How Much Does a Car Rise on a Hydraulic Lift?

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Homework Help Overview

The problem involves a hydraulic car lift with two cylindrical pipes and seeks to determine how much a car rises when a plunger is depressed. The context includes specific dimensions of the pipes and the displacement of the plunger.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the forces and displacements in the hydraulic system, referencing the work-energy principle. Some express confusion regarding the absence of mass in the calculations, questioning how to apply certain equations without it.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to relate the forces and displacements in the hydraulic lift. There is recognition of the need for clarification on the implications of missing mass information, and some guidance has been offered regarding the use of work equations.

Contextual Notes

Participants note the lack of mass information, which may affect the application of certain equations in the context of the hydraulic lift problem.

Calculuslover9
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Homework Statement


A hydraulic car lift has a reservoir of fluid connected to two cylindrical fluid filled pipes. The pipe directly below the car has a diameter of 1.8 m. the pipe o which the plunger acts has a diameter of 0.045m. the plunger is depressed a distance of 1.5m. How much does the car rise?
http://tinypic.com/r/wam5xd/5
If image won't show please look at link
http://tinypic.com/r/wam5xd/5

Homework Equations



Ain X ΔXin=Aout X Δout

The Attempt at a Solution


Since the piston movess downaward thorugh a displacement Δx1 equals the volume of liquid pushed up on the right as the right piston moves upward through a displacementhttp://tinypic.com/r/2pr8wvt/5

If the image did not show please look at below or look at link

http://tinypic.com/r/2pr8wvt/5

(π (〖0.0225〗^2 m)/4m)(1.5m)=(π (〖0.9〗^2 m)/1.5m)(x)
x=0.00094m

My attempt was wrong. But I am not sure why, any hints to what I can do?
 
Last edited:
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The work done by Force F1 should be the same as the work done by the force F2 such that

F1x1 = F2x2

From this you can find the ratio of F2/F1 and thus get x2
 
I understand your thought processes, however, no mass is given so how, can I use that equation?
 
rock.freak667 said:
The work done by Force F1 should be the same as the work done by the force F2 such that

F1x1 = F2x2

From this you can find the ratio of F2/F1 and thus get x2

I understand your thought processes, however, no mass is given so how, can I use that equation?
 
The change in volume of fluid in one piston is equal to the change in volume of the other.
 

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