# Video: Calculating the Upthrust Force on a Compound Object

A submarine under the sea has a lower density than the seawater around it, which is 1025 kg/m³, because most of the submarine’s volume consists of air that has a density of 1.225 kg/m³. If 90% of the submarine’s volume is air and 10% of its volume is steel of density 7700 kg/m³, how many times greater is the upthrust force that the water applies to the submarine than the submarine’s weight? Round your answer to one decimal place.

10:10

### Video Transcript

A submarine under the sea has a lower density than the seawater around it, which is 1025 kilograms per meter cubed, because most of the submarine’s volume consists of air that has a density of 1.225 kilograms per meter cubed. If 90 percent of the submarine’s volume is air and 10 percent of its volume is steel of density 7700 kilograms per meter cubed, how many times greater is the upthrust force that the water applies to the submarine than the submarine’s weight? Round your answer to one decimal place.

Okay, so in this question, we know that we’re looking at a submarine. And we know that the submarine has a lower density than the seawater around it. Now, we’re told that the seawater has a density of 1025 kilograms per meter cubed. We’re also told that the submarine’s volume mainly consists of air and air has a density of 1.225 kilograms per meter cubed. Specifically, we’re told that 90 percent of the submarine’s volume is air and 10 percent of the volume is steel. Now steel has a density of 7700 kilograms per meter cubed.

What we’ve been asked to do is to find out how many times greater the upthrust force is and the upthrust force is applied by the water to the submarine compared to the submarine’s weight. Finally, we know that we need to give our answer to one decimal place. Okay, so let’s start by labelling some quantities.

Firstly, we know that the density of seawater is 1025 kilograms per meter cubed. We’ll call this quantity 𝜚 sub 𝑤 — 𝜚 for density and 𝑤 for water. Secondly, we know that the submarine contains a lot of air. Now, the density of air is also something that we’ve been given. We’ll call this 𝜚 sub 𝑎 for air and we know that this quantity is 1.225 kilograms per meter cubed. The third density that we’ve been given in the question is the density of steel. We’re calling this 𝜚 sub 𝑠 for steel and it’s 7700 kilograms per meter cubed.

Now the rest of the information that we’ve been given in the question we’ll label on a diagram. So here’s a diagram of our submarine, where we know that 90 percent of it is air and we also know that the remaining 10 percent is steel. Now, let’s say that the submarine has a total volume 𝑣. We don’t actually know what 𝑣 is. But let’s not worry about that now. Instead what we can do is to label the volumes of the air and the steel that make up the submarine in terms of the volume 𝑣.

In other words, we can say that the volume of air 𝑣 sub 𝑎 is equal to 90 percent or 0.9 times the total volume 𝑣. And similarly, the volume of steel that makes up the submarine, which we’ll 𝑣 sub 𝑠, is equal to 10 percent or 0.1 times the total volume. This will prove useful later. Instead let’s look now at the forces acting on the submarine.

So because the question has asked us how many times greater is the upthrust force compared to the weight of the submarine, we know that we’re looking at two different forces acting on the submarine: firstly, the upthrust force which we’ll call 𝑇 and secondly, the weight of the submarine which we’ll call 𝑤. Now, we need to find how many times greater is 𝑇 than 𝑤. In other words, we need to find the ratio 𝑇 divided by 𝑤 because if 𝑇 is a certain number of times greater than 𝑤, then 𝑇 divided by 𝑤 will tell us how many times greater 𝑇 is.

So at this point, what we need to do is to start finding out what 𝑇 and 𝑤 are. Well, 𝑇 can be found using Archimedes’s principle. Archimedes’s principle tells us that the upthrust on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. In other words, our submarine is in the sea. Now, in order to be in that position in the sea, it’s displaced some amount of water. Specifically, it’s displaced water that has the same volume as the submarine.

In other words, let’s imagine a simpler scenario. Let’s say we’ve got a pond and we’ve got a ball inside a pond. Well, in order for that ball to be there, it had to displace some water that has the same volume as the ball. And it’s for this reason that the ball can occupy this volume here because the water that used to be there is no longer there. It’s been displaced by the ball.

Now, Archimedes’s principle tells us that the upthrust on the ball is equal to the weight of that displaced water. And similarly, with our submarine, the upthrust on the submarine is going to be equal to the weight of the water that the submarine has displaced. So what we need to do is to work at the weight of the water that would occupy the volume 𝑣 because that’s the volume of water that the submarine has displaced.

In order to do this, we need to recall two things: firstly, that the weight of an object 𝑤 is given by multiplying the mass of that object 𝑚 by the gravitational field strength of the Earth 𝑔 and secondly, that we can find out the mass of an object by multiplying the density of that object by the volume that it occupies. In other words, 𝑚 is equal to 𝜚𝑣.

