91Ó°ÊÓ

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

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

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,

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

(b)

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

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.