The piston of a hydraulic press with an area of 360 cm2 acts with a force of 18 kN.

The piston of a hydraulic press with an area of 360 cm2 acts with a force of 18 kN. The area of the small piston is 45 cm2. With what force does the small piston act on the oil in the press?

Given:

n1 = 360 cm2 (square centimeters) is the area of ​​the large piston of a hydraulic press;

n2 = 45 cm2 is the area of ​​the small piston of the hydraulic press;

P1 = 18 kN is the force with which the large piston of the hydraulic press acts on.

It is required to determine P2 (Newton) – the force with which the piston of a small area acts on the oil.

Since the values ​​will decrease among themselves, there is no need to convert the values ​​to the SI system. Then, to determine the strength, you must use the following formula:

P1 / n1 = P2 / n2;

P2 = P1 * n2 / n1;

P2 = 18 * 45/360 = 810/360 = 2.25 kN = 2250 Newtons.

Answer: A small piston acts on oil in a hydraulic press with a force equal to 2250 Newtons.