# Question #06104

Refer to the figure above. Tractor is parked on the incline and is at rest. To keep it in the state of equilibrium, downward force along the incline $m g \sin \theta$ should be balanced by the force of friction $f$ given as $\mu N$.
$\therefore f = m g \sin \theta$
where $m$ is mass of tractor, $g = 9.81 {\text{ ms}}^{-} 2$ is acceleration due to gravity.
$f = 2000 \times 9.81 \times \sin {25}^{\circ}$
$\implies f = 8292 \text{ N}$