Finding the Frictional Force Necessary to Prevent Skidding

A motorcycle with a mass of 150 kg is driven at a constant speed of 120 km/h on a circular track whose radius is 100 m. To keep the motorcycle from skidding, what frictional force must be exerted by the track on the tires?

Solution By Newton's Second Law of Motion, the force \(\mathbf{F}\) required to keep an object of mass \(m\) traveling along a curve traced out by \(\mathbf{r}=\mathbf{r}(t)\) is \(\mathbf{F}=m\mathbf{a}\). The magnitude of the frictional force exerted by the tires must therefore equal \begin{equation*} \left\Vert \mathbf{F}\right\Vert\;=\;m\left\Vert \mathbf{a}\right\Vert \end{equation*}

In this example, the motion is circular. So, using (2), we have \begin{eqnarray*} &&\left\Vert \mathbf{F}\right\Vert\;=\;m\left( \frac{v_{0}^{2}}{R}\right)\;=\;150\enspace \textrm{kg} \left[\dfrac{\left( 120 \enspace \textrm{km/h}\right)^{2}}{100\enspace \textrm{m}}\right]\quad\,\color{#0066A7}{\underset{{\hbox{$\uparrow$}} }{=}} ( 150) ( 144) \left(\dfrac{1000^{2}}{3600^{2}}\right) \dfrac{\textrm{kg}\,{\cdot}\,\textrm{m}}{\textrm{s}^{2}} \approx 1667\ \enspace \textrm{N}\\[-16.1pt] &&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \color{#0066A7}{\hbox{1 h = 3600s}} \end{eqnarray*}

The track must exert a force of magnitude 1667 \(N\) or greater to prevent skidding.