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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> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
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| <p> | | <p> |
| $4.2.14$ | | $4.2.14$ |
| Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal. | | Determine the pressure force of the logs of mass $m$ on the walls of the channel. The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal. |
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| loading="lazy" alt="4.2.14" width="300" /> | | loading="lazy" alt="4.2.14" width="300" /> |
| <figcaption> | | <figcaption> |
| For problem 4.2.14 | | For problem 4.2.14 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider the following figure | | Let's consider the following figure |
| </p> | | </p> |
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| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" alt="4.2.14" width="550" /> | | loading="lazy" alt="4.2.14" width="550" /> |
| <figcaption> | | <figcaption> |
| Force analysis | | Force analysis |
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| From right figure | | From right figure |
| $$\cos{\theta} = \frac{R}{2R} = 0.5$$ | | $$\cos{\theta} = \frac{R}{2R} = 0.5$$ |
| so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis, | | so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis, |
| $$N = N_t~\sin{\theta} \;(1)$$ | | $$N = N_t~\sin{\theta} \;(1)$$ |
| on $y$-axis | | on $y$-axis |
| $$F_{A1} = mg + N_t~\cos{\theta}$$ | | $$F_{A1} = mg + N_t~\cos{\theta}$$ |
| $$\rho_w~g~V = mg + N_t~\cos{\theta} \;(2)$$ | | $$\rho_w~g~V = mg + N_t~\cos{\theta} \;(2)$$ |
| For half-submerged trunk, on $x$-axis, | | For half-submerged trunk, on $x$-axis, |
| $$N = N_t~\sin{\theta}$$ | | $$N = N_t~\sin{\theta}$$ |
| on $y$-axis | | on $y$-axis |
| $$F_{A2} + N_t~\cos{\theta} = mg$$ | | $$F_{A2} + N_t~\cos{\theta} = mg$$ |
| $$\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \;(3)$$ | | $$\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \;(3)$$ |
| Substituting $(2)$ into $(3)$, it is obtained | | Substituting $(2)$ into $(3)$, it is obtained |
| $$N_t = \frac{mg}{3~\cos{\theta}} \;(4)$$ | | $$N_t = \frac{mg}{3~\cos{\theta}} \;(4)$$ |
| Putting $(4)$ into $(1)$ | | Putting $(4)$ into $(1)$ |
| $$N = \frac{mg}{3}~\tan{\theta}$$ | | $$N = \frac{mg}{3}~\tan{\theta}$$ |
| $$\boxed{N = \frac{mg}{\sqrt{3}}}$$ | | $$\boxed{N = \frac{mg}{\sqrt{3}}}$$ |
| Note: You can prove that both pressure forces generated by trunks are equals. | | Note: You can prove that both pressure forces generated by trunks are equals. |
| </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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