Calculae the total torque acting on the body shown in figure about the point O.

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Solution

$\Rightarrow T_{1} = - F r sin θ$
$= - (10) (0.04) sin 90 °$
$= - 0.4 N m$
$\Rightarrow T_{2} = F r sin θ$
$= (20) (0.04) sin 150 °$
$= 0.4 N m$
$\Rightarrow T_{3} = F r sin θ$
$= (5) (0.03) sin 0 °$
$= 0$
$∴$ Total torque $= T_{1} + T_{2} + T_{3} + T_{4}$
$= (- 0.4 + 0.4 + 0.54) N m$
$= 0.54 N m .$
Hence, the answer is $0.54 N m .$

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