Hair Elasticity

[alondra011]

Hello, My science Fair topic is, How does chemical lighting hair treatments affect the elasticity of hair? My teacher already approve it but I just don't know how or with what to test it. I was planning to use household or inexpensive lighting hair treatments because i don't want to spent a lot of money in the science fair. What are some things that i can test hair elasticity on?

[deleted-71588]

Your biggest problem is "How are you going to define the elasticity of hair?". Without a definition, you can't devise a way to measure it.

Were you thinking about how curls in hair make it behave like a spring?
Were you thinking about hair as something that will stretch and then return to its previous length when released?
Were you thinking about how much force it takes to stretch a hair and exceed its elastic point and cause it to break (in-elastic point)?
Something else?

[alondra011]

Yes, my answer is below one of your options :" how much force it takes to stretch a hair and exceed its elastic point and cause it to break (in-elastic point)?"
I couldn't find another method to measure elasticity other than pulling. My teacher told me to research if there was another method other than stretching but i was unable to find a different method. Any suggestions ?

[deleted-113017]

Actually what you are trying to find is called Ultimate Tensile Strength or the Ultimate Strength (the point at which if the material is ductile, it starts to "neck"i.e.start thinning out before breaking, or in case of brittle material, completely breaks).

The Elastic Limit is actually the point after which Hooke's Law (which says that the deformation will increase proportionally with increase in force) no longer applies and your material has permanent deformation (even if you remove the force).

Now that we have the technicalities out of the way, I can not understand why you would want to do anything other than pulling to find the ultimate strength, since "pulling" or tensile testing is what even professionals in the materials industry do to find the ultimate strength of a material.

What I would suggest you do is that you find a small pan, tie one end of the uncolored hair to it and tie the other end to a vertically standing hook and increase the weight in the pan incrementally until the hair breaks. Repeat for the colored hair and repeat multiple times to get a good average. You can find the force by using F=mass*gravity and if you are looking for the ultimate stress then it would be Stress=Force/Area. I leave it up-to you to find the cross-sectional area of a human hair. And don't forget to weigh the pan too.

Having said that, it is very important that when you are performing these tests on those hairs to make sure that they have the same cross-sectional area since Ultimate Strength depends a great deal on the cross sectional area of your material. I know it will be next to impossible but try your best to have those hairs of similar dimensions.

If you need some more info on these things, I would suggest you to Google Tensile stress and you can find a lot of information about it.

Hope this helps!

[Jacobie09]

Interesting project! Give us the results when done. I'd like to see them and good luck!

[alondra011]

Is there any diagram that could represent the process using the pan? and how would you measure gravity?

Actually what you are trying to find is called Ultimate Tensile Strength or the Ultimate Strength (the point at which if the material is ductile, it starts to "neck"i.e.start thinning out before breaking, or in case of brittle material, completely breaks).

The Elastic Limit is actually the point after which Hooke's Law (which says that the deformation will increase proportionally with increase in force) no longer applies and your material has permanent deformation (even if you remove the force).

Now that we have the technicalities out of the way, I can not understand why you would want to do anything other than pulling to find the ultimate strength, since "pulling" or tensile testing is what even professionals in the materials industry do to find the ultimate strength of a material.

What I would suggest you do is that you find a small pan, tie one end of the uncolored hair to it and tie the other end to a vertically standing hook and increase the weight in the pan incrementally until the hair breaks. Repeat for the colored hair and repeat multiple times to get a good average. You can find the force by using F=mass*gravity and if you are looking for the ultimate stress then it would be Stress=Force/Area. I leave it up-to you to find the cross-sectional area of a human hair. And don't forget to weigh the pan too.

Having said that, it is very important that when you are performing these tests on those hairs to make sure that they have the same cross-sectional area since Ultimate Strength depends a great deal on the cross sectional area of your material. I know it will be next to impossible but try your best to have those hairs of similar dimensions.

If you need some more info on these things, I would suggest you to Google Tensile stress and you can find a lot of information about it.

Hope this helps!

