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Archimedes principle and buoyant force

Explore Archimedes' principle and the buoyant force in fluid dynamics. Understand how pressure varies with depth and acts in all directions, leading to the upward net force on submerged objects. This principle explains why objects float and is key to fluid displacement and the concept of buoyancy. Created by Sal Khan.

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  • leaf green style avatar for user Veronica Schmidt Teron
    I am confused because I thought at the beginning of this lecture Sal said that the pressure around a submerged object was equal from all directions but at the end of his lecture he says that the pressure is greater underneath the object than on top and that is the buoyant force. Could someone help clear this up for me? What is Sal actually saying?
    (56 votes)
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    • starky sapling style avatar for user leonardo.ulian
      His first example was a point, which has a volume of zero. Around this point the pressure is equal because no volume of liquid is displaced. Also, a point has no "height", hence no difference between pressure up and bottom. In the cube example, it has volume, which generates difference in pressure up and bottom.
      (2 votes)
  • male robot donald style avatar for user Karamvir Singh
    does that mean the pressure in me helps me not to be squashed by the atmosphere
    (31 votes)
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  • spunky sam blue style avatar for user Bundi Bedu
    This question appeared in my physics final paper and there is confusion amongst students over its answer: "A wooden block is lying on the bottom of the tank sticking (with glue) to it. When water is poured into the tank, water does not enter below the block. Is there a buoyant force acting on the block? Explain."
    Please help.
    (13 votes)
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  • leaf green style avatar for user Pat
    Does an object (such as a hot air ballooon) float because it weighs less than a volume of normal air equivalent to the volume of space it takes up? Also, as the air is heated up by the balloon, it becomes less dense and it should float. But what exactly is pushing it up--is it the hot air itself pushing upwards on the inner walls of the balloon?
    (7 votes)
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    • male robot hal style avatar for user Satwik Pasani
      There are two explanations as to what is pushing up the balloon. One is that atmosphere (which is a fluid in static equilibrium) cannot distinguish between the balloon and an equivalent amount of normal air in its place. Therefore, it provides an upward force due to difference in hydrostatic pressure at the bottom of balloon and the top, which is equal in magnitude to the weight of normal air the size of the balloon. Had there been normal air there it would have been static as the upward force would have been equal to its weight but since the weight of the balloon is less than the upward force acting on it, it will move up. This is what concept of floating and upthrust is. The second explanation is considering the tendency of the entire system to lower its gravitational potential energy which can be done if the balloon were replaced with air (due to its greater mass) and therefore all elements of air above the balloon, try to reduce the net energy by coming down in place of the balloon and in the process providing an upward push on it.
      (3 votes)
  • piceratops ultimate style avatar for user Nathan Shapiro
    I've seen people write things like "Archimedes' principle says that the buoyant force acting on an object is equal to the weight of the liquid displaced. This simply means that if something is denser than the liquid, it will sink." I've tried figuring out how they came to that conclusion and did a considerable amount of research on it, but I could never figure it out. How did they come up with that?
    (3 votes)
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    • piceratops seed style avatar for user Tony Micer
      This is something difficult to visualize. But here is how to get there:

      The force of water above the object is given by rho*g*h, and the buoyant force underneath the object is equal to the (pressure at the bottom of the object)*(surface area of the bottom of the object). Let's look a little closer at that surface area. The surface area is related to the volume; generally, the greater the total surface area of an object, the greater the object's total volume. For example, an empty balloon has a much smaller surface area than a balloon filled with air. Why did the surface area change? Well, that's because we increased the balloon's volume!

      Now, if an object has a greater density, that means that, per amount of surface area, that object also has a greater mass for that given area. If that amount of mass on the surface of the object is greater than the mass of the area of the water (or any liquid) underneath it, then the gravitational force pulling downwards on the mass of the object will cause the object to "push aside" the liquid in its way.
      (7 votes)
  • blobby green style avatar for user PJ.Hearle
    if I exhale completely, I sink to the bottom of a pool. If my lungs are full of air, I float. Is this simply due to the fact that the volume of my body is greater when my lungs are full of air, and thus my overall density is less? Is there not some other buoyant property afforded by the fact that my lungs are full of a low-density gas?
    (3 votes)
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  • marcimus pink style avatar for user Markie
    At , to summarize Archimedes principle- for every submerged object the weight of water displacement equals the object's weight? Is that right? Thanks.
    (0 votes)
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    • starky sapling style avatar for user leonardo.ulian
      No. Nothing was talked about the object's weight. It summarizes saying that the buoyancy force acting on the submerged object is equal to weight of displaced liquid, which depends only on the volume of the object. That is not the only force acting in the object though, there is also the object's weight but it was not mentioned so far.
      (6 votes)
  • blobby green style avatar for user sathishkumar.kannan71
    how does he know that the pressure of the cube at the bottom is higher than the pressure on top?
    (3 votes)
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  • aqualine ultimate style avatar for user Gala Balaguer
    At , Sal says the net upward force of submerged objects equals the weight of liquid displaced. Would it be accurate to say this applies to gases too since we're talking about weight, and possible compression wouldn't change weight?
    (2 votes)
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  • orange juice squid orange style avatar for user Kai Shaikh
    How can water exert a force, creating a greater pressure on the bottom of an object than the top, if the object (take a cube for example) has been placed on the bottom of the container holding the water, so that there is no water below the object (the object is directly against the container), and thus the only force from the water (besides the pressure resulting from the water's force acting on the sides of the object) would be that acting downwards. If this were the case, even if the object's density were such that the object had the potential to float, i.e. if its density were less than that of water, making the force of gravity on the object lesser than that of the buoyant force that would be produced by the volume of the water displaced, how the water be able to fulfill this desire to displace the cube so that only part of it would be submerged. What I am trying to say is that, if an object that should float is placed on the bottom of a container, how could water create the buoyant force needed to accelerate the object, if the water is unable to create a larger pressure on the bottom of the object, due to a lack of contact between the bottom of the object and the water itself. Would the object still float, or would it remain at rest on the bottom, with the force of gravity on the object being cancelled by the normal force provided by the bottom of the container?
    (2 votes)
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    • leaf green style avatar for user Mark Zwald
      That's a good question. My initial response is if there was a vacuum between the cube and the bottom surface, then you would be correct, the cube would stay pinned to the bottom even though it is less dense than water. However, in reality, there is no vacuum there even with the cube touching the bottom surface and as a result there is still greater water pressure pushing up on the bottom surface of the cube than pushing down on the top surface causing the cube to float.

      Not sure if anyone has done this experiment but I would guess that if you take a very flat cube and place it against a very flat surface under water, that it would take more force to initially lift the cube when it is flat on the bottom versus when it is off the bottom.
      (2 votes)

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.