Question

A car is moving at 10 m/s on a horizontal road with friction on a dry day. The car can travel around a traffic circle with a minimum radius of 4.8 meters. It rains and the car around a traffic circle with a minimum radius of 11.8 meters. What is the percentage of the coefficient of static friction on the rainy day compared to the dry day

Answer 1

Answer:

$\mu_w=86\%$

Explanation:

From the question we are told that:

Velocity on Dry road $V_d=10m/s$

Radius Dry $r_d=4.8$

Radius wet $r_w=11.8$

Generally the equation for coefficient of static friction on the dry day is mathematically given by

$\mu mg=\frac{mv^2}{r_d}$

$\mu g=\frac{v^2}{r_d}$

$\mu_d 9.8=\frac{10^2}{4.8}$

$\mu_d=2.125$

Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by

$\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}$

$\frac{\mu_w}{2.125}=\frac{4.8}{11.8}$

$\mu_w=0.86$

Therefore

$\mu_w=86\%$