# Net Forces on a Car on a Turn

1. Dec 10, 2008

### pillanoid

1. The problem statement, all variables and given/known data
At a certain instant of time, a 1200 kg car traveling along a curve 250 m in radius is moving at a speed of 10 m/s but is slowing down at a rate of 2 m/s2. Ignoring air friction, what is the total static friction force on the car as a fraction of its weight at that instant?

2. Relevant equations
Fc = mv2/R
Ff = $$\mu$$FN = $$\mu$$mg

3. The attempt at a solution
I have an idea at how to attempt this...
mv2/R = Ff
(since the centripetal force is provided by the frictional force)
but I need clarification: I see it that the frictional force and the force due to the acceleration (deceleration) of the car act perpendicular to each other and therefore on different axes.
How would you account for this force due to deceleration in calculating the centripetal force? Is this just a matter of calculating the speed of the car at that instant, in which case the deceleration doesn't matter?

2. Dec 10, 2008

### Staff: Mentor

Careful. Friction provides all the force accelerating the car, not just the centripetal force. There's both a tangential and a centripetal component.

3. Dec 10, 2008

### pillanoid

Ah, okay. So the total frictional force is equal to mv^2/R - 2m = -1.6m?

4. Dec 10, 2008

### LowlyPion

Unfortunately it's not that simple.

The accelerations are vectors and they are not acting in the same direction.

Figure the sum of the vectors first because the slowing is tangential and the centripetal is radial.