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<title>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. </title>
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<title>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. </title>
$4.1.12^*.$ 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 $P$.
</p>
<center>
<figure>
<img src="4.1.12.png"
loading="lazy" width="230" />
<figcaption>
For problem $4.1.12^*$
</figcaption>
</figure>
</center>
<p>
</p>
<h3>Solution</h3>
<p>
</p>
<center>
<figure>
<img src="4.1.12_1.png"
loading="lazy" width="210" />
<figcaption>
Geometry of the problem
</figcaption>
</figure>
</center>
<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
$$h_0=\frac{H}{3}$$
</p>
<center>
<figure>
<img src="4.1.12_2.png"
loading="lazy" width="250" />
<figcaption>
A close look at the base of the pyramid
</figcaption>
</figure>
</center>
<p>
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):
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
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<meta name="description" content="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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<meta property="og:description" content="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. ">
<meta property="og:description" content="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. ">
<title>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. </title>
<title>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. </title>
$4.1.12^*.$ 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 $P$.
$4.1.12^*.$ 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 $P$.
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="4.1.12.png"
<img src="4.1.12.png"
loading="lazy" width="230" />
loading="lazy" width="230" />
<figcaption>
<figcaption>
For problem $4.1.12^*$
For problem $4.1.12^*$
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="4.1.12_1.png"
<img src="4.1.12_1.png"
loading="lazy" width="210" />
loading="lazy" width="210" />
<figcaption>
<figcaption>
Geometry of the problem
Geometry of the problem
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
From the drawing, we could obtain the following expressions
From the drawing, we could obtain the following expressions
$$l=a\sin(60^{\circ}) = \frac{\sqrt{3}}{2}a$$
$$l=a\sin(60^{\circ}) = \frac{\sqrt{3}}{2}a$$
Height of the center of mass above the base of the tetrahedron
Height of the center of mass above the base of the tetrahedron
$$h_0=\frac{H}{3}$$
$$h_0=\frac{H}{3}$$
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="4.1.12_2.png"
<img src="4.1.12_2.png"
loading="lazy" width="250" />
loading="lazy" width="250" />
<figcaption>
<figcaption>
A close look at the base of the pyramid
A close look at the base of the pyramid
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
After a close look at the basement of the tetrahedron, we could obtain the base area
After a close look at the basement of the tetrahedron, we could obtain the base area
$$S=\frac{\sqrt{3}}{4}a^2$$
$$S=\frac{\sqrt{3}}{4}a^2$$
On the other hand, Pythagoras' theorem
On the other hand, Pythagoras' theorem
$$H^2=l^2-x^2=\frac{2}{3}a^2$$
$$H^2=l^2-x^2=\frac{2}{3}a^2$$
Whence the height of the tetrahedron is equal to
Whence the height of the tetrahedron is equal to
$$H=\sqrt{\frac{2}{3}}a$$
$$H=\sqrt{\frac{2}{3}}a$$
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
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
$$OO' = h_0 = \frac{1}{3}\sqrt{\frac{2}{3}}a$$
$$OO' = h_0 = \frac{1}{3}\sqrt{\frac{2}{3}}a$$
In this case, the pressure difference between the surface of the water and the level of the center of mass is
In this case, the pressure difference between the surface of the water and the level of the center of mass is
$$p_0=\rho g (h-h_0)=\rho g \left(h-\frac{1}{3}\sqrt{\frac{2}{3}}a\right)$$
$$p_0=\rho g (h-h_0)=\rho g \left(h-\frac{1}{3}\sqrt{\frac{2}{3}}a\right)$$
Hence, the force acting on the side of the pyramid (force due to water pressure):
Hence, the force acting on the side of the pyramid (force due to water pressure):
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
