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Friction on an inclined plane

  1. Sep 26, 2006 #1
    I am suppose to find the initial velocity of a car before it skids to a stop.
    The car begins it's stop right as it is going up a road that is 20 degrees above the horizontal. The car makes a 15.2 m skid mark before it stops. The coefficient of kinetic friction is 0.60, and static friction is 0.80. The mass of the car and its driver is 1630kg.
    Now this is my assumption, that since the car left a skid mark, that would mean the wheels aren't turning. Which I don't think matters, since it would still be kinetic even if the wheels were turning (?) Since I am stopping, there is no other positive force working against the force of friction. My final velocity is 0 since I stop.

    First thing I should do is find the force of friction.
    Force of friction = (coef) (normal force)
    The normal force is the force parallel to the plane, so it is m*g*cos20
    F(k) = (0.60) (1630kg*9.8m/s^2) = 9010 N

    Next:
    F = ma; 9010N = 1630kg(a) = 5.53 m/s^2
    I think I can use kinematics now, and:
    vf^2 - vi^2 = 2a(x)
    vi^2 = 2(5.53 m/s^2)(15.2m)
    vi = 13.0 m/s

    Does that seem about right? Or am I suppose to use the static friction in there somehow?
     
  2. jcsd
  3. Sep 26, 2006 #2

    SGT

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    No, if the weels were turning you should use static friction. A weel turns because at each instant one of it´s points is static relative to the ground.
    This is why you break more fast if you don´t block your weels, since static friction is greater than the kinectic.
    No, normal is a synonim to perpendicular. The normal force is orthogonal to the plane. It is mg sin(20).
    Redo your calculations with the correct normal force.
     
  4. Sep 26, 2006 #3
    No, it's not (at least not when the x-axis is oriented parrallel to the incline :smile: ).
     
    Last edited: Sep 26, 2006
  5. Sep 26, 2006 #4

    Doc Al

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    Staff: Mentor

    The normal force is perpendicular to the plane, but m*g*cos20 is correct. What happened to the cos20?

    Friction is not the only force acting on the car.

    Kinetic friction applies here. See my comments: https://www.physicsforums.com/showpost.php?p=1093726&postcount=6
     
  6. Sep 26, 2006 #5
    oops, forgot to put it in the equation. But I did solve with it.
    F(k) = (0.60) (1630kg*9.8m/s^2)*cos20 = 9010 N

    Okay, I think I understand why it is kinetic friction for locked wheels now, and static for unlocked. Seemed odd at first, but it makes sense now.

    I believe I am missing the weight of the car on the incline? Now this force should be parallel, not the other, and is F= m*g*sin20.
    F(w) = (1620kg * 9.8m/s^2 * sin20)= 5430N.

    Both forces are against the car, so 5430N + 9010N = 14440N
    F=ma
    14440N = 1630kg * (a)
    a= 8.86 m/s^2
    vf^2 - vi^2 = 2a(x)
    vi^2 = 2(8.86 m/s^2)(15.2m)
    vi = 16.4 m/s

    Does that look better?
     
  7. Sep 26, 2006 #6

    Doc Al

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    Staff: Mentor

    Exactly.
    Isn't the mass 1630 kg?

    Much better. (Just check your calculations for errors. Don't round off until the last step.)
     
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