Speed at which a car on a banked curve will slide.

[SheldonG]

Homework Statement

If a car goes around a banked curve too fast, the car will slide out of the curve. Find an expression for the car speed v_{max} that puts the car on the verge of sliding out. What is the value for R=200, \theta = 10, \mu_s = 0.60?

Homework Equations

a = v^2/r, F = ma.

The Attempt at a Solution

Place the x-axis along the bank, and write the for equations for x and y:

\sum F_x = \frac{mv^2}{R}\cos\theta = mg\sin\theta + f_s
\sum F_y = 0 = F_n - \frac{mv^2}{R}\sin\theta - mg\cos\theta

It seems that the car should start to slide when
\frac{mv^2}{R}\cos\theta > mg\sin\theta + f_s

Using fs = \mu_s F_n

\frac{mv^2}{R}\cos\theta > mg\sin\theta + \mu_s\left(\frac{mv^2}{R}\sin\theta + mg\cos\theta\right)

Solving for v:

v > \sqrt{\frac{gR(\tan\theta + \mu_s/\tan\theta)}{1-\mu_s\tan\theta}

Unfortunately, the figure I get for v using R = 200, theta = 10, coefficient of static friction = 0.60 is much too high (88.6 m/s).

Could someone give me a hint about where I went wrong?

Thank you,
Sheldon

 

[Astronuc]

In this term \mu_s mg\cos\theta, if one divides by cos\theta, then one does not get a tan function, but simply

\mu_s mg


So one would obtain

v > \sqrt{\frac{gR(\tan\theta + \mu_s)}{1-\mu_s\tan\theta}

 

[SheldonG]

Thank you, Astronuc. I am in your debt. What a stupid error :-)

Sheldon