The large piston of a hydraulic press with an area of 180 cm2 acts with a force

The large piston of a hydraulic press with an area of 180 cm2 acts with a force of 18 kN. Small piston area 4 cm2. With what force does the small piston act on the oil in the press?

Initial data: S1 (area of the working surface of the larger piston) = 180 cm2; F1 (acting force of the larger piston) = 18 kN; S2 (smaller hydraulic press working surface area) = 4 cm2.

The force with which the small piston acts on the oil can be determined from the equality: P = F1 / S1 = F2 / S2, whence F2 = F1 * S2 / S1.

Let’s calculate: F2 = 18 * 4/180 = 0.4 kN = 400 N.

Answer: The small piston acts on the oil with a force of 400 N.