In a hydraulic lift, the diameteres of the pistons are 4.97 cm and 20.3 cm. A car weighing W=10.3 kN is to be?

Question

lifted by the force of the large piston. a). What force Fa must be applied to the small piston? b). When the small piston is pushed in by 10.5 cm, how far is the car lifted? c). Find the mechanical advantage of the lift, which is the ratio W/Fa.

Answer 1

This has to use pressure of liquids, where

Pressure = Force / Area.

Area of the small piston = Pi [(0.0497/2)^2] m^2

Area of the large piston = Pi [(0.203/2)^2] m^2

(a) Since pressure of liquid at the same level are equal, the hydraulic pressure of the smaller piston and the larger piston are the same.  

Pressure of small piston = pressure of large piston  

Fa / Pi [(0.0497/2)^2] = 10 300 / Pi [(0.203/2)^2]  

Fa = 617.39 N

(b) Volume of liquid in small piston = volume of liquid in large piston

Pi [(0.0497/2)^2] X 10.5 = Pi [(0.203/2)^2] X h

h = 0.629 cm (height lifted in large piston)

(c) Mechanical advantage = 10300 / 617.39

= 16.68 (ans)