Question Video: Pascal’s Principle | Nagwa Question Video: Pascal’s Principle | Nagwa

# Question Video: Pascal’s Principle Physics • Second Year of Secondary School

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A bag attached to an intravenous drip holds saline solution that has a density of 2,160 kg/m³. The bag is 15 cm in height and full to the top. The solution flows from the drip through a hole of area 0.785 cm² and passes through the tube into a cannula that has an opening of area 0.0225 cm². What is the magnitude of the force exerted by the saline solution at the hole at the base of the drip bag? Give your answer to two decimal places. What is the magnitude of the force exerted by the saline solution at the cannula? Give your answer to three decimal places.

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### Video Transcript

A bag attached to an intravenous drip holds saline solution that has a density of 2160 kilograms per cubic meter. The bag is 15 centimeters in height and full to the top. The solution flows from the drip through a hole of area 0.785 square centimeters and passes through the tube into a cannula that has an opening of area 0.0225 square centimeters. What is the magnitude of the force exerted by the saline solution at the hole at the base of the drip bag? Give your answer to two decimal places.

Okay, so, in this example, what we have is a saline solution in a bag. So, let’s say that this here is our bag. And we’re told that this bag is 15 centimeters in height and that it’s full to the top. This saline solution in the bag flows through this point here called the drip into a narrow tube. And in our statement, we’re told the cross-sectional area of this hole in the bag. We’ll refer to that area as 𝐴 sub one. And it’s given as 0.785 square centimeters.

So, after leaving the bag, our salt solution flows through this tube until it reaches what is called a cannula. Now, we may think, isn’t that the name of an Italian dessert? But actually, that’s a cannoli. This word cannula refers to the very small opening between this narrow tube and the needle that will go into the patient’s body. We’ll call the cross-sectional area of our cannula 𝐴 two. And we’re given that value as 0.0225 square centimeters. Knowing all this, our first question asks, what is the magnitude of the force exerted by the saline solution at the hole at the base of the drip bag? In other words, what’s the force acting on the cross-sectional area 𝐴 one as we’ve located it? We can call that force 𝐹 one. And we’ll clear a bit of space to solve for it.

When we think about what causes force 𝐹 one, we can see that it’s the saline solution piled up in this bag pressing down on that spot in the bag. In other words, the layers of this fluid lying on top of one another create a downward pressure thanks to their weight. And this pressure is applied across the area we’ve called 𝐴 one. At this point, we can recall the general relationship between pressure, area, and force. A pressure 𝑃 is equal to a force 𝐹 spread over an area 𝐴. In this case, it’s the force, what we’ve called 𝐹 one, that we want to solve for. But that will require knowing the area, what we’ve called 𝐴 one, which we do know since it’s given to us, as well as the pressure 𝑃 of the fluid at that point.

That’s not something we yet know. But notice that in our problem statement, we’re told what the density of our saline solution is. That can be a clue to us because, recall, we realize that the pressure at the bottom of the bag is due to the weight of the saline solution on itself. And we realize that the greater the density of this solution, the more weight a given volume of it has. In fact, there’s a mathematical relationship that connects pressure and fluid density. That relationship says that the pressure created by a fluid is equal to the density of the fluid multiplied by the acceleration due to gravity multiplied by the vertical height of that fluid.

And at this point, we can recall that not only are we given the fluid density of this solution, but we’re also given its height because we’re told that the bag is 15 centimeters tall and that its full to the top. So, that means we know 𝜌, the density, and ℎ, the height of our fluid. And 𝑔, the acceleration due to gravity, we can take to be exactly 9.8 meters per second squared. All this means that we have enough information to calculate the pressure of the fluid at the bottom of the drip bag. But remember that it’s not exactly pressure we want to solve for, but rather the force, what we’ve called 𝐹 one.

So, here’s what we will do. Let’s take this relationship here and rearrange it to solve for force. If we multiply both sides by the area 𝐴, that term cancels out from the right. And we can see that force is equal to pressure times area. This tells us that the force 𝐹 one at the bottom of the drip bag is equal to the pressure at that location, we’ll call it 𝑃 one, multiplied by the cross-sectional area of that opening, what we’ve called 𝐴 one. We’re given that area 𝐴 one. And 𝑃 one is equal to 𝜌, the density of our saline solution, times 𝑔 multiplied by the height of the solution in the bag. That is, we could write out our equation for 𝐹 one as 𝜌 times 𝑔 times ℎ times 𝐴 one.

