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Newton's Second Law: Rotatory Version

O is the cent...

Question

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

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

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

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F_{1} - F_{2}

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F_{1} + F_{2}

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Solution

The correct option is D $F_{1} + F_{2}$
The torques $F_{1} d$ and $F_{2} d$ of $F_{1}$ and $F_{2}$ respectively are counterclockwise. The torque $F_{3} d$ is clockwise. Applying condition for rotational equilibrium,
$F_{1} d + F_{2} d - F_{3} d = 0$ or $F_{3} = F_{1} + F_{2}$

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

The correct option is D F1+F2The torques F1d and F2d of F1 and F2 respectively are counterclockwise. The torque F3d is clockwise. Applying condition for rotational equilibrium,F1d+F2d−F3d=0 or F3=F1+F2

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

The correct option is D $F_{1} + F_{2}$
The torques $F_{1} d$ and $F_{2} d$ of $F_{1}$ and $F_{2}$ respectively are counterclockwise. The torque $F_{3} d$ is clockwise. Applying condition for rotational equilibrium,
$F_{1} d + F_{2} d - F_{3} d = 0$ or $F_{3} = F_{1} + F_{2}$