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  • Today we're going to continue with playing with liquids.

  • If I have an object that floats, a simple cylinder that floats in some liquid, the area

  • is A here, the mass of the cylinder is M. The density of the cylinder is rho and its

  • length is l and the surface area is A. So this is l.

  • And let the liquid line be here, and the fluid has a density rho fluid.

  • I call this level y1, this level y2. The separation is h, and right on top here,

  • there is the atmospheric pressure P2, which is the same as it is here on the liquid.

  • And here we have a pressure P1 in the liquid. For this object to float we need equilibrium

  • between, on the one hand, the force Mg and the buoyant force.

  • There is a force up here which I call F1, and there is a force down here which I call

  • F2-- barometric pressure.

  • The force is always perpendicular to the surface. There couldn't be any tangential component

  • because then the air starts to flow, and it's static.

  • And here we have F1, which contains the hydrostatic pressure.

  • So P1 minus P2-- as we learned last time from Pascal--

  • equals rho of the fluid g to the minus y2 minus y1, which is h.

  • So that's the difference between the pressure P1 and P2.

  • For this to be in equilibrium, F1 minus F2 minus Mg has to be zero, and this we call

  • the buoyant force. And "buoyant" is spelt in a very strange way:

  • b-u-o-y-a-n-t. I always have to think about that.

  • It's the buoyant force. F1 equals the area times P1 and F2 is the

  • area times P2, so it is the area times P1 minus P2, and that is rho fluids times g times

  • h. And when you look at this, this is exactly

  • the weight of the displaced fluid. The area times h is the volume of the fluid

  • which is displaced by this cylinder, and you multiply it by its density, that gives it

  • mass. Multiply it by g, that gives it weight.

  • So this is the weight of the displaced fluids. And this is a very special case of a general

  • principle which is called Archimedes' principle. Archimedes' principle is as follows: The buoyant

  • force on an immersed body has the same magnitude as the weight of the fluid which is displaced

  • by the body. According to legend Archimedes thought about

  • this while he was taking a bath, and I have a picture of that here--

  • I don't know from when that dates-- but you see him there in his bath, but what

  • you also see are there are two crowns. And there is a reason why those crowns are

  • there. Archimedes lived in the third century B.C.

  • Archimedes had been given the task to determine whether a crown that was made for King Hieron

  • II was pure gold. The problem for him was to determine the density

  • of this crown-- which is a very irregular-shaped object--

  • without destroying it. And the legend has it that as Archimedes was

  • taking a bath, he found the solution. He rushed naked through the streets of Syracuse

  • and he shouted, "Eureka! Eureka! Eureka!" which means, "I found it! I found it!" What

  • did he find? What did he think of? He had the great vision to do the following: You

  • take the crown and you weigh it in a normal way.

  • So the weight of the crown-- I call it W1--

  • is the volume of the crown times the density of which it is made.

  • If it is gold, it should be 19.3, I believe, and so this is the mass of the crown and this

  • is the weight of the crown. Now he takes the crown and he immerses it

  • in water. And he has a spring balance, and he weighs

  • it again. And he finds that the weight is less and so

  • now we have the weight immersed in water. So what you get is the weight of the crown

  • minus the buoyant force, which is the weight of the displaced fluid.

  • And the weight of the displaced fluid is the volume of the crown--

  • because the crown is where... the water has been removed where the crown

  • is-- times the density of the fluid--

  • which is water, which he knew very well-- times g.

  • And so this part here is weight loss. That's the loss of weight.

  • You can see that, you can measure that with a spring.

  • It's lost weight, because of the buoyant force. And so now what he does, he takes W1 and divides

  • that by the weight loss and that gives you this term divided by this term, which immediately

  • gives you rho of the crown divided by rho of the water.

  • And he knows rho of the water, so he can find rho of the crown.

  • It's an amazing idea; he was a genius. I don't know how the story ended, whether

  • it was gold or not. It probably was, because chances are that

  • if it hadn't been gold that the king would have killed him--

  • for no good reason, but that's the way these things worked in those days.

  • This method is also used to measure the percentage of fat in persons' bodies, so they immerse

  • them in water and then they weigh them and they compare that with their regular weight.

  • Let's look at an iceberg. Here is an iceberg.

  • Here is the water-- it's floating in water.

  • It has mass M, it has a total volume V total, and the density of the ice is rho ice, which

  • is 0.92 in grams per cubic centimeter. It's less than water.

  • This is floating, and so there's equilibrium between Mg and the buoyant force.

  • So Mg must be equal to the buoyant force. Now, Mg is the total volume times rho ice

  • times g, just like the crown. The buoyant force is the volume underwater,

  • which is this part, times the density of water, rho water, times g.

