Chapter 5, Problem 59P

Since the curve is designed for a speed of 85 km/h, traveling at that speed would mean no friction is needed to round the curve. From Example 5-15 in the textbook, the no-friction banking angle is given by

$\theta ={\tan }^{-1}\frac{v^2}{rg}={\tan }^{-1}\frac{{\left[(85\text{ km/h})\left(\frac{1\text{ m/s}}{3.6\text{ km/h}}\right)\right]}^2}{(68\text{ m})(9.80{\text{ m/s}}^2)}=39.91°$

Driving at a higher speed with the same radius means that more centripetal force will be required than is present by the normal force alone. That extra centripetal force will be supplied by a force of static friction, downward along the incline, as shown in the first free-body diagram for the car on the incline. Write Newton’s second law in both the x and y directions. 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_fr = μ_sF_N.

$\begin{array}{l}\sum F_y=F_{\text{N}}\cos \theta -mg-F_{\text{fr}}\sin \theta =0\to F_{\text{N}}\cos \theta -{\mu }_s{F}_{\text{N}}\sin \theta =mg\to \\ F_{\text{N}}=\frac{mg}{(\cos \theta -{\mu }_s\sin \theta )}\\ \sum F_x=F_{\text{N}}\sin \theta +F_{\text{fr}}\cos \theta =m{v}^2/r\to F_{\text{N}}\sin \theta +{\mu }_s{F}_{\text{N}}\cos \theta =m{v}^2/r\to \\ F_{\text{N}}=\frac{m{v}^2/r}{(\sin \theta +{\mu }_s\cos \theta )}\end{array}$

Equate the two expressions for the normal force, and solve for the speed.

$\begin{array}{l}\frac{m{v}^2/r}{(\sin \theta +{\mu }_s\cos \theta )}=\frac{mg}{(\cos \theta -{\mu }_s\sin \theta )}\to \\ v=\sqrt{rg\frac{(\sin \theta +{\mu }_s\cos \theta )}{(\cos \theta -{\mu }_s\sin \theta )}}=\sqrt{(68\text{ m})(9.80{\text{ m/s}}^2)\frac{(\sin 39.91°+0.30\cos 39.91°)}{(\cos 39.91°-0.30\sin 39.91°)}}=32\text{ m/s}\end{array}$Now for the slowest possible speed. Driving at a slower speed with the same radius means that less centripetal force will be required than that supplied by the normal force. That decline in centripetal force will be supplied by a force of static friction, upward along the incline, as shown in the second free-body diagram for the car on the incline. Write Newton’s second law in both the x and y directions. 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_fr = μ_sF_N.

(III) A curve of radius 68 m is banked for a design speed of 85 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or loo fast.]