Question

A hydraulic lift in a garage has two pistons: a small one of cross-sectional area 3.60cm2 and a large one of cross-sectional area 250cm2 .

a) If this lift is designed to raise a 3300-kg car, what minimum force must be applied to the small piston?

b)If the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?

Answer 1

one piston has 250/4 = 62.5 times the area of the other.

That means 1 Newton of force on the small piston is multiplied into 62.5 Newtons of force on the large piston.

First we must find the weight of the car:
W = m*g
where W is weight, m is mass, and g is gravity.
W = (3,500 kg) * (9.81 m/s^2)
W = 34,335 N

So the car pushes down with 34,335 N of force. Presumably it is on the large piston. The applied force with which we must push on the small piston to exactly balance the car is:
Fapplied = (34,335 N) / 62.5
Fapplied = 549.36 N

If the force is applied via gas pressure (or liquid pressure), we can convert this force into a pressure using the definition of pressure:
P = F / A
where P is pressure, F is force, and A is area.
P = (549.36 N) / (4 cm^2)
P = (549.36 N) / (0.0004 m^2)
P = 1,373,400 N/m^2
P = 1,373,400 Pa

The pressure on the small piston (which ends up being the same pressure on the large piston) must be 1,373,400 Pascals, or about 13.56 atmospheres differential pressure.