Coefficient of static friction with brakes

[conniechiwa]

On a level road with its brakes on, the shortest distance in which a car traveling with 83 km/hr can stop is 95 m. This shortest distance occurs when the driver uses anti-lock brakes which means that the car brakes without skidding.

What is the coefficient of static friction between the tires and the pavement?

I have determined that the acceleration is 2.80, but now I'm stuck.

 

[AlephZero]

Your acceleration of 2.80 m/s^2 looks right. (You should really give the units. The question uses m, km, and hr, so just "2.80" might be 2.80 km/hr^2 !)

The coefficient of friction = friction force / normal force.

The normal force is the weight of the car, so call the mass of the car m kg. Then you can find both forces in terms of m. The mass will cancel out when you find the friction coefficient.

 

[HallsofIvy]

However, the coefficient of static friction has nothing to do with this problem! The coefficient of static friction, multiplied by the normal force on an object, gives the force necessary to start the object moving from a stop. In this problem, you should be using sliiding friction.

 

[AlephZero]

There is no sliding between the tires and the road, assuming the anti-lock brakes are working perfectly.

But I agree that all you can really say is the coefficient of static friction is >= your calculated value.

And because of the anti-lock brakes, the coeff. of sliding friction could be LESS than the calculated value.

 

[conniechiwa]

i'm not really sure how to find the friction force

 

[AlephZero]

The friction force produces the (negative) acceleration that stops the car.

What does Newton's second law say about force, mass, and acceleration?

 

[conniechiwa]

F=ma

 

[learningphysics]

F=ma

Yes, and the normal force = mg

masses cancel, in your equations. and you can get coefficient of friction.

 

[conniechiwa]

thanks