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The actual problem:

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 on 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?

My attempt: 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 displacement

$\ π (\frac{0.0225m^2}{4}) (1.5m) = π (\frac{0.9m^2}{4})(x)$


My thoughts: My attempt is wrong but I am confused as to why. Since $(A1)(Δx1)=(A2)(Δx2)$ , I thought that if I used the depressed distance of 1.5m for $Δx1$, then I would be able to find the hight of the car. I understand the diameter of the pipe on the left is very small and unrealistic if you ask me (typical text book problem).

My question

Any hints to how I can look at this problem differently? Maybe I am looking at the problem wrong. Any hints would be appreciated!

  • $\begingroup$ Where does the 4 in the LHS denominator come from? $\endgroup$
    – DJohnM
    Nov 30, 2013 at 21:04
  • $\begingroup$ x=0.0009375m so what's your question. Volume=Volume or (cross multiply and divide). $\endgroup$ Nov 30, 2013 at 21:14
  • $\begingroup$ @User58220 I realized I made a mistake. Thanks for pointing that out. I copied my work down wrong.the 4 is from the surface area. The answer is still the same though. $\endgroup$
    – monkeyman
    Nov 30, 2013 at 21:25
  • 1
    $\begingroup$ Now, how do the two $4$'s appear? You've already changed the diameter to radius... $\endgroup$
    – DJohnM
    Nov 30, 2013 at 22:43

2 Answers 2


Actually, to the best of my knowledge, your solution is correct. Your logic is also very sound. The displacement is calculated from your formula. Note how the guys that send people to orbit also agree with you. And also, I'm not sure about what is the policy regarding Wikipedia links, but take a look at this article.

More specifically, the following quote:

Therefore, the small piston must be moved a large distance to get the large piston to move significantly.

Would you be able to perhaps post the solution from your book, so we can at least have a clue as to what the author's result may have been?

  • $\begingroup$ Unfortunately, I do not have a solution for this problem. But I do now believe my answer is right since it follows the logic. My professor may have marked my answer to this solution wrong by mistake. Thank you for taking a closer look at this problem. $\endgroup$
    – monkeyman
    Nov 30, 2013 at 21:27
  • $\begingroup$ @monkeyman Probably. You should check with him, you've done a pretty good job with your solution. $\endgroup$
    – gezibash
    Nov 30, 2013 at 21:36
  • $\begingroup$ It would appear that the OP's solution had two errors that just happened to cancel out. Do profs award marks for luck? $\endgroup$
    – DJohnM
    Nov 30, 2013 at 23:07
  • $\begingroup$ @User58220 I am using formula $(A1)(Δx1)=(A2)(Δx2)$ the 4 is apart of the surface area formula for A1 and A2 $\endgroup$
    – monkeyman
    Dec 1, 2013 at 23:19
  • 1
    $\begingroup$ A "$4$" appears in the area formula for a circle, if the diameter is the measurement used. By dividing the diameter by $2$ before you squared it, you have already divided by $4$. $\endgroup$
    – DJohnM
    Dec 2, 2013 at 1:55

The ratio of large-to-small piston diameter is $1.8:0.045$ or $40:1$

The ratio squared is $1600:1$

Since volume depends on diameter (or radius) squared, and the fluid volume is constant, the piston motion must scale as $1:1600$

Input plunger moves $1.5\text{ m}$, so output moves $1.5/1600$ or $0.0009375\text{ m}$


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