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Homework Help: The safely curve

  1. Mar 27, 2007 #1
    A car is safely negotiating an unbanked circular turn at a speed of 23 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static friction force by a factor of three. If the car is to continue safely around the curve, to what speed must the driver slow the car?

    i have it so where mu_s = v^2/r but I don't know if thats right, help please! thanks :-)
  2. jcsd
  3. Mar 27, 2007 #2


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    I think you mean mu_s(g) = v^2/r. But in any case, if mu_s is reduced by 3, then v^2 must be reduced by what factor to keep the car from slipping?
  4. Mar 27, 2007 #3
    3/r maybe?
  5. Mar 27, 2007 #4
    The radius of the turn is constant. If you divide [tex]\mu_s[/tex] by 3, what must you do to balance your equation?
  6. Mar 27, 2007 #5
    then you must divide the other side of the equation to balance it, so would it be ((v^2/r)/3)?
  7. Mar 27, 2007 #6
    Exactly. But the only thing you're allowed to change is v.
    What should you do to v to get the whole expression to equal [tex]{\frac{v^2}{3r}[/tex]?
  8. Mar 27, 2007 #7
    I'm not sure...not having the radius measurement is throwing me off
  9. Mar 27, 2007 #8
    It doesn't matter what the radius is. It doesn't change.
    You have [tex]\frac{v^2}{r}[/tex]
    You need to divide all of that by three, but you're NOT allowed to touch r. All you're allowed to change is v. What might you do to v so that

    [tex]\frac{v^2}{3r}=\frac{(v')^2}{r}[/tex] ?
  10. Mar 29, 2007 #9
    I tried to divide by 3 again...what is v prime??
  11. Mar 29, 2007 #10


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    The speed the driver must slow down to in order to continue along the curve safely. This is what you are trying to solve for.
  12. Apr 1, 2007 #11
    v prime is your new velocity. The one that makes the origianl exprssion (v^2/r) get divided by 3.
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