# Video: Finding the Magnitude of the Upthrust Force on an Object Partially Submerged in Water

A solid object is placed in water that is in a container. The object partially submerges, which displaces a volume of water into another container, as shown in the diagram. The water has a density of 1000 kg/m³. What is the magnitude of the upthrust force on the object?

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### Video Transcript

A solid object is placed in water that is in a container. The object partially submerges, which displaces a volume of water into another container, as shown in the diagram. The water has a density of 1000 kilograms per meter cubed. What is the magnitude of the upthrust force on the object?

Okay, so in this question, we’ve got a solid object which is placed in water — that is in a container. So here’s the solid object in the diagram. We’ve placed it in the water which is sort of bluey. And here is the container. The object partially submerges, which again has been shown in the diagram there’s a partially submerged bit. And this displaces a volume of water into another container, which is this water here. We’re also told that the water has a density of 1000 kilograms per meter cubed. We need to find the magnitude — the size of the upthrust force on the object and that’s this force labelled by the arrow.

So first of all, we can write down a few important quantities. Firstly, we’ve been told that the density of water, which we’re calling the Greek letter 𝜌, is one 1000 kilograms per meter cubed. Secondly, we’ve also been told the volume of the displaced water here which we’ll call capital 𝑉 is equal to 0.25 meters cubed. And we’ve been given that information in the diagram. So we need to find out the magnitude of the upthrust force. We’ll call this force 𝐹.

And we need to find out what 𝐹 actually is, to do this we can use something known as Archimedes’ principle. The principle states that the upthrust on an object immersed in a fluid is equal to the weight of the displaced fluid. In other words, the upthrust 𝐹 on the object in grey here is equal to the weight of the water that it displaced. And the water that it displaced is this water here. So we need to find out the weight of that water.

Let’s say that the displaced water has a weight 𝑊, then Archimedes’ principle is telling us that the upthrust 𝐹 is equal to the weight of the displaced water 𝑊, which means that if we can find out what 𝑊 is, then we know what 𝐹 is. So how do we find the weight of an object? Well, the weight of an object is found by multiplying the mass of the object 𝑚 by the gravitational field strength of the Earth which is 𝑔. So if we want to find out the weight of the displaced water, then we do this by multiplying the mass of the displaced water. We’ve said that the displaced water has a mass 𝑚. And we multiply the mass by the gravitational field strength of the Earth, which is 𝑔.

Now 𝑔 we already know. We know that 𝑔 is 9.8 meters per second squared. This is a constant value that we need to know. It’s the gravitational field strength of the Earth. But we don’t yet know what 𝑚 is. However, we do know the density of the water and the volume of the displaced water. So to work out the mass of the water, we can recall that density 𝜌 of an object is equal to the mass of an object divided by the volume it occupies.

In this case, the object that we’re talking about of course is the displaced water. And we’re trying to find out the value of the mass of the water having already known what the values of 𝜌 and 𝑉 are. This means we need to rearrange the equation. We can multiply both sides of the equation by 𝑉 and so the volumes on the right-hand side cancel, which leaves us with the density multiplied by the volume is equal to the mass.

Now since we’re trying to find out the mass of the water — the displaced water — and we need to use that to find out the weight of the displaced water, we can simply substitute this equation in. We’ll replace 𝑚 in the weight with 𝜌𝑉. Well, that leaves us with is 𝑊 is equal to 𝜌𝑉, which is the mass, multiplied by the gravitational field strength 𝑔. And remember we said earlier that this weight is equal to the upthrust force 𝐹.

So having done all of this mathematical manipulation, we finally found an equation that tells us the quantity that we’re looking for, which is 𝐹, in terms of three quantities that we know already 𝜌, 𝑉, and 𝑔. This means we can get rid of all of other working out and just focus on 𝐹 is equal to 𝜌𝑉𝑔. Now all that remains for us to do is to plug in all the values. 𝐹 is equal to 𝜌 which is 1000 kilograms per meter cubed multiplied by 𝑉 which is 0.25 meters cubed multiplied by 𝑔 which is 9.8 meters per second squared.

And since all of the quantities that we’ve used are already in their standard units, we know that the force will come out in its standard unit, which is newtons. So we can evaluate this expression to give us a value of 𝐹 of 2450 newtons. And that is our final answer. The magnitude of the upthrust force on the object is 2450 newtons.