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(B) Let $d$ be the perpendicular distance from the center $O$ to any side of the equilateral triangle.The torque $	au$ produced by a force is given b

Vedclass · Accepted Answer

$F_3 = F_1 + F_2$

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(B) Let $d$ be the perpendicular distance from the center $O$ to any side of the equilateral triangle.The torque $	au$ produced by a force is given by the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force.Taking the counter-clockwise direction as positive,the torques produced by $F_1$,$F_2$,and $F_3$ about point $O$ are:$	au_1 = F_1 	imes d$ (counter-clockwise)$	au_2 = F_2 	imes d$ (counter-clockwise)$	au_3 = F_3 	imes d$ (clockwise)Given that the total torque about $O$ is zero:$\sum 	au = 	au_1 + 	au_2 - 	au_3 = 0$$F_1 d + F_2 d - F_3 d = 0$Dividing by $d$ (since $d 
eq 0$):$F_1 + F_2 - F_3 = 0$$F_3 = F_1 + F_2$

The center of an equilateral triangle $ABC$ is $O$. Three forces $F_1$,$F_2$,and $F_3$ are applied along the sides $AB$,$BC$,and $AC$ respectively. If the total torque about $O$ is zero,what is the relationship between $F_1$,$F_2$,and $F_3$?

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