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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