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?
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.
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.
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