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.