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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#4.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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>
+<center>
+ <figure>
+ <img src="4.1.12.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $4.1.12^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="4.1.12_1.png"
+ loading="lazy" width="210" />
+ <figcaption>
+ Geometry of the problem
+ </figcaption>
+ </figure>
+</center>
+<p>
+From the drawing, we could obtain the following expressions
+$$l=a\sin(60^{\circ}) = \frac{\sqrt{3}}{2}a$$
+Height of the center of mass above the base of the tetrahedron
+$$h_0=\frac{H}{3}$$
+
+</p>
+<center>
+ <figure>
+ <img src="4.1.12_2.png"
+ loading="lazy" width="250" />
+ <figcaption>
+ A close look at the base of the pyramid
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+After a close look at the basement of the tetrahedron, we could obtain the base area
+$$S=\frac{\sqrt{3}}{4}a^2$$
+On the other hand, Pythagoras' theorem
+$$H^2=l^2-x^2=\frac{2}{3}a^2$$
+Whence the height of the tetrahedron is equal to
+$$H=\sqrt{\frac{2}{3}}a$$
+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
+$$OO' = h_0 = \frac{1}{3}\sqrt{\frac{2}{3}}a$$
+In this case, the pressure difference between the surface of the water and the level of the center of mass is
+$$p_0=\rho g (h-h_0)=\rho g \left(h-\frac{1}{3}\sqrt{\frac{2}{3}}a\right)$$
+Hence, the force acting on the side of the pyramid (force due to water pressure):
+$$F_0 = p_0S = \rho ga^2\left(\frac{\sqrt{3}}{4}h - \frac{\sqrt{2}}{12}a \right)$$
+After mathematical transformations
+$$F_0 = \frac{\rho g a^2}{12}\left(3\sqrt{3}h - \sqrt{2}a \right)$$
+Force acting on the side of the pyramid (force due to water pressure and atmospheric pressure)
+$$F=pS+F_0 = \frac{\sqrt{3}}{4}a^2p+F_0$$
+$$\boxed{F_0 = \frac{\rho g a^2}{12}\left(3\sqrt{3}h - \sqrt{2}a \right) + \frac{\sqrt{3}}{4} pa^2}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$F =\frac{1}{12}\rho ga^2(3\sqrt{3} h-\sqrt{2} a)+\frac{1}{4}Pa^2\sqrt{3}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
+ </p>
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