I have a question with which I am having trouble. A 71m radius curve is banked for a design speed of 91km/h. Given a coefficient of static friction of 0.32, what is the range of speeds in which a car can safely make the turn?

I'm confused. Do I not need to know the mass of the vehicle for this question, so that I can calculate the forces etc and find how much force I need/how much force is available?

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I could be wrong, but I believe you'll need the bank angle. –  Ataraxia Apr 13 '13 at 14:09
According to the formula in my textbook, the angle is given by tan theta = v^2/rg, yielding 42 degrees, I'm pretty sure. –  user23100 Apr 13 '13 at 14:10
Mass is needed so that net friction force can be calculated which could help in solving the problem –  Hash Apr 13 '13 at 15:03

Consider the centripetal force. There will be two forces acting on the vehicle: the normal force of the road, since the curve is inclined toward the center of the curve, and the static friction force from the traction of the wheels.

$$\frac{mv^2}{r}=f_{static}cos(\theta)+Nsin(\theta)$$

Since there is only acceleration in the radial direction, the forces in the direction perpendicular to the radial direction cancel one another:

$$Ncos(\theta)=f_{static}sin(\theta)+mg$$

Solving for m in the second equation and substituting it into the first, you get:

$$\frac{v^2(Ncos(\theta)-\mu_{static}Ncos(\theta))}{rg}= \mu_{static}Nsin(\theta)+Nsin(\theta)$$

Finally, solve for v, which will give you two different solutions. These are the minimum and maximum speeds required for the vehicle to remain on the curve. Notice that m cancels out. This is because all the forces involved, gravity, friction, and normal force, act relative to the mass of the vehicle.

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$v=\sqrt{\dfrac{rg(\sin \theta+ \mu_s\cos\theta}{\cos \theta-\mu_s \sin \theta}}$
Where $\mu_s, r$ is given. You can find $\theta$ using the formula you just posted in comments. And do refer here.