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?

Answer:

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?

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

Area of the big piston
A2 =πr2
=π(1)2
=πm2 

Pressure of the oil


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

According to Pascal’s Principle,

Example:

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?

Answer:

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