A cube of side a, made of a material of density
ho, is found on container's bottom that forms an angle lpha with the horizon. Find the force which the cube acts over bottom, if a liquid of density
ho_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.
<h3 id="back-link"><a href="/#4.1">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
<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 $g$.
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<img src="statement.png"
loading="lazy" alt="4.1.13" width="280" />
<figcaption>
For problem 4.1.13
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<h3>Solution</h3>
<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.
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<figure>
<img src="draw1.png"
loading="lazy" alt="4.1.13" width="450" />
<figcaption>
Pressure analysis at $x$-axis
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<p>
The pressures at points $A$ and $C$ are
$$P_A = P + \rho_0 g(h+a\sin{\alpha})$$
$$P_C = P + \rho_0 gh$$
If both equations are rested,
$$P_A-P_C = \rho ga\sin{\alpha} \\;(1)$$
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
$$F_x = \rho g a^3 \sin{\alpha} \\;(2)$$
Applying Newton's Second Law for $x$-axis, also considering friction force
$$mg\sin{\alpha} = f_r + F_x$$
As $m = \rho a^3$ and according $(2)$
$$\boxed{f_r = a^3 g (\rho-\rho_0)\sin{\alpha}}$$
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.
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<figure>
<img src="draw2.png"
loading="lazy" alt="4.1.13" width="480" />
<figcaption>
Pressure analysis at $y$-axis
</figcaption>
</figure>
</center>
<p>
Atmospheric-pressure force is
$$F_0 = \frac{Pa^2}{\cos{\alpha}}$$
but the cube doesn't "feel" that force, else a component of it ($F_{0y}$).
$$F_{0y} = F_0\cos{\alpha} = Pa^2 \\;(3)$$
The pressure exerted by water mass is
$$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \\;(4)$$
This volume of water $V_w$ is
$$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \\;(5)$$
Putting $(5)$ into $(4)$
$$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \\;(6)$$
The pressure force over superior side of cube is
$$F_p = F_{0y} + F_w \\;(7)$$
Substituting $(3)$ and $(6)$ into $(7)$
$$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \\;(8)$$
Applying Newton Second Law for $y$-axis
$$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \\;(9)$$
and according to $(8)$
$$\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$$
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 $F_{\perp}$.
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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