We already know that the pressure in a static fluid increases with depth, following the hydrostatic law $P = \rho g h$. But what happens when we need to find the total force acting on a submerged surface, like the wall of a water tank or a sluice gate? The pressure at the top of the gate is less than the pressure at the bottom. We can't simply multiply pressure by area, because the pressure isn't constant.

To find the total force, we need to sum up the forces acting on every part of the surface. We can imagine dividing the surface into infinitesimally small horizontal strips. On each tiny strip of area $dA$, the pressure $P$ is essentially constant. The small force $dF$ on this strip is therefore $dF = P \cdot dA$. Since $P = \rho g h$, we get $dF = (\rho g h) \cdot dA$.

The total resultant force, $F_R$, is the sum of all these tiny forces. In mathematical terms, this is an integral over the entire area A:

This integral can be cumbersome to solve directly. Fortunately, there's a much more elegant solution that uses a shape's geometric properties.

The total hydrostatic force on any plane surface submerged in a fluid is equal to the product of the area of the surface and the pressure at its centroid. The centroid is the geometric centre of an area, its average position. Think of it as the point where you could balance a cutout of the shape on a pin.

This simplifies our calculation dramatically. We just need to find the vertical depth of the centroid ($h_c$) from the free surface, calculate the pressure there, and multiply it by the total area.

For a rectangular plate of height $b$ submerged vertically with its top edge a distance $d$ below the surface, the centroid is at its centre. The vertical depth to the centroid is $h_c = d + b/2$.

So, we have the magnitude of the force. But where, exactly, does this resultant force act on the surface? It's tempting to think it acts at the centroid, but that's incorrect.

Because pressure increases with depth, the lower portion of the submerged surface experiences more force than the upper portion. This shifts the 'average' point of force application downwards. This point is called the centre of pressure ($C_P$).

To find the vertical location of the centre of pressure, $h_p$, we use the following formula:

The term $I_c$ is a purely geometric property that describes how an area's points are distributed with regard to an arbitrary axis. You can typically find these values in engineering handbooks.

What if the surface is not vertical, but inclined at an angle $\theta$ to the free surface? The good news is that the formula for the magnitude of the force, $F_R = (\rho g h_c)A$, remains exactly the same. We still just need the vertical depth to the centroid.

However, the location of the centre of pressure changes. If we measure distance along the inclined plate from the free surface, let's call it $y$, the position of the centre of pressure $y_p$ is:

This shows that for an inclined plate, just like a vertical one, the centre of pressure is always located further down the plate than the centroid.

Understanding these principles is fundamental for designing any structure that holds back a fluid. By correctly calculating both the magnitude of the hydrostatic force and its point of application, engineers can ensure the stability and safety of dams, gates, tanks, and ship hulls.