 We can all observe what happens when we release a helium balloon into the sky. And, like our comedies, we can observe that when we put something into water that floats, it will displace its own weight in the amount of water that it displaces. But let's see if we can quantify this concept of buoyancy by looking at it a little more closely. So we know that air is a gas and that water is a liquid. Both of these two states of matter can fall under a category known as fluids, which are basically distinguished from solids based on the characteristic of not having a coherent shape. A fluid will take the shape of whatever container it is put in. So when considering any amount of fluid, we can define an arbitrary shape of a desired size called a control volume. The idea of a control volume is that we're defining a shape or an area, or actually defining a volume in which we're containing our fluid. And that volume doesn't have to be actual physical volume, it can just be an area, sort of the boundaries that the fluid is in. So let's start with a control volume in the shape of a cylinder and consider a column of air in the shape of a cylinder that's filling the cylinder. Notice that could be just sort of floating out in space somewhere and we think of the air as being a column of air stacked on top. Well we know the air inside that control volume has a certain amount of weight and that the weight of the air in that cylinder would depend upon the size of the cylinder of the control volume that we define. Now let's actually consider two different control volumes of air. We'll consider a second control volume identical to the first one, but we won't consider it as being in a different space. Instead we're going to compare two volumes that are at different times. One volume that is a volume of the air before a certain event occurs and then one the same volume of air after we've taken part of that air and replaced it, for example, with a balloon in some portion of that volume. Now let's take a look at that a little more carefully with some physical examples. So now let's replace our theoretical air columns with something that's a little more visible. We're going to say that because air has weight we're going to represent it with some actual gram masses here and in two columns. Now notice these two columns are the same. So when we take the two columns and attempt to balance them we should notice that they indeed maintain balance. However, let's consider the idea of our balloon now. In the case of our balloon we have a one column of air. Let's look at the column of air here. In the case of our balloon we have one column of air that we're going to take some of the air here and replace it with something of similar volume. In this case we're going to compare the weight to a piece of styrofoam that's cut to approximately the same size. We're going to replace it with something of similar volume but significantly less density. Now the first thing we have to think about is the fact that the air is going to be displaced and therefore no longer there. Let me remove that. Here's the air that we're removing. It's no longer going to be there. In that case, notice that would definitely change the relative weight of the two things. However, just because it's not there doesn't mean that there isn't something there. There is something there. We're going to replace it with the helium or the other lighter than air material. Maybe there's some air on top of it. In which case there is still definitely a difference between the two. They are no longer balanced. This side is going to tend to want to go up. How much does it want to go up? Well, that amount depends on how much was removed and how much was added back. So our helium or our lighter material does affect it but it's not affecting it in the way you think. The helium is not making it go up. The helium is actually creating weight down. What's making it go up is the absence of the heavier material. So now let's go back again to our control volumes. We know that there's a relationship between the weight of the air in our control volume and the volume of the control volume itself. That relationship is quantified by something known as the specific weight. And that specific weight can be written and represented with the Greek letter gamma. And it's specifically related as the ratio between the weight, in this case the weight of the air inside our control volume or whatever fluid is inside our control volume and the volume of the control volume itself. Notice this is analogous to density, density which is mass per unit volume. And we can interchange these two things as long as we're near the surface of the earth and can easily interchange weight and mass. So if we take this relationship we can see that the weight of the air inside the control volume will be determined, doing a little bit of algebra, by taking the specific weight of our air or whatever fluid we're determining and multiplying it by the volume. Now let's consider our two control volumes. In the case of the first control volume we actually have weight of the air. We're going to consider that the old situation, the starting condition. And then we're going to create a new condition where we're going to take some of that air and we're going to replace some of that air with a balloon. And this is going to be a new weight. And that weight in that space is going to be the weight of something lighter than air. I'll abbreviate it L-T-A for lighter than air which could be helium or it could be hot air or something else that's lighter than air. Notice we're not going to replace the entire volume with the control volume. We're only going to replace some smaller portion of the volume here and we'll go ahead and let that be the volume that we're considering in both places. So now if we're interested in determining how much the weight changes between these two control volumes before and after, we're going to record that as the change in weight is the new weight or the weight lighter than air minus the old weight, which is the weight of the air that was there before. If I then recognize this relationship here between weight, specific weight, and volume, I can rewrite that as the volume times the specific weight of the lighter than air material minus the specific weight of the air, plugging that into both places and taking the volume out of both. Notice that volume is not the volume of the entire column but simply the volume of the piece that's being replaced. Now I recognize that weight, we define weight as a force that's acting down. If instead I want to think about this as being a force that's pulling up, all I need to do is reverse the sign and I'm going to do so by putting a negative sign out in front of this change in weight equation that we just created. In doing so, I'm thinking about this as being how much the force is pulling up and I'm going to define that as our net buoyant force, net buoyant force. Our net buoyant force FB, which if I rewrite it is the volume times, now notice I'm going to switch the signs here so I'm going to start with the specific weight of the air and I will subtract off the specific weight of whatever is lighter than air. A key point here, as we demonstrated with the gram weights, is to notice that our buoyant force, the positive part of our buoyant force is created by the absence of whatever fluid you're floating in but then you are replacing it with something that does have some weight but that contributes in a negative fashion. So your helium or your hot air is actually still acting in a downward direction. It's the absence of the cool air or just the air itself that creates a positive value for our net buoyant force. This relationship here for net buoyant force is the primary one we will use to solve problems involving buoyancy.