### Video Transcript

A cuboid-shaped object floats at rest in water of density 1000 kilograms per meter cubed. The top of the object is at a depth of 2.5 meters and its base is at a depth of 5.5 meters. How much greater is the pressure exerted by water at the object’s base than at its top? The horizontal faces of the object both have an area of 0.25 meters squared. How much greater is the force exerted by water at the object’s base than at its top?

Okay, so the first thing we’re told in this question is that we’ve got a cuboid-shaped object. Let’s draw a diagram of it. So here’s our cuboid-shaped object. And we know that this object is at rest in water of density 1000 kilograms per meter cubed. Now, we also know that the top of the object is that at depth of 2.5 meters and the base of the object is at a depth of 5.5 meters. So if we draw in to our diagram the surface of the water roughly about here, then we know that the top of the object which is this face here and we can represent this by this dotted line- we know that this face is at a depth of 2.5 meters. We also know something about the bottom of the object or the base, which is represented by this dotted line. Now the base is 5.5 meters below the surface.

So given all this information, we first need to work out how much greater the pressure exerted by water is at the object’s base compared to at its top. In other words, the water exerts both an upward and downward pressure on the object. However, the upward pressure is larger than the downward pressure. And what we need to do is to work out by how much. Now to do this, we’re going to use something known as Archimedes’s principle.

This principle tells us that the upthrust on an object is equal to the weight of the fluid displaced by the object. In other words, the total upward force on the object — in this case our cuboid-shaped object — is equal to the weight of the water — in this case displaced by the object.

Now, this net upward force is caused by a pressure imbalance between the top surface of the cuboid and the bottom surface of the cuboid. More specifically, the upward force exerted by the water on the bottom surface of the object is larger than the downward force exerted by the water on the top surface. And so the net result of this is going to be an upward force.

Now, you may have realized that since we’re talking about forces, we first need to work with forces before we can work with pressures. So let’s try and work out this net upthrust on the object using Archimedes’ principle — let’s call this net upthrust force 𝑇 for thrust — and also rather importantly, the surface that we’ve already shaded in in orange the top surface of the keyboard or block as well as the bottom surface which we can’t see in the diagram. But let’s say that those two surfaces have a surface area of 𝐴 that will become relevant later.

For now, let’s try and work out the thrust force. Now, we know that this thrust force according to Archimedes’s principle is equal to the weight of the water. So we’ll say 𝑊 sub water. And specifically, this is the weight of the water displaced by the cuboidal object.

And we can also recall that the weight of anything is given by multiplying the mass of that thing by the gravitational field strength of the Earth 𝑔. So therefore, we come back to this thrust equation and we say that the thrust is equal to the weight of the water which in itself is equal to the mass of the water 𝑚 sub water multiplied by the gravitational field strength of the Earth 𝑔.

Secondly, we can also recall that the mass of anything is given by multiplying the density of that thing 𝜌 by the volume it occupies 𝑉. And we can substitute this relationship into the mass of the water. Specifically, we find that this is equal to the density of the water 𝜌 sub water multiplied by the volume that it occupies which we’ll call 𝑉. But remember we’re looking at the volume of the water displaced by the cuboidal block. Therefore, this volume is the volume of the cuboidal block and we need to multiply this by the gravitational field strength of the Earth 𝑔.

So at this point, we know the density of water. We’ve been given this in the question. It’s 1000 kilograms per meters cubed. And we can recall that the gravitational field strength of the Earth is 9.8 meters per second squared. So the only thing that we don’t know is the volume of the block. This is where the area of the top and bottom surfaces of the block come in because we can recall that the volume of the block is given by multiplying the area of the top surface or the bottom surface by the thickness of that block, which in this case is this distance here. And that distance is simply 5.5 meters minus 2.5 meters.

