The area of the small piston of the hydraulic lift is S1 = 0.8 cm2, and the area of the large one is S2 = 40 cm2.

The area of the small piston of the hydraulic lift is S1 = 0.8 cm2, and the area of the large one is S2 = 40 cm2. What force must be applied to the small piston to lift a 7kN load?

The action of the force F1 on the piston with area S1 creates a pressure p under it equal to

p = F1 / S1.

This pressure acts on the piston of area S2, creating a force F2:

p = F2 / S2.

Let’s equate the two expressions and find the required value of the force F1:

F1 / S1 = F2 / S2;

F1 = F2 * S1 / S2;

F1 = 7 * 0.8 / 40 = 0.14 kN.

Thus, to lift a 7 kN load, a force of 0.14 kN must be applied to the small piston.