The coefficient of kinetic friction \(\mu_{ s }\) is given by
\(\mu_{r}=\frac{\text { frictional force }}{\text { normal reaction }}\) body of mass \(M\) sliding down an inclined plane with \(\theta\) as the angle of inclination has its coefficient of static friction as \(\mu_{ s }=\tan \theta\). This is proved below i.e.
(Free body diagram of the body for the body FBD)
Note: If the body slides down, the frictional force \((f r)\) act the opposition direction as shown so that
\(\Sigma F_{\text {net }}=m a-\) applying Newton's law
\(m g \sin \theta-f r=m a\)
\(f r=m g \sin \theta-m a\) \(f r=m(\sin \theta-a)\)
\(f r=m(g \sin \theta-a)\) but \(f r=\mu m g \cos \theta\)
but fr \(=\mu m g \cos \theta\)
\(\mu m g \cos \theta=m(g \sin \theta-a), a=0\)
\(\mu m g \cos \theta=m g \sin \theta\)
\(\mu \cos \theta=\sin \theta\)
\(\mu s=\frac{\sin \theta}{\cos \theta}=\tan \theta\)