Jonathan Osbourne

**PhD., University of Maryland**

Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

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In **flotation**, the buoyant force equals the weight of the floating object and the volume of the object is always greater than the volume of water displaced. **Floatation** can be calculated using Archimedes' Principle.

Alright so let's talk about flotation. Flotation is a major application of Archimedes Principle and it's very different from just the simple idea of buoyancy because when you have flotation, you always have the buoyant force equaling the weight. And that's true for a very simple reason, if something's floating then its weight got to be canceled by the buoyant force and remember what the buoyant force, it's just that net force that the fluid puts on the solid when you try to immerse it in the fluid. Alright, so when we have a floating object the amount of volume that's displaced is always going to be less than the total volume of the solid. Because if it was equal then the solid would be entirely immersed and I wouldn't be correct in calling it floating, and there's no way in which I could displace a greater amount of volume than the volume of the solid. I mean the solid is not going to carrying air with it you know and if it was then I would call that air part of the solid too.

Anyway so the volume displaced must always be less than the total volume of the solid. Okay so flotation problems usually are asking 1 or 2 things. What's the fraction of the volume that's submerged or what's the fraction of the volume that is visible. Alright so let's look at a sample problem. What fraction of the volume of an iceberg is visible above the ocean water? Alright now in order to solve this problem we're going to use that ratio idea that we've seen before but we can derive it again here. So we're going to say the buoyant force divided by the weight which of course buoyant force is equal the weight so this ratio got to be 1 right. So the buoyant force divided by the weight will be the density of the fluid, volume displaced times gravity divided by the density of the solid. The total volume times gravity.

And that gives us a very, very important relationship that we can always use when something is floating as long as it's a uniform substance. We have that the density of the fluid divided by the density of the solid is equal to the total volume over the volume displaced. So that means that the ratio of volume displaced to volume is the same as the ratio of density of the solid to the density of the fluid just by flipping that guy up-side-down. Alright, well I know the density of ice it's 917 kilograms per cubic meter for the density of the fluid we'll go ahead and use ocean water rather than pure water because it's a little bit different it's 1,025 alright. Now this is going to the fraction that is displaced, so that's the fraction underneath. But I didn't want the fraction underneath I wanted the part that was visible above, so we'll say well jeez the part that's visible above plus the part that's not visible that's beneath okay it's got to be 1. So that means that the answer, the volume visible divided by the total volume got to be equal to 1 minus density of the solid over density of the fluid.

And it turns out if you plug those numbers into a calculator that you'll get something like 10.6%. So that when you're driving your ocean liner through the Arctic cycle and you see an iceberg and you say well jeez that's a pretty big iceberg you should keep in mind that about 89.4% of that iceberg you can't see. Alright so there's one thing, let's go ahead and look at the next one. We've got a raft and it's 2.5 meters by 1 meter by 20 centimeters okay so that's giving us the volume of the raft and the weight of the raft is 30 Newtons. Now I've got an 800 Newton person who's going to go ahead and lay on that raft and I want to know how many centimeters of the raft are visible above the surface after I got this person laying down on the raft right?

And then after that I want to know what's the maximum weight that the raft can support without sinking alright let's see how this goes. So we've got our 800 Newton person and then we've got our 30 Newton raft. So that means the buoyant force, remember the buoyant force with flotation always equals the total weight it, has to. So this is going to be 830 Newtons, the weight of the person plus the weight of the raft. Alright, f buoyant is always equal to the density of the fluid times the volume displaced times gravity. So now I can just get the volume displaced. The volume displaced will be 830 Newtons divided by the density of the fluid which we'll take to be water times acceleration due to gravity which of course has 9.8 like it always is. So when we do this division we'll end up with 0.0847 cubic meters so that's how much of the volume of the raft is displaced by or is displacing the water, is actually underneath the water right?

But we want to know what height, how many centimeters of the raft alright? But we got to divide by the cross sectional area of the raft. So the raft looks something like this right, where this is 1 meter, this is 2.5 meters and this distance right here was 20 centimeters okay. So if we want to know how much, how many centimeters of the raft are under the water then we got to take the volume that's under the water and divide by this cross sectional area 2.5 meters by 1 meter. So we'll divide by 2.5 square meters and we'll have height displaced when we do that division we end up with 3.4 centimeters. So therefore, height visible, well most of it is visible 16.6 centimeters of the raft are visible above the water when you've got this 800 Newton person laying on it alright.

What about the maximum weight that this raft can support? Well at maximum the raft is going to go all the way down so that 20 centimeters will be displaced. So that means that my weight mg will have to equal the density of the fluid times the whole volume of the raft. Because now it's all the way submerged, with because this is the maximum times g and of course we can do that 10 to the third kilograms per cubic meter. The whole volume is nothing more than 2.5 times 1 times and I got to multiply by 20, no remember it's got to be SI unit so I have to express that in terms of meters so it'll be 0.2 so it'll be 2.5 times 0.2 cubic meters times 9.8 meters per second squared. And when you multiply all these out you'll end up with 4,900 Newtons. Now of course 30 Newtons of that is the weight of the raft, so that means that 4,870 Newtons which is a huge amount of weight can be supported by this raft. Alright so that's the way that, that problem goes.

Now I wanted to do one other problem that's just kind of a qualitative idea of the way that these things work. I really like these problems because they make you think about it but they don't have any numbers in them at all. You're just kind of looking at the way that it works. So let's say that we've got this raft right here and we're going to do something a little bit different with it. We're going to take this raft and we're going to put a lead ball on top of it and we're going to ask what happens to the level of water, well if we put that lead ball on top of the raft. That lead ball's weight is adding to the weight of the raft, but we said that the buoyant force has to equal the weight right it's still floating. So what that means is that this thing is going to displace more water it's going to go down a little bit further. And that's going to cause the water level to go up a little bit and that's because the raft is now displacing its own weight plus the weight of the lead ball.

Alright now what happens if we then take this lead ball and we drop it to the bottom of the pool? And we want to know now does the water level go up, go down or remain the same? Now you could go through and try to put in numbers and do all that that would be a lot of work. So let's just think about it in a really straight forward way. When the lead was on top of the raft, it was part of a floating object and that means that it was displacing an amount of water equal to its own weight. But now when I take that ball and I drop it into the pool so that it goes all the way down to the bottom now it's only displacing its own volume. Since the density of lead is greater than the density of water, when it's displacing its own volume it's not displacing as much water as when it's displacing its own weight. And that means that the water level will go down when I take the lead and I drop it into the water, because it's not displacing as much. Alright that's floatation.