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

Answer: (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 $F_1=3F_2$ ?(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 $\theta ={30}^o$ ) $\mu =\frac{1}{2}\tan {30}^o$ $\mu =\frac{1}{2\sqrt{3}}$