Question

A hydraulic lift in a garage has two pistons: a small one of cross-sectional area 4.50cm2 and a large one of cross-sectional area 220cm2 .

There are two parts to this question please give CORRECT answer.

Part A If this lift is designed to raise a 3400-kg car, what minimum force must be applied to the small piston?

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

Answer 1

Since the system operates at constant pressure, the force on each piston is proportional to its area. In this case, one piston has 220/4.5 = 48.89 times the area of the other. That means 1 Newton of force on the small piston is multiplied into 48.89 Newtons of force on the large piston.

A)

First we must find the weight of the car:
W = m*g
where W is weight, m is mass, and g is gravity.
W = (3,400 kg) * (9.81 m/s^2)
W = 33,320 N

So the car pushes down with 33,320 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:
F(applied) = (33,320 N) / 48.89
F(applied) = 681.53 N

B)

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 = (681.53 N) / (4.5 cm^2)
P = (681.53 N) / (0.00045 m^2)
P = 1,514,511.034 N/m^2
P = 1,514,511 Pa ~ 1514.511 KPa ~ 1.514 MPa

The pressure on the small piston (which ends up being the same pressure on the large piston) must be 1,514,511 Pascals, or 1.51*10^6 pa