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Stopping distance inclined plane

  1. Aug 3, 2009 #1
    If the stopping distance with kinetic to static friction is:

    d=vi2/2ukg

    Then,

    What is the stopping distance equation on an inclined plane?
     
  2. jcsd
  3. Aug 3, 2009 #2
    How did you get that forumula?

    Do the same thing, except you have to deal with kinetic energy being lost to friction and being converted to gravitational potential energy.
     
  4. Aug 3, 2009 #3
    Well,

    I have that:

    (initial velocity) vi=20m/s
    (kinetic coefficient) uk=0.180
    (angle) [tex]\Theta[/tex]=5*

    So,

    In that: d=vi2/2ukg

    The only addition would have to be of sin5 at the denominator?
     
  5. Aug 3, 2009 #4
    The formula is correct.
    It's derived from :

    The acceleration of the mass ( thing ) is determined by [tex]\frac{Fx}{m}[/tex].
    The force acting on the mass is only [tex]-Fk[/tex].
    So, the acceleration of the mass is [tex]-g*\mu k[/tex].....1
    So ( input the first equation ) :
    [tex]v' = v + a*t[/tex]
    [tex]t = \frac{V}{g*\mu k}[/tex]......2
    And
    [tex]s = vt +\frac{1}{2}at^2[/tex]
    use the first and the second equation
    [tex]s = \frac{v^2}{2g*\mu k}[/tex]

    Hope you can derive the equation of stopping distance on inclined plane !
     
  6. Aug 3, 2009 #5
    Oh, your killing me with all these edifying intimations. Yes, indeed, I will try! :rofl:

    Thank youu..
     
  7. Aug 3, 2009 #6
    s= vi2/2gsin[tex]\Theta[/tex]uk

    is that correct?
     
  8. Aug 3, 2009 #7
    That is only correct for a flat incline. For a complete answer, you would have to take into consideration the normal force and the net force relative to the incline.
    Remember:
    [tex]F_{friction}\equiv \mu_k N[/tex]
    And on an inclined plane, there is also a component of gravity opposite the direction of friction. You must take that into consideration as well, when calculating the net acceleration of the mass.

    From there, kinematics should suffice.

    A work-energy approach would work here as well, but that's outside your scope at the moment, right?
     
  9. Aug 3, 2009 #8
    Correct, out of the scope. I only need the accelration. Only problem, I totally agree with your remark on the normal force. But you need a mass to figure out this. In this case the skiers mass and force is omitted. I only was given the initial velocity, kinetic coefficient, and angle of descent.

    n = mgcos

    and this Fg opposite to the friction?

    well, this = mgsin

    However, note that in the above the mass drops out.

    Ergo, why I thought that the equation would be:

    s= vi2/2gsin[tex]\Theta[/tex]uk

    Then again... mgsin[tex]\Theta[/tex] - ffk = max=ma

    So, a = gsin[tex]\Theta[/tex]-ffk

    Correct?

    So,

    s= vi2/2gsin-ffk
    I think... maybe not, still.

    If I do not have the mass where as g is constant I am stuck, right?
     
  10. Aug 3, 2009 #9
    The mass is irrelevant. You are only asked about velocities and accelerations. As all of your accelerations are tied in to the gravitational acceleration, which is uniform regardless of mass, you can use a rearranged form of Newton's second law to find the acceleration.

    [tex]\vec a=\frac{\vec F}{m}[/tex]

    Also, you have a mistake in your very final line.

    Starting from [tex]v_f^2=v_0^2+2ad[/tex]
    Isolating [tex]d[/tex] provides us with:
    [tex]d=\frac{v_f^2-v_0^2}{2a}[/tex]

    Our final velocity is 0, since we're looking for the stopping distance, and the net acceleration of the mass can be found using Newton's second law:
    [tex]\vec a=\frac{\vec F}{m}[/tex]

    The net force in the direction of the slope is, as you've found:
    [tex]\Sigma \vec F=mg\sin{\theta}-F_{friction}[/tex]
    [tex]\Sigma \vec F=mg\sin{\theta}-\mu_k*N[/tex]

    Find [tex]N[/tex], and the problem should become trivial.
    Try drawing an FBD, and everything should become clear. Break gravity down into its components, and remember that the mass is at equilibrium in the direction perpendicular to the slope.
     
  11. Aug 3, 2009 #10
    [tex]\sum[/tex][tex]\textbf{F}[/tex]=mgsin[tex]\theta[/tex]-[tex]\mu[/tex]k*mgcos[tex]\theta[/tex]

    Okay, I see now... Thank you.
     
  12. Aug 3, 2009 #11
    [tex]a=gsin{\theta}-\mu_k*gcos{\theta}[/tex]

    thus,

    [tex]d=\frac{v_f^2-v_0^2}{2gsin{\theta}-\mu_k*gcos{\theta}}[/tex]


    ? :D
     
  13. Aug 4, 2009 #12
    Close enough, just don't forget that the 2 in the denominator encompasses both components of the acceleration.
     
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