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| <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> |
| $4.1.12^*.$ The bottom face of a regular tetrahedron with edge $a$, completely immersed in a liquid of density $\rho$, is at depth $h$. Determine the force exerted by the liquid on the side edge of the tetrahedron if the atmospheric pressure is $P$. | | $4.1.12^*.$ The bottom face of a regular tetrahedron with edge $a$, completely immersed in a liquid of density $\rho$, is at depth $h$. Determine the force exerted by the liquid on the side edge of the tetrahedron if the atmospheric pressure is $P$. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="4.1.12.png" | | <img src="4.1.12.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $4.1.12^*$ | | For problem $4.1.12^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="4.1.12_1.png" | | <img src="4.1.12_1.png" |
| loading="lazy" width="210" /> | | loading="lazy" width="210" /> |
| <figcaption> | | <figcaption> |
| Geometry of the problem | | Geometry of the problem |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| From the drawing, we could obtain the following expressions | | From the drawing, we could obtain the following expressions |
| $$l=a\sin(60^{\circ}) = \frac{\sqrt{3}}{2}a$$ | | $$l=a\sin(60^{\circ}) = \frac{\sqrt{3}}{2}a$$ |
| Height of the center of mass above the base of the tetrahedron | | Height of the center of mass above the base of the tetrahedron |
| $$h_0=\frac{H}{3}$$ | | $$h_0=\frac{H}{3}$$ |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="4.1.12_2.png" | | <img src="4.1.12_2.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| A close look at the base of the pyramid | | A close look at the base of the pyramid |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| After a close look at the basement of the tetrahedron, we could obtain the base area | | After a close look at the basement of the tetrahedron, we could obtain the base area |
| $$S=\frac{\sqrt{3}}{4}a^2$$ | | $$S=\frac{\sqrt{3}}{4}a^2$$ |
| On the other hand, Pythagoras' theorem | | On the other hand, Pythagoras' theorem |
| $$H^2=l^2-x^2=\frac{2}{3}a^2$$ | | $$H^2=l^2-x^2=\frac{2}{3}a^2$$ |
| Whence the height of the tetrahedron is equal to | | Whence the height of the tetrahedron is equal to |
| $$H=\sqrt{\frac{2}{3}}a$$ | | $$H=\sqrt{\frac{2}{3}}a$$ |
| Since the center of mass of the pyramid is located at the point $O'$, it is known that the pyramid its height above the base is | | Since the center of mass of the pyramid is located at the point $O'$, it is known that the pyramid its height above the base is |
| $$OO' = h_0 = \frac{1}{3}\sqrt{\frac{2}{3}}a$$ | | $$OO' = h_0 = \frac{1}{3}\sqrt{\frac{2}{3}}a$$ |
| In this case, the pressure difference between the surface of the water and the level of the center of mass is | | In this case, the pressure difference between the surface of the water and the level of the center of mass is |
| $$p_0=\rho g (h-h_0)=\rho g \left(h-\frac{1}{3}\sqrt{\frac{2}{3}}a\right)$$ | | $$p_0=\rho g (h-h_0)=\rho g \left(h-\frac{1}{3}\sqrt{\frac{2}{3}}a\right)$$ |
| Hence, the force acting on the side of the pyramid (force due to water pressure): | | Hence, the force acting on the side of the pyramid (force due to water pressure): |
| $$F_0 = p_0S = \rho ga^2\left(\frac{\sqrt{3}}{4}h - \frac{\sqrt{2}}{12}a \right)$$ | | $$F_0 = p_0S = \rho ga^2\left(\frac{\sqrt{3}}{4}h - \frac{\sqrt{2}}{12}a \right)$$ |
| After mathematical transformations | | After mathematical transformations |
