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			<h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3>

			<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.
			</p>
			<center>
			<figure>
			<img src="2.8.25.png"
			loading="lazy" width="200" />
			<figcaption>
			For problem $2.8.25^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="2.8.25_1.jpg"
			loading="lazy" width="350" />
			<figcaption>
			Forces acting on the system
			</figcaption>
			</figure>
			</center>
			<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: $2n-1$
			$$\boxed{F_n=\frac{(2n-1)mg}{\sqrt{3}}}$$
			</p>

			<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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	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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.		$2.8.25^*.$ 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.
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="2.8.25.png"		<img src="2.8.25.png"
	loading="lazy" width="200" />		loading="lazy" width="200" />
	<figcaption>		<figcaption>
	For problem $2.8.25^*$		For problem $2.8.25^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="2.8.25_1.jpg"		<img src="2.8.25_1.jpg"
	loading="lazy" width="350" />		loading="lazy" width="350" />
	<figcaption>		<figcaption>
	Forces acting on the system		Forces acting on the system
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

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

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

	And for the second rods		And for the second rods

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

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

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