# Cars on a banked road

#### Chokfull

A banked circular road is designed for traffic moving at 60 km/h. The radius of the curve is 200m. On a rainy day (I think this means to assume there is no friction in determining the bank angle) traffic moves at 40km/h. What is the minimum coefficient of friction between the tires and the road that will allow cars to negotiate the turn without sliding off the road?

I use the formula $$\displaystyle \theta = \tan^{-1} \frac {v^2} {gR}$$ to get the bank angle of the road, assuming theres no friction on the rainy day. I don't know what to do from here, though, to find the coefficient of static friction between tires and the road.

#### hollywood

There are three forces acting on the car - gravity, a normal force N by the road, and a frictional force by the road. Since the car is accelerating at $$\displaystyle \frac{v^2}{r}$$ (the centripetal acceleration), these must add up to a force $$\displaystyle \frac{mv^2}{r}$$ acting horizontally. Since we are looking for the minimum coefficient of friction, set the frictional force to be the maximum possible, given by $$\displaystyle \mu{N}$$ where $$\displaystyle \mu$$ is the coefficent of friction.

So we have:
$$\displaystyle mg$$ (straight down) plus
$$\displaystyle N$$ (perpendicular to the road, up and toward the center) plus
$$\displaystyle \mu{N}$$ (parallel to the road, up and away from the center) equals
$$\displaystyle \frac{mv^2}{r}$$ (horizontal toward the center)

First, calculate $$\displaystyle \frac{N}{m}$$ by looking at the components perpendicular to the road:
$$\displaystyle -mg\cos{\theta}$$ plus
$$\displaystyle N$$ plus
$$\displaystyle 0$$ equals
$$\displaystyle \frac{mv^2}{r}\sin{\theta}$$

Then calculate $$\displaystyle \mu\frac{N}{m}$$ (and then $$\displaystyle \mu$$) by looking at the components parallel to the road:
$$\displaystyle mg\sin{\theta}$$ plus
$$\displaystyle 0$$ plus
$$\displaystyle -\mu{N}$$ equals
$$\displaystyle \frac{mv^2}{r}\cos{\theta}$$

You should be able to take it from there. If you're still having trouble, go ahead and post again in this thread.

- Hollywood

#### Chokfull

There are three forces acting on the car - gravity, a normal force N by the road, and a frictional force by the road. Since the car is accelerating at $$\displaystyle \frac{v^2}{r}$$ (the centripetal acceleration), these must add up to a force $$\displaystyle \frac{mv^2}{r}$$ acting horizontally. Since we are looking for the minimum coefficient of friction, set the frictional force to be the maximum possible, given by $$\displaystyle \mu{N}$$ where $$\displaystyle \mu$$ is the coefficent of friction.

So we have:
$$\displaystyle mg$$ (straight down) plus
$$\displaystyle N$$ (perpendicular to the road, up and toward the center) plus
$$\displaystyle \mu{N}$$ (parallel to the road, up and away from the center) equals
$$\displaystyle \frac{mv^2}{r}$$ (horizontal toward the center)

First, calculate $$\displaystyle \frac{N}{m}$$ by looking at the components perpendicular to the road:
$$\displaystyle -mg\cos{\theta}$$ plus
$$\displaystyle N$$ plus
$$\displaystyle 0$$ equals
$$\displaystyle \frac{mv^2}{r}\sin{\theta}$$

Then calculate $$\displaystyle \mu\frac{N}{m}$$ (and then $$\displaystyle \mu$$) by looking at the components parallel to the road:
$$\displaystyle mg\sin{\theta}$$ plus
$$\displaystyle 0$$ plus
$$\displaystyle -\mu{N}$$ equals
$$\displaystyle \frac{mv^2}{r}\cos{\theta}$$

You should be able to take it from there. If you're still having trouble, go ahead and post again in this thread.

- Hollywood

OK, I don't understand your calculations fully but I haven't taken a lot of time to look at them yet. However, I did have trouble with some parts (hilighted in red). Firstly, you said there are three forces acting on the car, but you left out the force pushing the car to the center, but then you added it in later. I assume that was just a mistake.

2nd, $$\displaystyle \mu{N}$$ is towards the center; it is the frictional force, which helps keep cars on the road (and thus closer to the center), not pushing them off it.

3rd, your final equation had m in it, but if you look at my original post you see that I was not given the value for mass. This is a large reason why I am having trouble with the problem; it has to work for anywhere from a bicycle to an 18-wheeler.

MHF Helper

Chokfull

#### hollywood

I think skeeter's picture says it all, but here is my response to your questions.

OK, I don't understand your calculations fully but I haven't taken a lot of time to look at them yet. However, I did have trouble with some parts (hilighted in red). Firstly, you said there are three forces acting on the car, but you left out the force pushing the car to the center, but then you added it in later. I assume that was just a mistake.
Three forces acting on the car (gravity, normal force, and frictional force), and the sum of those forces causes the car to accelerate toward the center. The three forces combine to push the car toward the center.

2nd, $$\displaystyle \mu{N}$$ is towards the center; it is the frictional force, which helps keep cars on the road (and thus closer to the center), not pushing them off it.
I tend to try to make forces positive in all cases. In this case, since the car is going around the curve at a speed less than the designed speed, I figured that friction is holding the car up, keeping it from sliding down the road.

Frictional force is always parallel to the surface.

3rd, your final equation had m in it, but if you look at my original post you see that I was not given the value for mass. This is a large reason why I am having trouble with the problem; it has to work for anywhere from a bicycle to an 18-wheeler.
Yes, m cancels out of the equations. You can either assign a numerical value to m, knowing that the result will be the same regardless of what value you choose, or (my preferred method) do all the calculations with variables (including m), and when you get to the final answer, notice that m is not there.

