Centripetal Force and Static Friction??

[Caffine]

A car is driving on a level road at a contant speed 8m/s when it attempts to execute a turn about curve of effective radius 10m. For the following questions, we will assume the turn is successful, that is, the car performs the turns as the driver intends. The static coeff of friction between the tires and the road is .9, the kinetic coeff of friction is .7.

What force provides the centripetal force?
A. Gravity
B. The normal force
C. Static friction
D. Kinetic friction

Answer is C. I thought that it would be kinetic friction since the car is moving. The explanation for static friction is: Since the tires are not slipping on the road, the appropriate friction is static.

I'm still having a hard time understanding this. If the car is moving, how is the frictional force static. I always assumed that meant that the object wasn't moving. Could someone please elaborate on this and help me understand the difference between kinetic friction vs. static friction for a moving vehicle - or atleast why the answer isn't D.

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[kcola]

"It's the relative velocity between the surfaces that counts. As long as the tires aren't skidding, the patch of tire in contact with the road has zero speed with respect to the road. But a better way to say it is that friction always acts to oppose slipping between surfaces. And note that for static friction, there is no velocity between the surfaces."

http://www.physicsforums.com/archive/index.php/t-50276.html

 

[michaelfranklin]

I agree with kcola. When dealing with friction problems, you want to focus on the relative movement (sliding) between two surfaces. Ask yourself, are two surfaces sliding past one another?

With regards to your problem: the centripetal force is caused by (equal to) static friction because the two surfaces (tire - road interface) are not sliding past one another; the tire is rolling. Once the car's tires starts sliding on the road (as the driver does not intend), then kinetic friction comes into play.

So the equation for this problem would be something like:

UsN = (mv^2)/r​

...where Us is coefficient for static friction, N is the normal force (mg), m is the mass of the vehicle, v is the velocity of the vehicle, and r is the radius of curvature.

 

[Rabolisk]

Why are you reviving a topic that is 3 weeks old?

 

[michaelfranklin]

Because I just got an account and wanted to flex my posting muscles on a question that I felt was inadequately answered. Necessary? Probably not...just as useful as commenting on a post answering a question that's three weeks old. But I do apologize for wasting 5 seconds of your life.