A heavy box is to be dragged along a rough horizontal floor. To do… | JEE Main 2014 (19 Apr, Shift 1)

Q28JEE Main 2014 · 19 APR · SHIFT 1Difficult

Friction Basics

A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle 30° from the horizontal and requires a minimum force FA, while person B pulls the box at an angle 60° from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is $\frac{\sqrt{3}}{5}$ , the ratio $\frac{F_A}{F_B}$ is:

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Answer: (d)

Push at $30^\circ$ below horizontal: $F_A\cos 30^\circ = \mu(mg + F_A\sin 30^\circ) \Rightarrow F_A = \dfrac{\mu mg}{\cos 30^\circ - \mu\sin 30^\circ}$
Pull at $60^\circ$ above horizontal: $F_B = \dfrac{\mu mg}{\cos 60^\circ + \mu\sin 60^\circ}$
With $\mu = \dfrac{\sqrt{3}}{5}$ : $\cos 30^\circ - \mu\sin 30^\circ = \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{10} = \dfrac{2\sqrt{3}}{5}$ ; $\cos 60^\circ + \mu\sin 60^\circ = \dfrac{1}{2} + \dfrac{3}{10} = \dfrac{4}{5}$
$\dfrac{F_A}{F_B} = \dfrac{4/5}{2\sqrt{3}/5} = \dfrac{2}{\sqrt{3}}$
So the answer is (d) $\dfrac{2}{\sqrt{3}}$ .

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