Question

A hydraulic lift is designed to lift heavy objects of maximum mass 2000kg. The area of cross section of carrying the load is 2.25×10^-2 m^2. What is the maximum pressure the smaller piston would have to bear? ( Use g= 9.8 m/s^2)
Ans. 0.8711×10^6 N/m^2

Answer 1

Answer:

Pressure on the piston P=

A

F

Force F=m×a

=3000×9.8

=29400 N

Area of cross section A=425×10

−4

sqm

Therefore the pressure P=

425×10

−4

3000×9.8

=6.92×10

5

Pa