# Find the Coefficient of friction on a banked curve.

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1. Nov 29, 2017

### Motzu2098

1. The problem statement, all variables and given/known data

A car giving a turn on a curve with 88m of radius, traveling at a speed of 95km/h, the curve is perfectly banked for a car traveling at 75 km/h, meaning that the curve has an angle θ of 26.7º.
2. Relevant equations

μ = ?
r = 88m
v = 26.38 m/s
θ = 26.7º

Ff = μ · g · m
Fc = m (v^2/r)
Fg = m · g

3. The attempt at a solution

Hoping that The FBD is right, I came up with
ΣFy = FNy - mg - Ffy = 0
= FN cos θ - mg - FN μ f sin θ = 0
FN cos θ - FN μ f sin θ = mg
FN (cos θ - μ sin θ) =mg
FN = (mg)/(cos θ - μ sin θ)

then I used the FN = Fc, which I substituted to (mg)/(cos θ - μ sin θ) = m(v^2/r).
And then, for what I understand, masses cancel each other.
g/(cos θ - μ sin θ) = (v^2/r)
The problem is that I am stuck, I do not know how to transform the equation for solving for μ. nor if the procedure I've followed this far is right. (seems right to me)

2. Nov 29, 2017

### PeroK

I don't really understand your FBD. How many forces are acting on the car?

3. Nov 29, 2017

### BvU

Hello Motzu,

Your problem statement looks to me like a statement without a problem
What is it that is asked from you ?

The way I read it is that on a perfectly banked curve the sum of normal force and force from gravity should add up to the centripetal force exactly, so that $\mu$ can be zero ...

4. Nov 29, 2017

### PeroK

... so that when the car is travelling faster we must calculate the required coefficient of friction to support the extra speed.

5. Nov 29, 2017

### BvU

Ah, right. I should learn to read a bit further before responding .
For Motzu I gave it away a little bit: perfect means two forces, so for more speed the friction has to supply the remainder

6. Nov 29, 2017

### haruspex

I do not understand the label Ffy in your diagram. The name suggests it is the vertical component of the frictional force, but it is not drawn as vertical. If it is the whole frictional force then it should be normal to the FN, but it is not that either. Instead, it is shown as being a kind of reflection in the horizontal plane of the normal force.
However, you did get the right vertical force sum.
What directions are those two forces in?
Was that given or did you calculate it? in many such problems there is no need to determine the angle as so many degrees or radians. Your algebra will find a trig function of the angle, and that is all you will need for the next part.

7. Nov 29, 2017

### Motzu2098

I previously calculated it.

8. Nov 29, 2017

### jack action

First, this is wrong:

Ff = μ · g · m

The friction force depends on the normal force which is not necessarily $mg$.

For you free body diagram, you took a very complicated way to do it and I didn't bother to check if it was OK. Instead, you should analyze the forces acting parallel and perpendicular to the road. Because what you are interested in is the normal force acting on the road and the friction force parallel to the road. It should be clearer that way.

9. Nov 29, 2017

### PeroK

Let me show you how I would set up the problem. See whether you think this is clearer than what you wrote in your OP.

The track is set at an angle of $\theta = 26.7°$. The radius of the track is $r = 88m$. The car has a velocity of $v = 95 km/h$.

There are three forces acting on the car: gravity $F_g = mg$ where $g = 9.8 m/s^2$ acting vertically downward. The normal force from the track $F_N$. A frictional force acting tangentially down the track $F_f$.

The net force is ... (I'll leave that to you).

Then you need a FDB with these forces shown and how you are going to decompose the forces.

10. Dec 4, 2017

### PeterO

And be careful: The Normal Reaction force is larger when the car is doing 95 kph, compared to when it was doing 75 kph. The only thing the same at each speed is the direction (perpendicular to the surface; that is why it is called a Normal force. Not normal as in usual, but normal as in perpendicular)