A triangular lamina of area A and height h is immersed in a liquid of density ρ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is

The correct option is Aρgh/3 (Note: The provided option in the correct answer field "2Aρgh/3" corresponds to a triangular lamina with its vertex on the free surface and base parallel to it at depth h. For a triangular lamina with its base on the free surface and vertex pointing downwards at depth h, the depth of its centroid below the free surface is $\frac{h}{3}$, and thus the thrust is $F=PA=\rho g\overline{x}A=\frac{A\rho gh}{3}$).

Let us derive the thrust on the lamina step-by-step:

1. According to the principles of hydrostatics, the total thrust (hydrostatic force) $F$ acting on a plane surface immersed vertically in a liquid is given by the formula:
$F=\rho ·g·\overline{h}·A$
where:
• $\rho$ is the density of the liquid,
• $g$ is the acceleration due to gravity,
• $\overline{h}$ is the depth of the centroid (center of gravity) of the area from the free surface of the liquid,
• $A$ is the area of the lamina.

2. For a triangle of height $h$ with its base lying on the free surface of the liquid:
The centroid of a triangle lies at a distance of $\frac{h}{3}$ from its base.
Since the base of the triangular lamina is on the surface of the liquid, the depth of the centroid $\overline{h}$ from the liquid surface is:
$\overline{h}=\frac{h}{3}$

3. Substituting the depth of the centroid $\overline{h}$ into the hydrostatic force equation:
$F=\rho ·g·\left(\frac{h}{3}\right)·A$
$F=\frac{A\rho gh}{3}$

Therefore, the thrust on the lamina is $\frac{A\rho gh}{3}$.