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Homework Help: Frictional force/coeff of friction

  1. May 25, 2009 #1
    I have been working at this question for a while. I have to find the frictional force and the coefficient of friction. skier has a mass of 64kg, starts at top of a 18 degree slope and lets gravity take her downhill. slope is 65 m long and her speed is 15 m/s when she get to the bottom. how do I find the x component of weight as well? The answer for coefficient is 0.14N according to the book and frictional force is 83N.


    I came up with Fn=605N in relation to the y axis. a=3.03 m/s^2 (alteast what I got)
    I have tried a thousand different ways to do this and I keep getting myself more confused. The number I keep coming up with for coeff is .32 and I don't have a clue about frictional force. Any help would be great.
  2. jcsd
  3. May 25, 2009 #2


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    My first suggestion is to draw a free-body diagram if you have not done so yet. Next, write out the sum of the forces in the x and y directions.
  4. May 26, 2009 #3
    I have written out the free body diagram. the sum of the forces for x I have is 197N and for y it is 605. I am not sure if these are exactly correct. for y I used mg cos theta, and for x I used mg sin theta
  5. May 26, 2009 #4


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    Assuming you've got your coordinate axes oriented properly, you should find that the net force in the y direction is 0. As a suggestion, don't plug numbers in until you arrive at your final expression.

    Some help since I will be signing off for the night:

    Fy = F_N - mgcos(theta) = 0

    Fx = mgsin(theta) - F_f = ma

    Therefore, F_f = mgsin(theta) - ma

    Since F_f is of constant magnitude, the acceleration will be uniform, hence, V_f^2 = 2ax.

    Hopefully you can do the rest from here on out, good luck.

    P.S. I think the answer for the friction force should be 87 N, I'm not positive though, my math could be wrong.
  6. May 26, 2009 #5
    Thanks very much. I got 85 for my frictional force and .14 for coeff. The book says 84N for frictional and .14 for coeff. That should be close enough. I am off one for frictional force, but that could be rounding errors. Thanks again
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