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Homework Help: Buoyant Force of sand and steel

  1. Jan 28, 2012 #1
    I am just not too sure how to start this equation. The answer was given (i.e. Vc = 3698 cm^3), but I just need to know the steps on how to get to the answer.

    A sand core used to form the internal surfaces of a steel casting experiences a buoyancy force of 225.63 N. What is the volume of the sand core in cm cubed.

    Steel Density: 7.82 g/cm^3
    Sand Core Density: 1.6 g/cm^3

    Fb = Wm - Wc
    Wc = (Mc)(Gravity) & Wm = (Mm)(Gravity)
    D = M/V

    I have tried rearranging the equation: 225.63 N = Wm - Wc --> Wm = 225.63 + Wc
    Then substituting it into the equation: Wm = (Mm)(Gravity) --> 225.62 + Wc = (Mm)(Gravity)
    I then isolate the... I pretty much get lost after that.

    Thanks for the help
  2. jcsd
  3. Jan 28, 2012 #2

    Filip Larsen

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    Gold Member

    Welcome to PF!

    First, the definition is that the buoyancy force equals mass of object (sand core) minus mass of displaced fluid (steel), that is, the negative of what you wrote.

    Secondly, from that definition you can see that you also need to use the fact, that the mass of the displaced fluid is calculated from the volume of the object, that is with your terms you should use that Wc = Mcg = VDcg and Wm = VDmg. It should now be possible for you to relate the buoyancy force with the volume V, the difference in density Dm-Dc and the acceleration of gravity g and isolate for V.
  4. Jan 28, 2012 #3

    I substituted the equations for Wm and Wc into the Fb formula:
    225.63 N = (V * 7.82 g/cm^3 * 9.8 m/s^2) - (V * 1.6 g/cm^3 * 9.8 m/s^2)
    I came up with the answer 3.7 m^3. Can anyone confirm if this is right or whether I just got a lucky answer close to the real one? The answer I was given was 3698 cm^3.
  5. Jan 28, 2012 #4


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    Staff: Mentor

    Your method is okay. Keep a few more decimal places and use a more accurate value for g and you'll be good. g = 9.807 m/s2 is a good to three decimal places.
  6. Jan 28, 2012 #5

    Filip Larsen

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    Gold Member

    Looks good to me. Whoever calculated the answer of 3698 cm3 probably used a value of g around 9.81 m/s2 and rounded of the volume to nearest cm3.
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