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The volume of a metal peace with a hole on the inside (weight in air and water given)

  1. Sep 28, 2014 #1
    I've got this problem with my HW. Would appreciate the help. Thanks.

    1. The problem statement, all variables and given/known data

    A metal peace with the density of 10,4 g/cm3, has a hole on the inside. The metal weighs 8,2N in air and 7,1N in water. What is the volume of the hole?

    2. Relevant equations
    V = m / ρ
    m = Fg / g
    Maybe some else I don't know.

    3. The attempt at a solution
    So I have already managed to figure out that: V1 = (8,2N / 10m/s2) / 10400 kg/m3 = 0,0000788m3 = 78,8cm3; Which means the volume of the metal has to be 78,8cm3, since the hole inside is filled with air and there is air around it. Now I am probably suppose to figure out the volume of the meta peace in the water and then subtract the volume of the metal from it, to get the volume of the hole. But how can I calculate the volume of the metal peace in the water? Help?
     
  2. jcsd
  3. Sep 28, 2014 #2

    Orodruin

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    What does Archimedes' principle tell you?
     
  4. Sep 28, 2014 #3
    That any floating object displaces it's own weight of fluid. So technically the equation would be V = (Fg / g) = ρ;
    But that doesn't seem right...
     
  5. Sep 28, 2014 #4

    Orodruin

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    That is a special case for when an object floats and therefore has the same buoyancy as its weight. The more general statement is that the buoyancy force is equal to the weight of the displaced water. An object that sinks displaces a volume of water equal to its own volume.
     
  6. Sep 28, 2014 #5
    Would that mean that: Fb = Fg = ρ * g * V
    V = Fb / (ρ * g) ?

    But that can't be right, because it is smaller than the actual volume of the metal...
     
  7. Sep 28, 2014 #6

    Orodruin

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    What values are you putting in for Fb and rho?
     
  8. Sep 28, 2014 #7
    ρ = 10,4 g/cm3 = 10400kg/m3 (the density of the metal)
    Fb = 7,10N (the weight of the metal in the water)

    Is this not right?
     
  9. Sep 28, 2014 #8

    Orodruin

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    The total weight of the object in the water is the gravitational force on the object minus the buoyant force.

    Archimedes' principle states that the buoyant force is equal to the weight of the displaced *water*.
     
  10. Sep 28, 2014 #9
    So if I'm interpreting this right this means:
    Fb = ρ(water) * g
    Fb = 1000kg/m3 * 10m/s2 = 10000N

    F = Fg - Fb
    F = m * g - Fb
    m = (F + Fb) / g
    m = (7,1N + 10000N) / 10m/s2 = 1000,71kg

    V = m / ρ = 1000,71kg / 10400kg/m3 = 0,0962m3

    Which is not correct.... This is so hard....
     
  11. Sep 28, 2014 #10

    Orodruin

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    You are missing the displaced volume in your computation of Fb. A good practice is to always check that the units in your expressions are correct. The force on the object due to gravity you have from the weight outside of water.
     
  12. Sep 28, 2014 #11
    Fb = ρ * g * V1
    F = Fg - Fb

    F = Fg - ρ*g*V1
    V1 = (Fg - F) / (ρ*g) = (8,2N - 7,1N) / (1000kg/m3 * 10m/s2) = 0,00011m3 = 110cm3

    V= V1 - V2 = 110cm3 - 78,8cm3 = 31,2cm3

    And according to the solutions below 31,2cm3 is right! Thanks kind Sir from the internet. :)
     
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