Forces of Friction

1. Dec 6, 2011

jer_hall99

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

13. As a car skids with its wheels locked trying to stop on a road covered with ice and snow, the force of friction between the icy road and the tires will usually be: (Please note that “normal” and “perpendicular” have the same meaning)

A. greater than the normal force of the road times the coefficient of static friction
B. equal to the normal force of the road times the coefficient of static friction
C. less than the normal force of the road times the coefficient of kinetic friction
D. greater than the normal force of the road times the coefficient of kinetic friction
E. equal to the normal force of the road times the coefficient of kinetic friction.

2. Relevant equations

Fk=μkn

3. The attempt at a solution
I know that it is kinetic friction since the car is moving. I think that the answer is "C" because the force of friction would be less since the car is skidding. Not sure though, could somebody either confirm my answer or clear up my confusion please?

2. Dec 6, 2011

TaxOnFear

Is this a trick question?

You have the equation right in front of you:

FR = μKN

3. Dec 6, 2011

jer_hall99

It's really not a quick question... I am really just unsure of myself... I take that from your comment you are trying to tell me that the answer then is "E"? Please confirm if the answer is E then. Thank you.

4. Dec 6, 2011

TaxOnFear

I don't like this question. Do they mean the coefficient of kinetic friction while skidding or not? It could be either or depending on the context.

5. Dec 6, 2011

jer_hall99

Not sure. I am guessing it would be "E" since the more I think about it, I put "C" on the test and got it wrong.

6. Dec 6, 2011

7. Dec 6, 2011

TaxOnFear

Ahh yes it is E if you think about it, the frictional force is the same as if it wasn't skidding, however the decceleration due to braking is gone because the tyres are skidding, therefore it takes longer to stop.

8. Dec 6, 2011

Thank you.