Added 2.8.25

astrosander правка от 26 сент. 2024 г., 15:08

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		+	<title>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.</title>
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		+	<body style="">
		+	<header style="text-align:center;">
		+	<div id = "logo">
		+	<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
		+	</div>
		+	<p class="author">
		+	Solutions of Savchenko Problems in Physics <br>
		+	<i><b>knowledge must be free</b></i>
		+	</p>
		+	</header>
		+
		+	<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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			<meta name="description" content="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.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="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.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="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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			<title>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.</title>
			<link rel="stylesheet" href="https://savchenkosolutions.com/css/css-latex/style.css">
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			messageStyle: 'none'
			});
			</script>
			</head>
			<body style="">
			<header style="text-align:center;">
			<div id = "logo">
			<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
			</div>
			<p class="author">
			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
			</p>
			</header>

			<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>


			<footer class="row container">
			<br>
			<p>
			<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
			</p>
			<p>
			<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
			</p>
			</footer>
			</body>

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