Banked Turn: Navigating Car Dynamics on Curves

[Mr Davis 97]

Homework Statement

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

Homework Equations

Polar coordinates
Newton's laws

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).

 

[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.

 

[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.

 

[haruspex]

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.

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

 

[Mr Davis 97]

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

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

 

[haruspex]

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

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

 

[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?

 

[billy_joule]

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?

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.

 

[haruspex]

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?

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.

 

[billy_joule]

I see billy sees no advantage in using horizontal and vertical axes. Maybe.

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

 

[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...

 

[Mr Davis 97]

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

 

[haruspex]

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

Well done.