Updated en 2.2.37, 2.2.44, 2.2.36, 2.2.39

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			<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

			$$p=\Delta m v$$

		@@ -122,11 +122,12 @@
			From where, considering that $p'$ and $F'$ acts against $Ox$

			$$p'=-2\Delta m v\cos\frac{\alpha}{2}$$
		−	$$F'=-2F\cos\frac{\alpha}{2}$$
		+	$$F'=2F\cos\frac{\alpha}{2}$$
			Let's write Newton's second law
			$$\sum \vec{F}=0$$
		−	$$N-F'-p''=0$$
		+	$$N-F'-F''=0$$
			$$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$
		+	$$F''=\frac{p'}{\Delta t}$$
			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
			$$N\leq0$$
			$$F-\rho v^2\leq0$$
		@@ -142,7 +143,6 @@
			$${F\leq \rho v^2}$$
			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>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Almaskhan Arsen<br>
			</p>


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	<h2>Solutions of Savchenko Physics Textbook</h2>		<h2>Solutions of Savchenko Physics Textbook</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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?		$2.2.44^*.$ 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>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" width="150" />		loading="lazy" width="150" />
	<figcaption>		<figcaption>
	For problem $2.2.44^*$		For problem $2.2.44^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw1.jpg"		<img src="draw1.jpg"
	loading="lazy" width="180" />		loading="lazy" width="180" />
	<figcaption>		<figcaption>
	Forces acting on the system		Forces acting on the system
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

	$$		$$
	\rho = \frac{m}{l} = \frac{\Delta m}{\Delta l}		\rho = \frac{m}{l} = \frac{\Delta m}{\Delta l}
	$$		$$

	$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$		$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw2.jpg"		<img src="draw2.jpg"
	loading="lazy" width="130" />		loading="lazy" width="130" />
	<figcaption>		<figcaption>
	For problem $2.2.44^*$		For problem $2.2.44^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<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 $\vec{F}$:		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 $\vec{F}$:

	$$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$		$$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$
	$$F'=F\sqrt{2(1+\cos\alpha)}$$		$$F'=F\sqrt{2(1+\cos\alpha)}$$
	$$F'=2F\cos\frac{\alpha}{2}$$		$$F'=2F\cos\frac{\alpha}{2}$$


	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw3.jpg"		<img src="draw3.jpg"
	loading="lazy" width="130" />		loading="lazy" width="130" />
	<figcaption>		<figcaption>
	For problem $2.2.44^*$		For problem $2.2.44^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

	Similiarly,		Similiarly,

	$$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$		$$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$
	$$p'=p\sqrt{2(1+\cos\alpha)}$$		$$p'=p\sqrt{2(1+\cos\alpha)}$$
	$$p'=2p\cos\frac{\alpha}{2}$$		$$p'=2p\cos\frac{\alpha}{2}$$

	By definition of momentum		By definition of momentum

	$$p=\Delta m v$$		$$p=\Delta m v$$

	@@ -122,11 +122,12 @@
	From where, considering that $p'$ and $F'$ acts against $Ox$		From where, considering that $p'$ and $F'$ acts against $Ox$

	$$p'=-2\Delta m v\cos\frac{\alpha}{2}$$		$$p'=-2\Delta m v\cos\frac{\alpha}{2}$$
	$$F'=-2F\cos\frac{\alpha}{2}$$		$$F'=2F\cos\frac{\alpha}{2}$$
	Let's write Newton's second law		Let's write Newton's second law
	$$\sum \vec{F}=0$$		$$\sum \vec{F}=0$$
	$$N-F'-p''=0$$		$$N-F'-F''=0$$
	$$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$		$$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$
			$$F''=\frac{p'}{\Delta t}$$
	From where we obtain the force $\vec{N}$		From where we obtain the force $\vec{N}$

	$$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$		$$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$
	By definition of density		By definition of density
	$$\Delta m = \rho \Delta x$$		$$\Delta m = \rho \Delta x$$
	By definition of speed		By definition of speed
	$$ v = \frac{\Delta x}{\Delta t}$$		$$ v = \frac{\Delta x}{\Delta t}$$
	$$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$		$$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$
	$$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$		$$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$
	When the rope comes off		When the rope comes off
	$$N\leq0$$		$$N\leq0$$
	$$F-\rho v^2\leq0$$		$$F-\rho v^2\leq0$$
	@@ -142,7 +143,6 @@
	$${F\leq \rho v^2}$$		$${F\leq \rho v^2}$$
	From where we finally get		From where we finally get
	$$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$		$$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$
	$$p''=\frac{p'}{\Delta t}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$		$$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Almaskhan Arsen<br>		Almaskhan Arsen<br>
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