1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: The safely curve
  1. Slippery safe problem (Replies: 16)