What is the approximate force needed to apply to the small piston of a hydraulic

What is the approximate force needed to apply to the small piston of a hydraulic lift to lift a car weighing 800 kg, if the area of the small piston is 10 cm2, the area of the large one is 100 cm2?

Given:

m = 800 kilograms – vehicle weight;

g = 10 kg / m ^ 2 – acceleration of gravity;

S1 = 100 cm ^ 2 – area of ​​the large piston of the hydraulic lift;

S2 = 10 cm ^ 2 – the area of ​​the small piston of the hydraulic lift.

It is required to determine F2 (Newton) – the force that needs to be applied to the small piston of the hydraulic lift to lift the car.

The force acts on the large piston of the hydraulic lift:

F1 = m * g = 800 * 10 = 8000 Newtons.

Then:

F1 / S1 = F2 / S2;

F2 = F1 * S2 / S1 = 8000 * 10/100 = 8000/10 = 800 Newtons.

Answer: A force equal to 800 Newtons must be applied to the small piston of the hydraulic lift.