The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu $ is the coefficient of friction, the inclination of plane to the horizontal is
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Let the max friction be $\mathrm{f}$

case $1:$ body is just about to move up

friction and gravitational force will act in downward direction so

$2 F=m g \sin \theta+f$         $...(1)$

case $2 ;$ body is just prevented to slip

Friction will act upwards and gravitation will act downwards $F=m g \sin \theta-f$          $...(2)$

Using equation $1$ and $2$

$m g \sin \theta-3 f=0$

max friction $=\mu m g \cos \theta$

$\therefore m g \sin \theta=3 \mu m g \cos \theta$

$\therefore \tan \theta=3 \mu$

$\theta=\tan ^{-1} 3 \mu$

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