A force of 2500 N acts on the smaller piston of a hydraulic machine with an area of 5 cm2.

| March 17, 2021 | education

A force of 2500 N acts on the smaller piston of a hydraulic machine with an area of 5 cm2. What load does the machine lift with a large piston with an area of 200 cm2?

Given:

n1 = 5 cm2 (square centimeters) is the area of ​​the smaller piston of the hydraulic machine;

n2 = 200 cm2 is the area of ​​the large piston of the hydraulic machine;

P1 = 2500 Newtons – the force acting on the smaller piston.

It is required to determine m (kilogram) – the mass of the load that the hydraulic machine lifts.

Let’s find the weight of the load that the car is lifting:

P2 = P1 * n2 / n1 = 2500 * 200/5 = 500 * 200 = 100000 Newtons.

Then its mass will be equal to:

m = P2 / g, where g = 10 Newton / kilogram (approximate value);

m = 100000/10 = 10000 kilograms.

Answer: The large piston of the hydraulic machine lifts a load weighing 10,000 kilograms.



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