The piston of a hydraulic press with an area of 180 cm acts with a force of 18 kN on an area

The piston of a hydraulic press with an area of 180 cm acts with a force of 18 kN on an area of the small piston of 4 cm. With what force does the smaller piston act on the oil in the press?

180 cm2 = 0.018 m2;

4 cm2 = 0.00004 m2;

The pressure exerted by the large piston on the small one is equal to the pressure exerted by the small piston on the oil. Assuming that the pressure is equal to the ratio of the acting force to the surface area, we find the force with which the small piston presses on the oil:

F1 * S1 = F2 * S2;

F2 = F1 * S1 / S2 = 18000 * 0.018 / 0.00004 = 8100000 N = 8100 kN.

Answer: F2 = 8100 kN.