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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

2 Answers 2

up vote 2 down vote accepted

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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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.

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Even if correct a unexplained formula like this is not a particularly good answer, because it provides no basis for understanding nor any indication of how this is known. Nor does pointing to wikipedia help much because Stack Exchange sites seek to be repositories of answered questions and not mere link farms. –  dmckee Apr 14 '13 at 20:38
    
I will see to that, from next time I have my formula with explanation (or derivation). –  Inceptio Apr 15 '13 at 14:48

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