# Conflict normal force in inclined plane

## Homework Statement

The original problem was as follows:

A civil engineer wishes to redesign a curved roadway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a ramp is usually banked, meaning the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 11.2 m/s (25.1 mi/h) and the radius of the curve is 34 m. At what angle should the curve be banked?

I am getting a conflict if I do this problem in different ways:

## Homework Equations

I can resolve N into its vertical and horizontal components. Since the car is supposed to be in vertical equilibrium,
$$\begin{array}{l} N\cos \theta - mg = m{a_{ycar}} = 0\\ or\quad N\cos \theta = mg \end{array}$$
If I do the problem the second way:
I resolve mg into the component along the plane and the component perpendicular to the plane:

Then

$$\begin{array}{l} N - mg\cos \theta = m{a_{y'}} = 0\\ or\quad N = mg\cos \theta \end{array}$$

WHAT'S WRONG WITH MY REASONING!

## The Attempt at a Solution

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NascentOxygen
Staff Emeritus
You need to set centrifugal force = weight component down the banked track

If the car is travelling over a banked curve I don't think the height will remain constant; and your second solution isn't specific to your problem, the normal force always has that value on an inclined plane but that won't tell you what angle to bank your curve at.
You want the force pointing towards the center of the curve to be equal to
$$m\frac{V^2}{r}$$
So that your car moves in a circular path, this force can only come from gravity, which component of gravity points towards the center of the curve?

There is no component of gravity pointing towards the centre.
It's the component of the normal reaction force from the track that points towards the centre, and in doing so provides the centripetal force for the circular motion.

There is no component of gravity pointing towards the centre.
It's the component of the normal reaction force from the track that points towards the centre, and in doing so provides the centripetal force for the circular motion.
but even if you take the correct first solution where I resolve N into its components,if I find Nsin(theta) in the following way,

$$\begin{array}{l} N\cos \theta = mg\\ {N^2} - {N^2}{\cos ^2}\theta = {N^2}{\sin ^2}\theta = {N^2} - {\left( {mg} \right)^2}\\ N\sin (\theta ) = \sqrt {{N^2} - {{\left( {mg} \right)}^2}} = \sqrt {{{(mg\sec (\theta ))}^2} - {{(mg)}^2}} = mg\tan \theta \end{array}$$

there is still an 'mg' in the the expresssion for Nx

D H
Staff Emeritus
I can resolve N into its vertical and horizontal components.
You found the vertical component. What about the horizontal component? (See posts #2 and #3).

but even if you take the correct first solution where I resolve N into its components,if I find Nsin(theta) in the following way,

$$\begin{array}{l} N\cos \theta = mg\\ {N^2} - {N^2}{\cos ^2}\theta = {N^2}{\sin ^2}\theta = {N^2} - {\left( {mg} \right)^2}\\ N\sin (\theta ) = \sqrt {{N^2} - {{\left( {mg} \right)}^2}} = \sqrt {{{(mg\sec (\theta ))}^2} - {{(mg)}^2}} = mg\tan \theta \end{array}$$

there is still an 'mg' in the the expresssion for Nx
The vertical component of N equals mg as there is no vertical motion.
The horizontal component of N equals the centripetal force.
Combine these two and eliminate N and m
Obtain a formula relating the angle to v r and g

Doc Al
Mentor
I can resolve N into its vertical and horizontal components. Since the car is supposed to be in vertical equilibrium,
$$\begin{array}{l} N\cos \theta - mg = m{a_{ycar}} = 0\\ or\quad N\cos \theta = mg \end{array}$$
There's nothing wrong with this. Of course, as has been pointed out, you must also analyze the horizontal forces.

If I do the problem the second way:
I resolve mg into the component along the plane and the component perpendicular to the plane:

Then

$$\begin{array}{l} N - mg\cos \theta = m{a_{y'}} = 0\\ or\quad N = mg\cos \theta \end{array}$$
Since you are analyzing forces normal to the surface, the acceleration is not zero! The centripetal acceleration has a component normal to the surface.

The vertical component of N equals mg as there is no vertical motion.
The horizontal component of N equals the centripetal force.
Combine these two and eliminate N and m
Obtain a formula relating the angle to v r and g
The problem is not with the problem; I knew from the beginning that the solution was as follows

$$\begin{array}{l} {F_r} = {\rm{ - }}n{\rm{sin}}\theta = - \frac{{m{v^2}}}{r}\\ {\rm{and I know that}}\\ n\cos \theta = mg\\ \tan \theta = \frac{{{v^2}}}{{rg}}\\ \theta = \arctan \left( {\frac{{11.2{\kern 1pt} }}{{34 \times 9.8}}} \right) \end{array}$$

but what I don't understand is why isn't n=mgcos(θ). I understand why ncos(θ)=mg is correct.

To summarize, I don't understand what is wrong with n=mgcos(θ)

Doc Al
Mentor
To summarize, I don't understand what is wrong with n=mgcos(θ)
See my post above.

There's nothing wrong with this. Of course, as has been pointed out, you must also analyze the horizontal forces.

Since you are analyzing forces normal to the surface, the acceleration is not zero! The centripetal acceleration has a component normal to the surface.
What I meant was that in the plane x', y' that is inclined at an angle θ so that the axes are parallel to the inclined bank, the y' accleration is zero. Please correct me if I am wrong

Doc Al
Mentor
What I meant was that in the plane x', y' that is inclined at an angle θ so that the axes are parallel to the inclined bank, the y' accleration is zero. Please correct me if I am wrong
You are wrong. The acceleration is horizontal, so it will have components parallel and perpendicular to the inclined bank.

You are wrong. The acceleration is horizontal, so it will have components parallel and perpendicular to the inclined bank.
but if acceleration in the y' plane was not zero, then wouldn't the car have to go through the ramp.

my y' axis is perpendicular to the road. Accleration in the y' direction would mean jumping off the road or going into it.

Doc Al
Mentor
but if acceleration in the y' plane was not zero, then wouldn't the car have to go through the ramp.
No.
my y' axis is perpendicular to the road. Accleration in the y' direction would mean jumping off the road or going into it.
Again, no.

Remember that we are dealing with circular motion here where acceleration means a change in the direction of the velocity vector. Just because the acceleration acts towards the center doesn't mean that the car moves towards the center. And just because there's a component of acceleration perpendicular to the surface doesn't mean that the car moves in that direction.

D H
Staff Emeritus