Question

The minimum force required to move a body up an inclined plane of inclination $ 30{}^\circ $ , is found to be thrice the minimum force required to prevent it from sliding down the plane. The coefficient of friction between the body and the plane is:

• A. $ \frac{1}{\sqrt{3}} $
• B. $ \frac{1}{2\sqrt{3}} $
• C. $ \frac{1}{3\sqrt{3}} $
• D. $ \frac{1}{4\sqrt{3}} $

Answer 1

The Correct Option is B

From the relation, force required to move on the inclined plane is given by $ {{F}_{1}}=mg(sin\theta +\mu cos\theta ) $ ?(i) But, when the body slides down then force required to slide down is given by $ {{F}_{2}}=mg(sin\theta -\mu cos\theta ) $ ?(ii) (because force of friction acts opposite) The condition given $ {{F}_{1}}=3{{F}_{2}} $ ?(iii) Now, putting the values from (i) and (ii) Eqs. (iii), we obtain $ mg(sin\theta +\mu cos\theta )=3mg(sin\theta -\mu cos\theta ) $ $ \sin \theta +\mu \cos \theta =3\sin \theta -3\mu \cos \theta $ $ 2\sin \theta =4\mu \cos \theta $ $ \mu =\frac{1}{2}\tan \theta $ (given $ \theta ={{30}^{o}} $ ) $ \mu =\frac{1}{2}\tan {{30}^{o}} $ $ \mu =\frac{1}{2\sqrt{3}} $