# Coefficient of kinetic friction for lab

1. Jun 20, 2013

### catzmeow

Hello,

I was competing calculations for a simple machine lab, and I got a different mu_k for 3 separate inclines of 20,30 and 40 degrees.

But a question asks says ,"explain why the coefficient of kinetic friction should be independent of the inclination angle of the inclined plane."

How do I explain it if I got a different mu_k for each angle??

2. Jun 20, 2013

### barryj

How did you calculate the mu_k for each angle. Give an example of your work.

3. Jun 20, 2013

### tiny-tim

welcome to pf!

hello catzmeow! welcome to pf!
probably something wrong either with your experiment or your calculations

describe what you did

4. Jun 20, 2013

### catzmeow

Ha - no doubt something wrong with my calculations! I used mu_k= tan(theta) from mechanical advantage= 1/sin(theta) + mu_k(cos(theta)). I was supposed to derive how to calculate mu_k myself, so maybe that's my error.

Thanks so much :)

5. Jun 20, 2013

### tiny-tim

hi catzmeow!

that looks suspiciously like the formula for static friction

what exactly was your set-up?​

6. Jun 20, 2013

### catzmeow

Hi tiny Tim! Here's what this is all pertaining to and some of my work:

7. Jun 20, 2013

### tiny-tim

how did you measure the force (or the acceleration)?

(and your equations are difficult to read … can you please type them out for us?)

8. Jun 20, 2013

### catzmeow

Sure! Sorry :(

We measured the force needed to move the a wooden block (2.67 N) up the incline (also wood) for 20, 30 and 40 degrees. Acceleration is constant.

For an incline of 20, we used 1.96 N/200 g to move the block.
For 30, we used 2.21 N/225 g to move the block
For 40, we used 2.45 N/250 g to move the block.

The equations that are given are: Mechanical Advantage= 1/sin(theta)+ mu_k*cos(theta) and
efficiency= 1/ 1+ mu_k*cos(theta).

From either one of these, I'm supposed to derive mu_k and explain why it's independent of the inclination angle of the inclined plane.

Then I'm also supposed to show the "steps leading to" these equations. I guess that means show how they're derived?

Thanks so much- I really appreciate your help!

9. Jun 20, 2013

### tiny-tim

hi catzmeow!
first, you seem to be confusing g for gram with g for gravity

eg 1.96N is 0.2 g (g = gravity), ie the weight of 0.2 kg = 200 g (g = gram)

second, i don't see where your 0.2 etc comes into your calcuations

10. Jun 20, 2013

### barryj

I am confused as to what you are doing. To get a good understanding of what is happening I think you should draw a free body diagram of the block on the incline. You should get something like..
F = (mu)mgcos(theta) - mgsin(theta) where F is the force to move at a constant speed, and go from there.

11. Jun 20, 2013

### haruspex

Do you mean speed is constant? If not, how are you sure the acceleration is always the same?
That looks strange. Do you mean 1/(sin(theta)+ mu_k*cos(theta)) ?
The block is being pulled up the slope, so it's '+'.

12. Jun 20, 2013

### barryj

You are correct. I must have had a senior moment. The block is being pulled up the slope, so it's '+'.

13. Jun 21, 2013

### catzmeow

14. Jun 21, 2013

### catzmeow

My apologies- I mistyped. I meant that we used grams but I had already converted them to Newtons using the gravity constant. So my calculations are done with newtons. The .2 was the weight in kilograms of the force.

15. Jun 22, 2013

### tiny-tim

(just got up :zzz:)
yes, but i don't see where the 0.2 (or m) comes in your calculations

16. Jun 23, 2013

### catzmeow

yes, but i don't see where the 0.2 (or m) comes in your calculations [/QUOTE]

You're right- I didn't actually use these to calculate the mu_k, they were part of the experiment to calculate the ideal mechanical advantage and mechanical advantage. Sorry for the confusion :(

17. Jun 23, 2013

### tiny-tim

but the 0.2 etc are one of the forces in your force equation

your force equation should add all the forces together to get zero (along the slope), including the 0.2