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Rotational Inertia

O is the cent...

Question

$O$ is the centre of an equilateral triangle $A B C$ . $F_{1}, F_{2}$ and $F_{3}$ are three forces acting along the sides $A B, B C$ and $C A$ as shown in the figure. What should be the magnitude of $F_{3}$ so that the total torque about $O$ is zero?

\frac{F_{1} + F_{2}}{2}

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(F_{1} - F_{2})

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2 (F_{1} + F_{2})

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(F_{1} + F_{2})

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Solution

The correct option is C $(F_{1} + F_{2})$
All the forces are are at same perpendicular distance from the point O and let this distance be $r$ .
Torque acting due to force $τ = r \times F$
Torque by $F_{1}$ is, $τ_{1} = F_{1} r$ in the anti-clockwise direction.
Torque by $F_{2}$ is, $τ_{2} = F_{2} r$ in the anti-clockwise direction.
Torque by $F_{3}$ is, $τ_{3} = F_{3} r$ in the clockwise direction.
Since net torque acting about point O is zero.
$∴$ $τ = F_{1} r + F_{2} r - F_{3} r = 0$
$⟹ F_{3} = F_{1} + F_{2}$

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