Originally Posted by **leh3188** A car is safely negotiating an unbanked circular turn at a speed of 26 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-ninth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
I'm have no idea how to solve this problem. |

The sideways force that friction must resist comes from the centrifugal acceleration due to the turn.

$f_c = m \dfrac{v^2}{R}$

the frictional force is given by

$f_f = \mu m g$ where $\mu$ is the static friction coefficient

if the car is not to skid

$f_c \leq f_f$ so

$m \dfrac{v^2}{R} \leq \mu m g$

$v \leq \sqrt{\mu g R}$

thus $v$ varies as the square root of $\mu$

so if $\mu$ is reduced by 9, $v$ must be reduced by $\sqrt{9}=3$

and thus the speed must be reduced to $\dfrac{26}{3}~m/s$