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Hydraulic Lift

  1. Dec 3, 2004 #1
    In a hydraulic system the piston on the left has a diameter of 4.5 cm and a mass of 1.7 kg. The piston on the right has a diameter of 12 cm and a mass of 2.5 kg. If the density of the fluid is 750 kg/m3, what is the height difference h between the two pistons?
    can someone please help me with this?
    [tex]A_1 = (3.14)(0.045m/2)^2 = 0.000506m^2[/tex]
    [tex]A_2 = (3.14)(0.12m/2)^2 = 0.0036m^2[/tex]
    [tex]P_1 = mg/A_1 = (1.7kg)(9.8N/kg)/0.00506m^2 = 32924.9[/tex]
    [tex]P_2 = mg/A_2 = (2.5kg)(9.8N/kg)/0.0036m^2 = 6805.56[/tex]
    [tex]P_1 - P_2 = (\rho)(h)(g)[/tex]
    [tex]h = (P_1 - P_2)/(\rho)(g)[/tex]
    [tex]h = 3.55m[/tex]
  2. jcsd
  3. Dec 4, 2004 #2

    Andrew Mason

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    What help do you need? You have the right answer.

    I assume that the cylinders are connected at the bottom and the fluid is not flowing through that connection. This means that the pressure at the bottom of each cylinder is the same. Since the pressure at the bottom of each is:

    [tex]P = F/A = (mg + \rho Vg)/A[/tex] where V = volume = hA

    the condition for no flow is:

    [tex]P_1 = P_2[/tex] so:

    [tex]m_1g/A_1 + \rho h_1g = m_2g/A_2 + \rho h_2g[/tex]

    [tex]m_1/A_1 + \rho h_1 = m_2/A_2 + \rho h_2[/tex]

    [tex]m_1/A_1 - m_2/A_2 = \rho (h_2 - h_1)[/tex]

    [tex]\Delta h = (m_1/A_1 - m_2/A_2)/\rho)[/tex]

  4. Dec 4, 2004 #3
    are you sure cause i got the answer wrong so i thought i did something wrong. i think i made a wrong assumetion somewhere
    Last edited: Dec 4, 2004
  5. Dec 5, 2004 #4

    Andrew Mason

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    I was using your values for area of the pistons, but I see that you forgot to multiply by [itex]\pi[/itex]. So divide your answer by 3.14 (1.13 m).

  6. Dec 5, 2004 #5
    oh and i had it setup right too. thanks
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