The area of the large piston of the press is 375 cm2. What should be the area of the small

The area of the large piston of the press is 375 cm2. What should be the area of the small piston to lift a 12kN load using a force of 160N?

Given:

S1 = 375 cm ^ 2;

F1 = 12 kN;

F2 = 160 N.

Find S2 -?.

Let’s start by translating the meaning of force into the siz system:

F1 = 12 kN = 12,000 H.

A hydraulic press is two communicating cylindrical vessels of different diameters, with pistons whose areas S1 and S2 are different (S2> S1).

When unloaded, the pistons are flush. The force F1 acts on the piston S1, and a body is placed between the piston S2 and the upper support, which needs to be pressed and the force acts on the piston:

F2 = pS2 = F1S2 / S1.

We get equality:

F2 / F1 = S2 / S1.

S1 = (F2 * S1) / F1 = (160 * 375) / 12,000 = 5 cm ^ 2.