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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
$\sqrt 3 $
$\frac{5}{{\sqrt 3 }}$
$\sqrt {\frac{3}{2}} $
$\frac{2}{{\sqrt 3 }}$
Solution
$\begin{array}{l}
{F_A} = \frac{{\mu mg}}{{\sin \theta – \mu \cos \theta }}\\
similarly,\\
{F_B} = \frac{{\mu mg}}{{\sin \theta + \mu \cos \theta }}\\
\,\therefore \,\frac{{{F_A}}}{{{F_B}}} = \frac{{\frac{{\mu mg}}{{\sin \theta – \mu \cos \theta }}}}{{\frac{{\mu mg}}{{\sin \theta + \mu \cos \theta }}}}\,\,\,\,\,\\
\, = \frac{{\sin {{60}^ \circ } – \frac{\begin{array}{l}
\mu mg\\
\sqrt 3
\end{array}}{5}\cos \,{{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\begin{array}{l}
\mu mg\\
\sqrt 3
\end{array}}{5}\cos \,{{30}^ \circ }}}\,\,
\end{array}$
$\begin{array}{l}
\,\,\left[ {\mu = \frac{{\sqrt 3 }}{5}\,given} \right]\,\\
= \frac{{\sin \,{{30}^{ \circ \,}} + \frac{{\sqrt 3 }}{5} – \cos \,{{30}^ \circ }}}{{\sin \,{{60}^ \circ } – \frac{{\sqrt 3 }}{5}\cos \,{{60}^ \circ }}}\\
= \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} \times \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} – \frac{{\sqrt 3 }}{5} \times \frac{1}{2}}}
\end{array}$
$\begin{array}{l}
= \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} \right)}}{{\frac{{\sqrt 3 }}{5}\left( {1 – \frac{1}{5}} \right)}} = \frac{{\frac{1}{2} \times \frac{8}{5}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}}\\
= \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3 \times 4}}{{10}}}} = \frac{8}{{\sqrt 3 \times 4}} = \frac{2}{{\sqrt 3 }}
\end{array}$
