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.
<h3 id="back-link"><a href="/#4.2">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
<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.
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<figure>
<img src="statement.png"
loading="lazy" alt="4.2.13" width="300" />
<figcaption>
For problem 4.2.13
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<h3>Solution</h3>
<p>
Let's analyze the following force diagram
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<img src="diagram.png"
loading="lazy" alt="4.2.13" width="300" />
<figcaption>
Force diagram for each trunk
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<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$^3$,
$$\boxed{\rho=\frac{2}{3}\~\rm{g/cm^3}}$$
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<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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