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Prove of formula

  1. Oct 7, 2007 #1
    I am trying to prove that the skid to stop formula works with vehicles of different weight, with the use of kinetic energy. I think I have proven this, but if someone would like to check this over to make sure I am correct I would appreciate it.

    13000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7

    13000 * 21 * .7 = 191100 joules

    V = sqr ((2gKE)/w)
    V = sqr ((2* 9.81*191100))/13000)
    V = sqr (3749382/13000)
    V = sqr 288.414
    V = 16.98 m/s
    S = 16.98 / .2777
    S = 61.14 km/h

    2000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7

    2000 * 21 * .7 = 29400 joules

    V = sqr ((2gKE)/w)
    V = sqr ((2* 9.81*29400))/2000)
    V = sqr (576828/2000)
    V = sqr 288.414
    V = 16.98 m/s
    S = 16.98 / .2777
    S = 61.14 km/h

    This is the skid to stop formula.

    S = sqr (254 µ d ))
    S = sqr (254 * .7 * 21))
    S = sqr 3733.8
    S = 61.10 km/h
     
    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You're doing fine (except for mixing up weight and mass). But rather than plugging in specific numbers, try to prove it symbolically:

    The only force stopping the vehicle is friction, which equals: [itex]\mu mg[/itex]

    What's the acceleration due to that force? Does it depend on mass?

    Or you can use kinetic energy like you did. Set the work done by friction over a stopping distance D ([itex]W = FD = \mu mg D[/itex]) equal to the kinetic energy ([itex]{KE} = 1/2 m v^2[/itex]) and solve for D in terms of speed. Does it depend on mass?
     
  4. Oct 7, 2007 #3
    Thank you very much for your responce and your vote of encouragment.


    No, when I try it now I loose joules

    .5 * (13000 / 9.81 ) * V² = 191037.98 Joules

    I do not know why it is that I have now lost 62.02 joules

    w = FD = µmgD

    .7 *1325.178389 * 9.81 * 21 = 191100 joules ( now I get my joules back again)

    So answering your question, I do not think it depends on mass.
    But I am really confused now because I do not even know if this is what you wanted me to do??? This stuff is really hard for me to understand.
     
  5. Oct 7, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Why did you divide the mass by 9.81? Where did you get the speed?

    Where did you get the 1325.178389? The mass you are using is 13000 kg, right?

    I wanted you to set up the equation: Work = KE and solve for stopping distance (or speed).

    [tex]\mu mg D = 1/2 m v^2[/tex]

    If your math skills are up to it, you may see right away that mass (m) cancels out. So D does not depend on mass.
     
  6. Oct 8, 2007 #5
    I am sorry I thought you wanted me to change the weight to mass which is why I divided the weight by gravity.
    As for the velocity, I got it by using the skid to stop equation.

    S = sqr (254 µ d ))
    S = sqr (254 * .7 * 21))
    S = sqr 3733.8
    S = 61.10 km/h * .2777 = 16.96 m/s

    I got 1325.178389 from dividing the weight of 13000 kg / 9.81 to convert to mass.

    I do believe that I misunderstood what it was you were saying to me.

    µmgD = .5mV²

    Yes I do see that mass cancels out becuase it is the same on both sides
    µgD = 144 or
    .5V² = 144
    So mass in this instance is not relevant.


    So what was it you wanted me to do with the above statement? "Does it depent on mass"? The answer is No.
    Forgive my stupidity.
     
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