A hydraulic lift has two connected pistons with cross-sectional areas 20 cm2 and

Question

A hydraulic lift has two connected pistons with cross-sectional areas 20 cm2 and 640 cm2. It is filled with oil of density 650 kg/m3.

1) What mass must be placed on the small piston to support a car of mass 1200 kg at equal fluid levels? m= (in kg)

2) With the lift in balance with equal fluid levels, a person of mass 70 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons? h =

3) How much did the height of the car drop when the person got in the car? h =

Explanation / Answer

1) Pressure = constant

m1g/A1 = m2g/A2 => m1/A1 = m2/A2 => 1200/640 = m2/20 => m2 = 37.5 kg

2)

pressure by 37.5 kg mass + hydrostatic pressure by column of fluid = pressure by (70 kg person and 1200 kg car)
m2*g*A2 + d*g*H = m1*g*A1

H = (m1*A1 - m2*A2)/d = (1270*0.0640 - 37.5*0.0020)/650 = 0.125 m

3) volume displaces is the same in both pistons:
A1*h1 = A2*h2
A1*(H - h2) = A2*h2
A1*H = h2*(A1 + A2)
h2 = H*{A1/(A1+A2)}
h2 = 0.125 m (20 cm²/665 cm²)
h2 = 0.0038 m

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