Banked curves, coefficient of static friction

[mncyapntsi]

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 :)

 

[PeroK]

What is the angle of the bank?

 

[ergospherical]

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 $\theta$...].

 

[jack action]

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.

 

[PeroK]

What is the angle of the bank?

Is it 15 degrees?

 

[bob012345]

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.

 

[haruspex]

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.

 

[bob012345]

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.

 

[PeroK]

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?

 

[bob012345]

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.