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Kinetic Friction of sliding down a pole

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A firefighter whose weight is 812 N is sliding down a vertical pole, her speed increasing at the rate of 1.45 m/s2. Gravity and friction are the two significant forces acting on her. What is the magnitude of the frictional force?


    2. Relevant equations
    Fk = ukFn
    uk = coefficient of kinetic friction


    3. The attempt at a solution
    I'm not sure what the coefficient of KF should be so I just tried ..
    812*1.45 = 1177.4
    I'm not sure how to incorporate the gravity and friction into the equation.
     
  2. jcsd
  3. Oct 14, 2008 #2

    Doc Al

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    Staff: Mentor

    You're not given the coefficient of friction. Luckily you won't need it--you'll figure out the friction force another way.
    Gravity and friction are the two forces acting on the firefighter. You are given her acceleration. Apply Newton's 2nd law: ∑ F = ma

    (What's the firefighter's mass?)
     
  4. Oct 14, 2008 #3
    OK ...
    F = ma
    mass -> 812 N/9.8 = 82.8571 kg
    F = (82.8571)(1.45)
    F = 120.143 N

    Is it that simple?
     
  5. Oct 14, 2008 #4

    Doc Al

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    Staff: Mentor

    Good.
    Not that simple, but almost. The "F" in F = ma stands for the net force acting on the object. Express the net force in terms of the actual forces acting on the firefighter (weight and friction) by adding those forces up. (∑ F means the sum of the forces.) Then you can solve for the unknown friction force.

    Hint: Careful with signs.
     
  6. Oct 14, 2008 #5
    Would the next step be to multiple her weight and gravity? Then add that net force to 120.143 N?

    F = (82.8571)(9.8) = 812 N - but that's the weight already given??
    120+812 = 932.143
     
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