ABC is an equilateral triangle with O as its | vedclass

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(C) Let $x$ be the perpendicular distance from the center $O$ of the equilateral triangle to each of its sides.Since the triangle is equilateral,the d

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$({F_1} + {F_2})$

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(C) Let $x$ be the perpendicular distance from the center $O$ of the equilateral triangle to each of its sides.Since the triangle is equilateral,the distance from the center to each side is the same.The torque $	au$ produced by a force $F$ about point $O$ is given by $	au = F 	imes x$.Looking at the directions of the forces $\vec F_1, \vec F_2$,and $\vec F_3$ along the sides of the triangle,we can see that $\vec F_1$ and $\vec F_2$ produce torques in the same rotational sense (e.g.,clockwise) about $O$,while $\vec F_3$ produces a torque in the opposite sense (e.g.,counter-clockwise).For the total torque about $O$ to be zero,the sum of the torques must be zero:$	au_1 + 	au_2 - 	au_3 = 0$$F_1 x + F_2 x - F_3 x = 0$Dividing by $x$ (since $x 
eq 0$):$F_1 + F_2 - F_3 = 0$Therefore,$F_3 = F_1 + F_2$.

$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2$ and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero,then the magnitude of $\vec F_3$ is

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