A rectangular plane surface 2 m wide and 3 m deep lies in water in such a way that its plane makes an angle of 30° with the free surface of water. Determine the total pressure and position of centre of pressure when the upper edge is 1.5 m below the free water surface.

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Pressure on an Inclined Rectangular Surface

Problem Statement

A rectangular plane surface 2 m wide and 3 m deep lies in water in such a way that its plane makes an angle of 30° with the free surface of water. Determine the total pressure and position of centre of pressure when the upper edge is 1.5 m below the free water surface.

Given Data

  • Width of plane surface, \( b = 2 \, \text{m}\)
  • Depth (length) of plane surface, \( d = 3 \, \text{m}\)
  • Angle of inclination with free surface, \( \theta = 30^\circ \)
  • Vertical depth of upper edge, \( h_{top} = 1.5 \, \text{m}\)
  • Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

Diagram of the Inclined Surface

The setup showing the rectangular plane immersed at an angle in water.

Diagram of the inclined rectangular plane in water

Solution

(i) Total Pressure (Force)

First, calculate the area of the plate and the vertical depth to its centroid (\(\bar{h}\)).

$$ A = b \times d $$ $$ A = 2 \times 3 = 6 \, \text{m}^2 $$
$$ \bar{h} = h_{top} + \left(\frac{d}{2}\right) \sin\theta $$ $$ \bar{h} = 1.5 + \left(\frac{3}{2}\right) \sin 30^\circ $$ $$ \bar{h} = 1.5 + (1.5 \times 0.5) $$ $$ \bar{h} = 1.5 + 0.75 = 2.25 \, \text{m} $$

Now, calculate the total pressure force (F).

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 6 \times 2.25 $$ $$ F = 132435 \, \text{N} $$

(ii) Position of Centre of Pressure (\(h^*\))

First, find the moment of inertia (\(I_G\)) of the rectangular area about its centroidal axis.

$$ I_G = \frac{bd^3}{12} $$ $$ I_G = \frac{2 \times 3^3}{12} $$ $$ I_G = 4.5 \, \text{m}^4 $$

Now, use the formula for the vertical depth to the centre of pressure for an inclined plane.

$$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{4.5 \times (\sin 30^\circ)^2}{6 \times 2.25} + 2.25 $$ $$ h^* = \frac{4.5 \times (0.5)^2}{13.5} + 2.25 $$ $$ h^* = \frac{4.5 \times 0.25}{13.5} + 2.25 $$ $$ h^* = \frac{1.125}{13.5} + 2.25 $$ $$ h^* = 0.0833 + 2.25 $$ $$ h^* \approx 2.3333 \, \text{m} $$
Final Results:

Total Pressure (Force): \( F = 132435 \, \text{N} \)

Centre of Pressure (Vertical Depth): \( h^* \approx 2.3333 \, \text{m} \) below the free surface

Explanation of Concepts

Pressure on Inclined Surfaces: For a submerged plane that is not vertical, the pressure calculations must account for the angle. The total force still depends on the vertical depth to the centroid (\(\bar{h}\)), not the slant distance. The centre of pressure, which is the point where the resultant force acts, is also calculated based on its vertical depth (\(h^*\)). The term \(\sin^2\theta\) in the formula for \(h^*\) corrects the moment of inertia for the inclination of the plane, ensuring the final result is a vertical depth from the free surface.

Physical Meaning

The inclined rectangular surface is subjected to a total force of 132,435 N, which is equivalent to the weight of over 13.5 metric tons. This is the total load the plate must be designed to withstand without breaking or deforming.

The centre of pressure is located at a vertical depth of 2.3333 m from the water surface. This is slightly below the centroid's vertical depth of 2.25 m, which is always the case for submerged surfaces. This exact point is critical for engineering design; for example, if this were a hinged gate, any supports or opening mechanisms must be designed to handle the force applied at this specific location to prevent failure.

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