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