A load of mass m is suspended from a system of identical rods connected by hinges, as shown in the figure. Determine the force that stretches the nth upper horizontal rod.

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    <h3> Statement </h3>
    <p>
        $2.8.25^*.$ A load of mass $m$ is suspended from a system of identical rods connected by hinges, as shown in the figure. Determine the force that stretches the nth upper horizontal rod.

    <h3>Solution</h3>
    <p>

Let's consider forces acting on the 1st rod
$$\left{\begin{matrix} F_1=F_1'\cos(60^{\circ})=F_1'/2 \ T=F_1'\cos(30^{\circ})=\sqrt{3}F_1'/2 \end{matrix}\right.$$

$$\left{\begin{matrix} F_1'''\cos(30^{\circ})=F_1'\cos(30^{\circ}) \ F_1'=F_1'''\ F_1'''\cos(60^{\circ})+F_1'\cos(60^{\circ})=F_1'' \ F_1'=F_1''\ \end{matrix}\right.$$

And for the second rods

$$\left{\begin{matrix} F_2''\cos(30^{\circ})=F_1'''\cos(30^{\circ}) \ F_2'=F_1'''=F_1'\ F_2=F_2'\cos(60^{\circ})+F_1+F_1'''\cos(60^{\circ})\ F_2=3F_1'/2 \end{matrix}\right.$$

$$F_1'=\frac{2T}{\sqrt{3}}=\frac{2mg}{\sqrt{3}}$$
$$\boxed{F_1=\frac{mg}{\sqrt{3}} \to F_2=\frac{3mg}{\sqrt{3}}}$$
As we can see, the index of $F_2$ increased by two, leading to recurrence relation for $n^\text{th}$ index: $2n-1$
$$\boxed{F_n=\frac{(2n-1)mg}{\sqrt{3}}}$$

    <h4>Answer</h4>
    <p>
        $$T_n=(2n-1)mg/\sqrt{3}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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