# Question 31590

##### 1 Answer
Oct 24, 2016

Forces acting on the body are shown in the diagram above

Sum of the forces along the inclined plane:
${F}_{\text{net}} = m g \sin \theta - f$
where $m g$ is the weight of the body and $f$ is force of friction.

There is no acceleration along the inclined plane, ${F}_{\text{net}} = 0$, therefore
$m g \sin \theta = f$ ......(1)

Similarly in a direction perpendicular to the inclined plane:
${F}_{\text{net}} = N - m g \cos \theta$
where $N$ is normal reaction.

As per given condition above there is no acceleration in this direction either, ${F}_{\text{net}} = 0$, and therefore
$N = m g \cos \theta$ ......(2)

If ${\mu}_{k}$ is coefficient of kinetic friction, then
f = µ_kN#

Using (1) and (2) we get
$m g \sin \theta = {\mu}_{k} \times m g \cos \theta$

Solving for ${\mu}_{k}$
${\mu}_{k} = \tan \theta$
${\mu}_{k} = \tan {30}^{\circ} = \frac{1}{\sqrt{3}}$