Question

$ABC$ is an equilateral triangle with $O$ as its centre. $\vec{F_1},\vec{F_2}$ and $\vec{F_3}$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about O is zero then the magnitude of $\vec{F_3}$ is

• A. $F_1+F_2$
• B. $F_1-F_2$
• C. $\frac{F_1+F_2}{2}$
• D. $2(F_1+F_2)$

Answer 1

The Correct Option is A

 An equilateral triangle $ABC$ is one where all sides are equal. Let us understand the problem step by step to find the magnitude of the force $\vec{F_3}$.

Given that three forces $\vec{F_1}$, $\vec{F_2}$, and $\vec{F_3}$ act along the sides $AB$, $BC$, and $AC$, respectively.

The problem specifies that the total torque about the center $O$ of the triangle is zero. In an equilateral triangle, $O$ lies at the centroid, which divides each medians in the ratio 2:1.

The torque due to a force $\vec{F}$ about a point is given by: \(\tau = r \times F\), where $r$ is the perpendicular distance from the point to the line of action of the force.

For equilibrium (zero torque), the sum of torques produced by all forces is zero:

1. \(\tau_1 + \tau_2 + \tau_3 = 0$\)

For the equilateral triangle, the perpendicular distances from $O$ to the sides $AB$, $BC$, and $AC$ are equal due to symmetry. Hence, only the magnitudes of the forces need to be considered.

Thus, considering the directions of the forces we can write:

1. \(\vec{F_1} + \vec{F_2} + \vec{F_3} = 0$\)which simplifies to: \(F_3 = F_1 + F_2\), noting that directionality is considered such that clockwise and counterclockwise torques cancel out.

 

Therefore, the correct magnitude of $\vec{F3}$, ensuring zero net torque about $O$, is:

1. \(F_1 + F_2$\)

The correct answer is therefore \(F_1 + F_2\), which corresponds to the option given.