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In a hydraulic system, a force of 400 N is exerted on a piston with an area of 0.001 m2. The load-bearing piston in the system has an area of 0.2 m^2.

Required:
a. Calculate in kPa the pressure in the hydraulic fluid induced by the applied pressure.
b. What is the magnitude of the force exerted on the load bearing piston by the hydraulic fluid?

Respuesta :

abidemiokin

Answer:

Explanation:

Pressure on the hydraulic system is expressed as;

Pressure = Force/Area

Given

Force on the fluid = 400N

Area = 0.001m²

Pressure in the fluid = 400/0.001

Pressure in the fluid = 400,000N/m²

1N/m² = 0.001kPa

400,000N/m² = x

x = 400,000 × 0.001

x = 400kPa

Hence the pressure in kPa is 400kPa

b) Using the formula;

Pa = Pb

Fa/Aa = Fb/Ab

Pa = Fb/Ab

Fb = PaAb

Fb = 400,000(0.2)

Fb = 80,000N

Hence the magnitude of the force exerted on the load bearing piston by the hydraulic fluid is 80,000N

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