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Help finding coefficient of kinetic friction

  1. Feb 8, 2006 #1
    "A box slides down a 30.0 degree ramp with an aceleration of 1.20m/s^2. Determine the coefficient of kinetic friction between the box and the ramp."

    I know that the coefficient of kinetic friction is found by kinetic friction / normal force, but the only problems I've delt with in the past gave you the mass of the object. In those cases, I would find the force of gravity then use it to find the force upwards using the sin of the angle of the incline. From there, I would find the normal force subtracting force upwards from the force of gravity - then finally multiplying the mass times the acceleration to find the kinetic friction.

    I know how to work through it one way, but I can't figure out how to work backwards given the acceleration and no mass. Could someone please point me in the right direction?
  2. jcsd
  3. Feb 8, 2006 #2

    Doc Al

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    Staff: Mentor

    Call the mass "m" and work it out just as you would if you had the mass. You'll find that in every equation you'll need, the mass will drop out. Try it.
  4. Feb 8, 2006 #3
    So I have force of gravity = m (9.81m/s^2)

    Force upwards = m(9.81m/s^2)sin30
    Force upwards = m(4.905)

    Normal force = Fg - Fup
    Normal force = m(9.81m/s^2) - m(4.905)
    Normal force = m(4.905)

    then I lose myself - I'm not sure if I'm on the track that you mean...
  5. Feb 8, 2006 #4

    Do you think you could just forget about the numbers and just work out the math to the end, and you should end up with an expression for the coefficient of kinetic friction. I mean forget about numerical stuff untill the very end, just call gravity g, mass m, acceleration a etc etc...
  6. Feb 8, 2006 #5
    in a ramp, if there was no friction, then the accelration would be 9.8m/s^2?
    i would guess then..with friction
    [tex] \mu_{k} = \frac{F_{fr}}{F_n}
    \\ = \ \frac{(mg-ma}{mg-\sin(\theta)mg}

    is the above correct? im a HS student too..

    or...wait..i messed up the normal force...
    Last edited: Feb 8, 2006
  7. Feb 8, 2006 #6

    Doc Al

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    Staff: Mentor

    First tip: Don't plug numbers in until the end--use symbols.

    There are three forces acting on the block:
    (1) the weight (mg) which acts vertically down
    (2) the normal force (what is that?) which acts normal to the surface
    (3) the friction force ([itex]\mu N[/itex]) which acts up the ramp

    Find the components of the weight parallel and perpendicular to the ramp. Then use that to find the normal force. Then use Newton's 2nd law for force components parallel to the ramp. (That's when you'll write an equation and the mass with cancel out. You'll solve that equation for [itex]\mu[/itex].)
  8. Feb 8, 2006 #7
    Thanks for all of the help, Doc Al - I finally got an answer of .44!
    Last edited: Feb 8, 2006
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