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Find the kinetic friction coefficient

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A pig is sliding down a ramp that has an angle of 35 degrees from the horizontal plane. If the friction coefficient was 0 the ride would only take half the time. What is the kinetic friction coefficient between pig and ramp?


    2. Relevant equations
    Fnet = ma
    v² = u² + 2as
    a = [itex]\Delta[/itex]v/t
    fk = ukN = ukmgsin[itex]\theta[/itex]


    3. The attempt at a solution
    Since I wasn't given a length of the ramp, I decide it to be s=10 meters (the length shouldn't matter, I just use the value so I can perform calculations).
    u = 0 m/s
    s = 10 m
    a = gsin[itex]\theta[/itex] = 5.62 m/s²
    v = [itex]\sqrt{2as}[/itex] = [itex]\sqrt{2*5.62*10}[/itex] = 10.6 m/s
    t = (v-u)/a = 10.6/5.62 = 1.87 s
    This is the time it takes without any friction. Now let's double this time, and find the acceleration. Then, through the expressions of the forces, solve for uk:

    t = 3.74 s
    u = 0 m/s
    s = 10 m
    v = s/t = 2.67 m/s
    a = (v-u)/t = 0.71 m/s²

    F = ma = mgsin[itex]\theta[/itex] - ukgcos[itex]\theta[/itex]
    Rearrange and you'll get this
    uk = [itex]\frac{a + gsin\theta}{gcos\theta}[/itex] = [itex]\frac{0.71 + 9.8sin(35)}{9.8cos(35)}[/itex] = 0.787

    This answer isn't correct, but it's rather close (probably by coincidence). Anyhow, can anyone hint me in the right direction? Where am I going wrong?
    Thanks
     
  2. jcsd
  3. Jul 25, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    First, you should use more decimal places in your intermediate results. Maybe three or four. This will guard against rounding errors creeping into 'sensitive' calculations. Also, you might want to use g = 9.807 m/s2 for gravity.

    The formula that you arrived at for μk is not quite right; it produces a negative value. This is because the numerator should be negated.
     
  4. Jul 25, 2011 #3
    Ah you're right. I see the mistake.

    F = ma = mgsin[itex]\theta[/itex] - ukgcos[itex]\theta[/itex]
    Rearrange and you'll get this
    uk = [itex]\frac{a + gsin\theta}{gcos\theta}[/itex] = [itex]\frac{0.71 + 9.8sin(35)}{9.8cos(35)}[/itex] = 0.787

    should be

    F = ma = mgsin[itex]\theta[/itex] - ukmgcos[itex]\theta[/itex]
    Rearrange and you'll get this
    uk = [itex]\frac{a - gsin\theta}{-gcos\theta}[/itex] = [itex]\frac{0.71 - 9.8sin(35)}{-9.8cos(35)}[/itex] = 0.613

    When I got 0.613 as my final answer I used 5+ decimals and I used 9.807 for g. This is closer to the correct answer, but I'm quite sure I won't pass with this answer.

    The correct answer is 0.52
    Any ideas?
     
  5. Jul 25, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    I would change the approach to finding the acceleration for the friction case. Using an average velocity to find acceleration is not good.

    Instead, go directly from the kinematic formula s = (1/2)at2. You're given t and s, so find a.
     
  6. Jul 25, 2011 #5
    Thank you, I solved it!
    Not by finding acceleration in the case of friction first, but thanks to you reminding me of that what I calculated was the average velocity. These two lines messed it up:
    v = s/t = 2.67 m/s
    a = (v-u)/t = 0.71 m/s²
    Where it says (v-u), it should say 2*v. Since v = average velocity and u = 0, then 2*v is the change of velocity. And right we are!! 0.52 is the answer.

    Again thanks for your help! :-]
     
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