Let’s call that distance ℎ and we know that ℎ is equal to as we’ve already said 5.5 meters minus 2.5 meters or in other words is equal to three meters. And so the volume of the block is equal to the area of the top or bottom surface multiplied by ℎ. And we can substitute this in to here to tell us that the upthrust is equal to the density of the water multiplied by the area of the top surface 𝐴 multiplied by ℎ multiplied by the gravitational field strength of the Earth 𝑔.

Now, here’s the interesting bit. We can also recall that this upward thrust force is only caused by a pressure difference between the top surface and the bottom surface. In other words, the pressure at the bottom surface of the block is much larger than the pressure at the top surface of the block. And the difference in these two pressures results in a pressure upwards and that upward pressure corresponds to the thrust force.

Therefore, if we relate this thrust force to the amount of net upward pressure, then that pressure is exactly the same as the difference in pressure between the bottom surface and the top surface. Or to put it even more simply, if we find the pressure exerted by this thrust force, that pressure is not actually a pressure, that pressure is actually a pressure difference. It’s the difference between the pressure exerted on the bottom surface and the pressure exerted on the top surface, which coincidentally is exactly what we’re trying to find out in the first part of this question.

To do this, we can recall that pressure 𝑃 is defined as force per unit area. And so we’re going to say that the net upward pressure, which we’ll call 𝑃, is equal to the thrust force divided by the area 𝐴 because this thrust force is acting perpendicularly to the top surface area 𝐴. Now remember that this thrust force once again is calculated because of a difference in forces that the water exerts on the bottom surface and the top surface. Therefore, this pressure is also the pressure difference between the bottom surface pressure and the top surface pressure.

So let’s go back calculating that. We already have an expression for 𝑇. All we need to do is plug that in, at which point we find that the pressure is equal to 𝜌 sub water multiplied by 𝐴 multiplied by ℎ multiplied by 𝑔 divided by 𝐴. And the 𝐴s cancel. So what we’re simply left with is that the pressure difference between the bottom surface and the top surface is 𝜌 sub water multiplied by ℎ multiplied by 𝑔.

Now, we know all of these quantities. Once again, we know 𝜌 sub water 1000 kilograms per meters cubed, we know ℎ three meters, and we can recall that 𝑔 is 9.8 meters per second squared. So at this point, we can just plug in all the values, which look something like this. And when we evaluate the right-hand side of the equation, we find that the pressure difference is equal to 29400 pascals because pascal is the standard unit of pressure. And we know that it’s gonna be in the standard units because we use the standard units for density which is kilograms per meters cubed, the standard units for ℎ which is meters, and the standard units for 𝑔 which is meters per second squared.

So at this point, we have the answer to the first part of our question. The upward pressure on the bottom surface is 29400 pascals greater than the downward pressure on the top surface. And at this point, we can move on to the next part of the question.

The horizontal faces of the object both have an area of 0.25 meters squared. So now we’ve been told that the value of 𝐴 here is 0.25 meters squared because these are the horizontal faces: the top face and once again the bottom face which we can’t see in the diagram. And we’ve been asked to work out how much greater the force is that is exerted by the water at the object’s base than at its top.

But if we remember back to a few minutes ago, we said earlier that the difference between the two forces exerted on the bottom face and the top face is simply the upthrust 𝑇. And we already have an expression for the upthrust 𝑇. And this time, we know what the value of 𝐴 is. So we know everything on the right-hand side of the equation, which means we can just plug stuff in.

The upthrust therefore is equal to the density of water 1000 kilograms per meter cubed multiplied by the surface area which we now know to be 0.25 meters squared times the value of ℎ which is three meters and also multiplied by the gravitational field strength of the Earth, at which we can evaluate the right-hand side of the equation to give us an upthrust of 7350 newtons.

And once again, we know that this is in newtons because the newton is the standard unit of our force. And we’ve used standard units throughout our calculation for the density, for the area, for the value of ℎ, and for the value of 𝑔. So what we found out is that the upward force exerted by the water on the bottom face is 7350 newtons larger than the downward force exerted by the water on the top face, at which point we can safely say we’ve reached the end of the question.