Added 2.2.44; Updated 2.3.36 2.3.44

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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.2.44^*.$ A rope thrown over a smooth nail is dragged at a speed of $v$ through the gap. The friction force in the slot $F$, the mass of the rope length unit $\rho$. Determine the force exerted on the nail if the rope sections on opposite sides of the nail form an angle $\alpha$. At what speed will the rope move away from the nail?
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="150" />
+ <figcaption>
+ For problem $2.2.44^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ </p>
+<center>
+ <figure>
+ <img src="draw1.jpg"
+ loading="lazy" width="180" />
+ <figcaption>
+ Forces acting on the system
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+$$
+\rho = \frac{m}{l} = \frac{\Delta m}{\Delta l}
+$$
+
+$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$
+
+</p>
+<center>
+ <figure>
+ <img src="draw2.jpg"
+ loading="lazy" width="130" />
+ <figcaption>
+ For problem $2.2.44^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+For the thread to be stretched at a constant speed $\vec{v}$, there must be a force on the first half that is modulo the frictional force $\vec{F}$:
+
+$$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$
+$$F'=F\sqrt{2(1+\cos\alpha)}$$
+$$F'=2F\cos\frac{\alpha}{2}$$
+
+
+</p>
+<center>
+ <figure>
+ <img src="draw3.jpg"
+ loading="lazy" width="130" />
+ <figcaption>
+ For problem $2.2.44^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Similiarly,
+
+$$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$
+$$p'=p\sqrt{2(1+\cos\alpha)}$$
+$$p'=2p\cos\frac{\alpha}{2}$$
+
+By definition of momentum
+
+$$p=\Delta m v$$
+
+From where, considering that $p'$ and $F'$ acts against $Ox$
+
+$$p'=-2\Delta m v\cos\frac{\alpha}{2}$$
+$$F'=-2F\cos\frac{\alpha}{2}$$
+Let's write Newton's second law
+$$\sum \vec{F}=0$$
+$$N-F'-p''=0$$
+$$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$
+From where we obtain the force $\vec{N}$
+
+$$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$
+By definition of density
+$$\Delta m = \rho \Delta x$$
+By definition of speed
+$$ v = \frac{\Delta x}{\Delta t}$$
+$$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$
+$$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$
+When the rope comes off
+$$N\leq0$$
+$$F-\rho v^2\leq0$$
+$${F\leq \rho v^2}$$
+From where we finally get
+$$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$
+$$p''=\frac{p'}{\Delta t}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
+ </p>
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