Question

The small piston of a hydraulic lift has a crosssectional area of 2.23 cm2 and the large piston 297 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN? (In service stations, this force is usually exerted by compressed air.) Answer in units of N.

Answer 1

Answer:

The force that must be applied to the small piston = 18.02 N.

Explanation:

Hydraulic Press: This is a device that produce a very large force to compress something e.g printing press.

From pascal's principle,

In an hydraulic press,

f/a = F/A......................... Equation 1

Where f = force applied to the small piston, a = area of the small piston, F = force required to lift the load/ force produced at the large piston. A = Area of the large piston.

Making f the subject of the equation above,

f = F×a/A.............................. Equation 2

Given: F = 2.4 kN = 2400 N, a = 2.23 cm², A = 297 cm².

Substituting into  equation 2

f = 2400(2.23)/297

f = 18.02 N.

Thus the force that must be applied to the small piston = 18.02 N.