Question

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

Answer 1

Answer:

4.18

Explanation:

Givens  

The car's initial velocity  $v_{i}$= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.  

Knowns

We know that the maximum static friction force is given by:

$f_{s_max} =$μ_s*n                         (1)

Where μ_s is the coefficient of static friction and n is the normal force.  

Calculations  

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx=$v_{i} +\frac{1}{2} at^2$

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:  

$f_{y}=n-mg=0\\ n=mg\\f_{x}=F-f_{s,max} =0\\ f_{s,max}=F=ma$

Substituting (3) into (1), we get:

$f_{s,max}=$ μ_s*m*g

Equating this equation with (4), we get:

ma=  μ_s*m*g

 μ_s=a/g

      =4.18