# Centripetal Force (Horizontally Banked Question)

1. Nov 2, 2011

### McKeavey

1. The problem statement, all variables and given/known data
If a curve with a radius of 60m is properly banked for a car travelling at 60km/h (16.66m/s) with no friction, what must be the coefficient of friction if the car is not to skid when travelling at 90km/h (25m/s)

2. Relevant equations
mac = mv^2/r

3. The attempt at a solution

First we need to find the angle, so since it said that when the speed was 60km/h, there is No friction, we can make this equation.. and do the following calculations
Fgx = mac
mgsinɵ = mv^2/r
gsinɵ = v^2/r
9.8sinɵ = 16.66^2/60
sinɵ = 4.625/9.8
ɵ = 28.16

So now we have the angle..
We can now use the speed 90km/h with our new angle.
So the new equation is..
Ff + Fgx = mv^2/r
μFn + Fgx = mv^2/r
μmgcos28 + mgsin28 = mv^2/r
μ9.8cos28 + 9.8sin28 = (25)^2/60
μ(8.65) + 4.6 = 10.41
μ = 10.41 - 4.6/8.65
μ = 0.67

So I'm not sure where I went wrong..

2. Nov 2, 2011

### PeterO

You will be familiar with the idea that when you drive through a dip in the road, you feel heavier. This is because when you move through the dip, the reaction force is larger than just mg.
A similar thing happens when you drive round a banked turn. The reaction force is stronger than when the vehicle is parked on the banked turn.
The net result is that your second line : mgsinɵ = mv2/r
should actually have been : mgtanɵ = mv2/r

3. Nov 2, 2011

### McKeavey

Hmm I got angle as 25.24.
And then the resulting coefficient of friction as 0.59 :S

By the way thanks for the other thread, I got the right answer for it ^^
Now..To finish this.. :(

4. Nov 2, 2011

### PeterO

By rounding off your 60 kph conversion you get angle 25.24. I didn't round, and got 25.28657. Only a slight difference, but it may make a difference.

I hope you took into account the fact that at 90 kph, the Reaction force is even greater than it is at 60 kph.

Draw a free body diagram and you will [hopefully] see how much bigger.