surface tension calculation

[MTR1997]

I read the previous posts. I have conducted the experiment. I have done my research. The sites I found say the surface tension of water is 72 dynes/cm and 77 degrees F. My numbers are no where near that. What I have done wrong? Is there a different scale for the equation F=2sd? Is the number generated using this formual Newton meters or milliNewton meters? Is there a conversion scale to show dynes/cm?

I used the equation in the surface tension experiment.

Here is my math.
(Avg. # of pins for 10 trials x weight of one pin) x (gravitational pull number .00983) divided by 2(length of the needle used in meters)=s

For water this looks like (3.1x.0625)x.00983/.0076=.2506

Do I call this .2506 Nm or mNm?

Thank you for helping me understand the math here.

[kgudger]

Hello:

I'll try and help with the math, but I agree it is confusing! First, lets look at the number "72 dynes / cm". In Newtons/meter that is 72 * (.00001 / .01), or .072 N/m. I found another web site, (http://www.engineeringtoolbox.com/water ... d_597.html), from which I approximate the surface tension of water as .072 N/m at 25 degrees C (room temperature).

With regards to your numbers, what is (3.1 x 0.0625)? Is this the "weight"? And how is .0076 related to the length (times 2) of the needle in meters?

I made some assumptions. If I work backwards from the "0.072" number, and assume your needle was about 3 cm long, I get that you should be measuring about 0.44 g on your balance beam scale.
My equation would then be 0.44g * .00983 N/g / (2 * .03m) = 0.0721 N/m

Did you make sure that your balance beam was perfectly balanced before you submerged the needle? In other words, did you "Use a small piece of modeling clay as a counterbalance"?

Please let us know how you got your numbers. Thanks!
Keith

[MTR1997]

Hi
Here is my math.
(Avg. # of pins for 10 trials x weight of one pin=weight) x (gravitational pull number .00983) divided by 2x(length of the needle used in meters)=s

For water this looks like (3.1 (avg. number of pins) x.0625(weight of one pin))x.00983/(2 x .0038)=.2506

I calculated the length of the needle, in meters, incorrectly according to your math.
correcting it gives this equation (3.1pins x.0625g)x.00983/2x.038m=.025N/m

Yes used the counterweight and the straw looked balanced.
Are their any other issues that would affect my results? Would tap water make a difference in this experiment, rather than distilled water?

also I didn't count the last pin like it said in the directions.

thanks for helping me

[deleted-71588]

Would tap water make a difference in this experiment, rather than distilled water?

Yes, it WILL make a difference; however, I can't tell you how much difference your tap water might make. If your apparatus has any dish washing detergent (or other detergent or other surfactant) residue on it, that could easily explain a large difference.