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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<h3 id="back-link"><a href="../../#4.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>

			<p>
			$4.2.7$
			A body in water weighs three times less than in air. What is the density of the body?
			</p>

			<h3>Solution</h3>
			<p>
			Let's consider the following figure
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			<img src="draw.png"
			loading="lazy" alt="4.2.7" width="400" />
			<figcaption>
			Force analysis
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			<p>
			These normal forces $N_i$ is what the equipment measures as weight in each case.
		@@ -73,11 +73,11 @@
			From Newton Second Law for air
			$$N_a = mg$$
			Since $m=\rho~V$,
		−	$$N_a = \rho~V~g \;(1)$$
		+	$$N_a = \rho~V~g \quad(1)$$
			Applying Newton Second Law for water
			$$mg = N_w + F_A$$
			$$\rho~V~g = N_w + \rho_w~V~g$$
		−	$$N_w = (\rho-\rho_w)~V~g \;(2)$$
		+	$$N_w = (\rho-\rho_w)~V~g \quad(2)$$
			Dividing $(1)$ by $(2)$, taking account that $\frac{N_a}{N_w} = 3$
			$$\rho = \frac{3}{2}\rho_w$$
			as $\rho_w = 1$ g/cm$^3$
			$$\boxed{\rho = 1.5~\rm{g/cm^3}}$$
			</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>
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	<p class="author">		<p class="author">
	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
	</p>		</p>
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	<h3 id="back-link"><a href="../../#4.2">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#4.2">$\leftarrow$Back</a></h3>

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

	<p>		<p>
	$4.2.7$		$4.2.7$
	A body in water weighs three times less than in air. What is the density of the body?		A body in water weighs three times less than in air. What is the density of the body?
	</p>		</p>

	<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="4.2.7" width="400" />		loading="lazy" alt="4.2.7" width="400" />
	<figcaption>		<figcaption>
	Force analysis		Force analysis
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	These normal forces $N_i$ is what the equipment measures as weight in each case.		These normal forces $N_i$ is what the equipment measures as weight in each case.
	@@ -73,11 +73,11 @@
	From Newton Second Law for air		From Newton Second Law for air
	$$N_a = mg$$		$$N_a = mg$$
	Since $m=\rho~V$,		Since $m=\rho~V$,
	$$N_a = \rho~V~g \;(1)$$		$$N_a = \rho~V~g \quad(1)$$
	Applying Newton Second Law for water		Applying Newton Second Law for water
	$$mg = N_w + F_A$$		$$mg = N_w + F_A$$
	$$\rho~V~g = N_w + \rho_w~V~g$$		$$\rho~V~g = N_w + \rho_w~V~g$$
	$$N_w = (\rho-\rho_w)~V~g \;(2)$$		$$N_w = (\rho-\rho_w)~V~g \quad(2)$$
	Dividing $(1)$ by $(2)$, taking account that $\frac{N_a}{N_w} = 3$		Dividing $(1)$ by $(2)$, taking account that $\frac{N_a}{N_w} = 3$
	$$\rho = \frac{3}{2}\rho_w$$		$$\rho = \frac{3}{2}\rho_w$$
	as $\rho_w = 1$ g/cm$^3$		as $\rho_w = 1$ g/cm$^3$
	$$\boxed{\rho = 1.5~\rm{g/cm^3}}$$		$$\boxed{\rho = 1.5~\rm{g/cm^3}}$$
	</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>
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