# Cars on a banked road

• May 25th 2010, 09:21 AM
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 $\theta = \tan^{-1} \frac {v^2} {gR}$ 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.
• May 27th 2010, 09:05 AM
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 $\frac{v^2}{r}$ (the centripetal acceleration), these must add up to a force $\frac{mv^2}{r}$ acting horizontally. Since we are looking for the minimum coefficient of friction, set the frictional force to be the maximum possible, given by $\mu{N}$ where $\mu$ is the coefficent of friction.

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

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

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

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

- Hollywood
• May 31st 2010, 09:13 AM
Chokfull
Quote:

Originally Posted by 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 $\frac{v^2}{r}$ (the centripetal acceleration), these must add up to a force $\frac{mv^2}{r}$ acting horizontally. Since we are looking for the minimum coefficient of friction, set the frictional force to be the maximum possible, given by $\mu{N}$ where $\mu$ is the coefficent of friction.

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

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

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

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, $\mu{N}$ 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.
• May 31st 2010, 10:16 AM
skeeter
• Jun 1st 2010, 12:26 PM
hollywood
I think skeeter's picture says it all, but here is my response to your questions.

Quote:

Originally Posted by Chokfull
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.

Quote:

Originally Posted by Chokfull
2nd, $\mu{N}$ 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.

Quote:

Originally Posted by Chokfull
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
• Jun 3rd 2010, 08:34 AM
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 $\theta$ right? My calculator gives $\tan^{-1} \frac{(40*1000/3600)^2}{9.8*200} = .062905$, which is in radians I assume, so I multiply by 57.296 and get $3.604^o$, 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.
• Jun 3rd 2010, 12:22 PM
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 $\theta = \tan^{-1}( \frac {v^2} {gR})$. 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:

$
mg \sin(\theta) = \mu N = \mu mg \cos(\theta)
$

$
\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.
• Jun 3rd 2010, 02:44 PM
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 $15^o$, whereas in the diagram above it is shown tilted at $45^o$.
• Jun 4th 2010, 01:16 PM
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 $\theta=\tan^{-1}\frac{v^2}{Rg}=\tan^{-1}\frac{((60)(1000)/(3600))^2}{(200)(9.8)}=0.14$ 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:
$v=\frac{(40)(1000)}{3600}=$ 11.1 m/s
$\frac{mv^2}{r}\sin{\theta}=N-mg\cos{\theta}$
$\frac{v^2}{r}\sin{\theta}=\frac{N}{m}-g\cos{\theta}$
$\frac{N}{m}=\frac{v^2}{r}\sin{\theta}+g\cos{\theta }$
$\frac{N}{m}=\frac{11.1^2}{200}\sin{8.07}+9.8\cos{8 .07}=9.79$

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

so $\mu=0.203$

- Hollywood
• Jun 7th 2010, 08:41 AM
Chokfull
Quote:

Originally Posted by 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.

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 $\mu=.203$. 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.
• Jun 7th 2010, 09:22 AM
ebaines
A $v_1 =$ 60 Km/h you calculate the angle of the bank to give zero side forces on the car, and as previously noted you find that

$\tan \theta = \frac {v_1 ^2} {gR}$

Now when the car is moving only at $v_2 = 40$ Km/h, the balance of forces must equal out:

$
\frac {m v_2 ^2} R \cos \theta = mg sin \theta -\mu mgcos \theta
$

Solve for $\mu$ and you'll get the answer of $\mu = 0.0787$.
• Jun 8th 2010, 08:32 AM
Chokfull
Quote:

Originally Posted by ebaines
A $v_1 =$ 60 Km/h you calculate the angle of the bank to give zero side forces on the car, and as previously noted you find that

$\tan \theta = \frac {v_1 ^2} {gR}$

Now when the car is moving only at $v_2 = 40$ Km/h, the balance of forces must equal out:

$
\frac {m v_2 ^2} R \cos \theta = mg sin \theta -\mu mgcos \theta
$

Solve for $\mu$ and you'll get the answer of $\mu = 0.0787$.

Well, could you elaborate a bit? it makes sense up until http://www.mathhelpforum.com/math-he...29f4949e-1.gif. It looks like you combined two equations to get only the variables for which we had values, but I don't know what the second equation is.
• Jun 8th 2010, 09:21 AM
hollywood
Quote:

Originally Posted by Chokfull
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 $\mu=.203$. 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.

On the final calculation, I added instead of subtracting. I should have calculated:

$\mu(9.79)=9.8\sin{8.07}-\frac{11.1^2}{200}\cos{8.07}=$ 1.38 - 0.61= 0.76 (not 1.38 + 0.61 = 1.99)

so $\mu=$ 0.76/9.79 = 0.078

- Hollywood
• Jun 8th 2010, 09:39 AM
hollywood
Quote:

Originally Posted by ebaines
A $v_1 =$ 60 Km/h you calculate the angle of the bank to give zero side forces on the car, and as previously noted you find that

$\tan \theta = \frac {v_1 ^2} {gR}$

Now when the car is moving only at $v_2 = 40$ Km/h, the balance of forces must equal out:

$
\frac {m v_2 ^2} R \cos \theta = mg sin \theta -\mu mgcos \theta
$

Solve for $\mu$ and you'll get the answer of $\mu = 0.0787$.

The normal force is not just $mg\cos{\theta}$ - that is the component due to gravity only. There is also a component $\frac{mv^2}{r}\sin{\theta}$ because the car needs to accelerate horizontally (it is moving in a horizontal circle), so it has a small component perpendicular to the road. This other component adds about 0.9% to the normal force, so the coefficient of friction can be 0.9% smaller, or 0.078.

- Hollywood
• Jun 8th 2010, 11:53 AM
ebaines
Chokfull: Actually, I see that I made an error in my earlier post by not realizing that the normal force of the car against the ramp has two elements to it - one due to the weight of the car, and the other due to the centripetal acceleration of the car that is in the direction that presses the car down against the surface of the ramp. Look at the attached figure, and consider the forces that are acting in the direction along the ramp. They must balance out - otherwise the car would slide up or down the ramp. The force acting to pull the car down the ramp is the car's weight times $\sin \theta$. The force acting to push the car up the ramp is $\frac {mv^2} R cos \theta$. You know that the first force is bigger than the second, since the car is only going 40 Km/h instead of 60 Km/h, which is the value for which these two forces are balanced. Hence to keep the car from sliding down the ramp you need to consider a third force, which is due to friction, and works in a direction to keep the car up on the ramp. The magnitude of that force is $\mu N$, where $N$ is the normal force of the car against the ramp. the normal force has two components as decibed above - one element of it due to the car's weight pressing against the ramp is $mg \cos\theta$, and the other element due to the component of the centripetal acceleration of the car pressing it into the ramp is $\frac {mv_2^2} R \sin \theta$. Put all three forces into one equation:

$
mg \sin \theta = \frac {mv_2^2} R \cos \theta + \mu (mg \cos \theta+ \frac {mv_2^2} R \sin \theta)
$

Divide through by $m \cos \theta$:

$
g \tan \theta = \frac {v_2^2} R + \mu (g + \frac {v_2^2} R \tan \theta)
$

Rearrange:

$
\mu = \frac {g \tan \theta - \frac {v_2^2} R } {g + \frac {v_2^2} R \tan \theta }
$

You already know that $\tan \theta = \frac {v_1 ^2} {gR}
$

You should find that $\mu = 0.078$.

http://i789.photobucket.com/albums/y.../CarOnRamp.jpg