Now, what we can do is make life simpler for ourselves by substituting this 𝑚 into this equation here. And so 𝑤 is equal to 𝑚𝑔 becomes 𝑤 is equal to 𝜚𝑣𝑔. In other words, the weight of an object is equal to the density of that object multiplied by the volume that occupies multiplied by the gravitational field strength of the Earth. And we can use this information to work out the weight of the water displaced by the submarine. So we say that the weight of the displaced water- we’ll call this 𝑤 sub 𝑑 because we’ve already used 𝑤 to represent the weight of the submarine itself. So we say that 𝑤 sub 𝑑 is equal to the density of the water, which is 𝜚 sub 𝑤, multiplied by the volume initially occupied by the water and now occupied by the submarine, which is 𝑣, multiplied by the gravitational field strength of the Earth, which is 𝑔.

Now, at this point, we won’t actually plug any numbers in because remember we don’t know the value of 𝑣. So we’ll leave this expression as it is for now. However, we can then apply Archimedes’s principle to say that the weight of the displaced water is equal to the upthrust 𝑇. Therefore, we found an expression for the upthrust of the submarine: 𝑇 is equal to 𝜚 sub 𝑤 multiplied by 𝑣 multiplied by 𝑔.

Now, it’s important to remember that the weight of the water of course acts in a downward direction. However, the upthrust acts in an upward direction. But for us, this is not relevant. All we care about is the size of the upthrust. We already know the direction in which it acts plus the water is actually not there anymore because the submarine is occupying that volume. So it doesn’t really make sense to worry too much about the direction of the forces on the water. In simple words then, we’re basically just trying to find out the size or magnitude of the upthrust 𝑇. And we have an expression for this now. So let’s move on to finding out the weight of the submarine.

Firstly, we’ll write down on the side the expression for the upthrust that we’ve just found and then we’ll work out how to find out the weight of the submarine. Now, once again, we can use the fact that weight is equal to the mass of an object multiplied by the gravitational field strength of the Earth. And using this logic, we can work out the mass of the submarine.

Now, the mass of the submarine is simply going to be the mass of the air plus the mass of the steel that make up the submarine. In other words, the mass of the submarine, which we’ll call 𝑚 sub sub — that’s 𝑚 subscript submarine, is equal to the mass of the air, which we’ll call 𝑚 sub 𝑎, plus the mass of the steel, which we’ll call 𝑚 sub 𝑠.

So firstly, what is the mass of the air? Well, to work this out, we can once again recall that the mass of an object is given by multiplying the density of that object by the volume occupied by that object. So we say that 𝑚 sub sub is equal to the mass of the air, which is the density of air 𝜚 sub 𝑎, multiplied by the volume that it occupies, which is 0.9 𝑣 — because remember 𝑎 occupies 90 percent of the volume of the submarine — and to this, we add the mass of steel, which is the density of steel 𝜚 sub 𝑠 multiplied by the volume it occupies, which is 0.1 𝑣 — 10 percent of the entire volume of the submarine. And this is our expression for the mass of the submarine.

So to find out the weight of the submarine, which remember we’ve labelled 𝑤, we can say that 𝑤 is equal to the mass of the submarine 𝑚 sub sub multiplied by the gravitational field strength of the Earth 𝑔. In other words, we just need to take this expression and multiply it by 𝑔. Doing this gives us that 𝑤 is equal to 𝜚𝑎 multiplied by 0.9𝑣 plus 𝜚𝑠 multiplied by 0.1𝑣 all multiplied by 𝑔.

Now, at this point, we have an expression for the weight of the submarine. However, before we move on, let’s notice something here. In both of the terms in the parentheses, we’ve got a factor of 𝑣, which means that we can pull it out of the parentheses. We can factorize it. And that’s exactly what we’ve done here. We’ve pulled out a factor of 𝑣. So it’s no longer inside the parentheses, but rather it’s outside. And the reason for this will become clear in a moment.

However, at this point, we can see that we’ve got expressions for both 𝑇 and 𝑤. And remember in this question, we’re trying to find out 𝑇 divided by 𝑤 or how much larger 𝑇 is compared to 𝑤. So all we need to do is to say that 𝑇 divided by 𝑤 is equal to the expression for 𝑇 in the numerator divided by the expression for 𝑤 in the denominator.

And at this point, we can notice that both the numerator and the denominator have a factor of 𝑣𝑔. So these cancel with each other. And what we’re left with is simply an expression in terms of the densities of the water, air, and steel and the proportion in which the air and steel occupy the volume of the submarine. In other words, 𝑇 divided by 𝑤 is equal to 𝜚𝑤, the density of the water, divided by 0.9 times 𝜚𝑎, the density of air, plus 0.1 times 𝜚𝑠, the density of steel.

Now, we already know all of these quantities. So we can plug them into our equation. And that looks something like this: 1025 is the density of water, 1.225 is the density of air, and 7700 is the density of steel. So when we evaluate the right-hand side of the equation, we find that 𝑇 divided by 𝑤 is 1.329... But remember we need to give our answer to one decimal place. So we have to round this value here. Now, the number after the first decimal place is a two. This is less than five. So the first decimal place will stay the same. It will not round up.

And hence, the one decimal place 𝑇 divided by 𝑤 is equal to 1.3. In other words, the upthrust force on the submarine is 1.3 times greater than its weight.