A force of 40 N acts on the small piston, what force acts on the large piston of a hydraulic machine if it gives a 20 times the force gain?
P1 = 40 Newtons – the force acting on the small piston of the hydraulic machine;
n = 20 times – the coefficient of gain in the power of the hydraulic machine.
It is required to determine P2 (Newton) – the force acting on the large piston of the hydraulic machine.
According to the condition of the problem, we do not take into account the forces of resistance and friction. Then, to determine the magnitude of the force acting on the large piston, you must use the following formula:
P2 / P1 = n, from here we find that:
P2 = P1 * n = 40 * 20 = 800 Newton (0.8 kN).
Answer: A force equal to 800 Newtons (0.8 kN) acts on the large piston of the hydraulic machine.
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