Question

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 = ???

Answer 1

Answer:

(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