Pascal’s Principle – Example 1

Example 1:
In a hydraulic system the large piston has a cross-sectional area A2 = 200 cm² and the small piston has cross-sectional area A1 = 5 cm². If a force of 250 N is applied to the small piston, what is
a. the pressure exerted on the small piston
b. the force F, produced on the large piston?


a. Pressure exerted on the small piston

b. Pressure exerted on the large piston = Pressure exerted on the small piston

Example 2:
A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the small piston and large piston of the lift is 5cm and 1 m respectively,

  1. what gauge pressure in Pa must be applied to the oil?
  2. What is the magnitude of the force required on the small piston to lift the truck?

a. Weight of the truck,
W = mg
W = (5000)(10) = 50,000N

Area of the big piston
A2 =πr2

Pressure of the oil

Area of the small piston
A1 =πr2
=π (0.05)2
=0.0025π m2

According to Pascal’s Principle,


Figure above shows a hydraulic system. The area of surface X is 5 cm² and the area of surface Y is 100 cm². Piston X has been pushed down 10cm. what is the change of liquid level, h, at Piston Y?


Distance move by the piston-X, h1 = 10cm
Distance move by the piston-Y, h2 = h
Area of piston-X, A1 = 5 cm²
Area of Piston-Y, A2 = 100 cm²