With what force does a hydraulic press compress a part if the area of the large piston is 100

With what force does a hydraulic press compress a part if the area of the large piston is 100 times the area of the small one? a force of 50 N acts on the small piston.

Given:

F1 = 50 Newton – force acting on the small piston of the press;

s2 = 100 * s1 – the area of the large piston is 100 times the area of the small piston.

It is required to determine F2 (Newton) – with what force the large piston squeezes the part.

To determine the strength, you need to use the following formula:

F1 / s1 = F2 / s2;

F2 = F1 * s2 / s1;

F2 = F1 * 100 * s1 / s1;

F2 = 100 * F1 = 100 * 50 = 5000 Newton = 5 kN.

Answer: The large piston of the press compresses the part with a force of 5 kN.