# The hydraulic press must produce a force of 2.7 × 10 ^ 5 N. The diameter of the small piston is 3cm

**The hydraulic press must produce a force of 2.7 × 10 ^ 5 N. The diameter of the small piston is 3cm, the large one is 90cm. How much force should be applied to the small piston.**

Given:

F1 = 2.7 * 10 ^ 5 Newton – the force that the hydraulic press must produce;

d1 = 90 centimeters = 0.9 meters – the diameter of the large piston;

d2 = 3 centimeters = 0.03 meters – the diameter of the small piston;

pi = 3.14 is a geometric constant (Pythagoras number).

It is required to determine F2 (Newton) – the force that must be applied to the small piston.

Find the area of the large piston:

S1 = pi * d1 ^ 2/4 = 3.14 * 0.9 ^ 2/4 = 3.14 * 0.81 / 4 = 2.54 / 4 = 0.64 m2.

Find the area of the small piston:

S2 = pi * d2 ^ 2/4 = 3.14 * 0.03 ^ 2/4 = 3.14 * 0.0009 / 4 = 0.003 / 4 = 0.00075 m2.

Then:

F1 / S1 = F2 / S2;

F2 = F1 * S2 / S1 = 2.7 * 10 ^ 5 * 0.00075 / 0.64 = 202.5 / 0.64 = 316.4 Newton.

Answer: a force equal to 316.4 Newtons must be applied to the small piston.