Answer
Minimum coefficient of static friction needed between tires and road to prevent skidding
is , $$ u = 0.306 $$
Work Step by Step
According to figure , Mg is the weight of car , N is normal force on car .
Since , according to question road is banked such that for the car at speed 20 m/s
$$ u = 0 $$
here , u is the coefficient of static friction
now , the given velocity is greater than 20 m/s so the car will tend to skid upward so friction will be acting downward let it be written as F .
Step 1 :
we draw free - body diagram and mark forces as shown in figure
Step 2 :
we apply newton's laws of motion in vertical and horizontal motion -
Ncos$\theta$ - Fsin$\theta$ = Mg $: - Equation 1$
Nsin$\theta$ + Fcos$\theta$ = M$v^{2}$$/$R $: - Equation 2$ (Here , $v$ is velocity and R is radius )
on dividing $Equations$ 1 and 2
we get , $$\frac{v^{2}}{Rg} = \frac{sin\theta + ucos\theta}{cos\theta - usin\theta}$$
on dividing cos$\theta$ in numerator and denominator at RHS
we get , $$\frac{v^{2}}{Rg} = \frac{tan\theta + u}{1 - utan\theta} :- Equation 3$$
Step 3 :
finding tan$\theta$ when $u = 0$
using Equation 3 we get , $$ tan\theta = \frac{20^{2}}{140 \times 9.8} = 0.291$$
Step 4 :
Since , we know tan$\theta$ therefore using Equation 3 we calculate $u$ for velocity $v$ = 30 m/s
$$\frac{30^{2}}{140 \times 9.8}=\frac{0.291 + u}{1 - 0.291u}$$
Therefore , $$ u = 0.306 $$
