ABC is an equilateral triangle with O as its centre. $\vec{F}_1$, $\vec{F}_2$ and $\vec{F}_3$ represent three

Question

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:

Accepted Answer

Let $r$ be the perpendicular distance from the centre O of the equilateral triangle to each of its sides. The direction of torque is determined by the cross product of the position vector and the force vector. 
Forces $\vec{F}_1$ (acting along AB) and $\vec{F}_2$ (acting along BC) will produce torque in the same rotational sense (either both clockwise or both counter-clockwise) about the centre O. 
However, the force $\vec{F}_3$ is acting along AC (not CA, which would be the continuous cyclic order). Therefore, $\vec{F}_3$ will produce a torque in the opposite sense to that of $\vec{F}_1$ and $\vec{F}_2$.
The total torque about O is the algebraic sum of the individual torques:
$	au_{	ext{total}} = F_1 r + F_2 r - F_3 r$
Given that the total torque about O is zero:
$F_1 r + F_2 r - F_3 r = 0$
Dividing by $r$ (since $r 
eq 0$):
$F_3 = F_1 + F_2$

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