A car weighing 10.7kNand traveling at 13.4 m/swithout negative lift attempts to round an unbanked curve with a radius of 61.0 m. (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is 0.350, is the attempt at taking the curve successful?

Weight of car, w=10.7 kN .

Speed of car, v =13.4 m/s .

Radius of unbanked curve, r =61.0m .

In this problem, the car is making a turn on an unbanked curve. Friction is what provides the centripetal force needed for this circular motion. We will first draw the free body diagram of the car. Then, using the centripetal force and Newton’s second law, we can solve the given problem.

Formula:

${\mathbf{f}}_{\mathbf{s}}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{R}}$

(a)

The mass of the car is

$$\begin{gathered} m=\left(10700/9.80\right)\mathrm{kg} \\ =1.09\times {10}^3\mathrm{kg} \end{gathered}$$.

Choose "inward" (horizontally toward the center of the circular path) as the positive direction.

The normal force is $F_N=mg$in this situation, and the required frictional force is $f_s=m{v}^2/R$.

With a speed of v=13.4m/s and a radius R=61 m, Newton's second law leads to,

$$\begin{gathered} f_s=\frac{m{v}^2}{R} \\ \Rightarrow f_s=\frac{\left(1.09\times {10}^3\mathrm{kg}\right){\left(13.4\mathrm{m}/\mathrm{s}\right)}^2}{61.0\mathrm{m}} \\ \Rightarrow f_s=3.21\times {10}^3\mathrm{N} \end{gathered}$$

(b)

The maximum possible static friction is found to be
$$\begin{gathered} f_{s,m}={\mu }_{s}m \\ \Rightarrow f_{s,m}=\left(0.35\right)\left(10700\mathrm{N}\right) \\ \Rightarrow f_{s,m}=3.75\times {10}^3\mathrm{N} \end{gathered}$$

We see that the static friction found in part (a) is less than this, so the car rolls (no skidding) and successfully negotiates the curve.