# Net Force Dispute with Teacher

1. Oct 27, 2013

### Ninjamonkey474

1. A 1000 kg automobile initially moving 72 km/h jams the brakes and skids to a stop in a distance of 24 m on a dry pavement but stops in 81.6 m for a wet pavement. (Neglect reaction delay.)
A. Calculate the applied forces for both weather conditions.
B. What is the coefficient of friction for dry weather?
C. What is the coefficient in wet weather conditions

μ=Ffriction/Fnormal,
F=ma

3. I said this was impossible since both friction and applied are pointing in the same direction, you can only calculate the net force and not the applied, without which you cannot find either of the coefficients. She said it was possible but I didn't understand what she meant, can somebody explain to me why I'm wrong or why I'm right?

2. Oct 27, 2013

### tiny-tim

Hi Ninjamonkey474! Welcome to PF!
The friction (from the road) is the applied force.

(there are no external forces on the car except the weight and the friction from the road)

3. Oct 27, 2013

### Ninjamonkey474

So do I also use the combined brake-road force to calculate the coefficients of friction?

4. Oct 27, 2013

### Staff: Mentor

What combined force? The only force you need to worry about is friction from the road.

Or are you thinking that she meant the force that the driver "applied" to the brake? That's not what she meant. And if she did, you would not be able to calculate that. (Note that any force the driver applies to the brake pedal is internal to the system and has no direct affect on the slowing of the car.)

5. Oct 27, 2013

### Staff: Mentor

When your teacher is talking about the applied forces, what she is talking about is Fnormal and Ffriction. Why don't you start out by first calculating the acceleration rates (or in this example, the deceleration rates) under the two different kinds of road conditions? The next step after that will be to calculate the normal force. The only other force acting on the car besides that is the frictional force exerted by the road on the car, in the direction opposite to the car's motion.

Chet