The piston of a hydraulic press with an area of 150 centimeters squared acts with an apore of 15 kilo Newton.

The piston of a hydraulic press with an area of 150 centimeters squared acts with an apore of 15 kilo Newton. With what force does the oil in the press act on a small piston with an area of 4 centimeters squared?

Given:

s1 = 150 square centimeters – the area of ​​the large piston of the hydraulic press;

s2 = 4 square centimeters – the area of ​​the small piston of the hydraulic press;

F1 = 15 kN = 15000 Newtons – the force with which acts on the large piston of the hydraulic press.

It is required to determine F2 (Newton) – the force with which the oil will act on the small piston of the hydraulic press.

Since in the process of calculating the value of the area is reduced, there is no need to convert units from square centimeters to square meters. Then, according to the laws of hydraulics:

F1 / s1 = F2 / s2, from here we find:

F2 = F1 * s2 / s1 = 15000 * 4/150 = 400 Newtons.

Answer: the oil acts on a small piston of a hydraulic press with a force equal to 400 Newtons.