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Another Manometer question

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine the pressure required to make the top of the oil, with height H0=443 mm even with the hight of mercury. The oil has relative density.920 and mercury has relative density13.55. To which side is the pressure applied?

    2. Relevant equations
    Code (Text):

    |      o----
    |      o
    |      o        ] 443mm
    |      o
    o      o----
    o      o
    o are mercury and | is oil

    3. The attempt at a solution

    my question is how the helllllll do i know what its relative too... and how would i go about balencing them
  2. jcsd
  3. Nov 11, 2007 #2


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    There is a pressure being applied to one tube of the manometer. The pressure has to suffice to balance the difference in the weights of the contents of the two tubes. Think about force balance at the bottom of the manometer.
  4. Nov 11, 2007 #3
    but i dont know the force being applied at the top am i crazy or are they too many unknowns?
  5. Nov 11, 2007 #4
    The only unknown is the answer
  6. Nov 11, 2007 #5


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    Pressure is force per unit area. The mg force it needs to balance is g*density*volume=g*density*length*area. You can put in symbolic variables for area and length. You will find in the end you don't need to know them because they will cancel.
  7. Nov 12, 2007 #6
    hmmm maybe this might help better than my sketchy diagram(see attachment)....

    i still dont know how to calculate

    but how do i calculate "rho" if i only have the relative density

    Attached Files:

  8. Nov 12, 2007 #7


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    Relative density is referring to a multiple of the density of water. It's not all that 'relative'. It's usually called 'specific gravity'.
  9. Nov 12, 2007 #8
    Good question! I guess they mean "specific gravity" or the question cannot be answered. I'm sorry -- I translated "relative density" into "specific gravity" when reading the question.
  10. Nov 12, 2007 #9
    ok so pgh-Patm = P?

    please help me its 3am here... i just need someone to actually explain how to get to the answer, because my teacher is a dick. and i hate university with a passion
  11. Nov 12, 2007 #10


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    Weight/area is rho*g*h, if that's what you mean. And don't worry about Patm - you are just worried about a pressure difference. And the pressure difference has to support the weight difference between the two tubes.
  12. Nov 12, 2007 #11
    Careful -- look who's helping you!
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