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Static friction co-efficient

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    from the free body diagram of a block resting on a flat surface, I am trying to derive the coefficient of static friction.



    2. Relevant equations



    3. The attempt at a solution
    I have the definition of the coefficient as:
    us=Fn/Ffr

    But in my free body diagram i am pulling on the block, which resists with the Ffr
    I have
    [tex]\Sigma[/tex]Fx=0
    F-Ffrus=0
    us=F/Ffr

    My confusion is because if the definition of the coefficient is above, why don't I derive it when I observe my free body diagram of the block. What am I missing?
    The forces in the x direction are my pulling force and the friction force that resists it * us
    How do i get the weight of the block into my derived equation?
     
  2. jcsd
  3. Nov 11, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The weight of the block is Fn.

    (If that's not what you're looking for, please state the complete problem exactly as it was given.)
     
  4. Nov 11, 2008 #3
    It is not a problem as such. I am trying to determine the coefficient of static friction us experimentally. Then, I am trying to determine if it is a function of surface area or of mass.
    My experiment consists of a wooden block on a wooden flat surface. I am going to attach a spring scale and determine at what force the block overcomes the friction and moves.

    In trying to determine us from my free body diagram I am having a little difficulty as I tried to explain.
    From my free body diagram, if at equilibrium the forces in the x plane are zero, I get
    Fpull-Ffriction*us=0
    but I don't see where I am getting the mass of the block to become part of my derived equation.
     
  5. Nov 11, 2008 #4
    you also need to apply newton's second low in the Y-direction (vertical direction)

    ma_y = F_normal - mg = 0

    do you understand this equation ? Why is it equal to 0 ?

    ps a_y is the component of the acceleration in the vertical direction
    and F_normal is the normal force.

    In this case, this is not gonna help you much because the table is horizontal.

    marlon
     
  6. Nov 11, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    What you should get is Fpull - Ffriction = Fpull - μFnormal = 0.
     
  7. Nov 11, 2008 #6
    Thank you.
     
  8. Nov 11, 2008 #7
    Nope,

    Fpull=Fnormal*us
     
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