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Density of bar submerged in water

  1. Oct 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A bar suspended in air weighs 1.75 N
    The same bar weigs only 1.4 N when suspended in water.
    Calculate;
    a. upthrust
    b. density of the bar

    2. Relevant equations
    D = m/v
    F = m*a

    3. The attempt at a solution
    a
    .
    Upthrust is 1.75 N - 1.4 N = 0.35 N

    b.
    Assuming density of water is 1x103 kg m-3
    and weight of bar displaces that amount of water.
    So
    volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3

    Density of bar = m/v = 1.75 N / 1.78 m3 = 0.98 N m-3

    which has to be wrong, surely, or the bar would float . . .
    Help!
    Thank you.
     
  2. jcsd
  3. Oct 29, 2015 #2
    Remember the difference between weight and mass.
     
  4. Oct 29, 2015 #3

    SteamKing

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    Homework Helper

    It's not clear how you got 1.75 N / (9810) = 1.78 m3

    Your calculator must have suffered a nervous breakdown.

    Remember, a cubic meter of water weighs a tonne (1000 kg), literally.

    As Dr. Courtney pointed out, density is mass divided by volume, not weight divided by volume. You can't use newtons and kilograms interchangeably.
     
  5. Oct 30, 2015 #4

    PhanthomJay

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    Gold Member

    yes, and this called the buoyant force
    you are not applying Archimedes Principle correctly. The upward thrust or buoyant force is equal to the weight of water displaced. That weight of displaced water is not 1.75 N.
     
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