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Tires mark after braking

  1. Jul 27, 2005 #1
    Does anybody know the empirical formula that relates the initial velocity of a car in a street with the lenght of the mark on ground left by the tires when the brakes are set at maximum force, stopping the wheel ?
     
  2. jcsd
  3. Jul 27, 2005 #2

    krab

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    [tex]v=\sqrt{2ad}[/tex]
    a is somewhere between 0 (on ice) and 1 g (ideal conditions, really good rubber). A g is 32 ft/s/s. So for example if a is .75 g (sorta typical), d = 100 ft,
    [tex]v=\sqrt{2*.75*32*100}=69\,\mbox{ft/s}[/tex]
    or 47 mph.
     
  4. Jul 28, 2005 #3
    Now I realized that dimensional analysis could have helped...

    But is there any simple way to derive this expression ?
     
  5. Jul 28, 2005 #4

    Doc Al

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    It's just an example of uniformly accelerated motion. Are you familiar with the kinematics of uniform acceleration?
     
  6. Jul 28, 2005 #5
    Ok, I got it now. But the reason it seems to me as a little more envolved problem was the presence of gravity acceleration in the explanation and formula of Mr. Krab. Was it because the friction coeficient can reach some sort of "maximum value" in ideal situations? It still doesn't make sense to me.

    But the mention to Torricelli equation has cleared the way a lot.
     
  7. Jul 28, 2005 #6

    Doc Al

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    The force of friction depends on the coefficient of friction and the normal force. The normal force is just the weight = mg. Thus the acceleration [itex]a = F/m = ( \mu m g)/ m = \mu g[/itex].
     
  8. Jul 29, 2005 #7
    Ok. I am satisfied with the answer.
    Just another complemetary question regarding something Mr. Krab has said:

    [tex] v = \sqrt{2 a g} [/tex]

    a is somewhere between 0 (on ice) and 1 g (ideal conditions, really good rubber). A g is 32 ft/s/s.

    Is it true that [itex] \mu [/itex] is 1 in the friction between tires' rubber and the asphalt ?
     
    Last edited by a moderator: Jul 29, 2005
  9. Jul 29, 2005 #8

    Stingray

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    Yes, it is about 1-1.2 for good tires on hot asphalt. It's quite a bit less (~0.6) if the wheels are locked.
     
  10. Jul 29, 2005 #9
    Thank you so much....
    But friction coeficient can be much larger than that for other pairs of materials isn't it?
     
  11. Jul 29, 2005 #10

    Stingray

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    Well, racing tires can generate quite a bit more friction (as can regular tires on special surfaces), but [tex]\mu[/tex] still won't go much over 2 or 3. For most pairs of materials, it is less than 1.

    Also note that there are different types of friction, and the one being discussed here (Coulomb friction) is pretty limited. I'm only talking about cases where [tex]F_{\rm{max}}= \mu N[/tex] is a reasonable approximation. It's pushing it for street car tires, and not very good at all for certain types of racing tires.
     
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