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.

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    <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 $P$.

    <h3>Solution</h3>
    <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 $$h_0=\frac{H}{3}$$

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)
$$F=pS+F_0 = \frac{\sqrt{3}}{4}a^2p+F_0$$
$$\boxed{F = \frac{\rho g a^2}{12}\left(3\sqrt{3}h - \sqrt{2}a \right) + \frac{\sqrt{3}}{4} pa^2}$$

    <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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