## College Physics (4th Edition)

The road should be banked at an angle of $5.08^{\circ}$
We can assume that the frictional force is zero. Let $F_N$ be the normal force exerted by the road on the car. We can make an equation with the vertical components of the forces acting on the car: $\sum F_y = 0$ $F_N~cos~\theta -mg = 0$ $F_N = \frac{mg}{cos~\theta}$ The horizontal component of the normal force provides the centripetal force to keep the car moving around the curve: $F_N~sin~\theta = m~a_r$ $(\frac{mg}{cos~\theta})~sin~\theta = m~\frac{v^2}{r}$ $g~tan~\theta = \frac{v^2}{r}$ $tan~\theta = \frac{v^2}{g~r}$ $\theta = tan^{-1}(\frac{v^2}{g~r})$ $\theta = tan^{-1}\left[\frac{(26.8~m/s)^2}{(9.80~m/s^2)(825~m)}~\right]$ $\theta = 5.08^{\circ}$ The road should be banked at an angle of $5.08^{\circ}$
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