A body of mass $10$ kg slides along a rough horizontal surface. The coefficient of friction is $1/\sqrt 3 $. Taking $g = 10\,m/{s^2}$, the least force which acts at an angle of $30^o $ to the horizontal is ...... $N$
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(c) Let $P$ force is acting at an angle $30^o $ with the horizontal.
For the condition of motion $F = \mu \;R$
$P\cos 30^\circ = \mu \,(mg - P\sin 30^\circ )$
==> $P\frac{{\sqrt 3 }}{2} = \frac{1}{{\sqrt 3 }}\left( {100 - P\frac{1}{2}} \right)$
==>$\frac{{3P}}{2} = \left( {100 - \frac{P}{2}} \right)$
==> $2P = 100$ $\therefore \;\;P = 50\;N$
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