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

`\mu_w=86\%`

Step-by-step 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=(mv^2)/(r_d)`

`\mu g=(v^2)/(r_d)`

`\mu_d 9.8=(10^2)/(4.8)`

`\mu_d=2.125`

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

`(\mu_d)/(\mu_w)=(r_d)/(r_w)`

`(\mu_w)/(2.125)=(4.8)/(11.8)`

`\mu_w=0.86`

Therefore

`\mu_w=86\%`