Skeeter - that's a really cool diagram! Did you draw it yourself or find it on the internet?

- Hollywood

#### Chokfull

The centripital force is pushing the car off the road, and when the cars are moving at 60 km/h it keeps them from going off the road on the outside of the curve, where, if you notice, the guardrail is usually positioned. the banking of the road helps keep the cars on the road, rather than tipping them off. Therefore, the frictional force keeping them on the road must be towards the center. Also, I meant that the fourth force would be the turning of the car itself, wouldn't it? otherwise the car's going to head straight, and go off the road.

Also, is my equation for $$\displaystyle \theta$$ right? My calculator gives $$\displaystyle \tan^{-1} \frac{(40*1000/3600)^2}{9.8*200} = .062905$$, which is in radians I assume, so I multiply by 57.296 and get $$\displaystyle 3.604^o$$, which seems way too small. I multiplied the 40 by 1000 and divided by 3600 to convert to m/s, rather than km/h. I'm worried that I was incorrect in working under the assumption that on the rainy day there is no friction, but if this is incorrect this problem looks impossible.

#### ebaines

I think this problem is a lot easier than all this, but is so poorly worded that it's causing a lot of confusion. The bit about the cars going at 60 KMH on a rainy day I think is exactly what the OP surmised - this sets the bank of the road for 0 side force on the car as $$\displaystyle \theta = \tan^{-1}( \frac {v^2} {gR})$$. Then the question about friction to keep the car from siding down the track - I think this means on a dry day. If you consider that the slower the car goes the less sideways force there is keeping it from sliding down the track, then the worst case is when velocity = 0. This gives you the relatively simple formula:

$$\displaystyle mg \sin(\theta) = \mu N = \mu mg \cos(\theta)$$
$$\displaystyle \mu = tan(\theta) = tan(tan^{-1} (\frac {v^2} {gR} ) )= \frac {v^2} {gR}$$

Admittedly this ignores the possibiliity that the car can be traveling at a very high rate of speed and consequently slide up the track, but I bet that's not what they're looking for, as it would take an infinite amount of friction to prevent sliding at infinitely high speeds.

#### Chokfull

Umm...no. The question is about centripetal force. I.e. the problem is about the cars sliding up the track. The road should only be banked by about $$\displaystyle 15^o$$, whereas in the diagram above it is shown tilted at $$\displaystyle 45^o$$.

#### hollywood

The way I read the problem is this:

The curve is designed for zero sideways force at 60km/h. On one day in particular, the traffic is moving slower than expected (because of traffic) and the coefficient of friction is smaller (because of the rain). We're worried that the combination of those two factors will cause cars to slide off the road. How low does the coefficient of friction have to be for cars to start sliding off the road?

And since you're going less than 60km/h, you're talking about sliding down the slope. Or to look at it another way, the slope was designed for a car that is going faster, so it is banked more than it should be.

I get $$\displaystyle \theta=\tan^{-1}\frac{v^2}{Rg}=\tan^{-1}\frac{((60)(1000)/(3600))^2}{(200)(9.8)}=0.14$$ radians or 8.07 degrees, which is similar to your calculation except that I used 60km/h instead of 40km/h. The road was designed for 60km/h, so that's what we have to use to find out the angle of incline. And you're right - 8 degrees seems like a small angle.

Pressing on, we have:
$$\displaystyle v=\frac{(40)(1000)}{3600}=$$ 11.1 m/s
$$\displaystyle \frac{mv^2}{r}\sin{\theta}=N-mg\cos{\theta}$$
$$\displaystyle \frac{v^2}{r}\sin{\theta}=\frac{N}{m}-g\cos{\theta}$$
$$\displaystyle \frac{N}{m}=\frac{v^2}{r}\sin{\theta}+g\cos{\theta}$$
$$\displaystyle \frac{N}{m}=\frac{11.1^2}{200}\sin{8.07}+9.8\cos{8.07}=9.79$$

and
$$\displaystyle mg\sin{\theta}-\mu{N}=\frac{mv^2}{r}\cos{\theta}$$
$$\displaystyle g\sin{\theta}-\mu\frac{N}{m}=\frac{v^2}{r}\cos{\theta}$$
$$\displaystyle \mu\frac{N}{m}=g\sin{\theta}-\frac{v^2}{r}\cos{\theta}$$
$$\displaystyle \mu(9.79)=9.8\sin{8.07}-\frac{11.1^2}{200}\cos{8.07}=1.99$$

so $$\displaystyle \mu=0.203$$

- Hollywood

#### Chokfull

The way I read the problem is this:

The curve is designed for zero sideways force at 60km/h. On one day in particular, the traffic is moving slower than expected (because of traffic) and the coefficient of friction is smaller (because of the rain). We're worried that the combination of those two factors will cause cars to slide off the road. How low does the coefficient of friction have to be for cars to start sliding off the road?

And since you're going less than 60km/h, you're talking about sliding down the slope. Or to look at it another way, the slope was designed for a car that is going faster, so it is banked more than it should be.
That's an interesting take on the problem, I hadn't thought of it like that before. My book should have clarified, I guess.

However, your answer states that $$\displaystyle \mu=.203$$. The book gives me .078, so either you made a mistake in your calculations, I was right about the car going up, rather than down, the slope, or the book is wrong. But it may just be the angle. The formula for the angle came from an example earlier in the book for which there was no friction, and this is why I was saying that maybe it is supposed to be assumed that there is no friction on the rainy day, seeing as this was the only formula I was given for the angle.

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