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A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 390 cm2. It is filled with oil of density 530 kg/m3.

a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels?


HELP: The pressure is constant at any height.

b) 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?


c) How much did the height of the car drop when the person got in the car?


  • Physics -

    big piston area *p = car mass*g
    little piston area * p = little mass*g

    car mass/big area = little mass/little area

    little mass = (lttle area/big area ) car mass

    little mass =(25/390)*1500
    enough additional mass of oil must rise in the little tube to balance 70 kg on the big piston
    oil mass = (25/390)70
    = 530 * 25 * h
    solve for h
    get the additional volume in the little piston
    same volume down in big piston
    volume = 390 * drop in big piston

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