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Calculating the coefficient of kinetic friction

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem:
    A 10 kg box slides down a hill and has a constant velocity. The angle of the hill is 10° with respect to the horizontal. What is the coefficient of kinetic friction between the surface of the hill and the box?



    2. Relevant equations
    Fn = mg + ma
    ƒk = μk * Fn

    3. The attempt at a solution
    To calculate the coefficient of kinetic friction I used the formula ƒk = μk Fn and isolated μk to be
    μk = ƒk / Fn
    Where ƒk is kinetic frictional force, μk is the coefficient of kinetic friction, and Fn is the normal force.

    I don't have any of these variables, so I drew a free-body diagram and tried to find my normal force by doing:
    ∑Fy = Fn + mg cos 10° = 0
    Fn = mg cos 10°
    Fn = (10 kg) (9.80 m/s^2) cos 10° = 96.5 N


    My problem is I am not sure on how to determine my kinetic fricitonal force ƒk to plug it into the
    μk = ƒk / Fn. Someone please guide me?
     
  2. jcsd
  3. Mar 10, 2015 #2

    NascentOxygen

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    Staff: Mentor

    What you can say is that the friction force must exactly balance the opposing force due to that body's weight.

    Q: why can you say this?
     
  4. Mar 10, 2015 #3
    Newton's third law that states for every action, there is an equal and opposite reaction ?

    Therefore does fk = Fn? I'm sorry if this isn't correct, I struggle a lot with this course...
     
  5. Mar 10, 2015 #4

    NascentOxygen

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    Staff: Mentor

    No, those two forces are not in the same direction. fk opposes a force acting in the exact opposite direction to fk. If fk were not exactly cancelling its opposing force, then the body would be accelerating or decelerating, but we are told it is sliding with a fixed speed.

    You draw the diagram and show how the body's weight can be broken into orthogonal components, one along the slope, and the other normal to the slope.
     
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