Translated 2.1.1-2.1.7

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+ <meta name="description" content="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?">
+ <meta name="author" content="Aliaksandr Melnichenka">
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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?
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.5/statement.png"
+ loading="lazy" width="280" />
+ <figcaption>
+ For problem $2.1.5$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<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>
+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.
+$$ a\, dm = F(x+dx) - F(x)\;(1) $$
+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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