Question

Three forces each of magnitude $F$ are applied along the edges of a regular hexagon as shown in the figure. Each side of hexagon is $a$. What is the resultant moment (torque) of these three forces about centre $O$?

• A. $\frac{3\sqrt{3}}{2}aF$
• B. $\frac{1}{2}aF$
• C. $3aF$
• D. $\frac{\sqrt{3}}{2}aF$

Answer 1

The correct option is D $\frac{\sqrt{3}}{2}aF$

We know that, moment force $=rF$
In right angled $\Delta OAB$,
${OG}^2={OA}^2-{AG}^2$
$\Rightarrow {OG}^2=a^2-{\left(\frac{a}{2}\right)}^2$
[$\because OAB$ is an equilateral triangle with side '$a$', then $AG=\frac{AB}{2}=\frac{a}{2}$]
$\Rightarrow {OG}^2=\frac{3}{4}{a}^2$
$\Rightarrow OG=\frac{\sqrt{3}a}{2}$
Moment of force $AB$ about '$O$' is $\left(OG\right)AB$
i.e. $\frac{\sqrt{3}}{2}a\cdot \left(F\right)$
Moment of force $DC$ about '$O$' is $\left(OG\right)CD$, i.e. $\frac{\sqrt{3}}{2}a(-F)$
Moment of force $DE$ about '$O$' is $\left(OG\right)DE$, i.e. $\frac{\sqrt{3}}{2}a\left(F\right)$
Resultant moment of force about '$O$'
$=\left(\frac{\sqrt{3}}{2}a\right)F-\left(\frac{\sqrt{3}}{2}a\right)F+\left(\frac{\sqrt{3}}{2}a\right)F$
$=\left(\frac{\sqrt{3}}{2}a\right)F$.