In a hydraulic press, the area of the small piston is 5 cm squared, the area of the large one is 500 cm2.

In a hydraulic press, the area of the small piston is 5 cm squared, the area of the large one is 500 cm2. the force acting on this piston, 400 N, on the large-36 kN. What kind of strength gain does this press give?

The fluid pressure is the same for the cylinders and the ratio F1 / S1 = F2 / S2 is true.
F1 / F2 = S1 / S2, where S1 is the surface area of the large piston (S1 = 500 cm ^ 2 = 500 * 10 ^ -4 m ^ 2), S2 is the surface area of the small piston (S2 = 5 cm ^ 2 = 5 * 10 ^ -4 m ^ 2), F1 is the force acting on the larger piston (F1 = 3600 N), F2 is the force acting on the small piston (F2 = 400 N),
F1 / F2 = 3600/400 = 90.
S1 / S2 = (500 * 10 ^ -4) / (5 * 10 ^ -4) = 100.
Answer. The maximum gain in strength is 100. Due to frictional forces, the actual gain is 90.