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Questions on forces and coefficients

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data Hello, this is my first time posting, I've run into a bit of a snag in my physics lab. I am to figure out the coefficient of friction based on the values we found in the lab. Now I found them out however I just want to be sure that im doing this right. these values were determined using a simple,horizontal pulley system that ran off the end of a table



    2. Relevant equations

    F=MA
    coefficient of friction= Force of friction/force normal

    3. The attempt at a solution
    So I have the weight on the table at 1,045g and the weight required to pull this weight towards the end of the table was 170g.
    I figured the force of friction was the mass of the pulley (170g) X the acceleration of Gravity, so F=170g X 9.8 m/s squared which came out to be 1666N

    Now the force normal i figured to be the same way. The weight on top of the table was 1,045g X 9.8m/s squared which came out to be 10241N

    so now I have the force of friction to be 1666N and the force normal to be 10241N
    I then used the coefficient of friction equation, 1666N/10241N which came out to be .16

    So after all that my coefficient of friction was .16
    Could anyone please tell me if I am doing this right? I believe I am but would like to be sure.
     
  2. jcsd
  3. Oct 17, 2009 #2

    ideasrule

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    A Newton is a kg*m/s^2, not a g*m/s^2. So the gravity on the weight wouldn't be 1045g*9.8 m/s; it would be 1.045 kg *9.8 m/s. Otherwise, your work is right.
     
  4. Oct 17, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi Kingbaldur! Welcome to PF! :smile:

    (I take it that the table and the string are both exactly horizontal, and that you're measuring the coefficient of static friction?)

    Yes, your method is correct, but you could just have left g as g (instead of multiplying by 9.8 twice), and said 170g/1045g = .16 :wink:
     
  5. Oct 17, 2009 #4
    Hmm, I see, well thanks to both of you.
     
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