Translated 2.1.13-2.1.33

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+ <meta name="description" content="The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of F, as indicated in the figure. The coefficient of friction between loads of mass m_1 and m_2 is equal to \mu. Find the acceleration of all loads in the system.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of F, as indicated in the figure. The coefficient of friction between loads of mass m_1 and m_2 is equal to \mu. Find the acceleration of all loads in the system.</title>
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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>
+ $2.1.31.$ The cargo system shown in the figure is located on a smooth horizontal table. The lower right weight is pulled along the table with a force of $F$, as indicated in the figure. The coefficient of friction between loads of mass $m_1$ and $m_2$ is equal to $\mu$. Find the acceleration of all loads in the system.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.31/statement.png"
+ loading="lazy" width="250" />
+ <figcaption>
+ For problem $2.1.31$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>
+Let's renumber the weights as shown in the figure, and direct the $X$ axis to the right.
+</p>
+<br>
+
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.31/sol.png"
+loading="lazy" width="350" />
+<figcaption>
+Forces acting on the weights
+</figcaption>
+</figure>
+</center>
+
+<p>
+It is clear that then none of the weights can have negative acceleration.
+</p>
+<p>
+Let's prove that weights $3$ and $4$ move as a single whole. To do this, let's assume the opposite: let weight $3$ slide on weight $4$. Then a friction force arises between them
+$$ F_{fr}= \mu mg $$
+and elastic force arises in the thread
+$$ T> \mu mg $$
+In this case, the acceleration of load $2$ would be directed to the left, which is impossible. Therefore, the accelerations of loads $2$, $3$, and $4$ are the same.
+</p>
+<p>
+Let us denote the acceleration of these loads as $\bar{a}_{1}=\bar{a}_{2}$, and the acceleration of load $1$ as $\bar{a}_{1}$.
+</p>
+<p>
+Now consider two cases.
+</p>
+
+<p>
+Case 1. Let loads $1$ and $2$ be at relative rest and $\bar{a}_{1}=\bar{a}_{2}$.
+</p>
+<p>
+Let us denote the modulus of the static friction force between them as $F_1$, the modulus of the friction force between loads $3$ and $4$ as $F_2$, and the modulus of the elastic force of the thread as $T$ . </p>
+
+<p>
+Then: for cargo $1$
+$$ F-F_{1}=Ma_{1} $$
+for cargo $2$
+$$ F_{1}-T=ma_{2} $$
+for cargo $3$
+$$ T-F_{2}=ma_{2} $$
+for cargo $4$
+$$ F_{2}=Ma_{2} $$
+Solving this system of equations, we obtain:
+$$ F_{1}=\frac{2m+M}{2(M+m)}F,a_{1}=a_{2}=\frac{F}{2(M+m)} $$
+The same result can be obtained in another way. Since the friction between all surfaces is static friction, the system of loads moves as one body with mass $M=2(M+m)$.
+</p>
+<p>
+Therefore
+$$ \bar{F}=m \bar{a}_{1},\bar{a}_{1}=\bar{a}_{2}= \frac{\bar{F}}{2(M+m)} $$
+Case $2$.
+</p>
+<p>
+Let load $2$ slide on load $1$. Then the friction force acts on load $1$
+$$ F_{fr}^{\prime}= \mu mg $$
+and this load receives acceleration $a_{1}=\frac{F- \mu mg}{m}$.
+</p>
+<p>
+The system of loads $2$, $3$ and $4$ moves as one body, the mass of which is $M_{0}=2m+M$ with acceleration
+$$ a_{2}=\frac{\mu mg}{2m+M} $$
+The first case is realized if
+$$ F \geq \frac{2 \mu m (m+M)g}{2m+M} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ <p>
+For $F\leqslant\frac{2\mu m_{1}g(m_{1}+m_{2})}{m_{2}+2m_{1}}\equiv F_{0}$ we get $a_{1\text{sing}}=a_{1\text{right}}=a_{2\text{right}}=\frac{F}{2(m_{1}+m_{2})}$
+</p>
+<p>
+For $F\geqslant F_{0}$ we get $a_{2\text{right}}=\frac{F-\mu m_{1}g}{m_{2}}$, $a_{1\text{left}}=a_{1\text{right}}=a_{2\text{left}}=\frac{\mu m_{1}g}{m_{2}+2m_{1}}$ </p>
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