## College Physics (4th Edition)

The surface should be banked at an angle of $58.8^{\circ}$
We can assume that the frictional force is zero. Let $F_N$ be the normal force exerted by the surface on the bike. We can make an equation with the vertical components of the forces acting on the bike: $\sum F_y = 0$ $F_N~cos~\theta -mg = 0$ $F_N = \frac{mg}{cos~\theta}$ The horizontal component of the normal force provides the centripetal force to keep the bike moving around the curve: $F_N~sin~\theta = m~a_r$ $(\frac{mg}{cos~\theta})~sin~\theta = m~\frac{v^2}{r}$ $g~tan~\theta = \frac{v^2}{r}$ $tan~\theta = \frac{v^2}{g~r}$ $\theta = tan^{-1}(\frac{v^2}{g~r})$ $\theta = tan^{-1}\left[\frac{(18~m/s)^2}{(9.80~m/s^2)(20.0~m)}~\right]$ $\theta = 58.8^{\circ}$ The surface should be banked at an angle of $58.8^{\circ}$