‘O’ is the center of a square ABCD. $F_{1}, F_{2}, F_{3} and F_{4}$ are four forces acting along the sides AB, BC, CD and AD as shown in figure. What should be the magnitude of F₄ so that, the total torque about ‘O’ is zero?

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$\frac{F_{1} + F_{2} + F_{3}}{2}$

$F_{1} - F_{2} - F_{3}$

$F_{1} + F_{2} - F_{3}$

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

answer is C.

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Detailed Solution

The Torques for all three forces F₁, F₂, F₃ are in same direction,

So the magnitude of F₄ must be equal to sum of all three forces.

$F_{4} = F_{1} + F_{2} + F_{3}$

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