Question

A hydraulic press has pistons of area 0.006m' and 0.350m', respectively. A force of 25 N is applied to the smaller piston. (a) What is the lifting force on the bigger piston? (b) Find the pressure exerted on the smaller piston and on the big piston.

Answer 1

Given:

$A_1=0.006\;\rm m^2$

$A_2=0.350\;\rm m^2$

$F_1=25\:\rm N$

(a) The hydraulic press equation says

$$F_1/A_1=F_2/A_2$$

Hence

$$F_2=F_1 A_2/A_1=25*0.350/0.006=1460\: \rm N$$

(b) The pressure exerted on the smaller piston and on the big piston is the same

$$P_1=P_2=F/A=25/0.006=4170\:\rm Pa$$