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Physics: Buoyancy and Archimedes' Principle


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About this Lesson

  • Type: Video Tutorial
  • Length: 12:53
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 138 MB
  • Posted: 07/01/2009

This lesson is part of the following series:

Physics (147 lessons, $198.00)
Physics: Fluids (13 lessons, $13.86)
Physics: Fluid Statics (9 lessons, $6.93)

This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.

Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.

Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.

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~ mventura


I've got two little objects here. They weigh about the same. I'm going to put them both into this glass of colored water. And when I put this one in, it floats. And when I put this one in, it sinks. So, what's the story here? They weighed the same. Everybody has a pretty decent intuition about this, although you usually think that light things float and heavy things sink, and that's not what's going on here. What's going on is closely related to that. It's that the object, which was less dense, is floating and the object, which was more dense, is sinking.
So I'd like to quantify that, and in doing so, we're going to come up with a formula, which really describes the forces of fluids on objects in fluids. And that's really important in lots of physical applications if you want to understand the physics of boats or icebergs or any situation whatsoever where you have one object inside of a fluid.
So let's try to analyze what's going on here, and I'm just going to do it using Newton's laws. I've got some fluid. And let me imagine that I've got an object with mass, M sub object (M[O]), submerged entirely. And it's got height (h) and area (a). It's just a little cube. And let me think about the forces acting on it. It's in a fluid, so there's pressure. And the pressure on the sides is equal - and of course, I'm pointing my fingers. Pressure doesn't point. Forces point. The pressure is a number on these sides and the force is perpendicular to the object. So there are equal forces on the side. There are not equal forces on top and bottom - not arising from the fluid only - not just arising from the pressure. Because, remember, this object has weight.
And so let's just try to separate out the forces arising from the fluid and the forces arising from gravity. Those are two distinct sets of forces. So, what are the forces arising only from the fluid? Let me save gravity for later. We'll add that in. But right now, I'm just going to consider the upward force, which arises from the pressure below. And that's just pressure is force per area. So force is pressure times area. So we're going to get P[1]A down, where P[1] is the pressure at the top. And P[2]A up, where P[2] is the pressure at the bottom. So this is level 1 and the bottom is level 2. Now, I know how the pressure in a fluid is related to depth. Pressure at level 2 is pressure at level 1 plus times g times h. And let's be careful, because there are two densities in this problem.
What's the density here? It's not the density of the object. I'm talking about the pressure in the fluid. And so that's the density of the fluid. And let me be very careful when I write this down. And I have P[2] is P[1 ]plus the density of the fluid times g times h, and all that times area . So these are the two forces - the down force and the up force - from the fluid. So if I were just to add these as vectors - which really means taking this one and subtracting that one, because they're in opposite directions - I will have the net force of the fluid on this object. You'll notice, I have a P[1]A[ ]up and I'm subtracting this P[1]A down. So I end up with a net upward force from the fluid along - of [fluid] g. And then what's left over - h times area - that's just the volume of this object.
So let me give that net force of the fluid alone on the object a name. I'm going to call it the buoyant force, because the water, or the fluid, is somehow buoying up this object. Now I've got to be very careful when I write this formula down. This is what I just derived. And I derived it for an object, which was entirely submerged. If you want to deal with the most general case, you could have an object, which is partially submerged. And then, what's the appropriate volume? Just think through this proof and convince yourself it's only the volume of the object under the fluid level. So people called that V[displaced]. V[displaced] represents the amount - you know what? That's a funny word. In this case, if you stick an object with mass, M[O], and volume, V[O], in the fluid, fluids are incompressible. So you're shoving some water aside in order to put the object in there. You're displacing some of the water. And how much are you displacing? In this case, you are simply displacing the volume of the object. And if you only put it halfway under, you're only displacing half of the volume of the object. So I'm trying to be careful to clearly write the volume of water displaced or fluid displaced, rather than the volume of the object itself, because this is the correct formula.
This was first discovered over 2,000 years ago by Archimedes, in Greece - a very brilliant fellow. Archimedes phrased it in words rather than in this formula. He said the buoyant force on an object will be the weight of the fluid which is displaced. So that's the same thing, because what's the weight of something? It's the mass times g. So what Archimedes is saying is the mass of the fluid displaced is density of the fluid times the volume of the fluid displaced. So it's a neat idea. It's a really useful formula. Because any time you put an object into a fluid - look what's mattering here is not what the object is made of. It's what the fluid is. It's the fluid which is pushing on the object, and that's what this buoyant force is.
