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- Thread starter DaTario
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krab

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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.

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But is there any simple way to derive this expression ?

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Doc Al

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But the mention to Torricelli equation has cleared the way a lot.

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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].DaTario said: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.

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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 ?

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 ?

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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.DaTario said:Is it true that [tex] \mu [\itex] is 1 in the friction between tires' rubber and the asphalt ?

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But friction coeficient can be much larger than that for other pairs of materials isn't it?

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Stingray

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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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