Answer:
[tex]\mu_s \geq 0.27[/tex]
Explanation:
The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:
[tex]f_s=\frac{mv^{2}}{R}[/tex]
But we know that:
[tex]f_s\leq \mu_s N[/tex]
And the normal force is given by the sum of the forces in the vertical direction:
[tex]N-mg=0 \implies N=mg[/tex]
Finally, we have:
[tex]f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27[/tex]
So, the minimum value for the coefficient of friction is 0.27.
