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<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>
+ <h3 id="back-link"><a href="../">$\leftarrow$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 $g$.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="4.1.13" width="280" />
+ <figcaption>
+ For problem 4.1.13
+ </figcaption>
+ </figure>
+ </center>
+
+ <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.
+ </p>
<center>
<figure>
− <img src="statement.png"
− loading="lazy" alt="4.1.13" width="280" />
+ <img src="draw1.png"
+ loading="lazy" alt="4.1.13" width="450" />
<figcaption>
− For problem 4.1.13
+ Pressure analysis at $x$-axis
</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>
+ 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.
</p>
+
<center>
<figure>
− <img src="draw1.png"
− loading="lazy" alt="4.1.13" width="450" />
+ <img src="draw2.png"
+ loading="lazy" alt="4.1.13" width="480" />
<figcaption>
− Pressure analysis at $x$-axis
+ Pressure analysis at $y$-axis
</figcaption>
</figure>
</center>
− <p class="TxtSolutions">
− The pressures at points A and C are
+
+ <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>
− <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 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>
</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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