Chapter 5 Problem 58

Problem

If a curve with a radius of 85 m is properly banked for a car traveling 65 km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/h?

Solution

Since the curve is designed for 65 km/h, traveling at a higher speed with the same radius means that more centripetal force will be required. That extra centripetal force will be supplied by a force of static friction, downward along the incline.

The no-friction banking angle is given by:

$F_{N}\cos \theta -mg=0\rightarrow F_{N}={\frac {mg}{\cos \theta }}$

$F_{N}\sin \theta =m{\frac {v^{2}}{r}}\rightarrow {\frac {mg}{\cos \theta }}\sin \theta =m{\frac {v^{2}}{r}}$

$\tan \theta ={\frac {v^{2}}{rg}}$

$\theta =\tan ^{-1}{\frac {\left[\left(65{\text{km/hr}}\right)\left({\frac {1{\text{m/s}}}{3.6{\text{km/hr}}}}\right)\right]^{2}}{\left(9.80{\text{m/s}}^{2}\right)\left(85{\text{m}}\right)}}=21.4^{\circ }$

The car will have no acceleration in the y direction, and centripetal acceleration in the x direction. We also assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of $F_{\text{FR}}=\mu _{s}F_{N}$.

$\sum F_{y}=F_{N}\cos \theta -mg-F_{\text{FR}}\sin \theta =0$ $\rightarrow F_{N}\cos \theta -\mu _{s}F_{N}\sin \theta =mg$ $\rightarrow F_{N}={\frac {mg}{\cos \theta -\mu _{s}\sin \theta }}$

$\sum F_{x}=F_{N}\sin \theta +F_{\text{FR}}\cos \theta =F_{R}=m{\frac {v^{2}}{r}}$ $\rightarrow F_{N}\sin \theta +\mu _{s}F_{N}\cos \theta =m{\frac {v^{2}}{r}}$ $\rightarrow F_{N}={\frac {m{\frac {v^{2}}{r}}}{\sin \theta +\mu _{s}\cos \theta }}$

equate

${\frac {m{\frac {v^{2}}{r}}}{\sin \theta +\mu _{s}\cos \theta }}={\frac {mg}{\cos \theta -\mu _{s}\sin \theta }}$ $\rightarrow \mu _{s}={\frac {{\frac {v^{2}}{r}}\cos \theta -g\sin \theta }{g\cos \theta +{\frac {v^{2}}{r}}\sin \theta }}={\frac {{\frac {v^{2}}{r}}-g\tan \theta }{g+{\frac {v^{2}}{r}}\tan \theta }}$

since $v=\left(95{\text{km/hr}}\right)\left({\frac {1{\text{m/s}}}{3.6{\text{km/hr}}}}\right)=26.39{\text{m/s}}$

$\mu _{s}={\frac {\left({\frac {\left(26.39{\text{m/s}}\right)^{2}}{85{\text{m}}}}\left(9.80{\text{m/s}}^{2}\right)\tan 21.4^{\circ }\right)}{\left(9.80{\text{m/s}}^{2}+{\frac {\left(26.39{\text{m/s}}\right)^{2}}{85{\text{m}}}}\tan 21.4^{\circ }\right)}}=0.33$