How Does a Hydraulic Lift Work with Different Piston Sizes?

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SUMMARY

The discussion centers on the mechanics of a hydraulic lift involving two pistons with cross-sectional areas of 15 cm² and 430 cm², filled with oil of density 620 kg/m³. Participants calculated the mass required on the smaller piston to support a 1600 kg car, arriving at 55.8 kg, and determined the equilibrium height difference when an 80 kg person enters the car, which was found to be 3.02 m. The challenge arose in calculating the height the car dropped when the person entered, with initial attempts yielding incorrect results. The correct approach involves applying the principle of conservation of volume and the relationship between forces and distances in hydraulic systems.

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  • Understanding of Pascal's principle in hydraulics
  • Knowledge of fluid density and its implications in buoyancy
  • Familiarity with the concept of force equilibrium in hydraulic systems
  • Basic algebra for solving equations related to hydraulic mechanics
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Pat2666
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Okay, so I'm stuck on the last part of the problem :

A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 430 cm2. It is filled with oil of density 620 kg/m3.

--------------------------------------------------------------------------------

a) What mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels?

kg *
55.8 OK

b) With the lift in balance with equal fluid levels, a person of mass 80 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?

m *
3.02 OK

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

m

HELP: The fluid is incompressible, so volume is conserved.
HELP: Remember, one side will go up and one side will go down. The difference you calculated in part (b) was the sum of those two changes.


Okay, so I managed to find out the first two parts of the problem but I'm having trouble with Part C. Any way I attempt either comes out to the answer for B, or is just wrong. Since it mentioned that the volume is the same I thought that maybe the volumes of the cylindrical area displaced for each piston would be equal to one another (as you will see below). However, I came up with 0.109m as the height that the car moved, but it was wrong. I don't know what I'm doing wrong or what I might be missing completely. Any help would be greatly apprciated :)

My work :

http://img255.imageshack.us/img255/7595/workns5.jpg
 
Last edited by a moderator:
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Your result looks fine to me. Another approach is Work-in=Work-out.

F1d1=F2d2

(3.02)(55.8)(9.80)= X(1680)(9.8)

X= .1003 m
 

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