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+ <body style="">
+ <header style="text-align:center;">
<h2>Solutions of Savchenko Physics Textbook</h2>
<p class="author">
Aliaksandr Melnichenka <br/>
October 2023
</p>
</header>
− <main>
− <article>
− <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
−
− <h4>Statement</h4>
− <p>
− $4.2.13$
− Two equal trunks are placed as the figure shows. The lower trunk is linked to the vertical wall by wires that form an angle of 45$^\circ$ with it. The upper trunk is half submerged in water. Find the density of the trunks.
− </p>
− <center>
− <figure>
− <img src="statement.png"
− loading="lazy" alt="4.2.13" width="300" />
− <figcaption>
− For problem 4.2.13
− </figcaption>
− </figure>
− </center>
− <h4>Solution</h4>
− <p class="TxtSolutions">
− Let's analyze the following force diagram
− </p>
− <center>
− <figure>
− <img src="diagram.png"
− loading="lazy" alt="4.2.13" width="300" />
− <figcaption>
− Force diagram for each trunk
− </figcaption>
− </figure>
− </center>
− <p class ="TxtSolutions">
− Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction,
− </p>
− <center>
− $-N'\cos{45^\circ}+N = 0 \;(1)$
− </center>
− and on $y$-axis
− <center>
− $mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$
− </center>
− <p calss="TxtSolutions">
− as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$
− </p>
− <center>
− $\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$
− </center>
− <p class="TxtSolutions">
− Applying Newton Second Law for cylinder 2, on $x$-axis
− </p>
− <center>
− $N'\cos{45^\circ}-T\cos{45^\circ}=0$
− </center>
− <center>
− $N' = T \;(3)$
− </center>
− <p calss="TxtSolutions">
− and for $y$-axis
− </p>
− <center>
− $\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$
− </center>
− <p class="TxtSolutions">
− in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$,
− </p>
− <center>
− $\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$
− </center>
− <p class="TxtSolutions">
− Putting $(3)$ int $(4)$
− </p>
− <center>
− $(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$
− </center>
− <p class="TxtSolutions">
− Substituting $(5)$ into $(2)$ and developing algebraically
− </p>
− <center>
− $\rho = \frac{2}{3}\rho_0$
− </center>
− <p class="TxtSolutions">
− as the water's density is $\rho_0=1$ g/cm$^3$,
− </p>
− <center>
− $\boxed{\rho=\frac{2}{3}~\rm{g/cm^3}}$
− </center>
− <p class="TxtSolutions" 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>
− </article>
− </main>
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+ <p>
+ $4.2.13$
+ Two equal trunks are placed as the figure shows. The lower trunk is linked to the vertical wall by wires that form an angle of 45$^\circ$ with it. The upper trunk is half submerged in water. Find the density of the trunks.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="4.2.13" width="300" />
+ <figcaption>
+ For problem 4.2.13
+ </figcaption>
+ </figure>
+ </center>
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− <script id="MathJax-script" async src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"></script>
+ <h3>Solution</h3>
+ <p>
+ Let's analyze the following force diagram
+ </p>
+ <center>
+ <figure>
+ <img src="diagram.png"
+ loading="lazy" alt="4.2.13" width="300" />
+ <figcaption>
+ Force diagram for each trunk
+ </figcaption>
+ </figure>
+ </center>
+ <p>
+ Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction,
+ $$-N'\cos{45^\circ}+N = 0 \;(1)$$
+ and on $y$-axis
+ $$mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$$
+ as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$
+ $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$$
+ Applying Newton Second Law for cylinder 2, on $x$-axis
+ $$N'\cos{45^\circ}-T\cos{45^\circ}=0$$
+ $$N' = T \;(3)$$
+ and for $y$-axis
+ $$\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$$
+ in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$,
+ $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$$
+ Putting $(3)$ int $(4)$
+ $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$$
+ Substituting $(5)$ into $(2)$ and developing algebraically
+ $$\rho = \frac{2}{3}\rho_0$$
+ as the water's density is $\rho_0=1$ g/cm$^3$,
+ $$\boxed{\rho=\frac{2}{3}~\rm{g/cm^3}}$$
+ </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>
+ </p>
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