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Homework Help: Calculating the coefficient of kinetic friction homework

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a system that looks like this:

    http://dept.physics.upenn.edu/courses/gladney/phys1/homework/images/half_Atwood.gif [Broken]

    m2 mass is 0.07kg. The m1 mass is gradually increased. Basically, if you tap it, m2 starts moving by itself at constant speed when m1 is at 0.03kg. What is the coefficient of static friction?

    2. Relevant equations

    Fmax=mu * Fn

    3. The attempt at a solution

    First I drew a free body diagram.

    The mass on the table is being pulled to the right with a force of (9.8)(0.03kg) = 0.294N.

    So basically the static force has to be equal that going the other way. Then I did the F=ma for the mass on the table, and for it to move it would need F= mu * 0.686N.

    So we can solve for 0.294N = mu * 0.686N, giving a coefficient of friction of 0.428. Is this right? I am not sure if I am thinking of this the right way.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 24, 2008 #2


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    Hi Lokhtar,

    That looks right to me.
    Last edited by a moderator: May 3, 2017
  4. Oct 24, 2008 #3
    A followup question is "would the coefficient of friction be different if the situation looked like this (angle of 40 degrees)?":

    (Switch m1 and m2, sorry for the inconsistency in labeling).

    I think it would be different, because now you have to find the coefficient of friction only in the x direction. So you'd draw a right triangle and your new coefficient of friction would be:

    mu * cos (40)?

    Is that right?
  5. Oct 24, 2008 #4


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    This problem would be considerably different from the first. The normal force would be different, the weight have a component pulling against the rope, etc. So I would say just start over and do the problem from the beginning.
  6. Oct 24, 2008 #5
    The question is conceptual. There are no numbers given. If you had a certain coefficient of friction on a horizontal surface, and then you changed it to an incline, would the coefficient of friction change?
  7. Oct 24, 2008 #6


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    When you put the question like that, it sounds like a very different situation.

    The coefficient of friction is a property of the surfaces. So if you have steel on aluminum, for example, you look up the coefficient in a table and there it is. It (ideally) does not depend on angle or anything else. (Of course in real life it depends on how dirty the surfaces are, how rough, etc.)

    I'm a bit confused about what is being asked. In this post, it sounds like the situation is that you already have found the coefficient for a particular object on a particular surface when the surface is horizontal; will the coefficient change if the surface is then inclined. The answer to that would be no.

    But back in post #3, the way it was worded made it sound like you are "restarting" the problem, and that the surfaces in parts 1 and 2 are not related. If that is the case you would need to redo the problem to find the coefficient of friction for the new surface.

    Does this make sense?
  8. Oct 24, 2008 #7
    Yes, I think so. The question is a followup which simply asks if the coefficient of friction would change if that flat surface were now placed on an incline (weights and everything would remain the same, except this time there would obviously be no movement due to the incline). It doesn't ask us to calculate anything.

    I think you've answered my question. I thought you'd have to break down the friction into x and y components, but if it is a property of the surface itself, then you wouldn't do that, and thus it would remain the same.
  9. Oct 24, 2008 #8


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    That sounds right to me.

    However, when you say that there is no movement due to the incline, that would not necessarily be the case. Since m2 is larger than m1, when you raise the incline enough, m2 would move down the incline.

    In fact, at 40 degrees if there were no friction m2 would already by moving down. (The component of the weight of m2 down the incline is greater than the total weight of m1.) This could be important, because if you need to draw the force diagram at that angle, the frictional force is up the incline, towards m1.
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