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Question

$\begin{array}{l}
{F_A} = \frac{{\mu mg}}{{\sin 	heta  - \mu \cos 	heta }}\
similarly,\
{F_B} = \frac{{\mu mg}}{{\sin 	heta  + \mu \

Vedclass · Accepted Answer

$\frac{2}{{\sqrt 3 }}$

Vedclass · Answer

$\begin{array}{l}
{F_A} = \frac{{\mu mg}}{{\sin 	heta  - \mu \cos 	heta }}\
similarly,\
{F_B} = \frac{{\mu mg}}{{\sin 	heta  + \mu \cos 	heta }}\
\,	herefore \,\frac{{{F_A}}}{{{F_B}}} = \frac{{\frac{{\mu mg}}{{\sin 	heta  - \mu \cos 	heta }}}}{{\frac{{\mu mg}}{{\sin 	heta  + \mu \cos 	heta }}}}\,\,\,\,\,\
\, = \frac{{\sin {{60}^ \circ } - \frac{\begin{array}{l}
\mu mg\
\sqrt 3 
\end{array}}{5}\cos \,{{60}^ \circ }}}{{\sin {{30}^ \circ } + \frac{\begin{array}{l}
\mu mg\
\sqrt 3 
\end{array}}{5}\cos \,{{30}^ \circ }}}\,\,
\end{array}$

$\begin{array}{l}
\,\,\left[ {\mu  = \frac{{\sqrt 3 }}{5}\,given} ight]\,\
 = \frac{{\sin \,{{30}^{ \circ \,}} + \frac{{\sqrt 3 }}{5} - \cos \,{{30}^ \circ }}}{{\sin \,{{60}^ \circ } - \frac{{\sqrt 3 }}{5}\cos \,{{60}^ \circ }}}\
 = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{5} 	imes \frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{5} 	imes \frac{1}{2}}}
\end{array}$

$\begin{array}{l}
 = \frac{{\frac{1}{2}\left( {1 + \frac{3}{5}} ight)}}{{\frac{{\sqrt 3 }}{5}\left( {1 - \frac{1}{5}} ight)}} = \frac{{\frac{1}{2} 	imes \frac{8}{5}}}{{\frac{{\sqrt 3  	imes 4}}{{10}}}}\
 = \frac{{\frac{8}{{10}}}}{{\frac{{\sqrt 3  	imes 4}}{{10}}}} = \frac{8}{{\sqrt 3  	imes 4}} = \frac{2}{{\sqrt 3 }}
\end{array}$

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