Static friction has me stuck

liljediboi

a drag car's tires have some of the highest friction coefficients, so the book says.

it takes 6 s to go a quarter mile (about 402.5m) with constant acceleration, no skidding

what is the coefficient for the static friction?

A CORRECTION HAS BEEN MADE, IT IS 402.5m NOT 602.5

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is it 1.14?

deliveryman

You need more information to solve this problem, maybe the mass of the car. Atleast I think this problem is unsolvable based on the information you gave us.

agro

First, we need to know the amount of acceleration that makes the car travels 602.5 m in 6 s (starting from zero velocity, I presume). Using d = 0.5 at^2, we get a = 33.47 m/s/s.

If the car has mas M, then the force that makes the car accelerates as much as 33.47 m/s/s is Ma. But that Ma is the friction force (static) between the tires and the road (Since no other force could cause it). Since Ffriction = uN where N is the normal force and u is the coefficient of static friction, and since N equals to Mg where g is gravity, then

Ffriction= uN
Ma = u Mg
a = ug
u = a/g
u = 33.47/9.8
u = 3.41

Correct me if I'm wrong...

liljediboi

umm.. the equation is d=at, not d=at^2 thats for falling objects. isint it?

agro

It is true that d = 0.5*at^2 is the equation for falling objects with zero initial velocity. More generally, it is the equation of motion for objects with constant acceleration and zero initial velocity. Falling objects happens to undergo constant acceleration (the acceleration of gravity), and so does your constantly accelerated car!

it takes 6 s to go a quarter mile (about 402.5m) with constant acceleration, no skidding
the equation d = vt (not d = at) is for objects moving with a constant velocity (zero acceleration). Your car's velocity is not constant, it starts from zero and increases constantly (constant acceleration).

Anything else that still puzzles you?

Isn't it wonderful that a falling apple and an accelerating car have something in common :) ?

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