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Surface Tenstion in Water (Needle resting on water)

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A needle can be supported by the suface tension of water.

    Given the density of steel (7900kg/m^3) a needle with diameter 0.25mm, the surface tension of water (0.072 N/m) and acceleration due to gravity (9.8) show that this is reasonable.

    What is the largest diameter needle that could be supported by the surface tension.


    2. Relevant equations

    [tex]
    \gamma=\frac{F}{L} \\
    \gamma L = mg \\
    2 \gamma l = mg
    [/tex]
    That last equation is for the largest possible mass

    3. The attempt at a solution

    So far I have done.
    [tex]
    \gamma L = mg \\
    0.072 × 2.5 × 10^{-4} = 9.8m \\
    1.8×10^{-5}=9.8m
    [/tex]
    But if I carry that through I get some crazy huge number for the mass which cannot be correct. I suspect that L is not the diameter of the needle but the true length but without that given I dont know how to find it. I know what to do or rather how to use all the values given apart from the density of steel so maybe that is relevant to my first issue but not sure how.
     
  2. jcsd
  3. Nov 12, 2013 #2

    CWatters

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    Science Advisor
    Homework Helper

    The length affects both the surface tension force and the mass of the needle so the length should cancel...

    Tension force = 2γ * length
    Mass = density * area * length

    So the force eqn is..

    2γ * length = density * area * length * g

    Length cancel.
     
  4. Nov 12, 2013 #3
    OK thanks. So is it always 2yL or only in the case when its the largest possible mass?
     
  5. Nov 12, 2013 #4
    If I follow that through, I get confused, see if what I am doing is on the right path or not as its not working out (but I think it is not supposed to - im not sure)

    [tex]
    2\gamma L=\rho ALg \\
    2\gamma=\rho Ag \\
    2×0.072=7900×g×A \\
    0.144=77420A \\
    A=\frac{0.144}{77420}=1.86×10^{-6}
    \\
    \\
    A=\pi r^2 \\
    r=\sqrt{\frac{A}{\pi}}\\
    r=\sqrt{\frac{1.86×10^{-6}}{\pi}}=7.69× 10^{-4}m = 0.77mm
    [/tex]

    And the diameter would be 1.54mm when it is only supposed to be 0.25mm.

    I am really confused.
     
  6. Nov 13, 2013 #5

    CWatters

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    Science Advisor
    Homework Helper

    It looks like a two part question to me..

    1) Show that surface tension can support a needle of diameter 0.25mm (eg show surface tension is greater than force due to gravity for that size).
    2) Then calculate the maximum diameter that can be supported.
     
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