Farm Power

A tractor center of gravity is located $0.8\text{ m}$ above ground and tread width is $1.6\text{ m}$. If the tractor drives on a side slope of angle $\theta = 20^\circ$, the overturning factor $\tan\theta / (w / 2h)$ (where $w/2h = 1.0$) is $\tan 20^\circ \approx 0.364$. The margin to overturning $(1 - 0.364)$ is (rounded off to two decimal places):