<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>