Pig on a slide Friction Problem

  1. A slide loving pig slides down a certin 35 degree slide in twice the time it would take to slide down a frictionless 35 degree slide. What is the coefficient of kinetic friction between the pig and the slide? Here is what I have so far....
    [sum]Fx=ma=mgsin [the] -[mu]mgcos[the] I canceled out the mass on both sides and get
    [sum]Fx=a=gsin [the] -[mu]gcos[the]

    For the y forces I have
    [sum]y=N=mgcos[the]

    As far as the difference in the time with or with out friction I have x=mgsin[the]=(1/2)a[tex]t^2[/tex] for the slide with no friction and I have x=mgsin[the]=2a[tex]t^2[/tex]for the slide with friction
    This is all the info I have been able to come up with...I can't figure out how to put it all together. Thanks for helping. I've been working on this one for awhile.
     
  2. jcsd
  3. jamesrc

    jamesrc 477
    Science Advisor
    Gold Member

    Be careful about mixing up forces, accelerations and positions. For example, your second equation has forces equalling accelerations (not good). I'm assuming the 'x' direction is along the ramp and that the pig starts at the top of the ramp from rest (I'm told pigs like to rest). So lets call the distance down the ramp (same for friction/no friction) Δx. In either case,

    [itex] \Delta x = \frac{at^2}{2} [/itex]

    which is where (it seemed like) you were heading.

    The acceleration is different in each case and the time is different in each case. Let's call the time down the frictionless slide T, and the time down the slide with friction 2T. From your force balance, you found the acceleration with friction already, let's call it af:

    [itex] a_f = g\sin \theta - \mu g\cos \theta [/itex]

    without friction (anf) is simply:

    [itex] a_{nf} = g\sin\theta [/itex]

    To put it all together, equate Δxf with Δxnf:

    [itex] \frac{a_f(2T)^2}{2} = \frac{a_{nf}T^2}{2} [/itex]

    substitute in your expressions for af and anf and solve for μ

    [itex] 4\left(g\sin \theta - \mu g\cos \theta\right) = g\sin\theta [/itex]

    [itex] \vdots [/itex]

    [itex] \mu = 3\tan\theta [/itex]

    (I suggest you verify my algebra for yourself in case I made a careless error or two.)


    Edit:
    (Last line corrected for algebra)

    [itex] \mu = \frac{3\tan\theta}{4} [/itex]
     
    Last edited: Nov 17, 2003
  4. Thanks

    Funny how it seems so obvious when someone explains it. Thanks for your help.
     
  5. Algebra Corrections

    It turns out to be (3/4)Tan[the] I got the final answer to be [mu]=.53 Ahhhh, now I can sleep at night[zz)]
     
  6. jamesrc

    jamesrc 477
    Science Advisor
    Gold Member

    Good job. Glad I put in that disclaimor; maybe I should just add it to my signature. I guess I was so wrapped up in using the new [itex] \LaTeX [/itex] functionality that I forgot to distribute the 4.
     
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