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Homework Help: Pressure/force hydraulic lift problem

  1. Oct 4, 2005 #1
    A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 630 cm2. It is filled with oil of density 690 kg/m3.

    I found that about 63.5 kg of mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels.

    But then it asks...

    With the lift in balance with equal fluid levels, a person of mass 100 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?

    I dont know how to approach this one....any help would be great!
  2. jcsd
  3. Oct 4, 2005 #2


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    So a 100 kg person gets in a car of mass 1600 kg, so that mass becomes 1700 kg.

    The system was at a equilibrium but now the larger mass increases, so it must drop.

    Besides the mass on the smaller piston, what other mass is available to maintain the system into equilibrium?

    Think statics.
  4. Oct 4, 2005 #3
    well, i guess my problem is first of all what it is really asking...

    is it asking for the change in height of the piston (fluid) when the person get into the car?

    b/c that doesnt make sense since that's the next part of the problem...
    Last edited: Oct 4, 2005
  5. Oct 4, 2005 #4


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    As I understand what has been written, the first part of the problem wants you to find the mass (weight or force) on the smaller piston (25 cm2) which will equilibrate with the mass of the car (1600 kg) on the larger piston (630 cm2) at the same height.

    Basically this is a statics problem. How did you solve for the initial mass of 63.5 kg?

    In the second part, a person (100 kg) is added to the car for a total mass of 1700 kg. The large piston will drop and the smaller piston will rise in order to establish a new equilibrium.

    So what is the new height if one does not add additional mass to the small piston?

    Remember the pressure in the fluid must be the same at the same elevation in a static situation.

    Also remember Force = Pressure x Area or conversely Pressure = Force/Area.
  6. Oct 4, 2005 #5
    Here's how i got 63.5 kg in the very first part:

    we are in an enclosed cylinder...(well hydraulic lift), if i push down on piston 1 with
    F(1), it increases the pressure in that cylinder by :

    change in pressure = F(1)/ A(1) Eqa. 1

    by pascal's principle, the pressure in cylinder 2 increases by same amount, so the increased force F(2) = change in pressure x A(2)

    by substituting equ. 1 into the above, and so on...i have the equation:

    F(2) = { F(1) / A(1) } x A(2), ...i used .25 m^2 as my first piston, 6.3 m^2 as my second piston area.

    the F = mg, so i plug in everything and solve for m(1)

    do i used the sort-maybe same approach to solve for the second part?
    im not sure where to start...
    Last edited: Oct 5, 2005
  7. Oct 5, 2005 #6
    I've been told to do this :

    deltaP = rho*g*h

    solve for h

    (P=F/A, so solve for F/A on each side. The difference in these two is deltaP, then divide by rho*g)

    i tried that didnt work, or is it right but im just doing it wrong? :cry:
    Last edited: Oct 5, 2005
  8. Oct 6, 2005 #7
    ok, after much try, i got total height difference to be 2.3 m. now it's asking:

    How much did the height of the car drop when the person got in the car?

    which i think it's asking how much of that height difference is attributable to either side.

    Because the areas of the sides are different, each will contribute differently..

    i kinda got the following equation, h = H x {A(1) / A(2)}

    where H is the height difference and solve for h....

    except it's not correct, is it the right approach though?
  9. Oct 6, 2005 #8


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    The height, h, of 2.3 m is correct.

    Now to solve for H, the drop in height of the large piston, one may assume an incompressible fluid, and use conservation of mass, which implies conservation of volume.

    The volume displaced by H, H*A (where A is large piston area), must equal the volume displaced by h in the smaller piston of area, a. However, remember the starting point is at equilibrium, which means that while the small piston is not displaced by h above the equilibrium point, but by h-H. The height h is about the large piston which has fallen by H.

    The change volume is measure with respect to the equilibrium elevation.
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