Translated 2.1.34-2.1.47

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+ <meta name="description" content="Two bodies of the same mass connected by a thread slide along an inclined plane. Thread tension force T. There is no friction between one body and the board. Determine the friction force between the board and another body.">
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+ <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>
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+ </header>
+
+ <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.1.47^*.$ Two bodies of the same mass connected by a thread slide along an inclined plane. Thread tension force $T$. There is no friction between one body and the board. Determine the friction force between the board and another body.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/2/2.1.47/statement.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $2.1.47^*$
+ </figcaption>
+ </figure>
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+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.47/newton05.png"
+loading="lazy" width="250" />
+<figcaption>
+System of connected bodies
+</figcaption>
+</figure>
+</center>
+<p>
+Body of mass $m_3$ moves with acceleration of the center of mass of the system
+</p>
+<p>
+Considering the system as a whole (not taking into account internal forces)
+$$ \boxed{a_3 = \frac{F}{M+2m}} $$
+In this case, a force equivalent to double the tension force of the thread $2T$ acts on the body $m_3$.
+$$ 2T = M a_3 $$
+
+$$ T = F\frac{M}{2(M+2m)} $$
+The body $m_1$ is acted upon by two forces $\vec{T}$ and $\vec{F}$
+</p>
+<p>
+We write Newton's second law for the first body as
+$$ F-T = ma_1 $$
+
+$$ \boxed{a_1=\frac{F(M+4m)}{2m(M+2m)}} $$
+Similarly, the second body $m_2$ is acted upon only by $\vec{T}$
+$$ m_2 = T $$
+
+$$ \boxed{a_2=\frac{FM}{2m(M+2m)}} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ Acceleration of cargoes 1–3:
+$$ a_1=\frac{F(M+4m)}{2m(M+2m)} $$
+
+$$ a_2=\frac{FM}{2m(M+2m)} $$
+
+$$ a_3=\frac{F}{M+2m} $$
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