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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$2.1.5.$ What force acts in the cross-section of a homogeneous rod of length $l$ at a distance $x$ from the end to which the force $F$ is applied along the rod?
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			For problem $2.1.5$
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			<h3>Solution</h3>
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			<figure>
			<img src="https://savchenkosolutions.com/2/2.1.5/draw.png"
			loading="lazy" width="350" />
			<figcaption>
			Forces acting on a small area
			</figcaption>
			</figure>
			</center>
			<p>
			If we consider a rod of mass $m$ as a single whole, then it will move with acceleration
			$$ a = \frac{F}{m} $$
			Since the rod is inextensible, the acceleration of all its parts is the same and equal to $a$
			</p>
			<p>
		@@ -84,7 +84,7 @@
			Let's consider a small section of the rod of length $dx$ and mass $dm$. Since the rod is homogeneous
			$$ dm = m \frac{dx}{l} $$
			Let's write down Newton's second law for this section.
		−	$$ a\, dm = F(x+dx) - F(x)\;(1) $$
		+	$$ a\, dm = F(x+dx) - F(x)\quad(1) $$
			Where $F(x+dx)$ and $F(x)$ are the interaction forces with neighbors
			</p>
			<p>
			Let's integrate expression $(1)$ along the horizontal coordinate $x$:
			$$ \int_x^l am\frac{dx}{l} = \int_x^l dF $$

			$$ ma\frac{l-x}{l} $$

			$$ F(x) = \fbox{$F(1-\frac{x}{l})$} $$
			</p>
			</p>

			<h4>Answer</h4>
			<p>
			$$T = F(1 − x/l)$$
			</p>


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	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
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	<h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$2.1.5.$ What force acts in the cross-section of a homogeneous rod of length $l$ at a distance $x$ from the end to which the force $F$ is applied along the rod?		$2.1.5.$ What force acts in the cross-section of a homogeneous rod of length $l$ at a distance $x$ from the end to which the force $F$ is applied along the rod?
	</p>		</p>
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	<figure>		<figure>
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	loading="lazy" width="280" />		loading="lazy" width="280" />
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	For problem $2.1.5$		For problem $2.1.5$
	</figcaption>		</figcaption>
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	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
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	<center>		<center>
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	<img src="https://savchenkosolutions.com/2/2.1.5/draw.png"		<img src="https://savchenkosolutions.com/2/2.1.5/draw.png"
	loading="lazy" width="350" />		loading="lazy" width="350" />
	<figcaption>		<figcaption>
	Forces acting on a small area		Forces acting on a small area
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	If we consider a rod of mass $m$ as a single whole, then it will move with acceleration		If we consider a rod of mass $m$ as a single whole, then it will move with acceleration
	$$ a = \frac{F}{m} $$		$$ a = \frac{F}{m} $$
	Since the rod is inextensible, the acceleration of all its parts is the same and equal to $a$		Since the rod is inextensible, the acceleration of all its parts is the same and equal to $a$
	</p>		</p>
	<p>		<p>
	@@ -84,7 +84,7 @@
	Let's consider a small section of the rod of length $dx$ and mass $dm$. Since the rod is homogeneous		Let's consider a small section of the rod of length $dx$ and mass $dm$. Since the rod is homogeneous
	$$ dm = m \frac{dx}{l} $$		$$ dm = m \frac{dx}{l} $$
	Let's write down Newton's second law for this section.		Let's write down Newton's second law for this section.
	$$ a\, dm = F(x+dx) - F(x)\;(1) $$		$$ a\, dm = F(x+dx) - F(x)\quad(1) $$
	Where $F(x+dx)$ and $F(x)$ are the interaction forces with neighbors		Where $F(x+dx)$ and $F(x)$ are the interaction forces with neighbors
	</p>		</p>
	<p>		<p>
	Let's integrate expression $(1)$ along the horizontal coordinate $x$:		Let's integrate expression $(1)$ along the horizontal coordinate $x$:
	$$ \int_x^l am\frac{dx}{l} = \int_x^l dF $$		$$ \int_x^l am\frac{dx}{l} = \int_x^l dF $$

	$$ ma\frac{l-x}{l} $$		$$ ma\frac{l-x}{l} $$

	$$ F(x) = \fbox{$F(1-\frac{x}{l})$} $$		$$ F(x) = \fbox{$F(1-\frac{x}{l})$} $$
	</p>		</p>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$T = F(1 − x/l)$$		$$T = F(1 − x/l)$$
	</p>		</p>


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