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.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <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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