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For toppling of the regular hexagon shown. the coefficient of friction must be

\[\begin {array} {1 1} (a)\; > 0.21 & \quad (b)\; < 0.21 \\ (c)\;= 0.21 & \quad  (d)\; < 0.21 \end {array}\]

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At the instant of toppling normal force passes through the point P.
Taking moments about P
$F. 2a \cos 30 =mg \large\frac{a}{2}$
$Fa \times \sqrt 3=\large\frac{a}{2}$
$\therefore F= \large\frac{mg}{2 \sqrt 3}$
$f= \mu mg$
$ f > F$ for the object not to slide.
$\therefore \mu mg > \large\frac{mg}{2 \sqrt 3}$
$\mu > \large\frac{\sqrt 3}{6}$
$\mu > \large\frac{1.732}{6}$
$\mu > 0.21$
answered Dec 6, 2013 by meena.p

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