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Kinetic Friction on a slope

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Note - I have used (uk) as the symbol for co-efficient of kinetic friction (mew k?).

    You want to measure (uk) between snow and your snowboard. You measure the angle x the slope makes with the horizontal. You kick your snowboard so it slides up the slope, and then back down. You notice that it takes twice as long to complete the downward portion of it's slide than the upward portion. Determine (uk). The steps to do so are as follows:

    1) FBD of snowboard, using variables f and a together with i-hat in the direction up the slop, to represent the friction and acceleration vectors. j-hat is chosen perpendicular to the slope then. What are the signs of f and a on the upward/downward portion of the motion?

    2) Use newton's 2nd law in the i-hat and j-hat directions to solve for a (on the way up and on the way down, denoted (au) and (ad) respectively) in terms of (uk), x and g. Be careful about the sign of f.

    ...

    The question after this point I understand. My problem is coming up with equations for (au) and (ad) that make sense to me. Please help :)

    2. Relevant equations

    NetForce = mass*acceleration
    NetForce(i-hat) = mass*accel(i-hat)
    NetForce(j-hat) = mass*accel(j-hat)

    3. The attempt at a solution

    When I worked this out, I am saying that a is negative all the time (remember (i-hat) is in the direction up the slope), and f is negative on the way up, and positive on the way down. I obtained...

    Fnet = Normal*(j-hat) + f(i-hat) - mgsinx(i-hat) - mgcosx(j-hat)
    = m*accel = m*a(i-hat)

    (i-hat) direction: f-(mgsin x) = ma
    (j-hat) direction: N - (mgcos x) = 0

    f = (uk) * N = (uk) * mgcos x

    I make f negative to solve for (au), and get au = g(-(uk)cos x - sin x). I keep f positive to solve for ad = g( (uk)cos x - sin x). My problem is this. If (uk) happens to be greater than 1, and the angle x is less than 45 degrees, won't (uk)*cos x be greater than sin x. And then (ad) would have to be positive, which means the snowboard is accelerating "up" the slope...which cannot happen obviously. So I assume I have done something wrong. If anything is unclear, please let me know. I realize using (uk) and (ad) creates a lot of clutter. Thanks for your time,
     
  2. jcsd
  3. Oct 16, 2007 #2
    Bump...any suggestions?
     
  4. Oct 16, 2007 #3
    Another bump. Sorry to be persistent, but 39 views now, and no one has thoughts (I don't need the problem solved completely, I'm pretty sure my issue is with grasping the variables f and a)? If I'm being too vague/unclear, please let me know.
     
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