# Turning car

1. Sep 16, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
A car enters a turn whose radius is $R$. The road is banked at angle $\theta$, and the coefficient of friction between the wheels and the road is $\mu_s$. Find the minimum and maximum speeds of the car to stay on the road without skidding sideways

2. Relevant equations
Polar coordinates
Newton's laws

3. The attempt at a solution

So I am confused about how to get started on this problem. I know that there are three forces acting on the car (friction, gravity, and normal force). But I am confused about how to put this into equations that involve the radius and the centripetal force. I have tried using polar coordinates, but since the vehicle is on a banked turn, the forces don't really align in any way. How should I get started? (Once I have the equations it seems that the problem is not too hard).

2. Sep 16, 2016

### billy_joule

You don't need to use polar coordinates, Cartesian is easier.
Draw a free body diagram of the road cross section, it should look very similar to the block on a ramp problems you've done on the past.

3. Sep 16, 2016

### Mr Davis 97

All I get is that $F_{friction} = mg \sin \theta$ and $N = mg \cos \theta$. None of these relate to the velocity of the car, so I don't see how I can get a minimum or maximum.

4. Sep 16, 2016

### haruspex

Neither of those equations is correct.
What are the force components in the horizontal direction? What must the net force be in that direction?

5. Sep 16, 2016

### Mr Davis 97

So we're situating our coordinate system not perpendicular to the ramp but perpendicular to the ground?

6. Sep 16, 2016

### haruspex

I feel that will be less confusing in this case because it relates directly to the known accelerations.

7. Sep 16, 2016

### Mr Davis 97

But if I do that won't I end up with 2 equations with four forces, since each force has a component in each direction?

8. Sep 16, 2016

### billy_joule

Have you drawn a free body diagram yet?

The math is very similar regardless of your choice of coordinates. You will need to find vector components either way.
Don't get hung up on which coordinates to choose.
Finding the correct answer to the questions in #4 is much more important.

9. Sep 17, 2016

### haruspex

There are three forces, one of which is known, so two unknowns. You know the angle, so even if you take horizontal and vertical components there are still only two unknowns.
I see billy sees no advantage in using horizontal and vertical axes. Maybe.

10. Sep 17, 2016

### billy_joule

No I do agree with your suggestion in #6, just trying to point out that either coordinates will work fine.

11. Sep 17, 2016

### Mr Davis 97

So I do as you say haruspex, but I keep getting the answer $\displaystyle v_{min} = \sqrt{\frac{gR(1 - \mu)}{\mu \sin \theta + \cos \theta}}$, which is not the correct answer. The correct answer is $\displaystyle v_{min} = \sqrt{\frac{gR(\sin \theta - \mu \cos \theta)}{\mu \sin \theta + \cos \theta}}$. I cannot for the life of me figure out what I am doing wrong...

12. Sep 17, 2016

### Mr Davis 97

Wait, nevermind. I figured out what I was doing wrong. I now have the right answer.

13. Sep 17, 2016

Well done.