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Q.
Block $M$ slides down on frictionless incline as shown. The minimum friction coefficient so that $m$ does not slide with respect to $M$ would be :
Laws of Motion
Solution:
The acceleration of the blocks down the incline will be $g \sin \theta$ so $F B D$ of $m$ wr to $M$ in limiting state is
Sol
$ N =m g-m g \sin ^{2} \theta $
$ \mu N =m g \sin \theta \cos \theta $
$ \mu\left(m g-m g \sin ^{2} \theta\right) =m g \sin \theta \cos \theta $
$ \mu\left(1-\sin ^{2} 37^{\circ}\right) =\sin 37^{\circ} \cos 37^{\circ} $
$ \mu\left(1-\frac{9}{25}\right) =\frac{3}{5} \times \frac{4}{5} $
$ \mu(16) =12 $
$ \mu =\frac{3}{4} $
