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

1. Apr 29, 2008

### SheldonG

1. The problem statement, all variables and given/known data
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, $$\theta = 10$$, $$\mu_s = 0.60$$?

2. Relevant equations

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

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 29, 2008

### Staff: Mentor

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

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

Last edited: Apr 29, 2008
3. Apr 29, 2008

### SheldonG

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

Sheldon