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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="../">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>

			<p>
			$2.8.1$
			The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b.
			</p>
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" alt="2.8.1" width="400" />
			<figcaption>
			For problem 2.8.1
			</figcaption>
			</figure>
			</center>
			<h3>Solution</h3>
			<p>
			Let's consider the following figure
			</p>
			<center>
			<figure>
			<img src="draw.png"
			loading="lazy" alt="2.8.1" width="500" />
			<figcaption>
			Force analysis
			</figcaption>
			</figure>
			</center>
			<p>
			$a$. Applying Newton Second Law for $x$-direction
			$$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$
		@@ -76,7 +76,7 @@
			$$T_1=T_2 \;(1)$$
			for $y$-direction
			$$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$
		−	Let be $T_1=T_2=T$, so, bak to $(2)$
		+	Let be $T_1=T_2=T$, so, back to $(2)$
			$$2T~\sin{30^\circ}=mg$$
			$$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$
			$b$. According figure,
			$$F^2 = T^2+T^2 = 2T^2$$
			We consider the same tension $T$ because we suppose the threads are unextensible.
			$$F = \sqrt{2}~T \;(3)$$
			Applying Newton Second Law
			$$T = mg \;(4)$$
			Putting $(4)$ into $(3)$
			$$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$
			Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41.
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			BSc. Luis Daniel Fernández Quintana<br>
			Physics Department (FCNE)<br>
			Universidad de Oriente, Cuba<br>
			</p>



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	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	<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="../">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>

	<p>		<p>
	$2.8.1$		$2.8.1$
	The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b.		The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b.
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" alt="2.8.1" width="400" />		loading="lazy" alt="2.8.1" width="400" />
	<figcaption>		<figcaption>
	For problem 2.8.1		For problem 2.8.1
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	Let's consider the following figure		Let's consider the following figure
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw.png"		<img src="draw.png"
	loading="lazy" alt="2.8.1" width="500" />		loading="lazy" alt="2.8.1" width="500" />
	<figcaption>		<figcaption>
	Force analysis		Force analysis
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	$a$. Applying Newton Second Law for $x$-direction		$a$. Applying Newton Second Law for $x$-direction
	$$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$		$$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$
	@@ -76,7 +76,7 @@
	$$T_1=T_2 \;(1)$$		$$T_1=T_2 \;(1)$$
	for $y$-direction		for $y$-direction
	$$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$		$$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$
	Let be $T_1=T_2=T$, so, bak to $(2)$		Let be $T_1=T_2=T$, so, back to $(2)$
	$$2T~\sin{30^\circ}=mg$$		$$2T~\sin{30^\circ}=mg$$
	$$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$		$$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$
	$b$. According figure,		$b$. According figure,
	$$F^2 = T^2+T^2 = 2T^2$$		$$F^2 = T^2+T^2 = 2T^2$$
	We consider the same tension $T$ because we suppose the threads are unextensible.		We consider the same tension $T$ because we suppose the threads are unextensible.
	$$F = \sqrt{2}~T \;(3)$$		$$F = \sqrt{2}~T \;(3)$$
	Applying Newton Second Law		Applying Newton Second Law
	$$T = mg \;(4)$$		$$T = mg \;(4)$$
	Putting $(4)$ into $(3)$		Putting $(4)$ into $(3)$
	$$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$		$$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$
	Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41.		Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41.
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	BSc. Luis Daniel Fernández Quintana<br>		BSc. Luis Daniel Fernández Quintana<br>
	Physics Department (FCNE)<br>		Physics Department (FCNE)<br>
	Universidad de Oriente, Cuba<br>		Universidad de Oriente, Cuba<br>
	</p>		</p>



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