A regular hexagon is given and a particle of mass m at the center of it is acted upon by six forces each of magnitude F. The forces are directed towards the vertices of the hexagon. Consider that one of the forces is reversed in direction, the acceleration of the particle is :

$\frac{2F_1}{m}$

Let the forces be : $\vec{F_1}, \vec{F_2}, \vec{F_3}, \vec{F_4}, \vec{F_5}and \vec{F_6}$

Given that :

Initially : $\vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} + \vec{F_5} + \vec{F_6} = 0 $

$\vec{F_2} + \vec{F_3} + \vec{F_4} + \vec{F_5} + \vec{F_6} = - \vec{F_1} \text{ ... (A)}$

Finally : $ - \vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} + \vec{F_5} + \vec{F_6} = \vec{F}_{net} $

From eq : (A)

$\vec{F}_{net} = - \vec{F}_1 - \vec{F}_1$

$\Rightarrow \text{Acceleration } = \frac{2 F_1}{m}$