Cars rounding a slippery banked corner in the rain

[titansarus]

Homework Statement

The question is from Fundamental Physics 10th Edition, Walker Resnik Holiday:
A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 200 m. Traffic is moving along the highway at 40 km/h on a rainy day.What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Homework Equations

$N cos \theta = mg , N sin \theta = m v_1 ^2 /r$

$F_s = μ F_N ::: Friction = μ N$)

$\theta = arctan (v^2 / rg)$

The Attempt at a Solution

[/B]
I solved the problem like this: if we think the slope of the road is θ, and if we think when the road is rainy, the friction is zero and velocity is $v_1$ and when the road is not rainy the coefficient of friction is μ and velocity is $v_2$, we will have: (N is the force of the road on the object that is normal to the road)
Rainy:
$N cos \theta = mg , N sin \theta = m v_1 ^2 /r$
Dry:
$N sin \theta + N μ cos \theta = m v_2 ^2 / r$.

$N μ cos \theta$ is the horizontal part of the friction force. because the friction itself will be directed along the road, the horizontal part that is directed along the center of the circle is $N μ cos \theta$.

by using the rainy equations and dividing them by m, we get:

$v_1 ^2 / r + g μ = v_2 ^2 / r$ and from this by changing km/h to m/s we get $μ=0.078$

The answer of the book is the same but the solution is completely different. I want to know whether my solution is acceptable or not.

The book solution (the Eq 6-1 is $F_s = μ F_N ::: Friction = μ N$):

 

[Doc Al]

I solved the problem like this: if we think the slope of the road is θ, and if we think when the road is rainy, the friction is zero and velocity is $v_1$ and when the road is not rainy the coefficient of friction is μ and velocity is $v_2$, we will have: (N is the force of the road on the object that is normal to the road)

Despite the description of the road as "rainy", there is friction in that case. Since the road is designed for traffic moving at 60 km/h, at that speed friction is zero. But not at other speeds.

The first step is to find the banking angle of the road.

 

[CWatters]

I think your method is close/ok but the description/explanation of the two cases is wrong.

When they say the road is "designed for 60kph they mean the bank angle is set so that no frictional force is required at that speed under any conditions (wet or dry). You wrote equations to describe that situation but you said the reason there was no friction was because it was wet. In fact no friction is required at 60kph wet or dry.

Then you calculated the coefficient of friction needed at 40kph. You said this was in the dry but actually the road state doesn't matter (as long as it has that coefficient of friction it doesn't matter if its wet or dry).

Doc Al is right. The best procedure is probably to calculate the bank angle but I guess technically you don't need is as it disappears in the wash as you found out

 

[DaveE]

The rain is irrelevant in this solution. Yes rain may lower friction, but you are only interested in finding the critical value of friction that makes the car slide off the road. What causes that friction is a different question.

 

[titansarus]

I think your method is close/ok but the description/explanation of the two cases is wrong.

When they say the road is "designed for 60kph they mean the bank angle is set so that no frictional force is required at that speed under any conditions (wet or dry). You wrote equations to describe that situation but you said the reason there was no friction was because it was wet. In fact no friction is required at 60kph wet or dry.

Then you calculated the coefficient of friction needed at 40kph. You said this was in the dry but actually the road state doesn't matter (as long as it has that coefficient of friction it doesn't matter if its wet or dry).

Doc Al is right. The best procedure is probably to calculate the bank angle but I guess technically you don't need is as it disappears in the wash as you found out

If the 60 kph is okay for rain and without rain, then if we use 40kph, then it must not be a problem and I think we will not need to calculate any friction. (consider that you are moving very slowly in a circular road, then you will not need that banking. But for fast speeds, it is important), because of that, I think there is no friction in the rain and 40 kph is okay and at the same time with friction of dry road that adds to the centripetal force, it increases to 60 kph.

My problem is that my solution and the solution of the book are completely different and even the reasonings behind them are different (the reasoning of the book seems better) but both of them give the exact same answer.

 

[Doc Al]

If the 60 kph is okay for rain and without rain, then if we use 40kph, then it must not be a problem and I think we will not need to calculate any friction.(consider that you are moving very slowly in a circular road, then you will not need that banking. But for fast speeds, it is important)

Rethink that. If you go too slow, what happens? (assuming no friction to help you)

 

[titansarus]

Rethink that. If you go too slow, what happens? (assuming no friction to help you)

If there is no friction, then there is maybe no movement, even for a flat road. I know that. But I think the lower speed is probably okay even with a not banked road. (less angle). Whenever the speed is not more than 60, there shouldn't be problem for the car. Am I wrong?

 

[Doc Al]

Whenever the speed is not more than 60, there shouldn't be problem for the car. Am I wrong?

Yes, you are wrong.

You need to better understand what "banking angle" means. Only at the designed speed is friction not needed.

(Hint: What's the direction of the friction force?)

 

[Andrew Mason]

Hi Titansarus. Welcome to PF!

In my view, yours is a better solution because it is clearer and works it out from first principles. The reasoning in the book solution seems backward to me. The 8.1 degree angle seems to be the angle that a car, without friction, traveling at 60 km/hr would be able to make the turn. So the car traveling at 40 km/hr at that angle would slide down the curve unless there was a friction force preventing it from sliding down.

At 40 km/hr and no friction, the angle of 3.6 degrees would allow the car to make the curve. At 60 km/hr, a force of friction would be required to keep it from sliding up off the road.

I haven't worked it out but maybe the coefficient of static friction in both those cases has to be .078 and that is why your answers agree.

AM

 

[Doc Al]

The 8.1 degree angle seems to be the angle that a car, without friction, traveling at 60 km/hr would be able to make the turn. So the car traveling at 40 km/hr without at that angle would slide down the curve unless there was a friction force preventing it from sliding down.

That's correct and is exactly what the solution describes. Why is the book's reasoning backward?

 

[Andrew Mason]

That's correct and is exactly what the solution describes. Why is the book's reasoning backward?

Because at 8.1 degrees and no friction, the cars would be sliding down the slope at only 40 km/hr, which is contrary to the stated facts. At 3.6 degrees, the cars traveling at 40 km/hr would make it without friction (i.e. in the rain) but cars traveling at 60 km/hr would need a dry surface with friction to keep from flying off, which is consistent with the stated problem.

I assumed, like the OP, that the question was asking what the co-efficient of friction was between the dry road and the tires. If the question was asking what the co-efficient of friction between a wet road and the tires is then the book answer would be correct. Maybe that ambiguity in the question and the OP's assumption that there was no friction in the rain is the source of the confusion. I am not familiar with the text but if the statement that the road was "designed for traffic at 60 km/hr" means that this is the bank angle for no friction, then there would be no ambiguity and we have both misinterpreted the question.

AM

 

[Doc Al]

Because at 8.1 degrees and no friction, the cars would be sliding down the slope at only 40 km/hr, which is contrary to the stated facts.

You are asked to find the coefficient of friction so it doesn't slide. That's the problem!

 

[titansarus]

Yes, you are wrong.

You need to better understand what "banking angle" means. Only at the designed speed is friction not needed.

(Hint: What's the direction of the friction force?)

My instructor said that roads are made with an angle so that when cars move with a maximum speed, don't exit from the path. He said any speed lower than max speed is OK. But your reasoning of that the friction is maybe upward (because weight is downward) seems correct. It is because I asked this question; The reasoning of the book seems right but it is different with mine and both give the same answer.

 

[CWatters]

My instructor said that roads are made with an angle so that when cars move with a maximum speed, don't exit from the path. He said any speed lower than max speed is OK.

In normal conditions that is correct. However it is quite possible to slide down by going too slowly. I have actually done that.

Some years ago I tried to drive up a hill with a banked bend near the top. Whole lot covered in compacted snow/ice. I wanted to bail but there was too much traffic and police were waving people to go up. Due to the hill I wasn't able to go up fast as fast I wanted, so when I got to the banked bend I started to slide sideways down the banking.

 

[Doc Al]

I am not familiar with the text but if the statement that the road was "designed for traffic at 60 km/hr" means that this is the bank angle for no friction, then there would be no ambiguity and we have both misinterpreted the question.

That's exactly what the statement means. (I guess I'm too familiar with the text. )

 

[verty]

Using trigonometry, I get this formula for mu_k: $$\mu_k = \tan(\tan^{-1} {v_0^2 \over rg} - \tan^{-1}{v_1^2 \over rg})$$

@titansarus, you wrote $N \cos\theta = mg$ but that is not correct. $N \cos\theta < mg$. This is because friction is holding the car up as well, so the normal force has less of a job to do. At least, that's how I understand it.

 

[haruspex]

@titansarus, you wrote $N \cos\theta = mg$ but that is not correct. $N \cos\theta < mg$. This is because friction is holding the car up as well, so the normal force has less of a job to do. At least, that's how I understand it.

@titansarus was confusing rainy with zero friction, hence the equation.

 

[haruspex]

Using trigonometry, I get this formula for mu_k: $$\mu_k = \tan(\tan^{-1} {v_0^2 \over rg} - \tan^{-1}{v_1^2 \over rg})$$

Or perhaps more simply
$\frac{v_1^2-v_2^2}{(gr)^2+(v_1v_2)^2}gr$.
Either way, we see that swapping the two speeds simply switches the sign of the answer, which is equivalent to having friction act the other way.
This is why the OP's misinterpretation of the question still got the right answer.

 

[haruspex]

Maybe that ambiguity in the question

I see no ambiguity. It is quite clear that risk of sliding is at the lower speed.

 

[Andrew Mason]

I see no ambiguity. It is quite clear that risk of sliding is at the lower speed.

The OP understood the physics quite well. As I said, there is no ambiguity if you understand that "designed for traffic moving at 60 km/h" means "with no friction" ie. one has to understand that "designed" does not necessarily mean "properly designed". I live in Saskatchewan. If you designed a banked curve to handle traffic at a "design speed" with no friction, winter traffic which is slowed down due to icy roads, would have to speed up to get around icy curves. Not my idea of good design.

AM

 

[haruspex]

there is no ambiguity if you understand that "designed for traffic moving at 60 km/h" means "with no friction"

I was not depending on that. It mentions 60kph as being fine, whatever that means exactly, then implies that at 40kph there is a risk of sliding. Ergo, the risk is of sliding down.