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Fluid mechanics mercury problem

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Mercury is poured into a U-tube as in Figure P15.18a. [see the picture here: http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch15/Images/P15.18.gif] [Broken] The left arm of the tube has a cross-sectional area of A1 = 10.0 cm2, and the right arm has a cross-sectional area of A2 = 5.00 cm2. One hundred grams of water are then poured into the right arm as in Figure P15.18b.

    (a) Determine the length of the water column in the right arm of the U-tube.

    (b) Given that the density of mercury is 13.6 g/cm3, what distance, h, does the mercury rise in the left arm?

    2. Relevant equations

    [tex]\rho*gh_{i} = \rho*gh_{f}[/tex] (at same level)
    [tex]\rho = \frac{m}{v}[/tex]

    3. The attempt at a solution

    I figured out a):
    [tex]V = Ah[/tex]
    [tex]\rho = \frac{m}{v}[/tex]
    [tex]\rho = \frac{m}{Ah}[/tex]
    [tex] 1000 = \frac{0.1}{(5*10^{-4}*h}[/tex]
    [tex]h = 0.2 m = 20 cm[/tex]

    I couldn't do part b), so I finally went to Google the problem and found a solution on this website: http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch15/Ch15.html

    Their solution says this:

    [tex]\rho*gl_{1} = \rho*gl_{2}[/tex]
    This equation is taken at the point where the water on the right hand side meets the mercury as shown in the picture on the site. 1s is the distance from the top of the mercury on the left hand side to this point and l2 is the distance from the top of the water to that point.
    [tex](13.6)(9.8)(l_{1}) = (1)(20)(9.8)[/tex]
    [tex]l_{1} = 1.47 cm[/tex]

    I understand this part. This is the part of the solution I do not understand:

    [tex]A_{1}h = A_{2}(1.47 - h)[/tex]
    [tex]10h = 5(1.47 - h)[/tex]
    [tex]h = 0.49 cm[/tex]

    Basically I don't understand why you would know to set the two volumes equal to each other? How do you know the volumes are the same?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 25, 2009 #2


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    Homework Helper

    The water on one side must have the same weight or mass as the Mercury on the other side (above the bottom of the water in both cases).
    mass on left = mass on right
    pV = 100 g
    pAh = 100 g
    solve for h. Looks like a little more than 0.7 cm.
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