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Physics LibreTexts

16.5: Pressure on a Vertical Surface

Figure XVI.5 shows a vertical surface from the side and face-on. The pressure increases at greater depths. I show a strip of the surface at depth \( z\).

Suppose the area of that strip is \( dA\). The pressure at depth \( z\) is \( \rho gz\), so the force on the strip is \( \rho gzdA\). The force on the entire area is \( \rho g\int zdA\), and that, by definition of the centroid (see Chapter 1), is \( \rho g\overline{z}A\) where \( \overline{z}\) is the depth of the centroid. The same result will be obtained for an inclined surface. 

Therefore:

The total force on a submerged vertical or inclined plane surface is equal to the area of the surface times the depth of the centroid.

Example \(\PageIndex{1}\)

Figure XVI.6 shows a triangular area. The uppermost side of the triangle is parallel to the surface at a depth \( z\). The depth of the centroid is \( z+\frac{1}{3}h\), so the pressure at the centroid is \( \rho g\left(z+\frac{1}{3}h\right)\). The area of the triangle is \( \frac{1}{2}bh\) so the total force on the triangle is \( \frac{1}{2}\rho gbh\left(z+\frac{1}{3}h\right)\).

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