Added Luis's English solution of 4.1.13

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+ <title>Savchenko Solutions</title>
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+ content="A website with solutions to physics problems from Savenko Textbook" />
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+ <header>
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <main>
+ <article>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h4>Statement</h4>
+ <p>
+ $4.1.13^*$
+ 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 $g$.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="4.1.13" width="280" />
+ <figcaption>
+ For problem 4.1.13
+ </figcaption>
+ </figure>
+ </center>
+ <h4>Solution</h4>
+ <p class="TxtSolutions">
+ 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>
+ <center>
+ <figure>
+ <img src="draw1.png"
+ loading="lazy" alt="4.1.13" width="450" />
+ <figcaption>
+ Pressure analysis at $x$-axis
+ </figcaption>
+ </figure>
+ </center>
+ <p class="TxtSolutions">
+ The pressures at points A and C are
+ </p>
+ <center>
+ $P_A = P + \rho_0 g(h+a\sin{\alpha})$
+ </center>
+ <center>
+ $P_C = P + \rho_0 gh$
+ </center>
+ <p class="TxtSolutions">
+ If both equations are rested,
+ </p>
+ <center>
+ $P_A-P_C = \rho ga\sin{\alpha} \;(1)$
+ </center>
+ <p class="TxtSolutions">
+ 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
+ </p>
+ <center>
+ $F_x = \rho g a^3 \sin{\alpha} \;(2)$
+ </center>
+ <p class="TxtSolutions">
+ Applying Newton's Second Law for $x$-axis, also considering friction force
+ </p>
+ <center>
+ $mg\sin{\alpha} = f_r + F_x$
+ </center>
+ <p class="TxtSolutions">
+ As $m = \rho a^3$ and according $(2)$
+ </p>
+ <center>
+ $\boxed{f_r = a^3 g (\rho-\rho_0)\sin{\alpha}}$
+ </center>
+ <p class="TxtSolutions">
+ 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>
+ <center>
+ <figure>
+ <img src="draw2.png"
+ loading="lazy" alt="4.1.13" width="480" />
+ <figcaption>
+ Pressure analysis at $y$-axis
+ </figcaption>
+ </figure>
+ </center>
+ <p class="TxtSolutions">
+ Atmospheric-pressure force is
+ </p>
+ <center>
+ $F_0 = \frac{Pa^2}{\cos{\alpha}}$
+ </center>
+ <p class="TxtSolutions">
+ but the cube doesn't "feel" that force, else a component of it ($F_{0y}$).
+ </p>
+ <center>
+ $F_{0y} = F_0\cos{\alpha} = Pa^2 \;(3)$
+ </center>
+ <p class="TxtSolutions">
+ The pressure exerted by water mass is
+ </p>
+ <center>
+ $F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \;(4)$
+ </center>
+ <p class="TxtSolutions">
+ This volume of water $V_w$ is
+ </p>
+ <center>
+ $V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \;(5)$
+ </center>
+ <p class="TxtSolutions">
+ Putting $(5)$ into $(4)$
+ </p>
+ <center>
+ $F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(6)$
+ </center>
+ <p class="TxtSolutions">
+ The pressure force over superior side of cube is
+ </p>
+ <center>
+ $F_p = F_{0y} + F_w \;(7)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(3)$ and $(6)$ into $(7)$
+ </p>
+ <center>
+ $F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(8)$
+ </center>
+ <p class="TxtSolutions">
+ Applying Newton Second Law for $y$-axis
+ </p>
+ <center>
+ $N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \;(9)$
+ </center>
+ <p class="TxtSolutions">
+ and according to $(8)$
+ </p>
+ <center>
+ $\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$
+ </center>
+ <p class="TxtSolutions">
+ This normal force is directed in opposed direction of $y$-axis.
+ </p>
+ <p class="TxtSolutions">
+ Friction force $f_r$ is the parallel reaction force $F_{||}$ and normal force is the perpendicular one $F_{\perp}$.
+ </p>
+ <p class="TxtSolutions" style="text-align: center; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
+ </p>
+ </article>
+ </main>
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