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Force Couple

The center of...

Question

The center of an equilateral triangle is O. Three forces $F_{1}$ , $F_{2}$ and $F_{3}$ are applied along AB, BC and AC respectively. The magnitude of $F_{3}$ so that the total torque about O should be zero is:

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

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Solution

The correct option is A $(F_{1} + F_{2})$
Both $F_{1}$ and $F_{2}$ exert counter-clockwise torque on the on the triangle about point O while $F_{3}$ exerts clockwise torque about point O. The perpendicular distances from point O on the forces are equal by symmetry.
Let the distance be l.
$∴ F_{1} l + F_{2} l - F_{3} l = 0$
$\Rightarrow F_{1} + F_{2} = F_{3}$

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The center of an equilateral triangle is O. Three forces F1, F2 and F3 are applied along AB, BC and AC respectively. The magnitude of F3 so that the total torque about O should be zero is:

The correct option is A (F1+F2)Both F1 and F2 exert counter-clockwise torque on the on the triangle about point O while F3 exerts clockwise torque about point O. The perpendicular distances from point O on the forces are equal by symmetry.Let the distance be l.∴F1l+F2l−F3l=0⇒F1+F2=F3

The center of an equilateral triangle is O. Three forces $F_{1}$ , $F_{2}$ and $F_{3}$ are applied along AB, BC and AC respectively. The magnitude of $F_{3}$ so that the total torque about O should be zero is: