# Trying to derive the equation for an ideally banked curve

1. Oct 28, 2014

### Jazz

I'm confused trying to understand the previous steps that lead to the formula for the angle of an ideally banked curve. In absence of friction, this is the angle where the Centripetal Force is provided by the horizontal component of the Normal Force.

I understand that the Normal Force is perpendicular to the surface where an object is being supported. In an incline, this is the component of the weight that is perpendicular to the slope:

$N = mgcos(\theta) \hspace{20mm} (1)$

And this says the textbook I’m reading :) :
The figure and the source of this quote is here:

I’ve also drawn the figure but only with the forces (I think the car makes the diagram a little messy):

This would mean that I can find $N$ just by dividing both sides by $cos(\theta)$:

$N = \frac{mg}{cos(\theta)} \hspace{20mm} (2)$

Right from the get go this equation looks quite different from equation $(1)$ (it's like the other way around), and so the results they give should be different.

Probably the following isn't necessary but let suppose there is a car that has a mass of $5\ kg$ in an incline that makes $30º$ with the horizontal. If I try to find $N$ with $(1)$ I get:

$N = mgcos(\theta)$

$N = (5\ kg)(9.8\ m/s)cos(30º)$

$N = 42.2\ N$

Then with $(2)$:

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

$N = \frac{(5\ kg)(9.8\ m/s)}{ cos(30º)}$

$N = \frac{(5\ kg)(9.8\ m/s)}{ cos(30º)}$

$N = 56.6\ N$

What am I doing wrong?

It is suppose that the 2nd equation in the quote above is written in terms of $N$ in order to plug it into the other one, rearrange in terms of $\theta$ and finally come up with the formula, but since I cannot make sense of the previous steps I cannot understand the rest of the derivation either.

Thanks!!

2. Oct 28, 2014

### PhanthomJay

You get two different values for N depending upon whethe or not the object is accelerating centripetaly. Each value is correct for the described situation. For the incline problem without the centripetal acceleration, can you explain why the normal force is mgcostheta? Does this explanation apply to the centripetal acceleration case?

3. Oct 29, 2014

### Jazz

I think I’ve understood what is going on.

When a car is taking a curve, there is an extra weight supplied by the vertical component of the static friction, preventing it from being thrown away in a direction tangential to the curve. The ‘new weight’ is this component plus the weight of the car. And that, I think, is the reason why the Normal Force is no longer equal to the component of the weight alone, but that of the ‘new weight’ (at rest in an incline, there is no such downward friction but upward instead). Here is a diagram:

Source: me.

Thus, the Centripetal Force is the component in the horizontal axis of this Normal Force plus the horizontal component of the static friction. Then the maximum velocity without slipping out of the curve is:

$v_{max} = \sqrt{\frac{rg(\mu cos(\theta)+ sin(\theta)}{cos(\theta) - \mu sin(\theta)}}$

If it is a banked, frictionless curve, then the maximum velocity is:

$v_{max} = \sqrt{rg tan(\theta)}$

Here I write this equation in terms of $\theta$ and I get what I was looking for.

But, do I get the explanation right?

Does that extra weight have a more technical name?

Last edited: Oct 29, 2014
4. Oct 29, 2014

### PhanthomJay

You are thinking along the right path , but your reasoning is not quite correct. Let's keep friction out of the picture for now , for simplification and for consistency with the original problem that excludes friction and depends only upon the horizontal component of the normal force to provide the centripetal acceleration. In your free body diagram for the vehicle rounding the curve , but without friction, realize that there is no acceleration in the vertical direction, and hence, using Newton's first law, the vertical component of the normal force, Ncostheta , must be equal to the vehicle's weight , that is ,
N = w/cos theta. The normal force will always be greater than the weight. The weight itself never changes . Now if you were to look at the inclined plane problem without friction and without centripetal acceleration, you now have no acceleration in the direction perpendicular to the incline, which is unlike the centripetal case which has no acceleration in the vertical direction. And in which case, N = wcostheta from Newton 1 in the direction perpendicular to the incline, N always being less than the weight. Now looking at the centripetal case again from another perspective, if by chance you wanted to sum forces in the direction perpendicular to the banked road surface, you now have a centripetal acceleration component in that direction, thus N - wcostheta is not 0 here, as Newton 2 applies in this direction.

5. Oct 30, 2014

### dean barry

Can i add this :
Considering only the primary forces ( centripetal and gravity )

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6. Oct 30, 2014

### Jazz

Now I think I’m beginning to understand this.

I can see that have brought friction into the picture was not necessary.

Could I try to set a sort of analysis in order to see if I get this right?

When the car is negotiating an inclined, curved path, it can happen one of these situations:

1) The car doesn’t slip. There is an $a_c$ as expected. And it also means that $a_c$ is right for the angle $\theta$ so that $Ncos(\theta)=w$.

2) The car’s speed is greater than in 1). The $a_c$ here is greater than in 1) and hence $Ncos(\theta) > w$. The car starts to slip upward.

I think it can also be understood by means of this equation: $tan(\theta) = \frac{v^2}{gr}$. $g$ is fixed and if the car’s speed is considered to be the same along the curve, and we know that $\theta$ doesn’t change, the centripetal acceleration will tend to increase the radius $r$ measured from the center of rotation to the car in order to make this equation hold.

3) The car’s speed is slower than in 1): The car slips downward: That’s because $Ncos(\theta) < w$. There is no centripetal acceleration and the car slips with a net force equal to $wsin(\theta)$.

Are these assumptions correct?

7. Oct 30, 2014

### Jazz

In that case, F1 is the reaction force of the car against $F_c$, F2 is weight and F3 is the Normal Force, right?

That reation force occurs because of the tendency of the car to keep going in a straight line while the curve is preventing it from doing so, right? (maybe it’s a silly question I’m still struggling to understand this without making wrong assumptions).

Thanks!

8. Oct 30, 2014

### PhanthomJay

assumptions 1 and 2 look good! Assumption 3 is true only if the car has no speed. If it has any speed less than the design speed, it will still have centripetal acceleration but still slip down the bank. You don't want to negotiate a curve with no friction, banked or unbanked.

9. Oct 31, 2014

### dean barry

I dont consider reactions at all : F1 is the centripetal force ( m * v² ) / r , F2 is the gravitational force ( m * g ) and F3 is the resultant of F1 & F2.
For no side forces, set the banking normal to this resultant.
Note : the result is the same regardless of the mass

10. Oct 31, 2014

### Jazz

Without the aid of friction I should say :)

Thanks for helping me to see the light. Three days trying to understand this issue isn't that bad.

Ah okey. I got somewhat confused following the direction of the arrows. That's why I thought F1 as a reaction force.

Last edited: Oct 31, 2014
11. Oct 31, 2014

### PhanthomJay

not bad at all, and you have done very well with it.
A comment on Dean's diagram: the horizontal force shown is the fictitious centrifugal force, and the diagram does also not show the real normal force. Although the design embankment angle determined from the sketch is correct , it is not intuitively clear why this is so.