A heavy box is to dragged along a rough horizontal floor. To do so, person $A$ pushes it at an angle $30^o$ from the horizontal and requires a minimum force $F_A$, while person $B$ pulls the box at an angle $60^o$ from the horizontal and needs minimum force $F_B$. 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 MAIN 2014, Difficult
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${F_A} = \frac{{\mu mg}}{{\sin \theta - \mu \cos \theta }}$
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{{\sin {{60}^ \circ } - \frac{\mu mg$
$\sqrt 3
}{5}\cos {{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\mu mg$
$\sqrt 3}{5}\cos {{30}^ \circ }}}$
$[ {\mu = \frac{{\sqrt 3 }}{5}}$ given $]$
$= \frac{{\sin {{30}^{ \circ }} + \frac{{\sqrt 3 }}{5} - \cos {{30}^ \circ }}}{{\sin {{60}^ \circ } - \frac{{\sqrt 3 }}{5}\cos {{60}^ \circ }}}$
$= \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} \times \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{5} \times \frac{1}{2}}}$
$= \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} \right)}}{{\frac{{\sqrt 3 }}{5}\left( {1 - \frac{1}{5}} \right)}} = \frac{{\frac{1}{2} \times \frac{8}{5}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}$
$= \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}$
$= \frac{8}{{\sqrt 3 \times 4}}$
$= \frac{2}{{\sqrt 3 }}$
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{{\sin {{60}^ \circ } - \frac{\mu mg$
$\sqrt 3
}{5}\cos {{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\mu mg$
$\sqrt 3}{5}\cos {{30}^ \circ }}}$
$[ {\mu = \frac{{\sqrt 3 }}{5}}$ given $]$
$= \frac{{\sin {{30}^{ \circ }} + \frac{{\sqrt 3 }}{5} - \cos {{30}^ \circ }}}{{\sin {{60}^ \circ } - \frac{{\sqrt 3 }}{5}\cos {{60}^ \circ }}}$
$= \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} \times \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{5} \times \frac{1}{2}}}$
$= \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} \right)}}{{\frac{{\sqrt 3 }}{5}\left( {1 - \frac{1}{5}} \right)}} = \frac{{\frac{1}{2} \times \frac{8}{5}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}$
$= \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}$
$= \frac{8}{{\sqrt 3 \times 4}}$
$= \frac{2}{{\sqrt 3 }}$
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