Help Proove Physics Teacher Wrong

DRock9

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Aug 22, 2006
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My physics teacher feels very strongly that you will not increase the force, due to friction, required to turn a wheel if you deflate the tire. I asked her, well why then is it harder to pedal a bike when the tires are flat, and she did not have an answer for me. I also ask, well why then do they say that inflating your tires increases your gas mileage, and also why do you deflate your tires when you go to the track?

She claims then when you deflate, or inflate your tires the change in surface area does not have any effect on friction what-so-ever.

This goes against anything I have ever read, can anyone clarify this?

Thanks!
 
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Ok well it does have an effect on friction because depending oh how much your tire is inflated you can see different contact patterns with the road. For instance an over inflated tire wears out the center of the tire and not the edges. An under inflated tire wears the outer edge of tread and not the center. I would ask her to show you the formula that proves her statement. I remember my physics teacher coming up with some strange things when I was taking it. Conservation of Force??? I can't remember but it had to do with if you jump the earth has to move up to u a little?? I laughed my a$$ off. I asked if we got every human on the earth to move to one country and we all jumped at the same time could we pull the earth out of orbit? It seemed silly to me but after all they are the smartest dumb people I have met. ( actually maybe engineers are)
 
She claims then when you deflate, or inflate your tires the change in surface area does not have any effect on friction what-so-ever.

I found this info, hope it helps..

If the coefficient of friction is held constant, increasing surface area does not increase the force of friction. That's technically true, but in practice:

There are exceptions: Friction and area of contact

The coefficient of friction depends upon surface roughness, molecular adhesion, and deformation effects. A fully inflated tire has greater resistance to deformation than a softer tire. In other words, the coefficient of friction is different for a flat tire than a fully inflated tire.
 
The coefficient of friction depends upon surface roughness, molecular adhesion, and deformation effects. A fully inflated tire has greater resistance to deformation than a softer tire. In other words, the coefficient of friction is different for a flat tire than a fully inflated tire.[/QUOTE]

Exactly....thats the difference right there.
 
A simple demonstration/experiment for your teacher. Deflate one rear tire to 10 lbs and leave the other side at 35 lbs. Now go for a leisurely 50 mph drive on a long, safe road. After 30 minutes, have her get out and place her hand on each tire. Which one is warmer? Where did that heat come from??
Disclaimer: This experiment is hypothetical and no-one should be driving around on a half-flat tire.
 
"The coefficient of friction depends upon surface roughness, molecular adhesion, and deformation effects. A fully inflated tire has greater resistance to deformation than a softer tire. In other words, the coefficient of friction is different for a flat tire than a fully inflated tire."

Thank you! That is pretty much what I was looking for right there.

She claimed that if you deflate your tires it will not help you get out of sand or help you on the ice, I do not know about the ice part but it helps in my friends 4x4 in the sand and on rocks I can tell ya that much.
 
do this...get a car..put it in natural and push...do the same but with all 4 tires flat...there you proved him wrong... same thing if your tire are deflated you lost some MPG....
 
Ok well it does have an effect on friction because depending oh how much your tire is inflated you can see different contact patterns with the road. For instance an over inflated tire wears out the center of the tire and not the edges. An under inflated tire wears the outer edge of tread and not the center. I would ask her to show you the formula that proves her statement. I remember my physics teacher coming up with some strange things when I was taking it. Conservation of Force??? I can't remember but it had to do with if you jump the earth has to move up to u a little?? I laughed my a$$ off. I asked if we got every human on the earth to move to one country and we all jumped at the same time could we pull the earth out of orbit? It seemed silly to me but after all they are the smartest dumb people I have met. ( actually maybe engineers are)

Thank an engineer everytime you sit in that car of yours :nice:
 
Thank an engineer everytime you sit in that car of yours :nice:

LOL yah yah. I know thats why I say they are the Smartest dumb people. I know they are very smart but from my experience common sence eludes them. Have you ever worked with an "engineered" sytem or thing and when you go to use it you say hey if they did this it would be easier to install or use?
 
Hahah yeah like working on cars you need to be houdini to reach some spots.....eh I'll be an engineer soon and everyone will be whining about my crap. I'll be working for GM this summer so it'll be hard to screw up and not actually make an improvement. :D
 
I'll give contribute this (which is probably what the instructor is getting at)...

fs < us n

fk = uk n

fs = magitude of static frictional forces
fk = magnitude of kinetic frictional forces
us = static coefficient of friction (usually this is a 'look-up' value)
uk = kinetic coefficient of friction (usually a 'look-up' value)
n = normal force of the object. For example the Force of tires on the concrete, or Force of a brake pad on a rotor.

There is no allowance for the surface area of the contact area, when calculating the Frictional Forces.

The above works on the assumption that the 2 surfaces are 'hard'





The following explains a little as to how 'soft' surfaces are different...
http://www.school-for-champions.com/science/friction_equation.htm
Coefficient when surfaces not hard and sliding

In the case where a surface is soft, there is molecular adhesion, and in rolling and fluid friction, the coefficient of friction is not a simple number. The coefficient may be dependent on the area of the surfaces, the amount of deformation, the amount of adhesion, the shape of the surfaces, the radius of the wheel or the viscosity of the fluid.
What this means is that although the standard friction equation holds in these cases, the coefficient of friction will only hold for a specific configuration. In other words, you can't accurately give something like the coefficient of rolling friction for a rubber tire on pavement without stating the type of rubber, area on the pavement, inflation of the tire, and its tread pattern.

Basically, hard surfaces are simpler to work with mathematically, which is why these equations are used in 101 level physics classes.

Soft surfaces are much tougher to work with.


jason
 
That site is pretty cool.

They also had this
http://www.school-for-champions.com/science/frictionrolling.htm
Coefficient of rolling friction
Rolling friction is not as straightforward as sliding friction. But still, there is a relationship between the rolling friction and the normal force, similar to that in sliding friction. Often the coefficient of rolling friction is determined by assuming a form of the Standard Friction Equation holds:

FR = μR*W

where

FR is the resistive force of rolling friction,
μR is the coefficient of rolling friction for the two surfaces (Greek letter "mu" sub R), and
W is the weight of the wheel (in the case of a wheel rolling on the ground surface).
Other factors in coefficient
The coefficient of rolling friction, μR, isn't a simple number as in the case of sliding friction for hard materials. Instead, it is proportional to the width of the wheel and inversely proportional to the radius. In the case of soft wheels and tires, μR the surface area of the tire on the ground is a factor.

Determining the coefficient
A way that μR is determined is by rolling a wheel at an initial velocity, v, and then measuring the time, t, that it takes to stop.

μR = v/(t*g)

where g is the acceleration of gravity 9.8 m/s2 or 32 ft/s2.

This method and equation can be used with hard or soft wheels. Although it does not require the weight or size of the wheel to be known, it only applies for a wheel similar to that being measured.