Question

The diameter of the large piston of a hydraulic press is 20 cm, and the area of the small piston is 0.50 cm2. If a force of 400 N is applied to the small piston, (a) what is the resulting piston? (c) Underneath the large piston?

Answer 1

a)

$\frac{F_1}{S_1}=\frac{F_2}{S_2},$

$F_1=F_2\frac{\pi d_1^2}{4S_2}=251.2~kN,$

b)

$p_1=p_2=\frac FS=8~MPa.$