Question Video: Comparing Pressure Changes on the Pistons of a Hydraulic Lift | Nagwa Question Video: Comparing Pressure Changes on the Pistons of a Hydraulic Lift | Nagwa

Question Video: Comparing Pressure Changes on the Pistons of a Hydraulic Lift Physics • Second Year of Secondary School

The ratio between the cross section of the large piston to that of the small piston of a hydraulic lift is 3. By how much should the pressure acting on the large piston increase to maintain the two pistons at the same horizontal level if the pressure acting on the small piston increased by Δ𝑃? [A] 3Δ𝑃 [B] Δ𝑃 [C] Δ𝑃/3 [D] 2Δ𝑃

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Video Transcript

The ratio between the cross section of the large piston to that of the small piston of a hydraulic lift is three. By how much should the pressure acting on the large piston increase to maintain the two pistons at the same horizontal level if the pressure acting on the small piston increased by Δ𝑃? (A) Three Δ𝑃, (B) Δ𝑃, (C) Δ𝑃 over three, (D) two Δ𝑃.

Here, we are told that there is a hydraulic lift and the ratio between the area of the large piston to the small piston is three. We are then asked, if the pressure acting on the small piston increases by Δ𝑃, by how much should the pressure acting on the large piston increase so that the two pistons stay at the same horizontal level?

Let’s remind ourselves about Pascal’s principle, information about the pressure fluids exert, and some general properties of hydraulics. Pascal’s principle states that at any point in a fluid, the pressure exerted by the fluid at that point is equal in all directions. As well, remember that the pressure exerted by a fluid at a point is due to the weight of the water directly above that point. This means that if we have two points that are at equal depths, they will be at the same pressure as long as the only force on them is the weight of the fluid above.

A hydraulic lift, like we see in this diagram, is constructed to utilize these properties of fluids. When a force is applied to the small piston, a force is transferred through the fluid and pushes up on the larger piston. We can find the pressure applying to each of the pistons using this equation. The pressure 𝑃 exerted over an area 𝐴 is equal to the force applied 𝐹 divided by that area. This means that the pressure on the base of the small piston is equal to the force being applied divided by the area of the small piston. The same equation is used for the large piston as well. The pressure on the base of the large piston is equal to the force of the fluid pushing up divided by the area of the large piston.

Now, because the fluid contained between the two pistons is completely enclosed and the same, we can set the pressure on the base of these two pistons to be the same as well. This allows us to relate the magnitudes of these forces to the cross-sectional areas of the pistons, like so. The force acting on the small piston divided by the area of its face is equal to the force acting on the large piston divided by its area. Notice that the left-hand side of this equation is equal to the pressure over the area of the large piston and the right-hand side is equal to the pressure over the area of the small piston.

Now, since the pressures must be equal to each other, if the pressure on the small piston increases by an amount Δ𝑃, in order for the pistons to stay at the same horizontal level, and for this equation to stay balanced, the pressure on the larger piston must also increase by the same amount. Therefore, the larger piston will also need an increase in pressure of Δ𝑃. The difference in area between the two pistons doesn’t matter here, since they’re at the same horizontal level. So, the second option, Δ𝑃, is the correct answer.

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