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Mass on an inclined plane with friction as a function of distance

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A small body starts sliding down an inclined plane. The angle between the plane and the horizon is [tex]\alpha[/tex]. The coefficient of friction depends on the distance [tex]s[/tex] that the body travels according to [tex]\mu = \beta s[/tex].
    Find the distance that the body travels until coming to a stop and its maximum speed.

    2. The attempt at a solution

    The force equation for the plane axis which I'm pretty sure I got correctly is
    [tex] g\sin{\alpha} - \beta s g\cos{\alpha} = a [/tex]

    That is after simplifying and substituting what is required. This is a differential equation which I'm not sure how to solve. First, is there any problem with treating the acceleration as the second derivative of the distance (rather than of the position, which appears the same in this case). Second when I do try to find a homogenous solution I get an imaginary root which I'm not sure is ok. Am I even on the right track ? Or is there something I'm totally missing ?
     
  2. jcsd
  3. Nov 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Welcome to PF! :smile:

    (have an alpha: α and a beta: β and a mu: µ :wink:)
    Yes, essentially you have s'' = A - Bs,

    and if you put r = s - A/B, r'' = s'', that's r'' = -Br, which is simple harmonic motion, r = C1cos(√B)t + C2sin(√B)t (except, of course, that it gets stuck when it reaches maximum "amplitude" :wink:)
    Yes, that's normal for simple harmonic motion … imaginary roots correspond to cos and sin, while real roots correspond to exp. :smile:
     
  4. Nov 24, 2009 #3
    O.K. Thank you :)

    This clears up things a bit :smile: I need to look into that and harmonic motion more carefully.. (it is something which I actually haven't studied yet)
     
  5. Nov 25, 2009 #4
    I got a clue from somewhere which seems to hint that I can do it another way
    The clue says to multiply both sides of the equation by [tex]\frac{ds}{dt}[/tex] but I have no clue what this clue gives me :). That is, how to continue from there..
     
  6. Nov 25, 2009 #5

    rl.bhat

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    You have given the equation of acceleration.
    g*sinα - β*s*g*cosα = a.------(1)
    Now a = dv/dt = dv/ds*ds/dt = (v*dv)*(1/ds)
    Put this value in eq(1) and simplify. You get
    g*sinα*ds - β*g*cosα*s*ds = v*dv.
    Find the integration on the both side and put v = 0 to find s.
     
  7. Nov 25, 2009 #6

    tiny-tim

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    Yes, if s'' = f(s), you can multiply both sides by s', giving:

    s's'' = f(s)s'

    The LHS is the derivative of 1/2 (s')2 (incidentally, that's exactly where the 1/2 mv2 in KE comes from :wink:), and RHS is the derivative of ∫f(s)ds. :smile:
     
  8. Nov 25, 2009 #7
    So the point of this method is to move from a second order (and non-separable?) DE to a separable DE ?
     
  9. Nov 25, 2009 #8

    tiny-tim

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    No :confused:, the point is that you start with a second-order equation which needs integrating twice, and first you integrate it once. :smile:
     
  10. Nov 25, 2009 #9
    How do I integrate this part: β*g*cosα*s*ds
    I need to find an expression which when differentiated with respect to s gives s
    I thought it might be (β*g*cosα*s^2)/2 but then according to the chain rule when differentiating it should be multiplied by s' . I'm sorry but I'm really weak with DE ..
     
  11. Nov 25, 2009 #10

    tiny-tim

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    Sorry, but you're not making sense :redface:

    when you differentiate wrt s, there's no chain, is there? :wink:
     
  12. Nov 25, 2009 #11
    True. In that case I was just weak with making sense. Thank you :)
     
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