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Hydraulic lift inquiry big problem

  • Thread starter mrjoe2
  • Start date
  • #1
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Hi all,

I have quite a perfect understanding of hydraulic lift, yet there is one huge problem that is contradicting my understanding. for a hydraulic lift where you push on on side of diameter d1 piston, and the object on a piston of diameter d2, you get the equation:

F1/A1 = F2/A2 just to support the object (car) without lifting it. but when you want to lift the car a certain height, you get F1/A1 = F2/A2 + pgh where p is the density of the fluid. this all makes sense.

but why when you do calculations to find the change in pressure force to lift the car a certain height h, it doesnt work. what im doing is:

F1' = A1(F2/A2 + pgh) minus F1 = A1(F2/A2) to get the change in pressure force to life the car a certain height. so Delta F = A1pgh right?
But subbing in values does not get the correct change in pressure. they use some question DeltaF=pg(A1 + A2)h which is something i have never seen before. I dont get the flaw in my logic, please someone help im about to cry.
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi mrjoe2,

Hi all,

I have quite a perfect understanding of hydraulic lift, yet there is one huge problem that is contradicting my understanding. for a hydraulic lift where you push on on side of diameter d1 piston, and the object on a piston of diameter d2, you get the equation:

F1/A1 = F2/A2 just to support the object (car) without lifting it. but when you want to lift the car a certain height, you get F1/A1 = F2/A2 + pgh where p is the density of the fluid. this all makes sense.

but why when you do calculations to find the change in pressure force to lift the car a certain height h, it doesnt work. what im doing is:

F1' = A1(F2/A2 + pgh) minus F1 = A1(F2/A2) to get the change in pressure force to life the car a certain height. so Delta F = A1pgh right?
But subbing in values does not get the correct change in pressure. they use some question DeltaF=pg(A1 + A2)h which is something i have never seen before. I dont get the flaw in my logic, please someone help im about to cry.
It's not so much a flaw in your logic; it appears that you are just not considering the correct height to use in your equation.

For example, to start, let the car be initially only supported by the hydraulic lift, with the pistons at the same vertical level. The car is sitting on the large piston with area A, and the applied force is at the small piston with area a.

Now you want to raise the car up a height H. But here to make the large piston go up a height H, the small piston is going to go down a height h. (And so the height in your original formula is the sum of these heights.) In other words the force difference is given by:

[tex]
\Delta F = a \rho g (H+h)
[/tex]

You next can get eliminate h in terms of H. When the large piston rises, it is because a certain amount of fluid is moving to that side of the lift; that same volume of fluid is leaving the side of the lift where the small piston moves down. By setting those volumes equal you can eliminate h and get the equation you have in your post (which in the variables I am using is:

[tex]
\Delta F = \rho g (a+A) H
[/tex]

Do you get that equation?
 

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