1. ## Banked Curve

Physics problem here.

A curve with a $140 m$ radius on a level road is banked at the correct angle for a speed of $20 {\rm m/s}$.

If an automobile rounds this curve at $30 {\rm m/s}$, what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

A friend got me started: $\tan \theta=\frac{20^2}{(9.8)(140)}$, $\theta=16.25$ but I'm not sure why this is important or what to do with it to find $\mu_s$ (friction coefficient).

2. Oh, and one more if you have the time. I have no idea where to begin with this one.

It's wanting to know the velocity of the bus.

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.530 kg suspended from the ceiling of the bus by a string of length 1.79 m is found to hang at rest relative to the bus when the string makes an angle of 34.0º with the vertical. In this position the lunch box is a distance 54.0 m from the center of curvature of the curve.

3. Originally Posted by cinder
Physics problem here.

A curve with a $140 m$ radius on a level road is banked at the correct angle for a speed of $20 {\rm m/s}$.

If an automobile rounds this curve at $30 {\rm m/s}$, what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

A friend got me started: $\tan \theta=\frac{20^2}{(9.8)(140)}$, $\theta=16.25$ but I'm not sure why this is important or what to do with it to find $\mu_s$ (friction coefficient).
Make a quick sketch of the situation looking at a cross-section of the road. My sketch has the road banking at a slope of angle $\theta$ with the horizontal up and to the right. (So the center of the curve is to the left in this diagram.)

If the road is banked correctly then a vehicle going around the curve will not need any friction in order to not slide down the bank. So assume we have a vehicle traveling at that speed around this curve.

What forces do we have on this vehicle? There is a weight w acting straight down and there is a normal force N acting perpendicular (and upward) from the plane of the bank. (Remember we are assuming no friction for this case.) Do NOT put a centripetal force in your Free-Body Diagram!!

I am going to choose what might at first seem to be a bizarre coordinate system: I'm going to choose a +y axis straight upward and a +x axis in the direction of the center of the circle. (Some would call this a +r axis since it changes as the vehicle moves around the curve.) We need to find components of the forces in this coordinate system.

Well w is easy: it's pointing straight down. What about N? If you work at the geometry for a few minutes, you will find that $N_x = Nsin(\theta)$ and $N_y = Ncos(\theta)$.

So let's do Newton's 2nd in the coordinate directions. I'm going to do the y direction first since we are going to need to know the normal force for the x direction.
$\sum F_y = Ncos(\theta) - w = 0$ <-- The vehicle can't accelerate in this direction if the curve is banked properly.

$N = \frac{w}{cos(\theta)} = \frac{mg}{cos(\theta)}$

Now in the x direction the net force is what? The centripetal force, of course! (This is why I told you not to put the centripetal force in the FBD: centripetal force is a net force, not necessarily a single force acting on an object.)
$\sum F_x = Nsin(\theta) = F_c$
(Remember the weight has no component in this direction.)

So
$Nsin(\theta) = \frac{mv^2}{r}$
where r is the radius of the curve.

Inserting our value of N into this:
$\frac{mg}{cos(\theta)} \cdot sin(\theta) = \frac{mv^2}{r}$

So
$\frac{sin(\theta)}{cos(\theta)} = \frac{v^2}{gr}$

$tan(\theta) = \frac{v^2}{gr}$

This is the equation your friend gave you.

Now, this gives you the angle of the bank given the radius of the curve and the "correct" speed to go around it. We need to know the minimum static friction coefficient required to keep a car from sliding when it is going at a higher speed around the same curve.

So we've got a number for $\theta$. Let's redo this problem assuming that we've got a static friction force in play.

We have the same diagram as before, but now we've got a static friction force. What direction is it in? Well, the car is going faster than the recommended speed, so the tendency of the car would be to slide up the bank. Thus static friction is acting against this: down the bank.

We need to find components of this force $f_s$ in our coordinate system. By looking at the geometry again I get that
$f_{s,x} = f_scos(\theta)$
and
$f_{s,y} = f_ssin(\theta)$

Now, we are looking for the minimum coefficient of static friction we need, so we are looking at the maximum possible value of static friction. (Think over this one for a bit and make sure you understand why.) Thus $f_s = \mu _s N$.

Now do your sum of forces in the y direction to get a value for the normal force (in terms of $\mu _s$), and use that value in the sum of forces in the x direction to get an equation for $\mu _s$ in terms of the speed of the car. Remember the net force in the y direction will be 0 and the net force in the x direction is the centripetal force.

There is a lot of algebra involved here in solving for $\mu _s$ so for reference I will give you the x direction equation and the final equation. Make sure you can derive them yourself.

For the sum of forces in the x direction I get:
$\left ( \frac{mg}{cos(\theta) - \mu_s sin(\theta)} \right ) sin(\theta) + \mu _s \left ( \frac{mg}{cos(\theta) - \mu_s sin(\theta)} \right ) cos(\theta) = \frac{mv^2}{r}$

For the solution of $\mu _s$ I get:
$\mu _s = \frac{\left ( \frac{v^2}{gr} \right ) cos(\theta) - sin(\theta)}{\left ( \frac{v^2}{gr} \right ) sin(\theta) + cos(\theta)}$

-Dan