In a hydraulic machine, the cross-sectional areas are 1:10. A body weighing 5 kg was placed on a small piston.

| July 28, 2021 | education

In a hydraulic machine, the cross-sectional areas are 1:10. A body weighing 5 kg was placed on a small piston. How much pressure must be applied to the larger piston to keep the system in balance.

To find the value of the force with which it is required to act on the larger piston of the specified hydraulic machine, we use the proportion: Fbp / Fmp = Fbp / (m * g) = Sbp / Smp, from where we express: Fbp = m * g * Sbp / Smp.

Variables and constants: m – body weight on a small piston (m = 5 kg); g – acceleration due to gravity (g ≈ 10 m / s2); Sbp / Smp – ratio of piston areas (Sbp / Smp = 10).

Let’s perform the calculation: Fbp = m * g * Sbp / Smp = 5 * 10 * 10 = 500 N.

Answer: The larger piston of the specified hydraulic machine must be pressed with a force of 500 N.



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