# Question #06104

##### 1 Answer
Jan 28, 2018

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.

Inserting given values we get

$f = 2000 \times 9.81 \times \sin {25}^{\circ}$
$\implies f = 8292 \text{ N}$