# Video: Calculating the Force a Piston Applies at One End due to a Force Applied at the Other End

A hydraulic pump has a thin shaft with an area of 0.15 m² and a thick shaft with an area of 1.2 m², as shown in the diagram. At the tops of the shafts are pistons that can be pushed. A force 𝐹₁ = 85 N is applied to the piston in the thin shaft and the pressure of the hydraulic fluid applies a force 𝐹₂ to the piston in the thick shaft. Find the magnitude of 𝐹₂.

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### Video Transcript

A hydraulic pump has a thin shaft with an area of 0.15 square meters and a thick shaft with an area of 1.2 square meters, as shown in the diagram. At the tops of the shafts are pistons that can be pushed. A force, 𝐹 one equals 85 newtons, is applied to the piston in the thin shaft. And the pressure of the hydraulic fluid applies a force, 𝐹 two, to the piston in the thick shaft. Find the magnitude of 𝐹 two.

Okay, taking a look at our diagram, we see on the left-hand side this thin shaft right here and on the right-hand side the thick shaft over here. Our diagram includes labels for the areas of those respective shafts. The thin one has a cross-sectional area of 0.15 square meters, and the thick one a cross-sectional area of 1.2 square meters.

On top of these shafts are pistons that we see marked in green. These pistons seal off either end of this container, which we’re shown encloses a fluid. And because we’re told that this is a hydraulic fluid, we can assume that it’s incompressible. That is, even if we put pressure on it, the volume of the fluid remains the same.

Into this system we apply a force. And that force is called 𝐹 one. Its magnitude is 85 newtons. And it’s a force pressing down on the small-area piston.

Now if we think about that force, 𝐹 one, spread over the area of the thin shaft, that force divided by this area will create a pressure on the enclosed fluid. And that increase in pressure due to the force 𝐹 one will be transmitted all throughout the fluid, as Pascal’s principle tells us. It’s this pressure spread over the area of our large-area piston on the right that creates the upward acting force 𝐹 two. And it’s the magnitude of that force, 𝐹 two, that we want to solve for.

Before we clear some space on screen to do that, let’s record the magnitude of the force 𝐹 one. We’re told that that force is 85 newtons. And then as we saw, we want to solve for the magnitude of the force 𝐹 two pushing up on the large-area piston. To do that, we can start over on the left-hand side with our force 𝐹 one. We said that this force spread over the area of the thin shaft creates a pressure. That’s because pressure, 𝑃, is related to force and area like this. Pressure is equal to force divided by area. Or in this case, we could say the pressure created by force 𝐹 one — and we’ll call that pressure 𝑃 one — is equal to the force 𝐹 one spread over the area of the thin shaft.

So 𝑃 one, the pressure in the thin shaft, is equal to 𝐹 one, the force whose magnitude we’re given, divided by what we’ll call 𝐴 one, where 𝐴 one is equal to 0.15 square meters. Now 𝑃 one, that pressure created by the force, is a change in pressure. Because at first, we weren’t exerting a pressure and now we are. And Pascal’s principle tells us that when we’re working with an incompressible fluid, like we are in this case. A change in pressure at one location in the fluid is transmitted to all other points in that fluid. This means that the pressure 𝑃 one experienced in the thin shaft of our container is the same pressure that’s experienced over in the thick shaft on the right. And the same pressure that’s exerted on the underside of our large-area piston.

If we call the name of the pressure on the underside of the large-area piston 𝑃 two, then that’s equal to 𝐹 two, the force we want to solve for, divided by what we can call 𝐴 two, which is the cross-sectional area of the thick shaft. As we’ve just seen though, Pascal’s principle connects these two equations. It says that 𝑃 one is equal to 𝑃 two. Or in particular, the pressure on the underside of the small-area Piston is the same as the pressure just beneath the large-area piston.

And then if 𝑃 one is equal to 𝑃 two, then that must mean 𝐹 two over 𝐴 two is equal to 𝐹 one over 𝐴 one. And then if we focus on this last equality, 𝐹 two over 𝐴 two equals 𝐹 one over 𝐴 one, we can see that we’re given 𝐹 one, the applied force. We know 𝐴 one and 𝐴 two. And therefore, we have all the information we need to rearrange and solve for 𝐹 two.

To isolate 𝐹 two on one side of this equation, we can multiply both sides by the area 𝐴 two, causing that term to cancel out on the left. And we see that 𝐹 two is equal to 𝐹 one multiplied by this ratio of areas. 𝐴 two, the cross-sectional area of the large shaft, divided by 𝐴 one, that of the small shaft. 𝐹 one is given to us as 85 newtons. 𝐴 one is the area of the thinner shaft, 0.15 meters squared. And then 𝐴 two is the cross-sectional area of the thicker shaft, 1.2 meters squared.

So we’re now ready to substitute in for these three values on the right. With those values plugged in, notice what happens to the units of meters squared. Since they’re in numerator and denominator, they cancel out. And we’re left with units of newtons, that is, units of force. When we multiply 85 by 1.2 and divide that result by 0.15, we find a result of 680 newtons. That’s the magnitude of the force acting on the large-area piston.