# Conflict normal force in inclined plane

1. Dec 6, 2011

1. The problem statement, all variables and given/known data
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:
2. Relevant 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2011

### Staff: Mentor

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

3. Dec 6, 2011

### JHamm

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?

4. Dec 6, 2011

### Stonebridge

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.

5. Dec 6, 2011

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

6. Dec 6, 2011

### D H

Staff Emeritus
You found the vertical component. What about the horizontal component? (See posts #2 and #3).

7. Dec 6, 2011

### Stonebridge

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

8. Dec 6, 2011

### Staff: Mentor

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.

9. Dec 6, 2011

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)

10. Dec 6, 2011

### Staff: Mentor

See my post above.

11. Dec 6, 2011

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

12. Dec 6, 2011

### Staff: Mentor

You are wrong. The acceleration is horizontal, so it will have components parallel and perpendicular to the inclined bank.

13. Dec 6, 2011

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.

14. Dec 6, 2011

### Staff: Mentor

No.
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.

15. Dec 6, 2011

### D H

Staff Emeritus
vadiraja, it will help if you look at this problem from both a dynamics and a kinematics perspective.

From dynamics, there are only two forces acting on the car: gravitation and the normal force. You already know at the gravitational force. The normal force has magnitude Fn and is directed away from the vertical by by some angle θ. Note that both Fn and θ are unknown at this point. Drawing a free-body diagram will help you get a picture of the problem.

From kinematics, you know that the car has zero vertical acceleration and has a horizontal acceleration given by the assumption that the car is undergoing uniform circular motion. You need to use these kinematics results in conjunction with the dynamics result to solve for θ.

You can also solve for Fn, but you can solve for θ without having to determine the magnitude of the normal force.