You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a force on a smaller piston of not more than 550N. a) What should be the specifications on the dimensions of the pistons? Asmall piston/Alarge piston = ? b) How far down will you need to push the piston in order to lift the car 50cm? h = ?

(a) Area (small piston) / Area (large piston) = 0.037

(b) h = 1336.36 cm = 13.36 m

Explanation:

(a)

The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

Stress (small piston) = Stress (large piston)

Force (small piston) / Area (small piston) = Force (Large Piston) / Area (Large Piston)

Area (small piston) / Area (large piston) = Force (small piston) / Force (Large piston)

Area (small piston) / Area (large piston) = 550 N / (1500 kg) (9.8 m/s²)

Area (small piston) / Area (large piston) = 0.037

(b)

The work is also transmitted equally to the large piston. So,

Work (small piston) = Work (Large Piston)

Force (small piston). Displacement (small piston) = Force (large piston). Displacement (small piston)

(550 N) (h) = (1500 kg) (9.8 m/s²) (50 cm)

h = 735000 N. cm/550 N

h = 1336.36 cm = 13.36 m