Figure shows three forces $\overrightarrow{\mathrm{F}}_{1}, \overrightarrow{\mathrm{F}}_{2}$ and…
Figure shows three forces $\overrightarrow{\mathrm{F}}_{1}, \overrightarrow{\mathrm{F}}_{2}$ and $\overrightarrow{\mathrm{F}}_{3}$ acting along the sides of an equilateral
triangle. If the total torque acting at point ' 0 ' (centre of the triangle) is zero then the
magnitude of $\overrightarrow{\mathrm{F}}_{3}$ is
$\frac{\mathrm{F}_{1}-\mathrm{F}_{2}}{2}$
$\mathrm{~F}_{1}-\mathrm{F}_{2}$
$\mathrm{~F}_{1}+\mathrm{F}_{2}$
$\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}$
Solution
The perpendicular distance of point $\mathrm{O}$ from the sides $\mathrm{AB}$ or $\mathrm{BC}$ or $\mathrm{AC}$ will be same (say r). Torque will be zero if total moment of force about point $\mathrm{O}$ is zero Hence
$\mathrm{rF}_{1}+\mathrm{rF}_{2}-\mathrm{rF}_{3}=0$
$\mathrm{F}_{3}=\mathrm{F}_{1}+\mathrm{F}_{2}=(4+2) \mathrm{N}=6 \mathrm{~N}$
