Surface Tension Help?

[Razghar]

The equation F=2sd, where i convert grams to force, in the equation.

I don't quite know how to do algebra, and I saw a similar question like mine, answered by Craig.
If 1 gram of water equals .00983 Newtons, then 2 grams, I would multiply .00983 by 2 of course.

But after that, i don't know how to find out how much force (N) it took to pull the needle out of the water.

Help please?
Thanks

[deleted-71417]

Hi,

Here is the science buddies writeup on determining surface tension:

https://www.sciencebuddies.org/mentorin ... p012.shtml

Try reading this writeup very carefully to see if it answers your question. I am having trouble understanding exactly what you are trying to ask. The writeup explains the equation fairly well, but does assume you know how to rearrange the equation to solve for the surface tension..

So if F=2sd, if you solve for s the result is s=F/(2d), where s is the surface tension, F is the force needed to pull the pin free, and d is the length of the pin, where all units are self consistent, as explained in the writeup.

If I have not answered the question, please ask again in a way that will help us understand what you need.

Good luck with the project and science fair!

Barrett Tomlinson

[Razghar]

Thank you.

If I solve for s, in the equation s=F/(2d), Do i have to multiply 2 by d, or the length of the needle. And how do I find out the Force, or Newtons?

Thanks.

[deleted-71417]

Hi,

If you carefully study step 4. Measureing Surface Tension in the writeup in the link on my first answer, the process of determining F is described step by step. Pay particular attention to steps 4d and 4h.

To compute s you divide F(in Newtons) by two times the length of the pin(in meters).

Barrett Tomlinson

[Razghar]

Thanks.

2 grams of water equals .01966 Newtons.

I wanted to solve for s, so I found out how many meters the needle was.

I then did: F (newtons) / length

i got .00333203103448276

Surface tension is measured in J/m^2. How do i make that number i got into a surface tension measurment?

Thanks.

[kgudger]

According to the writeup mentioned in Barrett Tomlinson's posting, the surface tension is in newtons/meter (N/m). This is the same number as Joules per meter squared (A Newton is a Joule / meter).

I see how you got .01966 N for the force. You need to divide that number by 2 times the length of the needle. If I do the math backwards, that means your needle was 2.95 meters long! I don't think that can be right. I just measured a couple of sewing needles, and they were 4 to 5 cm long. A cm is 1/100 of a meter. If your needle was 4 cm long, and your force is .01966 N, then the surface tension would be:
s = (.01966) / ( 2 * (.04)) which equals 0.24575 N/m.

I found out that the surface tension of water is 72.8 dynes / cm. To convert to N/m, multiply by .001. This means the surface tension of water should be about .0728 N/m. With the force you measured, this number says that the length of your needle was around 13.5 cm long.

If you could post all of your data, I would be happy to help with the equation.
Keith

[Razghar]

Thanks. My needle was 5.8 cm long.

That would be 0.057999999999999996 meters?

If i multiply 2 by that, it equals .116.

Then would I do Newtons / .116?

.01966 / .116 = approximately .16948
Correct? Not .03 or what I had posted before.

Thanks.

[deleted-71417]

Hi,

Your math checks out, although I suggest you pay careful attention to your use of significant figures when presenting your results for your science fair. The way your posts read you should only report 1 or two significant figures.

Another thought: Have you tried to replicate your measurement multiple times? The writeup suggests doing it at least 5 times (more might be better with a measurement this delicate). The average value of multiple measurements should give you a better estimate of the true surface tension value, and the standard deviation would give you an idea of the error in your measurement.

But congratulations, you have already gotten fairly close to a plausible result for a very delicate measurement!

Good luck at the fair!

Barrett Tomlinson