Now, let’s check to make sure we have all these values accounted for. The density of the fluid 𝜌 is given as 2160 kilograms per cubic meter. 𝑔 is a constant, 9.8 meters per second squared. ℎ is the height of the saline solution in our bag, given as 15 centimeters. And 𝐴 one is the area of the hole at the bottom of the bag, given as 0.785 square centimeters. So, we’re ready to substitute in for these values. With all our values plugged in, let’s take a moment to look at the units of these terms.

For the density, we have units of kilograms per cubic meter. For 𝑔, we have units of meters per second squared. For the height ℎ, our units are centimeters. And for our area 𝐴 one, they’re centimeters squared. We can see then that we’re not quite ready to multiply all these values together because we have different units for length. Two of these terms, the first two, use meters, while the last two use centimeters. We’ll want to put them all on the same footing.

We could decide to change the meters into centimeters or the centimeters into meters. Either method would work. But let’s choose the second option, converting centimeters into meters. We can recall that 100 centimeters is equal to one meter. Or another way of saying the same thing, one centimeter is equal to one one 100th of a meter. This means that wherever we see centimeters in our expression, we can replace that with one divided by 100 times a meter. When we do that, 15 centimeters becomes 15 times one divided by 100 meters and 0.785 square centimeters becomes 0.785 times one divided by 100 meters quantity squared.

So, then, 15 centimeters is equal to 15 divided by 100 meters, or put another way, 0.15 meters. And then, being careful to square both one divided by 100 as well as the units meters, we get a result for our area 𝐴 one of 0.785 divided by 10000 square meters. The reason we have 10000 in the denominator is that’s what one divided by 100 times one divided by 100 is. With that conversion done, our units are all in agreement. And we’re ready to calculate 𝐹 one. When we do, to two decimal places, we get a result of 0.25 newtons. That’s the force acting on the hole at the bottom of the drip bag. Now, let’s look at the second question in this exercise.

This question asks, what is the magnitude of the force exerted by the saline solution at the cannula? Give your answer to three decimal places.

Looking at our sketch, we can recall that the cannula is where the thin tube from the drip bag connects into the needle that goes into the patient. We’ve called the cross-sectional area of this cannula 𝐴 two. And in our problem statement, that value is given to us as 0.0225 square centimeters. Getting back to our question, we can see that, in this case, as before, we want to solve for a force magnitude. Let’s call this force the force of the saline solution at the cannula 𝐹 two. And this force 𝐹 two will be equal to the pressure, we can call it 𝑃 two, at the cannula multiplied by its cross-sectional area.

We’re given this area, 𝐴 two. But what about the pressure 𝑃 two? That is, what’s the pressure that acts on the cannula at the end of this long tube coming from the drip bag? Well, because this area in this thin tube from the bottom of the drip bag up to the cannula is a closed system of an incompressible fluid, that means the pressure of the saline solution all throughout this tube and up to the cannula will be the same. We could think of it as a stretch of pipe with fluid running through it. Since the pipe is very thin and has a constant diameter, fluid pressure is constant throughout.

This is helpful to us because it means that 𝑃 two, the pressure at the cannula, is the same in magnitude as 𝑃 one, the pressure at the bottom of the drip bag. And that pressure, we saw, was equal to 𝜌, the fluid density, times the acceleration due to gravity times the height of the fluid in the bag. So, we can use those same values from before to indicate the pressure at the cannula.

Plugging in for 𝜌, 𝑔, ℎ, as well as 𝐴 two, we see that we’ve run into the same issue as earlier where we have different units, meters and centimeters, for lengths. Once more, we can resolve that difference by converting centimeters to meters. We’ve seen that 15 centimeters is equal to 0.15 meters. And further, 0.0225 square centimeters is equal to 0.0225 divided by 10000 meter squared. We’re now ready to calculate the magnitude of the force 𝐹 two. To three decimal places, it’s 0.007 newtons. That’s the magnitude of the force exerted by the saline solution at the cannula.

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