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Archimede's principle

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.00 kg; the mass of the liquid is 1.80 kg. Balance D reads 3.50 kg, and balance E reads 7.50 kg. The volume of block A is 3.80 × 10-3 m3.

    (a) What is the density of the liquid?

    (b) What will each balance read if block A is pulled up out of the liquid?

    For figure see : http://session.masteringphysics.com/problemAsset/1026707/4/yg.13.46.jpg

    2. Relevant equations

    Wdisplaced fluid = density x gV

    3. The attempt at a solution

    Okay.... this has me stumped!

    a) V x density fluid + m block + m fluid = 7.5 kg
    from here density of fluid = 1237kg/m^3

    b) From archimede's principle we find the weight of the displaced fluid = weight of block = 46.1kg

    Then I am unsure of what to do next... assuming what I have done so far is correct :confused:
     
  2. jcsd
  3. May 28, 2010 #2

    collinsmark

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    I think you mean
    V x density fluid + m beaker + m fluid = 7.5 kg

    But anyway, I agree with your 1237 kg/m3 figure. :approve:

    You totally lost me on that. :uhh::rolleyes:

    There's a couple of ways to solve the problem from here. But since this is an Archimedes' principle problem, I suggest using that approach.

    Your goal here is to find the weight (thus mass) of the block. You know the mass/weight of the displaced liquid (you know its volume and density so you can calculate its mass). Then use the measured value of the spring D, as part of the equation.

    When making your equation, consider this. What value would spring D measure if the block and displaced fluid had the same densities (i.e. both had the same mass), such that the block was just floating? How dense/massive does the block need to be before the spring D will measure a positive, non-zero value? (Hint: you need to form your equation such that it represents the difference of something :wink:).
     
  4. May 28, 2010 #3
    Whoops.... yes I meant beaker :redface:

    okay, not sure I am getting this but:

    Wdisplaced fluid + T in spring - Wblock = 0

    46.1 + 3.5 - Wblock = 0

    W block = 49.6kg

    so if the block is removed from the fluid, D would read 49.6kg???

    E would read 1 + 1.8 = 2.8kg
     
  5. May 29, 2010 #4

    collinsmark

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    If the block was floating, D would would read 0 kg. Only when the density of the block becomes greater than the density of the fluid does D read a positive value. D is 3.50 kg, So 3.50 kg = m block minus m displaced fluid. (since you have already calculated the density of the displaced fluid, and you know the volume of the displaced fluid [same as the volume of the block] you can calculate its mass).

    There's another way to solve this problem too, which is useful to double check your answer. The mass measured by balance D plus the mass measured by balance E must be the total mass of everything; the beaker, the block, and the fluid.
     
  6. May 29, 2010 #5
    So I was on the right track.

    I used weight instead of mass tho, change them and it would be right.

    Mdisplaced fluid + T in spring - Mblock = 0

    4.7 + 3.5 -Mblock = 0

    Mblock = 8.2kg

    Thank you so much for your help!

    I can't believe I didn't think of the last way you talk about (to check answer) so simple yet so far from my mind :tongue:
     
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