The larger piston of the hydraulic machine lifts a 450 kg load. In this case, a force of 150N acts on the smaller piston.

The larger piston of the hydraulic machine lifts a 450 kg load. In this case, a force of 150N acts on the smaller piston. Find the area of the smaller piston if the area is larger than 90 cm ^ 2.

F1 / S1 = F2 / S2, where F1 is the force of pressure of the load on the larger piston (F1 = Fт = m * g, where m is the mass of the load being lifted (m = 450 kg), g is the acceleration of gravity (we assume g = 10 m / s2)), S1 is the surface area of the larger piston (S1 = 90 cm2), F2 is the pressure force produced by the smaller piston (F2 = 150 N), S2 is the surface area of the smaller piston (cm2).

S2 = F2 * S1 / (m * g) = 150 * 90 / (450 * 10) = 3 cm2.

Answer: The area of the smaller piston is 3 cm2.