# In a hydraulic press, the area of the large piston is 300 cm2, the area of the small piston is 30 cm.

**In a hydraulic press, the area of the large piston is 300 cm2, the area of the small piston is 30 cm. A force acts on the small piston, the modulus of which is 500N. Find the modulus of force acting on the large piston. Do not take into account friction.**

Because the fluid pressure in the pistons is the same, then the ratio is true:

F1 / S1 = F2 / S2; F1 is the pressure force on the larger piston (N), S1 is the surface area of the larger piston (S1 = 300 cm ^ 2 = 300 * 10 ^ -4 m ^ 2), F2 is the pressure force on the smaller piston (F2 = 500 N), S2 is the surface area of the smaller piston (S2 = 30 cm ^ 2 = 30 * 10 ^ -4 m ^ 2).

Let us express and calculate the force of pressure on the larger piston:

F1 = F1 * S1 / S2 = 500 * (300 * 10 ^ -4) / (30 * 10 ^ -4) = 5000 N = 5 kN.

Answer: The force acting on the larger piston is 5 kN.