I want to complete our discussion of fluid statics by looking at the final topic of buoyancy and stability. And this arises where we have an object in a fluid where the object is less dense than the fluid in which it is immersed. For example, a hot air balloon in the atmosphere or a piece of wood, or an iceberg floating in water. The questions that we want to address is, what is the net upward of buoyancy force which is acting? What is its magnitude, and where does it act? And what is the condition for an object which is fully immersed or floating to be stable? To answer the first question, the magnitude, I'll consider some arbitrary volume. And it's shown here like a sphere, but it could be any object of any shape. And this is immersed in some fluid. It's fully immersed. So to analyze this I'll imagine a prism and I can imagine this by taking a drill and drilling say vertically downwards through this object, and then putting the prism back in there and finding the forces on that prism that I've removed. So let's suppose that the height of that prism is h and the cross sectional area is dA, and the forces which are acting on it are the hydrostatic pressures which is P1, which is pushing down on the bottom, and P2 which is pushing upwards on the bottom. So, the net upward force on the prism is just the upward force due to P2 minus the downward force due to P1, in other words P2 minus P1 multiplied by the area dA. However, we can get now the total force, which is acting on the entire volume, is the summation of the forces on these elemental prisms, or in the limit the integral over the whole object of P2 minus P1 dA. And this is the net output force or the buoyancy force acting. Now from hydrostatics, we know that the pressure difference between the top and the bottom here due to the hydrostatic pressure distribution P2 minus P1 is equal to gamma, the specific weight of the fluid times h. Therefore, the buoyancy force is equal to the integral over the volume of gamma hdA, and because gamma is constant I can take that out of the integral and determine the integral hdA is the volume of that prism. Therefore, the integral of that expression over the whole volume Is just the volume of the object, V. So therefore, the upward force is equal to gamma times V, but gamma times V is just the weight of the fluid which is displaced by the body. Therefore, the final conclusion is that the buoyancy force is equal to the weight of the fluid, which is displaced by the body which is of course, Archimedes principle, and it acts through the Centroid of the body. If we have a floating object, then the buoyancy force is equal to the weight of the volume which is displaced. In other words, it's the weight of the volume which is displaced by the submerged portion of the volume, and it acts through the Centroid of that displaced volume. Not the Centroid of the volume of the whole, but the Centroid of the volume which is displaced. Now, the next question I want to look at briefly is the stability of an object. If the object is of uniform density, the Centroid of the volume, and the center of gravity are in the same place. The Centroid is a purely geometrical thing, it only depends on the shape of the object. So, if it's of uniform density, the Centroid and the center of gravity are at the same place. But what if it isn't? What if the object doesn't have uniform density? For example, let's suppose this was a wood block here, and a lead weight, let's suppose, on the bottom. Now the center of gravity is not coincidence with the Centroid, and the center of gravity is below the centroid. And the Centroid though is still in the same place because that only depends on the shape of the object. Now, in this case, if I displace this slightly, if I rotate it a little bit from its vertical position, now the forces which are acting on this object are like this. The buoyancy force is still acting upwards through the Centroid. The gravity force acts downwards through the center of gravity, that's the definition of center of gravity. So you can see that we have a restoring force, or a couple on this object which is rotating in this direction, which is trying to pull this back to its equilibrium position. In other words, this situation is stable. If we displace it a little bit from its vertical equilibrium position, a force acts on it to bring it back to its original position so it's stable. So, therefore, if the Centroid is above the center of gravity for a fully submerged object, that object is stable. On the other hand though, you can see the converse of this. If the center of gravity is above the entroid, then that is unstable. It cannot remain in that position. But what about floating objects like a ship? Floating objects, the center of gravity is often above the Centroid. So does that mean it's unstable? Well, no because if we displace the ship from it's vertical equilibrium position, then the buoyancy force is equal to the weight displaced here [COUGH] which is this volume. But the center of gravity stays in the same place, assuming that the cargo doesnt move around at all. But the center buoyancy moves because the center buoyancy is the Centroid of this volume, which you can see will now shift somewhat to the right. So in this case again, there's a restoring force acting on the submerged object which is trying to bring it back to its original equilibrium position. Now ,let me conclude this with a numerical example. And the example is that we have an ocean buoy, which is a sphere of 1.5 meters diameter, it weighs 8.5 kilonewtons, it's tethered to the seabed by a cable. If the specific weight of the sea water, which is surrounding this is 10.1 kilonewtons per cubic meter, the tension in the cable is most nearly which? Okay. So the solution, the first thing we do is to draw a free body diagram showing the forces acting on the buoy. And the forces that are acting are the tension in the cable T, which is holding it down, and the forces of buoyancy, which is acting upwards through the Centroid, and the weight, which is acting downwards also through the Centroid, assuming that the buoys are of uniform material. So a simple force balance for equilibrium says that a tension in the cable is equal to the buoyancy force minus the weight in air. The buoyancy force is equal to the specific weight of the sea water, gamma s, multiplied by the volume of the buoy minus W. The volume of a sphere is four thirds pi R cubed, so that is gamma times four thirds pi R cubed minus W. Plugging in the numbers, the specific weight of sea water is 10.1 kilonewtons per cubic meter, the diameter is 1.5 meters, and the weight of the buoy we're given in air is 8.5 kilonewtons. So the answer is 9.35 kilonewtons, and the closest answer, the best one, is A, 9.4 kilonewtons. And this completes our discussion of fluid statics
