# Circular Motion and Friction of car turning corner

1. Jun 16, 2010

### haroldtreen

1. The problem statement, all variables and given/known data
"A 1500kg car rounds a curve of radius 75m at a speed of 25m/s. The curve is banked at an angle of 22 degrees to the horizontal. Calculate:

a) The magnitude of the frictional force.
b) The Coefficient of Kinetic Friction."

2. Relevant equations
F=mv$$^{2}$$/r

a) 61000N
b) 0.33

3. The attempt at a solution

What makes sense to me is that the centripetal force is countered by another force. I believe this other force is a combination of the horizontal frictional force & the horizontal normal force (cause from the banked curve resisting the centripetal force).

The centripetal force = 12500N

So,

12500N = Force Friction x + Force Normal x = Force Friction x + Force Normal*Sin$$\phi$$

Force Normal = mg/cos$$\phi$$

This however have been unable to get me the provided answers :S.
It seems like a pretty simple question and I have been playing around with it for a bit and keep ending up with 6560N for the frictional force :(.

If anyone could help it would be great...I'm studying for exams! :P

2. Jun 17, 2010

### mo_0820

Now, please make the force analysis again.
How many forces act on the car? And what is the effect?

3. Jun 17, 2010

### inky

Actually your answer given for frictional force is not 61000N. Could you check this ? I think anwer is 6100 N. You need to use g =9.8 ms^-2. You can get 6083.08 N. It is around 6100N.
Then if use coefficent of friction formula, you can get mu value 0.33. Could you try again?

4. Jun 18, 2010

### ehild

This is not true.The force of friction has both horizontal and vertical components.

By the way, the answer for the frictional force is wrong, it is rather 6100 N.

ehild

5. Jun 18, 2010

### inky

Use summation F(x)=(mv^2)/r
summation F(y)=0
Both normal force and friction force have 2 components.