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Coefficient of friction

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A book has a mass of 400 g. When you slide the book against the floor with 5 N, it accelerated at the rate of -1.5 m/s^2. What would the coefficient of friction between the book and the floor be?


    2. Relevant equations
    Ff = μFN

    3. The attempt at a solution
    I am trying to learn physics on my own solely through whatever reliable resources i may stumble upon on the internet. So bear with me if my answer or my method is not correct. Anything I know about physics is self taught(trying to prepare for college)

    I tried it a few different ways
    Ff = μFN
    5N=μ(.4x9.8)
    5N=μ(3.92)
    5N/3.92=μ
    1.276=μ

    5N=μ(.4x-1.5)
    5N=μ(2.666[STRIKE]6[/STRIKE])
    5N/-.26666[STRIKE]6[/STRIKE]=μ
    8.75=μ (i know this cant be right)
     
  2. jcsd
  3. Apr 16, 2010 #2

    PhanthomJay

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    Neither solution is correct, but the problem is not all that clear. Presumably, the book is being pushed along a horizontal floor with an applied horizontal force of 5N acting to the left. Use Newton's 2nd law to calculate the net force acting on the book in the horizontal direction. The net force consists of the applied force in one direction and the friction force in the other direction.. You must calculate the friction force, then use your relevant equation to solve for the friction coefficient. The result is a rather high value, so I'm unsure if the problem is worded correctly. Wecome to PF!
     
  4. Apr 16, 2010 #3
    OK, well you have the correct equation, but you seem to lack understanding about Newtons's 2nd Law. Which ultimately yields F=ma. Secondly it's important to remember that forces and accelerations are vector quantities. Now the problem isn't very clear, but I'm nearly positive the 5N is applied horizontally to cause the book to slide.

    The first step, is to sum the force vectors in the y dimension, and then the ones in the x dimension.
    [tex]
    \sum F_{y} = F_{N} - F_{g} = ma_{y} = 0
    [/tex]
    The reason this equals 0 is because there is no acceleration in the y direction.
    Then you can conclude that:
    [tex]
    F_{g} = F_{n}
    [/tex]
    I think you already understood this, but I just want to be clear.
    Next you work on the x direction
    [tex]
    \sum F_{x} = F_{k} - F_{p} = ma_{x}
    [/tex]
    You already know what to sub in for Fk, and Fp is the force of the push. I think it's pretty easy from there. I hope I haven't said too much, but I wanted to show you this process. It's pretty much identical on mechanics problems.

    Additional, the reason I said Fk was positive and the Fp was negative is because the problem states that the acceleration is negative.

    Good luck!
     
  5. Apr 16, 2010 #4
    Yes. And the FM=-mg is an action reaction pair which is Newton's 3rd Law.

    Also just a tip from my experience. I can tell by the way you worded the question: "Calculate then...put in" that you are going to solve for it then move on. Many times it's better to just work with the symbols until you've solved to the desired variable, then put numbers in. I think this helps you to understand physics better conceptually, not to mention it helps your algebra skills. You don't have to, but just a thought.

    EDIT: haha, I guess you figured it out then.
     
  6. Apr 16, 2010 #5
    when i find the friction force(FN=-mg or Fn= (-.4kg x-9.8m/s2 i divide the net force by this value?

    Edit: i think i did lol thx
    also i think i forgot a one in the second answer of mine its supposed to be (18.75)
     
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