Pascal's Law And Hydraulics

Question

In a hydraulic lift the small piston has an area of $5\,	ext{cm}^2$ and the large piston an area of $250\,	ext{cm}^2$. What downward force on the small piston is needed to support a car of weight 10000 N resting on the large piston?

RankGuru · Accepted Answer

A hydraulic lift relies on Pascal's law, which states that a pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and the container walls. Because the pressure is the same on both pistons, $\frac{F_1}{A_1} = \frac{F_2}{A_2}$, so a small force on the narrow piston balances a large force on the wide one. Solving for the input force, $F_1 = F_2\frac{A_1}{A_2} = 10000	imes\frac{5}{250} = 10000	imes\frac{1}{50} = 200\,	ext{N}$. The value 100 N would require an area ratio of 1:100. The value 400 N doubles the correct force. The value 500 N uses an incorrect ratio of 1:20. This is the standard NCERT application of Pascal's law that explains the force-multiplying action of hydraulic machines. As a check, the area ratio of 50 means the device multiplies force fifty-fold, so lifting a 10000 N car with only 200 N input is exactly the expected mechanical advantage.

Pascal's Law And Hydraulics

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