Banked curves, coefficient of static friction

In summary: Slipping equations without the term: \dfrac{F_x}{R} = -kx## and \dfrac{F_x}{R} = -k(1-x)##.Slipping equations with the term: \dfrac{F_x}{R} = -k(1+x)##.In summary, the term does not have a significant effect on the results.Just comparing @ergospherical 's slipping equations with and without the ##X## term at 20##\frac{km}{hr}## and a presumed 15 degree bank angle.In summary, @ergospherical 's equations show that the term doesn't have a significant effect on the results.
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Homework Statement
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a problem on icy mountain roads). (a) Calculate the ideal speed to take a 100.0 m radius curve banked at . (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
Relevant Equations
tan(theta)-(v^2)/(rg) = us
where us is coefficient of static friction
The solution in my textbook says that for b, us = 0.234. However when I use the formula above I get 0.2364 which I feel like is too far off. Something must have gone wrong...
Any help would be much appreciated!
Thanks :)
 
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  • #2
What is the angle of the bank?
 
  • #3
In any case, your equation doesn't look correct. Can you show your work? You can write two equations by applying Newton's second law vertically and horizontally respectively, putting ##F = \mu N## in this limiting case.

[You should end up with an equation of the form ##\mu = \left(\tan{\theta} - \dfrac{v^2}{rg} \right)/X##, where ##X \sim 1## is a term depending on ##v,r,g## and ##\theta##...].
 
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  • #4
What you did is that you decomposed the acceleration due to gravity into components normal and perpendicular to the road, but you didn't do the same with the centripetal acceleration.
 
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  • #5
PeroK said:
What is the angle of the bank?
Is it 15 degrees?
 
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In my understanding of how this works, when a student gives a relevant equation it is a given in the problem. I would like to establish if that is true here or if the student derived it first.
 
  • #7
bob012345 said:
In my understanding of how this works, when a student gives a relevant equation it is a given in the problem. I would like to establish if that is true here or if the student derived it first.
At a guess, the student miscopied it.
 
  • #8
haruspex said:
At a guess, the student miscopied it.
Perhaps, but it is a reasonable simplification valid at low speeds as shown by @ergospherical in post#3. It could explain the slight numerical discrepancy.
 
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  • #9
bob012345 said:
Perhaps, but it is a reasonable simplification valid at low speeds as shown by @ergospherical in post#3. It could explain the slight numerical discrepancy.
Have you been thinking about a relativistic solution, then?
 
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  • #10
PeroK said:
Have you been thinking about a relativistic solution, then?
Just comparing @ergospherical 's slipping equations with and without the ##X## term at 20##\frac{km}{hr}## and a presumed 15 degree bank angle.
 
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1. What is a banked curve?

A banked curve is a curved section of a road or track that is angled or tilted towards the inside of the curve. This allows vehicles to travel at higher speeds without slipping or losing control.

2. How does a banked curve affect the coefficient of static friction?

A banked curve increases the coefficient of static friction, which is the measure of the amount of friction between two surfaces that are not moving relative to each other. This increased friction helps to keep a vehicle from sliding or slipping off the road or track while navigating the curve.

3. What factors affect the coefficient of static friction on a banked curve?

The coefficient of static friction on a banked curve is affected by the angle of the bank, the weight and speed of the vehicle, and the surface material of the road or track. A higher bank angle, heavier vehicle, and higher speed will result in a higher coefficient of static friction.

4. How does the coefficient of static friction impact a vehicle's ability to navigate a banked curve?

A higher coefficient of static friction means that the vehicle has a better grip on the road or track, allowing it to navigate the curve at higher speeds without slipping or losing control. However, if the coefficient of static friction is too low, the vehicle may slide or skid off the road or track.

5. Can the coefficient of static friction be manipulated to improve a vehicle's performance on a banked curve?

Yes, the coefficient of static friction can be manipulated by adjusting the bank angle, using different surface materials, or by adding friction-increasing substances such as sand or salt to the road or track. This can help improve a vehicle's performance and safety while navigating a banked curve.

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