So having this formula, now we can answer the question that I started off with. How do you decide if an object is going to sink or not? If it's going to sink, the force of gravity on it, which I've been neglecting up to now, is going to be bigger than the buoyant force. There are only two forces. Buoyant force is the net force up. Gravity is the net force down. And so if gravity wins, it's going to sink. Now let's just think about these formulas. The force of gravity on the object is the mass of the object times g. And the mass of the object, M[O], is the density of the object times the volume of the object times g. So if it's going to sink, that's got to be greater than the buoyant force, which is the density of the fluid times g times the volume displaced. And if we're sinking, how much water are we displacing - how much fluid? Well, if it's sinking, then it's totally underwater, and so it's displacing the entire volume of the object. So in this case, it is safe to go ahead and use V[displaced] = V[O]. And then we see the g's cancel and the V[O] cancels. The condition for sinking is density of object has to be greater than the density of the fluid in which it's immersed - very simple criterion for whether something sinks or floats. And it is not a question of who's heavier. You can be heavier, but still have less density. That depends on both your weight and your - I should be careful. Density is not weight. It's mass per volume. So you can have a massive object, and as long as it's got a big volume, mass per volume can be small and it will float.
A lovely example, which everybody loves - a helium balloon - why is this floating? Well, the simple answer is because the density of the balloon must be less than the density of the air and so the buoyant force on the balloon is larger than the downward force. If you thought about this without using Archimedes' principle, you might use the following logic. A balloon is made of rubber, and before it's inflated, the rubber is massive and gravity pulls it down to the table. Rubber of a balloon doesn't just float. Now I fill it up with helium. And helium is massive. It's got a density, so I'm just adding mass to that balloon. So I started off with something that was massive. I added mass. Why the heck does it float? And the answer is, yes, indeed, there's plenty of mass here. But there's less mass in this volume than there would be mass of air in this volume. And so up it goes.
Archimedes invented this principle in the context of a very famous problem and the problem was this. Let's just focus your attention on the picture here. There's a crown. Archimedes was in Greece and he was working with a king of a local city-state leader who had received a crown as a gift and wanted to know whether it was gold or whether they were lying to him and just pretending it was gold. And he didn't want to melt it down. He wanted to figure out, without destroying this beautiful crown, whether it was made of gold. This is 2,200 years ago - primitive technology. So how are you going to figure out whether something's made of gold? You know what the density of gold is - even 2,200 years ago. So if you could just figure out the mass and the volume, you divide them, and you'll know what the density is.
Well, it's easy to find the mass. You just hang it from a scale. So that's little force diagram is here to show you. Here's the crown. There's a tension in the string up and the weight of the crown down and g - that's it. And so T[1] - Mg = 0. So this little scale here is going to read the weight of the crown, because the scale reads the tension in the rope. So it's easy to figure out the weight, and you divide by g. So you do know the mass of the crown. How are you going to find the volume of a funny-shaped crown? This was the puzzle, 2,200 years ago, which Archimedes solved. The story goes that he sat down into his tub one day, pensing over this little problem, and the water spilled out, because it was filled to the brim. And he realized that this was a clever way of measuring volume. If you displace water, you're going to be able to figure out the volume of the object you've put in the water by just looking to see how much is the volume of the water you've displaced. So you could have done that with the crown - just put it into a pot filled to the brim with water and seen how much volume of water spilled out and that would tell you the volume of the crown. In principle, that's all you need to do.
In practice, that's a tough experiment, because there's this funny meniscus thing with water and it's got to be filled to the brim and then it's dribbling all over the place. It's just hard to measure that volume of water displaced. So Archimedes came up with a really clever idea, and what he did was he just weighed it and then he immersed it in a bucket of water. It's now totally underwater. So the force diagram has changed. There's still Mg down. You can't hide from gravity, even when you're underwater. There's still going to be some tension, but it's less. P[2] - the new tension in the string - is less, because there's a new force in the problem. You have an object submerged in water. There's a buoyant force. F[B]. And so if you measure the tension - that's easy; that's the scale reading - and you already knew Mg, you can deduce the buoyant force. And remember, what's the formula for the buoyant force? Density of fluid, which we know, times g times the volume displaced - that's the volume of the crown. So this is actually a very accurate measure of the volume of the crown and now you can test for the density.
The story goes that Archimedes was so excited by coming up with this idea, that he jumped out of the tub stark naked and ran down the streets yelling, "Eureka! Eureka!" I've found it. I've found it. Archimedes principle. It's old. It's derivable, by Newton's second law. And it's very useful. It tells you what is the force of a fluid on an object? We don't have to draw all these pressures and forces and everything anymore. We just know immediately what's the buoyant force. We can use this formula.
Let me just show you one last demo which I just think is remarkable. We take a bucket of water here. And I'm going to grab an ordinary 10-pound bowling ball. This is a real bowling ball. And what do you suppose is going to happen if I put it in the water? The whole question is: Is the density of a bowling ball greater than or less than the density of water? And the amazing thing is that water - 1000 kg/m^3 - is very dense. In fact, it's denser than a bowling ball. Isn't it just so great? Bowling balls float because they're less dense than water.
Fluid Statics
Buoyancy and Archimedes' Principle Page [3 of 3]

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