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Hydraulic Lift

  • Thread starter BeiW
  • Start date
  • #1
7
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Homework Statement


A hydraulic lift is used to jack a 970 kg car .12m off the floor. The diameter of the output piston is .18 m, and the input force is 270 N.

If the input piston moves .13 m in each stroke, how high does the car move up for each stroke?

2-3. Relevant equations and attempt at solving
Fin=Fout
That becomes pghA=pghA
p = density
g=gravity
h=height
A=area

So I get 270=p(9.8)(.13)(A) which is the input force.
I could set that equal to the output force:
270=p(9.8)(.13)(A1)=(p)(9.8)(h2)(.02545)
.02545 is the area of a circle when I plug in .09 for radius.

The p's would cancel, and I'm left with the variables A1 and h2.

I know I can use the system of equations to solve this, but I want to know if I set up the equations correctly. And also, what do I do with the .12 m (that the car is lifted off the ground)??
 
Last edited:

Answers and Replies

  • #2
64
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i believe the .12m is given so that you can find the initial height of the input piston since they are not in equilibrium
 
  • #3
7
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i believe the .12m is given so that you can find the initial height of the input piston since they are not in equilibrium
In that case, would I add the .12m to the .13 for h1?
 
  • #4
64
0
.12m refers to the car or height of the output piston, the input piston moves by .13m, the distance the car moves is related by the force/piston area ratios of the input and output
 

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