  • You lose your g, and so you find that the volume underwater divided by the total volume

  • equals rho ice divided by rho of water, which is 0.92.

  • That means 92% of the iceberg is underwater, and this explains something about the tragedy

  • on April 15, 1912, when the Titanic hit an iceberg.

  • When you encounter an iceberg, you literally only see the tip of the iceberg.

  • That's where the expression comes from. 92% is underwater.

  • I want to return now to my cylinder, and I want to ask myself the question, when does

  • that cylinder float? What is the condition for floating? Well, clearly, for that cylinder

  • to float the buoyant force must be Mg, and the buoyant force is the area times h--

  • that's the volume underwater-- multiplied by the density of the fluid times

  • g must be the total volume of the cylinder, which is the area times l, because that was

  • the length of the cylinder, times the density of the object itself times g.

  • I lose my A, I lose my g, but I know that h must be less than l; otherwise it wouldn't

  • be floating, right? The part below the water has to be smaller than the length of the cylinder.

  • And if h is less than l, that means that the density of the fluid must be larger than the

  • density of the object, and this is a necessary condition for floating.

  • And therefore, if an object sinks then the density of the object is larger than the density

  • of the fluid. And the amazing thing is that this is completely

  • independent of the dimensions of the object. The only thing that matters is the density.

  • If you take a pebble and you throw it in the water, it sinks, because the density of a

  • pebble is higher than water. If you take a piece of wood, which has a density

  • lower than water, and you throw it on water, it floats independent of its shape.

  • Whether it sinks or whether it floats, the buoyant force is always identical to the weight

  • of the displaced fluid. And this brings up one of my favorite questions

  • that I have for you that I want you to think about.

  • And if you have a full understanding now of Archimedes' principle, you will be able to

  • answer it, so concentrate on what I am going to present you with.

  • I am in a swimming pool, and I'm in a boat. Here is the swimming pool and here is the

  • boat, and I am sitting in the boat and I have a rock here in my boat.

  • I'm sitting in the swimming pool, nice rock in my boat.

  • I mark the waterline of the swimming pool very carefully.

  • I take the rock and I throw it overboard. Will the waterline go up, or will the waterline

  • go down, or maybe the waterline will stay the same? Now, use your intuition--

  • don't mind being wrong. At home you have some time to think about

  • it, and I am sure you will come up with the right answer.

  • Who thinks that the waterline will go up the swimming pool? Who thinks that the waterline

  • will go down? Who thinks that it will make no difference, that the waterline stays the

  • same? Amazing--

  • okay. Well, the waterline will change, but you figure

  • it out. Okay, you apply Archimedes' principle and

  • you'll get the answer. I want to talk about stability, particularly

  • stability of ships, which is a very important thing--

  • they float. Suppose I have an object here which is floating

  • in water. Here is the waterline, and let here be the

  • center of mass of that object. Could be way off center.

  • It could be an iceberg, it could be boulders, it could be rocks in there, right? It doesn't

  • have to be uniform density. The center of mass could be off the center...

  • of the geometric center. So if this object has a certain mass, then

  • this is the gravitational force. But now look at the center of mass fluid that

  • is displaced. That's clearly more here, somewhere here,

  • the displaced fluid. That is where the buoyant force acts.

  • And so now what you have... You have a torque on this object relative

  • to any point that you choose. It doesn't matter where you pick a point,

  • you have a torque. And so what's going to happen, this object

  • is clearly going to rotate in this direction. And the torque will only be zero when the

  • buoyant force and the gravitational force are on one line.

  • Then the torque becomes zero, and then it is completely happy.

  • Now, there are two ways that you can get them on one line.

  • We discussed that earlier in a different context. You can either have the center of mass of

  • the object below the center of mass of the displaced fluid or above.

  • In both cases would they be on one line. However, in one case, there would be stable

  • equilibrium. In the other, there would not be a stable

  • equilibrium. I have here an object which has its center

  • of mass very low. You can't tell that--

  • no way of knowing. All you know is that the weight of the displaced

  • fluid that you see here is the same as the weight of the object.

  • That's all you know. If I took this object and I tilt it a little

  • with the center of mass very low-- so here is Mg and here is somewhere the waterline--

  • so the center of mass of the displaced fluid is somewhere here, so Fb is here, the buoyant

  • force, you can see what's going to happen. It's going to rotate towards the right--

  • it's a restoring torque, and so it's completely stable.

  • I can wobble it back and forth and it is stable. If I would turn it over, then it's not stable,

  • because now I would have the center of mass somewhere here, high up, so now I have Mg.

  • And the center of the buoyant force, the displaced water, is about here, so now I have the buoyant

  • force up, and now you see what's going to happen.

