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Surface tension question

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data
    First part of the question was "Why is it possible to float a wax-coated stainless steel needle on the surface of water?" and I've already answered that, but I'm having trouble with the second part of the question, which is:

    If the surface tension at the interface is 0.073 N/m and the density of the steel needle is 7.8x10^3 kg/m^3, what is the maximum diameter of a needle that can float this way? (Hint: start by finding the weight of the needle. You may assume that the needle is cylindrical in cross-section)


    2. Relevant equations
    Density = mass / volume
    h = (2[tex]\sigma[/tex]cos[tex]\Psi[/tex]) / ([tex]\rho[/tex]g[tex]\alpha[/tex])
    V of needle = width x (2[tex]\Pi[/tex]r^2)
    I'm actually not too sure which equations I need...



    3. The attempt at a solution
    [tex]\sigma[/tex] = 0.073N/m
    [tex]\rho[/tex] of stainless steel needle = 7.8x10^3 kg/m^3
    Well it says to start by finding the weight of the needle, and I wasn't sure how to approach that in the first place, so I wasn't able to make a start.
     
  2. jcsd
  3. Jul 23, 2010 #2
  4. Jul 23, 2010 #3
    [tex]\Sigma[/tex]F = 0
    -W + ([tex]\sigma[/tex]L)cos[tex]\vartheta[/tex] + ([tex]\sigma[/tex]L)cos[tex]\vartheta[/tex] = 0
    W = 2([tex]\sigma[/tex]L)cos[tex]\vartheta[/tex]
    W = 2 x 0.073 x L x 1
    W = 0.146L N

    Is this correct?
    I have absolutely no idea what to do next.
    Just guessing, please correct me if I'm wrong...
    Mass of needle = 0.146L/9.8 kg
    Volume of needle = ([tex]\Pi[/tex]r^2)*L
    Volume of needle = 1/([tex]\rho[/tex]m) = 1/(7.8x10^3 x 0.015L) = 8.61x10^-3L m^3
    So 8.61x10^-3L = ([tex]\Pi[/tex]r^2)*L
    8.61x10^-3 = [tex]\Pi[/tex]r^2
    r = 0.05 m

    I've found a radius of some sort, but I don't think it's right.
    The question says I need to find the maximum diameter of a needle that can float.
    How do I find this???
     
    Last edited: Jul 23, 2010
  5. Jul 23, 2010 #4
    Hold on...
    I have another theory! Please tell me which one is wrong or right or both are wrong! :S
    [tex]\Sigma[/tex]F = 0
    -W + ([tex]\sigma[/tex]Lcos[tex]\theta[/tex]) + ([tex]\sigma[/tex]Lcos[tex]\theta[/tex]) = 0
    W = 2([tex]\sigma[/tex]L)cos[tex]\theta[/tex]
    W = 2 x 0.073 x L x 1
    W = 0.146L N

    And using another formula:
    P = (2[tex]\sigma[/tex]) / r
    P = 101325 Pa
    So, r = (2 x 0.073) / 101325
    r = 1.44x10^-6 m

    Or is this formula only used for bubbles?

    anyway, my other answer to this problem being
    d=2r
    d=2.88x10^-6m

    Please tell me where I went wrong and how I should tackle this problem to solve it.
    Please give me directions!
    Thank you!
     
  6. Jul 23, 2010 #5
    [tex]8.61\times 10^-3L = \pi r^2L[/tex]
    You have gone this far and there is just one more minor step to the result :wink: L is canceled as it appears in both sides, so there is only 1 unknown left :biggrin: By the way, please check your numerical values, because I'm too lazy to check them :biggrin:

    So there is no need for the weight, right? :yuck:
    The effect is due to surface tension, not ambient pressure.

    EDIT: I've just read again your post and found you made some corrections. You were on the right track at post #3. Now I'll organize it a bit:
    [tex]W=2\sigma Lcos\theta[/tex]
    [tex]\rho g \pi r^2L=2\sigma Lcos\theta[/tex]
    [tex]r=\sqrt{\frac{2\sigma cos\theta}{\rho g \pi}}[/tex]
    As you can see, r is max when cos(theta) is max, and max of cos(theta) is 1. You will find that [tex]r_{max} = 0.78mm[/tex], not so large as 0.05m that you got earlier.
     
    Last edited: Jul 23, 2010
  7. Jul 23, 2010 #6
    Thank you
     
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