. Let me say that it's some liquid. It doesn't have to be water, but some arbitrary liquid. It could be water. That's the surface of it. We've already learned that the pressure at any point within this liquid is dependent on how deep we go into the liquid. One point I want to make before we move on, and I touched on this a little bit before, is that the pressure at some point isn't just acting downwards, or it isn't just acting in one direction. It's acting in all directions on that point. So although how far we go down determines how much pressure there is, the pressure is actually acting in all directions, including up. The reason why that makes sense is because I'm assuming that this is a static system, or that the fluids in this liquid are stationary, or you even could imagine an object down here, and it's stationary. The fact that it's stationary tells us that the pressure in every direction must be equal. Let's think about a molecule of water. A molecule of water, let's say it's roughly a sphere. If the pressure were different in one direction or if the pressure down were greater than the pressure up, then the object would start accelerating downwards, because its surface area pointing upwards is the same as the surface area pointing downwards, so the force upwards would be more. It would start accelerating downwards. Even though the pressure is a function of how far down we go, at that point, the pressure is acting in every direction. Let's remember that, and now let's keep that in mind to learn a little bit about Archimedes' principle. Let's say I submerge a cube into this liquid, and let's say this cube has dimensions d, so every side is d. What I want to do is I want to figure out if there's any force or what is the net force acting on this cube due to the water? Let's think about what the pressure on this cube is at different points. At the depths along the side of the cube, we know that the pressures are equal, because we know at this depth right here, the pressure is going to be the same as at that depth, and they're going to offset each other, and so these are going to be the same. But one thing we do know, just based on the fact that pressure is a function of depth, is that at this point the pressure is going to be higher-- I don't know how much higher-- than at this point, because this point is deeper into the water. Let's call this P1. Let's call that pressure on top, PT, and let's call this point down here PD. No, pressure on the bottom, PB. What's going to be the net force on this cube? The net force-- let's call that F sub N-- is going to be equal to the force acting upwards on this object. What's the force acting upwards on the object? It's going to be this pressure at the bottom of the object times the surface area at the bottom of the object. What's the surface area at the bottom of the object? That's just d squared. Any surface of a cube is d squared, so the bottom is going to be d squared minus-- I'm doing this because I actually know that the pressure down here is higher than the pressure here, so this is going to be a larger quantity, and that the net force is actually going to be upwards, so that's why I can do the minus confidently up here-- the pressure at the top. What's the force at the top? The force at the top is going to be the pressure on the top times the surface area of the top of the cube, right, times d squared. We can even separate out the d squared already at that point, so the net force is equal to the pressure of the bottom minus the pressure of the top, or the difference in pressure times the surface area of either the top or the bottom or really any of the sides of the cube. Let's see if we can figure what these are. Let's say the cube is submerged h units or h meters into the water. So what's the pressure at the top? The pressure at the top is going to be equal to the density of the liquid-- I keep saying water, but it could be any liquid-- times how far down we are. So we're h units down, or maybe h meters, times gravity. And what's the pressure the bottom? The pressure at the bottom similarly would be the density of the liquid times the depth, so what's the depth? It would be this h and then we're another d down. It's h plus d-- that's our total depth-- times gravity. Let's just substitute both of those back into our net force. Let me switch colors to keep from getting monotonous. I get the net force is equal to the pressure at the bottom, which is this. Let's just multiply it out, so we get p times h times g plus d times p times g. I just distributed this out, multiplied this out. That's the pressure at the bottom, then minus the pressure at the top, minus phg, and then we learned it's all of that times d squared. Immediately, we see something cancels out. phg, phg subtract. It cancels out, so we're just left with-- what's the net force? The net force is equal to dpg times d squared, or that equals d cubed times the density of the liquid times gravity. Let me ask you a question: What is d cubed? d cubed is the volume of this cube. And what else is it? It's also the volume of the water displaced. If I stick this cube into the water, and the cube isn't shrinking or anything-- you can even imagine it being empty, but it doesn't have to be empty-- but that amount of water has to be moved out of the way in order for that cube to go in. This is the volume of the water displaced. It's also the volume of the cube. This is the density-- I keep saying water, but it could be any liquid-- of the liquid. This is the gravity. So what is this? Volume times density is the mass of the liquid displaced, so the net force is also equal to the mass of liquid displaced. Let's just say mass times gravity, or we could say that the net force acting on this object is-- what's the mass of the liquid displaced times gravity? That's just the weight of liquid displaced. That's a pretty interesting thing. If I submerge anything, the net force acting upwards on it, or the amount that I'm lighter by, is equal to the weight of the water being displaced. That's actually called Archimedes' principle. That net upward force due to the fact that there's more pressure on the bottom than there is on the top, that's called the buoyant force. That's what makes things float. I'll leave you there to just to ponder that, and we'll use this concept in the next couple of videos to actually solve some problems. I'll see you soon.