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Coefficient of kinetic friction for lab

  1. Jun 20, 2013 #1
    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??

    Thanks in advance!!
     
  2. jcsd
  3. Jun 20, 2013 #2
    How did you calculate the mu_k for each angle. Give an example of your work.
     
  4. Jun 20, 2013 #3

    tiny-tim

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    welcome to pf!

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

    describe what you did :smile:
     
  5. Jun 20, 2013 #4
    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 :)
     
  6. Jun 20, 2013 #5

    tiny-tim

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    hi catzmeow! :smile:

    that looks suspiciously like the formula for static friction :redface:

    what exactly was your set-up?​
     
  7. Jun 20, 2013 #6
    Hi tiny Tim! Here's what this is all pertaining to and some of my work: ImageUploadedByPhysics Forums1371762294.737052.jpg ImageUploadedByPhysics Forums1371762361.034172.jpg
     
  8. Jun 20, 2013 #7

    tiny-tim

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

    (and your equations are difficult to read … can you please type them out for us?)
     
  9. Jun 20, 2013 #8
    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!
     
  10. Jun 20, 2013 #9

    tiny-tim

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    hi catzmeow! :smile:
    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) :wink:

    second, i don't see where your 0.2 etc comes into your calcuations :confused:
     
  11. Jun 20, 2013 #10
    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.
     
  12. Jun 20, 2013 #11

    haruspex

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    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 '+'.
     
  13. Jun 20, 2013 #12
    You are correct. I must have had a senior moment. The block is being pulled up the slope, so it's '+'.
     
  14. Jun 21, 2013 #13
     
  15. Jun 21, 2013 #14
    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.
     
  16. Jun 22, 2013 #15

    tiny-tim

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    (just got up :zzz:)
    yes, but i don't see where the 0.2 (or m) comes in your calculations :confused:
     
  17. Jun 23, 2013 #16
    yes, but i don't see where the 0.2 (or m) comes in your calculations :confused:[/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 :(
     
  18. Jun 23, 2013 #17

    tiny-tim

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    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
     
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