Maximum Floating Needle Diameter Calculation

[hcho88]

Homework Statement

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)

Homework Equations

Density = mass / volume
h = (2\sigmacos\Psi) / (\rhog\alpha)
V of needle = width x (2\Pir^2)
I'm actually not too sure which equations I need...

The Attempt at a Solution

\sigma = 0.073N/m
\rho 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.

 

[hikaru1221]

Have a look at this site: http://en.wikipedia.org/wiki/Surface_tension#Basic_physics See the first picture and you may know what's next to do

 

[hcho88]

Have a look at this site: http://en.wikipedia.org/wiki/Surface_tension#Basic_physics See the first picture and you may know what's next to do

\SigmaF = 0
-W + (\sigmaL)cos\vartheta + (\sigmaL)cos\vartheta = 0
W = 2(\sigmaL)cos\vartheta
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 = (\Pir^2)*L
Volume of needle = 1/(\rhom) = 1/(7.8x10^3 x 0.015L) = 8.61x10^-3L m^3
So 8.61x10^-3L = (\Pir^2)*L
8.61x10^-3 = \Pir^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?

 

[hcho88]

Hold on...
I have another theory! Please tell me which one is wrong or right or both are wrong! :S
\SigmaF = 0
-W + (\sigmaLcos\theta) + (\sigmaLcos\theta) = 0
W = 2(\sigmaL)cos\theta
W = 2 x 0.073 x L x 1
W = 0.146L N

And using another formula:
P = (2\sigma) / 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!

 

[hikaru1221]

8.61\times 10^-3L = \pi r^2L
You have gone this far and there is just one more minor step to the result L is canceled as it appears in both sides, so there is only 1 unknown left By the way, please check your numerical values, because I'm too lazy to check them

P = (2LaTeX Code: \\sigma ) / r
P = 101325 Pa
So, r = (2 x 0.073) / 101325
r = 1.44x10^-6 m

So there is no need for the weight, right?
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:
W=2\sigma Lcos\theta
\rho g \pi r^2L=2\sigma Lcos\theta
r=\sqrt{\frac{2\sigma cos\theta}{\rho g \pi}}
As you can see, r is max when cos(theta) is max, and max of cos(theta) is 1. You will find that r_{max} = 0.78mm, not so large as 0.05m that you got earlier.

 

[hcho88]

Thank you