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Pascals principle and hydraulic lift

  1. Feb 8, 2012 #1
    says that the pressure is distributed evenly around a fluid.

    What I'm having a hard time understanding is the hydraulic lift, so I thought I would think of a simple system that could represent a fluid and then use that for intuition.
    Regrettably I did not get the same conclusions with that system, which is attached on the picture below.

    The idea is that we have a piston with one area A1 connected via a rod to a piston with a larger area A2=3A1, and we want to figure out the force we have to exert on the other piston to keep the system in equilibrium.

    It is not hard to see, that the force you have to exert one both pistons has to be the same. But unfortunately that is not the conclusion from using Pascals principle on a fluid, which tells us that the two pressures have to be equal, which in terms of forces would mean:

    F2 = F1*A2/A1

    I'm really having a hard time getting intuition for this. Why is it that the fluid, on which we uses Pascals principle can't just be replaced with a rigid rod, like in my system?

    Attached Files:

  2. jcsd
  3. Feb 8, 2012 #2


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    hi zezima1! :smile:

    if the rod isn't there, and the liquid is in equilibrium, then the forces on the left must equal the forces on the right

    but you have to include the forces on the upper and lower part of the left, not just the centre part :wink:

    (alternatively, if you consider only the central tube of fluid, the forces are still the same!)
  4. Feb 8, 2012 #3
    Yes okay, I sort of got to that conclusion after thinking about it. But I also think, I now know more clearly what puzzles me:
    It is how you can exert a force and get back a force much bigger. I know you can't call force a conserved quantity but it just doesn't fit with my notion of Newtons 3rd law.
  5. Feb 8, 2012 #4


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    hi zezima1! :smile:
    same as with a lever :wink:
    Newtons 3rd law is local

    eg, on a lever, it applies between each bit of the lever and the next bit, not from one end to the other

    Newtons 3rd law does apply between any two colliding molecules of the fluid :smile:
  6. Feb 8, 2012 #5
    Okay yes, I have seen levers and all such, so I guess you're right.

    But could you try to make the lever principle more intuitive, and explain how exactly it applies here? :)
  7. Feb 8, 2012 #6
    Hello zezima,

    To understand this in terms of a conservation law (conservation of energy) try this:

    In your diagram

    The volume of fluid displaced when you push piston1 = Distance travelled by piston x Area = d1A1.

    This fluid must be the same volume as 'pushed ' into the second chamber to move the second piston and so = d2A2

    Thus d1A1 = d2A2

    since in your example A2 = 3A1

    d1 = 3d2 or d2 = d1/3 (note the 3 factor is the other way round)

    Now work (energy) = force times distance and F2 = 3 F1

    So F2d2 = 3F1 * d1/3 = F1d1

    That is energy is conserved.

    go well
  8. Feb 8, 2012 #7


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    are you familiar with the principle of work done?

    W = F.d (work = force . displacement) …

    since work in = work out, the product F.d must be the same at both ends

    eg, if one end of the lever is displaced twice as much as the other end, then the force at the other end must be half as much :smile:

    similarly, when the tube gets narrower, the piston must move further, and so the force must be less :wink:
  9. Feb 8, 2012 #8
    ahh yes, I think I get what you mean :)

    But what if the liquid was confined? Then doesn't the principle still hold? And can you apply conservation of energy then?
  10. Feb 8, 2012 #9
    Can you now see that with a fluid each piston travels a different distance,

    Whilst with the rod connecting them in your original question each piston travel the same distance?

    BTW Hello, Tim
  11. Feb 8, 2012 #10


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    Hello, Studiot! :smile:

    Two pistons! :biggrin:
  12. Feb 8, 2012 #11
    Yeah I'm two pistol Pete.

  13. Feb 8, 2012 #12
    Okay guys I understand it now, in the case, where you can apply conservation of energy =)

    But.. What about if I had the system like on the picture:
    The pressure is equal so the system must be in equilibrium. But surely you now can't proof that the two are in equilbrium using work and energy since, there is no work done. How can you explain this?

    Attached Files:

  14. Feb 8, 2012 #13
    In equilibrium situations we usually find it more convenient to work in terms of the balance of forces (and moments) being zero as the conditions for equilibrium.

    Can you do this in this situation?

    Hint apply Newton's third law to each of the masses.

    I can, however, tell you that there are energy theorems that are not on your syllabus which you could use but it would be a sledgehammer to crack a nut.

    So we use the easy way.

  15. Feb 8, 2012 #14
    I can try. Since both masses are at rest, the upwards force from the fluid must be mg and 2mg. But since the area is twice as big for the other we actually do get:
    P1 = P2
    But that is under the assumption, that the two masses are initially at rest, when you put them on them piston simultaneously. And why can you assume that?
    Tell me the name of these - I know a bit about lagrangian/hamiltonian mechanics..
  16. Feb 8, 2012 #15
    Try this

    Let the 2m mass suffer a virtual downward displacement δy so that the m mass suffers an upward virtual displacement βδy, where is a parameter to be determined

    Then the total virual work done equals zero in equilibrium

    mgβ-(δy) + 2mgδy = 0

    β = 2
  17. Feb 8, 2012 #16
    virtual? :eek: I know there's something called virtual work in more advanced mechanics but I don't know what it is sorry. I just meant that I have heard about analytical mechanics, not that I have any idea how to apply it.
    Can't you explain it in terms of newtonian? Or was that what you were trying to? :)

    But I certainly can see the idea of what you try to say with the virtual work, and while the idea is neat, it to me seems like assuming that out of pure statistical fluctuation the one mass goes down a little bit, which is not really something that would happen in the mechanics I know - though I might be wrong since I don't know what you mean by virtual displacement.

    If this can't be rigourosly explained with Newtonian mechanics please tell me.
  18. Feb 9, 2012 #17
    Studiot :)

    can you please in terms of forces try to describe how the piston gets acted on by double the force as the other, when initially you push down on the piston with area A. Please don't use energy conservation :(
  19. Feb 9, 2012 #18
    I'm not quite sure what you mean.

    I assume you are referring to the hydraulic lift in post#12.

    So the situation is that the piston with mass m and area A exerts a downward force on the fluid =mg.
    In exerting this downward force the piston creates a pressure =mg/A in the fluid.

    By Newton's third law the fluid therefore exerts an equal upward force on the piston.

    Similarly the downward force at the second piston is 2mg and the upward force equals this.

    Now let us push down on the smaller piston with and additional force F.

    The piston will move down in accordance with Newton's first and second laws. You are now exerting a force mg + 2F on the fluid

    So the pressure created in the fluid now equals (mg+2F)/A = mg/A + F/A.

    I have split the pressure into the original pressure and the increase (F/A) due to the extra force you are now exerting.

    This extra pressure pushes up on the second piston with a extra force equal to (F/A) * (2A) = 2F

    The second piston will therfore move upwards in accordance with Newton's first and second laws since it now experiences a net upward force of 2F.

    So long as you maintain the downward foce F on the first piston it will continue to move down and the second piston will continue to move up.

    Does this help?
  20. Feb 9, 2012 #19
    hmm no I just find it hard to see how you can exert an external force and get a force bigger from that through internal action. I know that it's possible, but I've always found that hard to understand. The most simple example is a pulley in which you have an extra rope. The conservation of energy is still true since you are just pulling over twice the length.
    So something equal to that must be applyable here. I want to know how in terms of forces the individual particles in the column with the smallet area pushes on the other particles such that their net force on the bigger area is twice the external force
    hope I made myself clear now :) and thank you
  21. Feb 9, 2012 #20
    Yes, I thought we has already covered that.

    The small piston moves twice as far as the big one.
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