A piece of iron weighs 9.8 N in water. Determine its volume. The density of iron is 7.8 \cdot 10^3 m^3.
<h3 id="back-link"><a href="/#4.2">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
<p>
$4.2.6$
A piece of iron weighs 9.8 N in water. Determine its volume. The density of iron is 7.8 $\cdot$ 10$^3$ m$^3$.
</p>
<h3>Solution</h3>
<p>
Let's consider the following figure
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="4.2.6" width="300" />
<figcaption>
Force analysis
</figcaption>
</figure>
</center>
<p>
The normal force $N$ is what the equipment measures as apparent weight.
From Newton Second Law
$$N = mg-F_A$$
$$N = \rho\~g\~V-\rho_w\~g\~V$$
$$N = (\rho-\rho_w)\~g\~V$$
and taking $\rho_w = 1000$ kg/m$^3$
$$\boxed{V = \frac{N}{(\rho-\rho_w)\~g}=147 \rm{cm}^3}$$
</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>
</p>
<footer class="row container">
<br>
<p>
<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
</p>
<p>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
</p>
</footer>
Savchenko Solutions