• question_answer A heavy box is to 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 ${{F}_{A}},$ 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                [JEE ONLINE 19-04-2014] A) $\sqrt{3}$ B) $\frac{5}{\sqrt{3}}$ C) $\sqrt{\frac{3}{2}}$ D) $\frac{2}{\sqrt{3}}$

 [d] ${{F}_{A}}=\frac{\mu mg}{\sin \theta -\mu \cos \theta }$Similarly, ${{F}_{B}}=\frac{\mu mg}{\sin \theta +\mu \cos \theta }$ $\therefore$ $\frac{{{F}_{A}}}{{{F}_{B}}}=\frac{\frac{\mu mg}{\sin \theta -\mu \cos \theta }}{\frac{\mu mg}{\sin \theta +\mu \cos \theta }}$ $\frac{\frac{\mu mg}{\sin {{60}^{o}}-\frac{\sqrt{3}}{5}\cos {{60}^{o}}}}{\frac{\mu mg}{\sin {{30}^{o}}+\frac{\sqrt{3}}{5}\cos {{30}^{o}}}}$ $\left[ \mu =\frac{\sqrt{3}}{5}\text{given} \right]$ $=\frac{\sin {{30}^{o}}+\frac{\sqrt{3}}{5}\cos {{30}^{o}}}{\sin {{60}^{o}}-\frac{\sqrt{3}}{5}\cos {{60}^{o}}}$ $=\frac{\frac{1}{2}+\frac{\sqrt{3}}{5}\times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{5}\times \frac{1}{2}}$ $=\frac{\frac{1}{2}\left( 1+\frac{3}{5} \right)}{\frac{\sqrt{3}}{5}\left( 1-\frac{1}{5} \right)}=\frac{\frac{1}{2}\times \frac{8}{5}}{\frac{\sqrt{3}\times 4}{10}}$ $=\frac{\frac{8}{10}}{\frac{\sqrt{3}\times 4}{10}}=\frac{8}{\sqrt{3}\times 4}=\frac{2}{\sqrt{3}}$