11-18-08

The past week in Mr. Mannings first period class we have been discussing and drawing friction forces. Today we will be learning about the **coefficent of friction forces**.

The objectives for todays class include: Briefly before Mr. Manning began with the daily question, the class quickly went over the lab done on the previous day. He demonstrated the the lab in front of the class. He used a scale to pull the object first to make it start moving and then to keep it at a constant speed. To show the class exactly what was going on, he at the same time had his computer draw a graph of static friction vs. the kinetic friction. When this was all said and done, the key to this lab was that is take more force to get movement started than it does to keep the object at a constant speed.
 * __Objectives__**
 * 1) Daily Question
 * 2) Test/Reading Guide due Friday
 * 3) Discussion
 * 4) Promlems
 * __Daily Question__**

Anti-lock brakes keep and automobile's wheels from locking and skidding during a sudden stop. Apart from issues of steering. What is the advantage of preventing the wheels from skidding (sliding) on the pavement? (Hint: A tollding object receives static friction. Skidding objects receive kinetic friction)

The class with a bit of help from Mr. Manning came to the conclusion that this happens because when you skid you have kinetic friction under the tires which you don't want. You want your tired to freely rotate with static friction force to move along ithe road betting especially in icy conditions.


 * __Coefficient of Friction__**

The coefficient of friction symbolized as µ (myou) is the amount of pulling force. This can be defined as the ratio:

µ s = Ft(pulling force) / Fg(weight)

But since Fs = Ft and Fg= Fn,

µ s = Ft(pulling force) / Fg(weight)= Fs(static friction force) / Fn(normal force)

Therefore,

µ s = Fs/Fn

Which is also the equation for kinetic friction, you just replace the s with k:

µ k = Fk/Fn

Then we moved on to actually putting these equations to use. We did the first problem together:
 * 1) A tension force of 120 N pulls an 8kg box with an acceleration of 7m/s2 across a floor. How much friction must be acting on the box?

[|P1.bmp] £F= m x a £F = Ft + Fk £F= 8kg x 7m/s2 £F = 120 N + Fk £F= 56 N 56N = 120N + Fk Fk= - 64 N