# Question Video: Pascal’s Principle Physics • 9th Grade

A solid object falls through water of density 1000 kg/m³. At the instant that the top of the object is 25 cm below the water’s surface, the water exerts a pressure on the top, bottom, and side of the object, as shown in the diagram. Find 𝑃₁, 𝑃₂, and 𝑃₃.

07:16

### Video Transcript

A solid object falls through water of density 1000 kilograms per cubic meter. At the instant that the top of the object is 25 centimeters below the water’s surface, the water exerts a pressure on the top, bottom, and side of the object, as shown in the diagram. Find 𝑃 one. Find 𝑃 two. Find 𝑃 three.

Looking at our diagram, we see this object submerged a complete distance of 25 centimeters below the surface of this water. Along with this, we can see these three pressures exerted on the top, side, and bottom of this object respectively, 𝑃 one on the top, 𝑃 two on the side, and 𝑃 three on the bottom. It’s these three pressures that we want to calculate. To do this, it will be helpful to recall a relationship telling us the pressure created by a liquid a certain depth below the surface of that liquid.

If we have a liquid of density 𝜌 that’s filling a container and we want to know the pressure 𝑃 created by the liquid at some point within the liquid, then the way we do that is we multiply the liquid density 𝜌 by the acceleration due to gravity 𝑔. And then, if this point where we want to solve for the pressure is a distance, we can call it ℎ, below the surface of the liquid. Then we multiply the liquid density and gravity by that term as well. The pressure a distance ℎ below the surface of a liquid of density 𝜌 is equal to 𝜌 times 𝑔 times ℎ.

Now, in our scenario, the liquid we’re working with is water. And we’re told its density, 1000 kilograms per cubic meter. Knowing the density 𝜌, we can now recall what 𝑔, the acceleration due to gravity near the surface of the Earth, is. We can approximate that acceleration as 9.8 meters per second squared. Now, that we know density 𝜌 and the gravitational acceleration 𝑔, all we need to know is ℎ the distance below the surface of the liquid to solve for the pressure at that point.

This is where our diagram helps us. We know that we want to calculate 𝑃 one, which is the pressure on the top side of our object, the pressure right here at this depth below the surface of our liquid. And our diagram shows us what that depth is. It’s 25 centimeters. This means we can write an equation for the pressure on the top of our object 𝑃 one. It’s equal to the density of our liquid multiplied by 𝑔 multiplied by the depth at that point, 25 centimeters below surface level. And it’s at this point that we can plug in values for 𝜌 and 𝑔. We know that the density is 1000 kilograms per cubic meter and that 𝑔 is 9.8 meters per second squared.

At this point, we can notice that the units of distance in our expressions for 𝜌 and in our expression for 𝑔 are meters but for our ℎ value, we have a distance measured in centimeters. So, we’ll want to convert that to meters so our units are consistent all through. If we recall that 100 centimeters is equal to one meter. That means, to convert 25 centimeters to meters, we’ll move the decimal point two spots to the left. And in doing so, we find that 25 centimeters is equal to 0.25 meters.

Before we multiply these three terms together and solve for 𝑃 one, take a look again at the units involved in these terms. We have kilograms per meter cubed multiplied by meters per second squared multiplied by meters. Considering our distance units, we have two powers of the unit meter in our numerator, here and here. And we have three powers in our denominator here. This means that two of those meters units cancel out from both top and bottom, numerator and denominator. And when we do this cancellation, we find that in our denominator we simply have units of meters.

And again, this is after we multiply all three of these values together. If we gathered all the remaining units and put them to the far right of this expression, we would see our final units end up being kilograms per meter second squared. If we were to rearrange these units, we would find that this is equivalent to a newton per square meter. That’s the same thing as a kilogram per meter second squared.

But then, if we consider what a newton per meter squared is, that’s equivalent to the SI unit for pressure, the pascal. Since a newton per square meter is a pascal, and a newton per square meter is the same as the units we’ll end up with in our expression. What we’re saying is the overall units we’ll have, once we calculate 𝑃 one, are pascals, the SI base unit of pressure. All that being said, let’s now multiply these three numbers together and figure out just how many pascals of pressure 𝑃 one is. When we do this, we find a result of 2450 pascals. That’s the pressure being exerted on the top side of our object.

Next, looking back at our diagram, we want to solve for the pressure 𝑃 two, the pressure exerted on the side of the object. Just like before, we’ll write an equation for this pressure, In this case 𝑃 two, in terms of density, acceleration due to gravity, and the height of this particular point below the surface level. And since this height is different from our previous height, let’s give it a subscript. We’ll call it a ℎ two.

We already know what density and 𝑔 are; they’re the same as they were before. But it’s ℎ two that we want to solve for. Taking a look at our diagram, we see that 𝑃 two, this pressure exerted on the side of our object, is exerted exactly 7.5 centimeters below the elevation of 𝑃 one. In other words, if we take the depth value of our pressure point 𝑃 one, 25 centimeters, and add to it seven and a half centimeters, then we’ll have the accurate height, or, in this case, depth, of this pressure point 𝑃 two. So, ℎ two is 25 centimeters plus 7.5 centimeters, or, in other words, 32.5 centimeters.

Now, just like before, we’ll want to convert this value from units of centimeters to units of meters. And to do that, we’ll shift the decimal point two spots to the left and find that this depth of pressure point 𝑃 two in units of meters is 0.325 meters. Plugging in again for density and 𝑔, when we multiply these three numbers together, once again, we’ll get a result in units of pascals. This time, though, we calculate a pressure of 3185 pascals. The fact that this pressure 𝑃 two is greater than the pressure we calculated for 𝑃 one makes sense because we’re now at a lower point below the surface than 𝑃 one.

And now, having solved for the pressures 𝑃 one and 𝑃 two on the top and sides of our object, respectively, we’ll solve for the pressure on the bottom of the object, 𝑃 three. Once again, we’ll do this by multiplying the density of our liquid, the water, by the acceleration due to gravity multiplied by the height below the surface of this point 𝑃 three.

Looking at our diagram, we see that the underside of our object is located a distance of 7.5 centimeters below the side of the object where we calculated the depth of 𝑃 two. So, the depth of the bottom of our object is equal to 25 centimeters plus seven and a half centimeters plus seven and a half centimeters again. Or, in other words, 25 centimeters plus 15 centimeters, which adds up to 40 centimeters. And then, just like we’ve done before, we’ll convert this from a number in centimeters to a number in meters. It’s equal to 0.40 meters.

Lastly, we’ll once more plug in our values for 𝜌 and 𝑔. And having done that, we’ll multiply these three numbers together. When we do, the result we find is 3920 pascals. That’s the pressure we calculate acting on the bottom of this object. And having solved for that, we’ve now calculated the pressure on the top, side, and bottom of our object as shown in the diagram.