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Buoyant Forces

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    The "spirit-in-glass thermometer", invented in Florence, Italy, around 1654, consists of a tube of liquid (the spirit) containing a number of submerged glass spheres with slightly different masses (see the figure below). At sufficiently low temperatures all the spheres float, but as the temperature rises, the spheres sink one after the other. The device is a crude but interesting tool for measuring temperature. Suppose that the tube is filled with ethyl alcohol, whose density is 0.78945 g/cm3 at 20.0° and decreases to 0.78097 g/cm3 at 30.0°C.

    If one of the spheres has a radius of 1.400 cm and is in equilibrium halfway up the tube at 20.0°C, determine its mass?

    When the temperature increases to 30.0°C, what mass must a second sphere of the same radius have in order to be in equilibrium at the halfway point?

    At 30.0°C the first sphere has fallen to the bottom of the tube. What upward force does the bottom of the tube exert on this sphere
    2. Relevant equations
    buoyant force = mg

    3. The attempt at a solution
    So I go the first 2 correct.

    For the first one I had: M_sphere*g = p_alcohol_at_20 * g * 4/3 pi r^3
    solving for M_sphere, I got 9.07 grams

    For the second one I had M_sphere2*g = p_alcohol_at_30 * g * 4/3 pi r^3
    solving for M_sphere2, I got 8.98 grams

    For the last part, I set up this equation:
    buoyant force + normal force = mg_sphere
    p_alcohol_at_30*V_sphere*g + N = M_sphere*g
    But this equation has 2 unknowns, N (the normal force) and p_sphere, the spheres density. I'm not even a 100% sure if 1.4cm is the radius of this sphere so I may not even know the volume meaning 3 unknowns... What am I missing?
    Thanks in advance
     
  2. jcsd
  3. Nov 3, 2014 #2

    BvU

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    With 'first sphere', they mean the one for which you calculated the mass of 9.07 grams. You also have the volume for it.
     
  4. Nov 3, 2014 #3

    andrevdh

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  5. Nov 3, 2014 #4

    BvU

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    No reason to expect that. And no need:
    Just write a force balance for the sphere on the bottom.
     
  6. Nov 3, 2014 #5

    andrevdh

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    Ok. I was referring to the problem where it sat halfway up in the liquid.
    Abid did set a force equation up for the last problem though.
     
  7. Nov 3, 2014 #6
    Ok guys after some experimentation I found what was wrong. When I found the force, I found my units had grams, and the answer wanted Newtons which has kilograms... :rolleyes: So I set up the force equation and then divided the result by 1000 and the answer worked. Thank you guys!
     
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