# Surface tension question

1. Jul 23, 2010

### hcho88

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$$\sigma$$cos$$\Psi$$) / ($$\rho$$g$$\alpha$$)
V of needle = width x (2$$\Pi$$r^2)
I'm actually not too sure which equations I need...

3. 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.

2. Jul 23, 2010

### hikaru1221

3. Jul 23, 2010

### hcho88

$$\Sigma$$F = 0
-W + ($$\sigma$$L)cos$$\vartheta$$ + ($$\sigma$$L)cos$$\vartheta$$ = 0
W = 2($$\sigma$$L)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 = ($$\Pi$$r^2)*L
Volume of needle = 1/($$\rho$$m) = 1/(7.8x10^3 x 0.015L) = 8.61x10^-3L m^3
So 8.61x10^-3L = ($$\Pi$$r^2)*L
8.61x10^-3 = $$\Pi$$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
4. Jul 23, 2010

### hcho88

Hold on...
I have another theory! Please tell me which one is wrong or right or both are wrong! :S
$$\Sigma$$F = 0
-W + ($$\sigma$$Lcos$$\theta$$) + ($$\sigma$$Lcos$$\theta$$) = 0
W = 2($$\sigma$$L)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.
Thank you!

5. Jul 23, 2010

### 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

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:
$$W=2\sigma Lcos\theta$$
$$\rho g \pi r^2L=2\sigma Lcos\theta$$
$$r=\sqrt{\frac{2\sigma cos\theta}{\rho g \pi}}$$
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.

Last edited: Jul 23, 2010
6. Jul 23, 2010

Thank you