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Fluid Statics in U-Tubes

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Mercury is poured into a U-tube. The left arm has a cross-sectional area A1=.001 m and the right has A2=.0005m. One hundred grams of water are then poured into the right arm. Determine the length of the water column in the right arm of the U-tube; given the density of mercury is 13.6 g/cm^3, what distance h does the mercury rise in the left arm?

    2. Relevant equations

    Just volumes of cylinder and pressure variations with depth:

    [tex]P=P_o + ρgh[/tex]

    3. The attempt at a solution

    I’m not sure how to deal with this problem. I found the height in the column fairly easily since we know the density and mass and cross sectional area, it’s pretty easy to solve for h of the water column, it is .2 meters.

    I’m not sure how to set up equations to solve for the distance the mercury has risen though. I don’t know how to apply the principles of static fluids here since there are two different types of fluid.

    Pressures are not equal at equal heights I presume. Both surfaces of the water and the mercury must be at the same pressure, atmospheric pressure, yet are at different heights.

    However I do think the pressures at points A and B (labeled in the crude scetch) are equal because below the horizontal line I draw there is only mercury.

    Attached Files:

  2. jcsd
  3. Jan 31, 2012 #2


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    Hi AdkinsJr! :smile:
    that's right! :smile:

    now combine that with :wink:
    (or alternatively, the principle that if you start at the bottom, and deduce the pressure at the top of each tube, they have to come out the same)
  4. Jan 31, 2012 #3
    Ok; I think I'm making progress here. I just need to create another length "d" in the drawing and express the height of the mercury column as a sum h+d...


    [tex]P_A=P_{atm}+ρ_{h_2O}g(.2 m)[/tex]


    The three equations above give:

    [tex]ρ_{h_2O}(.2 m )=ρ_{Hg}(h+d)[/tex]

    So the 2 unknowns are h and d. So if I just come up with another equation with them it should get things moving.

    I think the weights of the columns have to be equal (above B for the mercury and above A for the water). So the mass of the mercury above the point B should be the same as the mass of the water poured in. I have the mass and the density, and thus the volume of mercury above B... which I can write as:


    Since the areas are given I think this should work; two equations two unknowns.

    Attached Files:

  5. Jan 31, 2012 #4
    Actually I think I'm wrong about the weights; they shouldn't be equal because of that principal of the hydraulic press...different cross sectional areas.

    The pressures are equal, so I'll have to say that F1/A1=F2/A2 and this should give the mass of the mercury column. Then I can find that volume and use V=(h+d)A1
  6. Jan 31, 2012 #5


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    that looks better! :smile:
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