Question

A cubical block rests on an inclined plane of coefficient of friction $ \mu =\frac{1}{\sqrt{3}} $ . What should be the angle of inclination so that the block just slides down the inclined plane?

• A. $ 30^{\circ} $
• B. $ 60^{\circ} $
• C. $ 45^{\circ} $
• D. $ 90^{\circ} $

Answer 1

Answer: (A)

$ \tan \,\,\theta =\mu $ $ \tan \theta =\frac{1}{\sqrt{3}} $
$ \tan \theta =\tan \,30^{\circ} $
Angle of inclination, $ \theta =30^{\circ} $