[kgudger]

Hi:

I've attempted to draw a diagram of how to test the hair tensile strength. Let us know if you have any questions.

hairpan.pdf

picture didn't upload

(9.46 KiB) Downloaded 561 times

Keith

[deleted-113017]

Thanks Keith for the diagram.

Now gravity on planet earth is a constant. For the purpose of this experiment using 9.81 meters/second/second should do (I strongly suggest working in SI units (meters,kilograms,seconds,etc) for this experiment).

Let me also just add a note here, you want to have the hair break at the center and not near the ends where you are tying the knots as it would give you a wrong reading. Looping the hair around the hooks should fortify it a bit and should ensure that the hair breaks in between the hook and the pan. Still I would recommend multiple runs and then taking an average to even out the errors.

[alondra011]

Thank You So much, it's helping me a lot. I'm still confused on the part about measuring gravity. If Using the formula F=Mass*Gravity is the most effective way I know that I will be finding the force after the mass (the objects/weights in the pan + the weight of the pan) times the Gravity. If gravity is 9.81 meter per second. How am I supposed to find the seconds? The time it will take for me to apply each weight? or the time it will take the hair to break? I'm really unsure on that part.
Thanks A lot to you all, I really Appreciated it.

[deleted-93346]

...If Using the formula F=Mass*Gravity is the most effective way I know that I will be finding the force after the mass (the objects/weights in the pan + the weight of the pan) times the Gravity. If gravity is 9.81 meter per second. How am I supposed to find the seconds? ...

That is a very perceptive question. It is a very good idea to check that the units match up correctly when making a calculation. Now for some quantities, like the "m" in F=ma, the unit is obvious (kgm using the preferred SI system of units), but for others, like the vector "F", you need to use a basic formula like F=ma in order to figure out what the units of F are. This probably seems circular reasoning, but it is not. Every time a new quantity such as force is defined, the units of that quantity become determined from the defining relationship. That defining relationship for force is found from setting the units of force to be the units of the quantity "ma". Thus we have m[kgm] x a[m/s/s] to get F[kgm m / s^2]. Now we know that whenever we encounter a force, the basic unit of that force will be kgm m/s^2, and we define 1 newton of force = 1 kgm m/s^2. You can see now that Newton's Law, F=ma is both 1) a description of how nature works, that is that the acceleration of an object, an observable, can be found using a scalar, m, that is characteristic of the body and will be fixed for a given body even as the acceleration may change this way and that, and a vector, which determined by the circumstances existing around that body, for example a spring pushing on the body in some direction, and 2) a definition of the force vector. I'm sure that seems a bit opaque the way I have tried to explain it -- the exact, "dotting the i's and crossing the t's" description of the simultaneous roles of description of a physical law and definition of a new quantity can be made very elegantly, but it takes several pages of rather abstract discussion to be both precise and clear. Anyway, the rule is clear -- when new quantities like force or electric charge, or pressure, or whatever, are first introduced, the equations in which they first come up can be used to deduce the units.

So for your problem we have F = mg with units of newtons [N] for left hand side and [kgm] x 9.81 [m/s^2] on the right hand side thus we are equating [N] with [kgm][m/s^2] which is correct since the newton was defined as [kgm m/s^2] above.

[alondra011]

Thank you. I undestand what i'm doing now.

How do I find the Cross Sectional Area of the Hair I will be testing, In order to get the right tensile strength?

[deleted-93346]

According to Wikipedia (18 Nov 2012)

"The diameter of human hair varies from 17 to 180 micrometers (0.00067 to 0.0071 in).[6]"

You could measure the diameter using diffraction of light from a laser, as described here:

http://tinyurl.com/d5j4f8d

You could make a bundle of 100 hairs (tedious) and measure the diameter of the bundle (about half a mm maybe?). From that the area of each hair is 0.01 of the area of the bundle ( pi*(diam/2)^2 ).

You might be able to use a micrometer on a hair.

You could look at the hair and a reticle (a kind of ruler for use under a microscope) with a microscope.

[alondra011]

Thank you, but the problem is that I don't have access to a microscope . And i was planning to just test a strand of hair not 100. If i can't find the cross sectional area of a hair then it will affect the whole tensile strength results...

