big piston area *p = car mass*g
little piston area * p = little mass*g
thus
car mass/big area = little mass/little area
little mass = (lttle area/big area ) car mass
little mass =(25/390)*1500
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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
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get the additional volume in the little piston
25*h
same volume down in big piston
volume = 390 * drop in big piston
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?
kg
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?
m
c) How much did the height of the car drop when the person got in the car?
m
1 answer
1 answer