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| <header style="text-align:center;"> | | <header style="text-align:center;"> |
| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#4.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#4.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| | | |
| <p> | | <p> |
| $4.1.13^*$ | | $4.1.13^*$ |
| A cube of side $a$, made of a material of density $\rho$, is found on container's bottom that forms an angle $\alpha$ with the horizon. Find the force which the cube acts over bottom, if a liquid of density $\rho_0$ fills the container. The superior side of cube is at distance $h$ from the liquid's surface. Between the container's bottom and the cube there is no liquid. The atmospheric pressure is $P$ and the acceleration due to gravity is $g$. | | A cube of side $a$, made of a material of density $\rho$, is found on container's bottom that forms an angle $\alpha$ with the horizon. Find the force which the cube acts over bottom, if a liquid of density $\rho_0$ fills the container. The superior side of cube is at distance $h$ from the liquid's surface. Between the container's bottom and the cube there is no liquid. The atmospheric pressure is $P$ and the acceleration due to gravity is $g$. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="4.1.13" width="280" /> | | loading="lazy" alt="4.1.13" width="280" /> |
| <figcaption> | | <figcaption> |
| For problem 4.1.13 | | For problem 4.1.13 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's define the $y$-axis perpendicular to bottom and the $x$-axis parallel to it. At $y$-axis there is no buoyancy force because between bottom and the cube there is no liquid. | | Let's define the $y$-axis perpendicular to bottom and the $x$-axis parallel to it. At $y$-axis there is no buoyancy force because between bottom and the cube there is no liquid. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw1.png" | | <img src="draw1.png" |
| loading="lazy" alt="4.1.13" width="450" /> | | loading="lazy" alt="4.1.13" width="450" /> |
| <figcaption> | | <figcaption> |
| Pressure analysis at $x$-axis | | Pressure analysis at $x$-axis |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| The pressures at points $A$ and $C$ are | | The pressures at points $A$ and $C$ are |
| $$P_A = P + \rho_0 g(h+a\sin{\alpha})$$ | | $$P_A = P + \rho_0 g(h+a\sin{\alpha})$$ |
| $$P_C = P + \rho_0 gh$$ | | $$P_C = P + \rho_0 gh$$ |
| If both equations are rested, | | If both equations are rested, |
| $$P_A-P_C = \rho ga\sin{\alpha} \;(1)$$ | | $$P_A-P_C = \rho ga\sin{\alpha} \;(1)$$ |
| Note that difference of pressures between any point over side $AB$ and any point over $CD$ is expressed by $(1)$ (for any two points at same $y$ and at $a$ distance one from the other). If $(1)$ is multiplied by the side area $a^2$, buoyancy force can be found for $x$-axis | | Note that difference of pressures between any point over side $AB$ and any point over $CD$ is expressed by $(1)$ (for any two points at same $y$ and at $a$ distance one from the other). If $(1)$ is multiplied by the side area $a^2$, buoyancy force can be found for $x$-axis |
| $$F_x = \rho g a^3 \sin{\alpha} \;(2)$$ | | $$F_x = \rho g a^3 \sin{\alpha} \;(2)$$ |
| Applying Newton's Second Law for $x$-axis, also considering friction force | | Applying Newton's Second Law for $x$-axis, also considering friction force |
| $$mg\sin{\alpha} = f_r + F_x$$ | | $$mg\sin{\alpha} = f_r + F_x$$ |
| As $m = \rho a^3$ and according $(2)$ | | As $m = \rho a^3$ and according $(2)$ |
| $$\boxed{f_r = a^3 g (\rho-\rho_0)\sin{\alpha}}$$ | | $$\boxed{f_r = a^3 g (\rho-\rho_0)\sin{\alpha}}$$ |
| This force is "felt" by bottom in opposed direction of $x$-axis. Now, pressure force $F_p$ which acts in $y$-axis, generated by above water mass and the atmospher, must be found. | | This force is "felt" by bottom in opposed direction of $x$-axis. Now, pressure force $F_p$ which acts in $y$-axis, generated by above water mass and the atmospher, must be found. |
| </p> | | </p> |
| | | |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw2.png" | | <img src="draw2.png" |
| loading="lazy" alt="4.1.13" width="480" /> | | loading="lazy" alt="4.1.13" width="480" /> |
| <figcaption> | | <figcaption> |
| Pressure analysis at $y$-axis | | Pressure analysis at $y$-axis |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <p> | | <p> |
| Atmospheric-pressure force is | | Atmospheric-pressure force is |
| $$F_0 = \frac{Pa^2}{\cos{\alpha}}$$ | | $$F_0 = \frac{Pa^2}{\cos{\alpha}}$$ |
| but the cube doesn't "feel" that force, else a component of it ($F_{0y}$). | | but the cube doesn't "feel" that force, else a component of it ($F_{0y}$). |
| $$F_{0y} = F_0\cos{\alpha} = Pa^2 \;(3)$$ | | $$F_{0y} = F_0\cos{\alpha} = Pa^2 \;(3)$$ |
| The pressure exerted by water mass is | | The pressure exerted by water mass is |
| $$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \;(4)$$ | | $$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \;(4)$$ |
| This volume of water $V_w$ is | | This volume of water $V_w$ is |
| $$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \;(5)$$ | | $$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \;(5)$$ |
| Putting $(5)$ into $(4)$ | | Putting $(5)$ into $(4)$ |
| $$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(6)$$ | | $$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(6)$$ |
| The pressure force over superior side of cube is | | The pressure force over superior side of cube is |
| $$F_p = F_{0y} + F_w \;(7)$$ | | $$F_p = F_{0y} + F_w \;(7)$$ |
| Substituting $(3)$ and $(6)$ into $(7)$ | | Substituting $(3)$ and $(6)$ into $(7)$ |
| $$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(8)$$ | | $$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(8)$$ |
| Applying Newton Second Law for $y$-axis | | Applying Newton Second Law for $y$-axis |
| $$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \;(9)$$ | | $$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \;(9)$$ |
| and according to $(8)$ | | and according to $(8)$ |
| $$\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$$ | | $$\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$$ |
| This normal force is directed in opposed direction of $y$-axis. | | This normal force is directed in opposed direction of $y$-axis. |
| Friction force $f_r$ is the parallel reaction force $F_{||}$ and normal force is the perpendicular one $F_{\perp}$. | | Friction force $f_r$ is the parallel reaction force $F_{||}$ and normal force is the perpendicular one $F_{\perp}$. |
| </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> |
| </p> | | </p> |
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