Solved Problem on Static Equilibrium

Solved Problem on 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

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

In one of the triangles in which the solid is divided, we draw an auxiliary line r from the center of the hexagon and perpendicular to the side of this, making an angle of 90°. As the triangles are equilateral triangles this line divides the central angle of the triangle into two equal angles (bisect the angel) and also divides the side of the solid into two equal segments (median), we obtain the height of the triangle that is the distance h from the center to where the force is applied (Figure 2).

Figure 2

Using the Pythagorean Theorem we determine the distance from the force to the center of the solid, the height of the triangle

\[ \begin{gather} L^{2}=h^{2}+\left(\frac{L}{2}\right)^{2}\\[5pt] L^{2}=h^{2}+\frac{L^{2}}{4}\\[5pt] h^{2}=L^{2}-\frac{L^{2}}{4} \end{gather} \]

multiplying and dividing the first term of the right-hand side by 4

\[ \begin{gather} h^{2}=\frac{4}{4}\times L^{2}-\frac{L^{2}}{4}\\[5pt] h^{2}=\frac{4L^{2}-L^{2}}{4}\\[5pt] h^{2}=\frac{3L^{2}}{4}\\[5pt] h=\sqrt{\,\frac{3L^{2}}{4}\,}\\[5pt] h=L\,\frac{\sqrt{3\,}}{2} \end{gather} \]

Note: We could obtain the same result by calculating the cosine of the angle of 30° in Figure 1

\[ \begin{gather} \cos 30°=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{h}{L} \end{gather} \]

From Trigonometry

\[ \begin{gather} \cos 30°=\frac{\sqrt{3\,}}{2} \end{gather} \]

\[ \begin{gather} \frac{\sqrt{3\,}}{2}=\frac{h}{L}\\[5pt] h=\frac{\sqrt{3\,}}{2}L \end{gather} \]

At the midpoint on the hexagon side, the force is perpendicular to the distance to the center, the torque of force will be given by the expression (i), where \( d=h=L\,\frac{\sqrt{3\,}}{2} \)

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

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.

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

substituting expression (II) into expression (III)

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

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

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