Question

The ratio of the radius of the output piston and that of the input piston in a hydraulic jack is 5:2.
How much load will be raised when an effort of 200N is applied at the input piston?

Answer 1

The hydraulic jack equation says

$$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$

Hence, the output force

$$F_2=F_1\frac{A_2}{A_1}=F_1\left(\frac{d_2}{d_1}\right)^2$$

$$F_2=200\:\rm N\times \left(\frac{5}{2}\right)^2=1250\: N$$