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Translational Equilibrium Revisited

O is the cent...

Question

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

F_{1} + F_{2}

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F_{1} - F_{2}

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

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\frac{(F_{1} - F_{2})}{2}

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Solution

The correct option is A $F_{1} + F_{2}$
Net torque about point $O$ is,

$τ = τ_{1} + τ_{2} + τ_{3}$

Assuming clockwise torque is positive.

$∴ τ = - F_{1} d - F_{2} d + F_{3} d$

For net torque to be zero,

$F_{3} d = F_{1} d + F_{2} d$

$∴ F_{3} = F_{1} + F_{2}$

Hence, $(A)$ is the correct answer.

Why this question?
When an object is in rotational equilibrium, make sure that net torque on the object is zero.

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O is the centre of an equilateral triangle ABC. F1, F2 and F3 are three forces acting along the sides AB, BC and AC as shown in the figure. What should be the magnitude of F3, so that the total torque about O is zero ?

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