  • I tilt it to the side, and it will rotate even further.

  • This torque will drive it away from the vertical. And that's very important, therefore, with

  • ships, that you always build the ship such that the center of mass of the ship is as

  • low as you can get it. That gives you the most stable configuration.

  • If you bring the center of mass of ships very high--

  • in the 17th century, they had these very massive cannons which were very high on the deck--

  • then the ship can capsize, and it has happened many times because the center of mass was

  • just too high. So here... the center of mass is somewhere

  • here. Very heavy, this part.

  • And so now, if I lower it in the water notice it goes into the water to the same depth,

  • because the buoyant force is, of course, the same, so the amount of displaced water is

  • the same in both cases. But now the center of mass is high and this

  • is very unstable. When I let it go, it flips over.

  • So the center of mass of the object was higher than the center of mass of the displaced fluid.

  • And so with ships, you have to be very careful about that.

  • Let's talk a little bit about balloons. If I have a balloon, the situation is not

  • too dissimilar from having an object floating in a liquid.

  • Let the balloon have a mass M. That is the mass of the gas in the balloon

  • plus all the rest, and what I mean by "all the rest"...

  • That is the material of the balloon and the string--

  • everything else that makes up the mass. It has a certain volume v, and so there is

  • a certain rho of the gas inside and there is rho of air outside.

  • And I want to evaluate what the criterion is for this balloon to rise.

  • Well, for it to rise, the buoyant force will have to be larger than Mg.

  • What is the buoyant force? That is the weight of the displaced fluid.

  • The fluid, in this case, is air. So the weight of the displaced fluid is the

  • volume times the density of the air-- that's the fluid in which it is now--

  • times g, that is the buoyant force. That's... the weight of the displaced fluid

  • has to be larger than Mg. Now, Mg is the mass of the gas, which is the

  • volume of the gas times the density of the gas.

  • That's the mass times g-- because we have to convert it to a force--

  • plus all the rest, times g. I lose my g, and what you see...

  • that this, of course, is always larger than zero.

  • There's always some mass associated with the skin and in this case with the string.

  • But you see, the only way that this balloon can rise is that the density of the gas must

  • be smaller than the density of air. Density of the gas must be less than the density

  • of the air. This is a necessary condition for this to

  • hold. It is not a sufficient condition, because

  • I can take a balloon, put a little bit of helium in there--

  • so the density of the gas is lower than the density of air--

  • but it may not rise, and that's because of this term.

  • But it is a necessary condition but not a sufficient condition.

  • Now I'm going to make you see a demonstration which is extremely nonintuitive, and I will

  • try, step by step, to explain to you why you see what you see.

  • What you're going to see, very nonintuitive, so try to follow closely why you see what

  • you will see. I have here a pendulum with an apple, and

  • here I have a balloon filled with helium. I cut this string and I cut this string.

  • Gravity is in this direction. The apple will fall, the balloon will rise.

  • The balloon goes in the opposite direction than the gravitational acceleration.

  • If there were no gravity, this balloon would not rise and the apple would not fall.

  • Do we agree so far? Without gravity, apple would not fall, balloon would not rise.

  • Now we go in outer space. Here is a compartment and here is an apple.

  • I'm here as well. None of us have weight, there's no gravity,

  • and here is a helium-filled object, a balloon, and there's air inside.

  • We're in outer space, there's no gravity. Nothing has any weight.

  • We're all floating. Now I'm going to accelerate.

  • I have a rocket-- I'm going to accelerate it in this direction

  • with acceleration a. We all perceive, now, a perceived gravity

  • in this direction. I call it g.

  • So the apple will fall. I'm standing there, I see this apple fall.

  • I'm in this compartment, closed compartment. I see the apple go down.

  • A little later, the apple will be here. I myself fall; a little later, I'm there.

  • I can put a bathroom scale here and weigh myself on the bathroom scale.

  • My weight will be M times this a, M being my mass, a being this acceleration.

  • I really think that it is gravity in this direction.

  • The air wants to fall, but the balloon wants to go against gravity.

  • The balloon will rise. The air wants to fall, so inside here you

  • create a differential pressure between the bottom, P1, and the top of the air, P2, inside

  • here. Just like the atmosphere on earth--

  • the atmosphere is pushing down on us-- the pressure is here higher than there.

  • So you get P1 is higher than P2. So you create yourself an atmosphere, and

  • the balloon will rise. The balloon goes in the opposite direction

  • of gravity. If there were no air in there, then clearly

  • all of us would fall: The apple would fall, I would fall, and the helium balloon would

  • fall. The only reason why the helium balloon rises

  • is because the air is there and because you build up this differential pressure.

  • Now comes my question to you: Instead of accelerating it upwards and creating perceived gravity