Pascal’s Principle

The pressure exerted on a fluid in a container is transmitted undiminished all throughout the fluid and to the walls of the container containing the fluid. The pressure is acting to the fluid in all directions and in the direction perpendicular to the container’s walls.

This one of the most important and useful properties of fluids was known as the Pascal’s principle which is named after the French mathematician, scientist, and philosopher Blaise Pascal, who discovered this property of fluids.

One of the important applications of the Pascal’s principle is the hydraulic lift which is commonly used to lift heavy objects. In a hydraulic lift, a pressure is created by a downward force, F1 on the small piston of area A1, which is transmitted throughout the hydraulic fluid and acts upward one the larger piston. The larger piston has greater area, A2 than that of the smaller piston, and because of this, it is possible that the upward force, F2 on it can be many times the downward force on the smaller piston. This creates a significant mechanical advantage. This relationship can be expressed mathematically as,

\(p_1 = p_2\)

\(\mathbf{{F_1 \over A_1} = {F_2 \over A_2}}\)


The radius of the large piston is 18 cm. If the radius of the small piston is 3.0 cm, how much downward force must be exerted on the small piston that would lift a 2, 300 kg car?


mcar = 2, 300 kg

radius of large piston, rL = 18 cm

radius of small piston, rs = 3.0 cm


  1. We first solve for area, A1 of the small piston and area, A2 of the large piston using their given radius and the formula for area, \(\pi r^2\).

\(A_1 = \pi (3.0\;cm)^2 = 28.27\;cm^2\)

\(A_2 = \pi (18\;cm)^2 = 1,017.88\;cm^2\)

  1. Solve for F2, which is the force used to lift the car, using the formula F2=mg.

\(F_2 = (2,300\;kg)(9.8\;N/kg) = 2.25 \times 10^4\;N\)

  1. We can now solve for the downward force, F1 on the small piston using the equation \({F_1 \over A_1} = {F_2 \over A_2}\).

\({F_1 \over A_1} = {F_2 \over A_2} \implies F_1 = {A_1F_2 \over A_2}\)

Substitute the values for each quantity,

\(F_1 = {(28.27\;cm^2)( 2.25 \times 10^4\;N) \over 1,017.88\;cm^2} = 624.90\;N\)