Tutorial on Friction Coefficient Formulas & Applications

[rculley1970]

Is there a link someone can provide that has a good tutorial on friction coefficient formulas and when to apply them. I seem to be having problems with this topic.

Everyone knows the old M1 and M2 connected by a string and pully question. That is on a level surface. But if the surface is tilted backward so the corner now makes a 53 degree angle instead of 90 how do I go about finding the acceleration. Is it the same way as before but with something else added into it besides the kinetic friction?

I have been working on physics all day and my brain is about to boil over.

Thank you.

 

[rculley1970]

problem:

a car is traveling at 65.0km/h on a flat highway. coefficient of friction between road and tires is 0.15, what is the minimum distance in which the car will stop.

I am assuming that I start out with:
.15(9.8)=a which = 1.47m/s^2

plug that into:
V^2=V0^2+2a(deltaX) and solve for deltaX.

is this a fairly accurate way to solve it?

 

[rculley1970]

I come up with:

a=1.47m/s^2
deltaX = 1437 m

 

[topsquark]

problem:

a car is traveling at 65.0km/h on a flat highway. coefficient of friction between road and tires is 0.15, what is the minimum distance in which the car will stop.

I am assuming that I start out with:
.15(9.8)=a which = 1.47m/s^2

plug that into:
V^2=V0^2+2a(deltaX) and solve for deltaX.

is this a fairly accurate way to solve it?

Wait a minute! Hold the Ferrari! The unit police are here!

In order to calculate the force due to friction you need a normal force. You said:

I am assuming that I start out with:
.15(9.8)=a which = 1.47m/s^2

I presume the 9.8 is g, but g has units of acceleration, not force! The normal force in this case, all things like wind resistance etc. being neglected, will be mg. Try using a round figure for the mass, such as 1500 kg maybe? (My estimates for masses are never great, but I think that's reasonable.)

Other than that, you're golden.

-Dan

 

[rculley1970]

The problem does not show any mass. The equation I get out of the book is:

(coefficient of k)(g)=a

Therefore, from the only given data from the problem:

(.15)(9.8) = a = 1.47m/s^2

 

[rculley1970]

Is that the correct answer for the problem? I have been working it out but putting it into the online answer form keeps saying it is wrong. For the friction of .15 I am getting a stopping distance of 1438 m. The problem asks for the answer in 3 significant figures only. I can't figure this out.

Please help.

 

[andrewchang]

i don't know if it's the right answer, but if you only need 3 significant figures, you're going to need to get rid of the 8. Round to the nearest 10 and you get 1440 m. (which has 3 sigfigs)

 

[andrewchang]

actually, it's the wrong answer. notice that the speed of the car is given in kilometers per hour, while your calculations are in meters per second. so, you should first convert the kilometers per hour to meters per second, and then do the problem

 

[rculley1970]

no, that didn't work. I think it's the answer. I am doing something wrong but can't figure it out. I have to submit the work in 30 minutes so i guess I will leave it blank and find out the hard way.

 

[andrewchang]

did you see my post just now?

 

[rculley1970]

I just did, recalculated, and it worked. Thank you very much andrew. I can't believe it was such a small detail that turned it into such a huge problem. Once again, thank you very much.