[deleted-93346]

In that case you may be stuck with relative measurements, i.e. comparing similar strands of hair that have been treated different ways. You would need many hairs (nearly 100, perhaps a 1 mm diameter tress) from the same spot on the same person. You would then need 5-10 repeat cases to estimate the variation between different hairs treated similarly, and hence the reliability of differences in the average strength found in groups treated differently. Something like

Treatment sample # average RMS variation
1 2 3 . . . 10
--------values------

none x1 x2 x3 . . .x10 <x> sigma_x
A y1 y2 y3 . . .y10 <y> sigma_y
B . . . , .
. . . . , .
. . . . , .
. . . . , .
F . . . , .

where <x> is the average of x, sigma_x is the RMS variation of the samples
of x and so forth. Sorry about the appearance of the table, the text input
widget won't let me put in spaces the way I wanted.
You could then estimate the difference in strength of B vs untreated as
(<y>+-sigma_y)-(<x>+-sigma_x)

Hope that is clear.

[alondra011]

According to Wikipedia (18 Nov 2012)

"The diameter of human hair varies from 17 to 180 micrometers (0.00067 to 0.0071 in).[6]"

You could measure the diameter using diffraction of light from a laser, as described here:

http://tinyurl.com/d5j4f8d

You could make a bundle of 100 hairs (tedious) and measure the diameter of the bundle (about half a mm maybe?). From that the area of each hair is 0.01 of the area of the bundle ( pi*(diam/2)^2 ).

You might be able to use a micrometer on a hair.

You could look at the hair and a reticle (a kind of ruler for use under a microscope) with a microscope.

Yes this would be a great aproxximation of the diameter of a hair strand. So what I did was that I made a Bundle of 100 hairs and measure it. It was 1mm so I divided 1/100 = 0.01 . I need to find area in meters so I converted 0.01mm to a meter which equals to 0.00001m since we divided 0.01/1000.
0.00001m is the diameter but we need to find the radius to calculate area so it will equal to 0.000005m.
Am I correcT?

[alondra011]

So in order to find stress i would Divided Force (Newtons ) by Area ( 0.0000000000785 m) will give me stress in pascals.. If 1 Pascal = 1 Pa = 1 N/m2 is the area 0.0000000000785 m or m2?

[deleted-93346]

I lost you at "I need to find area in meters" since area would be meter^2, not meters. A bundle of 100 hairs will have 100 times the area of one hair, and since area goes as diameter squared, the bundle of 100 will have 10 times the diameter of one hair. Thus diameter of one hair = (1/10) diameter of bundle of 100 = (1/10) * 1 mm = 0.1 mm = 1E-4 m = 100 micrometers.

[deleted-93346]

So in order to find stress i would Divided Force (Newtons ) by Area ( 0.0000000000785 m) will give me stress in pascals.. If 1 Pascal = 1 Pa = 1 N/m2 is the area 0.0000000000785 m or m2?

If you compute using scientific notation you are less likely to make arithmetical errors. Myself, I cannot decipher expressions like "0.0000000000785".

[alondra011]

Thank you so much for everybody's help and support. My science fair came out super great I'm just waiting for y results . Thank you so much

[deleted-71588]

So what I did was that I made a Bundle of 100 hairs and measure it. It was 1mm so I divided 1/100 = 0.01 . I need to find area in meters so I converted 0.01mm to a meter which equals to 0.00001m since we divided 0.01/1000.
0.00001m is the diameter but we need to find the radius to calculate area so it will equal to 0.000005m.
Am I correcT?

What justification do you have for dividing by 100 this early in the calculations?
If the diameter of a bundle of 100 hairs is 1mm, the radius of the bundle is 0.5 mm (radius = 1/2 the diameter), and the area of the bundle is 2*pi*(0.5*0.5) = aproximately 1.57 sq mm.
Now you get to divide this area by 100 hairs to get the cross section area of the average hair in the bundle = 0.0157 sq mm which is 0.0000000157 sq m (1.57 x 10**-8 m).
This cross sectional area in sq m is what you need in the formula.