Translated 2.1.1-2.1.7

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+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>Two bodies of mass m_1 and m_2 are connected by a thread that can withstand the tension force T. Forces F_1 = \alpha t and F_2 = 2 \alpha t are applied to the bodies, where \alpha is a constant coefficient, and t is the time of action of the force. Determine at what point in time the thread will break.</title>
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+ <header style="text-align:center;">
+ <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>
+ $2.1.6.$ Two bodies of mass $m_1$ and $m_2$ are connected by a thread that can withstand the tension force $T$. Forces $F_1 = \alpha t$ and $F_2 = 2 \alpha t$ are applied to the bodies, where $\alpha$ is a constant coefficient, and $t$ is the time of action of the force. Determine at what point in time the thread will break.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.6/statement.png"
+ loading="lazy" width="280" />
+ <figcaption>
+ For problem $2.1.6$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.6/draw.png"
+loading="lazy" width="320" />
+<figcaption>
+Forces acting on the system of bars
+</figcaption>
+</figure>
+</center>
+<p>
+Let's write Newton's second law for both bars along the horizontal axis
+$$ \left\{\begin{matrix}
+m_2 a = F_2 - T\\
+m_1 a = T - F_1
+\end{matrix}\right. $$
+</p>
+<p>
+Reduce for acceleration $a$
+$$ \frac{m_2}{m_1} = \frac{F_2 - T}{T - F_1} $$
+
+$$ \frac{m_2}{m_1} = \frac{2\alpha t - T}{T - \alpha t} $$
+Express $T$
+$$ T = \alpha t \frac{2m_1+m_2}{m_1+m_2} $$
+Where does the time until the thread breaks $T$ come from?
+$$ \fbox{$t = \frac{T}{\alpha}\frac{m_1 + m_2}{(2m_1 + m_2)}$} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$t = \frac{T}{\alpha}\frac{m_1 + m_2}{(2m_1 + m_2)}$$
+ </p>
+
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