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| <meta name="date" content="2023-10" scheme="YYYY-MM"> | | <meta name="date" content="2023-10" scheme="YYYY-MM"> |
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| <title>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.</title> | | <title>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.</title> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
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| <p> | | <p> |
| $4.2.13$ | | $4.2.13$ |
| 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. | | 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> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="4.2.13" width="300" /> | | loading="lazy" alt="4.2.13" width="300" /> |
| <figcaption> | | <figcaption> |
| For problem 4.2.13 | | For problem 4.2.13 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's analyze the following force diagram | | Let's analyze the following force diagram |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="diagram.png" | | <img src="diagram.png" |
| loading="lazy" alt="4.2.13" width="300" /> | | loading="lazy" alt="4.2.13" width="300" /> |
| <figcaption> | | <figcaption> |
| Force diagram for each trunk | | Force diagram for each trunk |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
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| <p> | | <p> |
| Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction, | | Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction, |
| $$-N'\cos{45^\circ}+N = 0 \;(1)$$ | | $$-N'\cos{45^\circ}+N = 0 \;(1)$$ |
| and on $y$-axis | | and on $y$-axis |
| $$mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$$ | | $$mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$$ |
| as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$ | | as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$ |
| $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$$ | | $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$$ |
| Applying Newton Second Law for cylinder 2, on $x$-axis | | Applying Newton Second Law for cylinder 2, on $x$-axis |
| $$N'\cos{45^\circ}-T\cos{45^\circ}=0$$ | | $$N'\cos{45^\circ}-T\cos{45^\circ}=0$$ |
| $$N' = T \;(3)$$ | | $$N' = T \;(3)$$ |
| and for $y$-axis | | and for $y$-axis |
| $$\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$$ | | $$\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$$ |
| in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$, | | in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$, |
| $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$$ | | $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$$ |
| Putting $(3)$ int $(4)$ | | Putting $(3)$ int $(4)$ |
| $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$$ | | $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$$ |
| Substituting $(5)$ into $(2)$ and developing algebraically | | Substituting $(5)$ into $(2)$ and developing algebraically |
| $$\rho = \frac{2}{3}\rho_0$$ | | $$\rho = \frac{2}{3}\rho_0$$ |
| as the water's density is $\rho_0=1$ g/cm$^3$, | | as the water's density is $\rho_0=1$ g/cm$^3$, |
| $$\boxed{\rho=\frac{2}{3}~\rm{g/cm^3}}$$ | | $$\boxed{\rho=\frac{2}{3}~\rm{g/cm^3}}$$ |
| </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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