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The safely curve

  • #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 :-)
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
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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 :-)
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?
 
  • #3
3/r maybe?
 
  • #4
486
1
The radius of the turn is constant. If you divide [tex]\mu_s[/tex] by 3, what must you do to balance your equation?
 
  • #5
then you must divide the other side of the equation to balance it, so would it be ((v^2/r)/3)?
 
  • #6
486
1
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]?
 
  • #7
I'm not sure...not having the radius measurement is throwing me off
 
  • #8
486
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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] ?
 
  • #9
I tried to divide by 3 again...what is v prime??
 
  • #10
hage567
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I tried to divide by 3 again...what is v prime??
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.
 
  • #11
486
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I tried to divide by 3 again...what is v prime??
v prime is your new velocity. The one that makes the origianl exprssion (v^2/r) get divided by 3.
 

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