Solved Problem on Static Equilibrium

Six forces of the same magnitude F act on a solid on the sides of a regular hexagon of side L. Calculate the torque of these forces relative to the axis passing through the center and perpendicular to the solid.

Problem data:

Length of the solid side: L;
Magnitude of force that acts in the solid: F.

Problem diagram:

We choose a reference frame on the axis passing through the center of the solid and perpendicular to this with a positive counterclockwise direction (Figure 1-A).
As the solid is a regular hexagon, it can be divided into 6 equal triangles of sides L and angles θ (Figure 1-B).

Figure 1

As the sum of the interior angles of a triangle is equal to 180°, then an angle measures (Figure 1-C)

\[ \begin{gather} \theta +\theta +\theta =180°\Rightarrow 3\theta=180°\Rightarrow \theta =\frac{180°}{3}\Rightarrow\theta =60° \end{gather} \]

Solution

The moment of a force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{\Large \tau}=Fd} \tag{I} \end{gather} \]

Consider the force \( \vec{F} \) applied to a vertex of the hexagon, this force can be decomposed into two components, a parallel component \( {\vec F}_{P} \) to the L segment, that goes from the center of the hexagon to the vertex, and another perpendicular or normal component \( {\vec F}_{N} \), makes a 90° angle. Only the normal component contributes to the torque of the solid. The angle between the \( \vec{F} \) force and the parallel component \( {\vec F}_{P} \) is 60°, which is the same angle of the triangle inside the hexagon (Figure 2), they are vertically opposite angles.

Figure 2

The normal component is given by

\[ \begin{gather} F_{N}=F\sin 60° \end{gather} \]

From the Trigonometry

\[ \begin{gather} \sin 60°=\frac{\sqrt{3\,}}{2} \end{gather} \]

\[ \begin{gather} F_{N}=\frac{\sqrt{3\,}}{2}F \tag{II} \end{gather} \]

The torque of the normal force be given by expression (I)

\[ \begin{gather} {\Large\tau}_{F_{N}}=F_{N}d \tag{III} \end{gather} \]

substituting expression (II) into expression (III) and the distance from the force to the central axis equal to L

\[ \begin{gather} {\Large\tau}_{F_{N}}=\frac{\sqrt{3\,}}{2}FL \tag{IV} \end{gather} \]

The total torque is given by the sum of the torques of the six forces that act in the body

\[ \begin{gather} {\Large\tau}=\sum_{k=1}^{6}{\Large\tau}_{F_{N}k}\\[5pt] {\Large\tau}={\Large\tau}_{F_{N}1}+{\Large\tau}_{F_{N}2}+{\Large\tau}_{F_{N}3}+{\Large\tau}_{F_{N}4}+{\Large\tau}_{F_{N}5}+{\Large\tau}_{F_{N}6} \end{gather} \]

as the solid is symmetrical and all forces are of equal magnitude, their torques are equal.

\[ \begin{gather} {\Large\tau}=6\,{\Large\tau}_{F_{N}} \tag{V} \end{gather} \]

substituting expression (IV) into expression (V)

\[ \begin{gather} {\Large\tau}=6\times\frac{\sqrt{3\,}}{2}FL \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {{\Large\tau}=3\,\sqrt{3\,}\,FL} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .