A circular plate of diameter 3 m is immersed in water in such a way that its least and greatest depth from the free surface of water are 1 m and 3 m respectively. For the front side of the plate, find (i) total force exerted by water and (ii) the position of centre of pressure.

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Inclined Circular Plate Problem

Problem Statement

A circular plate of diameter 3 m is immersed in water in such a way that its least and greatest depth from the free surface of water are 1 m and 3 m respectively. For the front side of the plate, find (i) total force exerted by water and (ii) the position of centre of pressure.

Given Data

  • Plate Diameter, \( D = 3 \, \text{m} \)
  • Least depth (top edge), \( h_{top} = 1 \, \text{m} \)
  • Greatest depth (bottom edge), \( h_{bottom} = 3 \, \text{m} \)
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Solution

1. Geometric Properties of the Plate

First, we find the depth of the plate's centroid (\(\bar{h}\)) and the angle of inclination (\(\theta\)).

$$ \bar{h} = \frac{h_{top} + h_{bottom}}{2} $$ $$ \bar{h} = \frac{1 + 3}{2} $$ $$ \bar{h} = 2 \, \text{m} $$
$$ \sin \theta = \frac{h_{bottom} - h_{top}}{D} $$ $$ \sin \theta = \frac{3 - 1}{3} $$ $$ \sin \theta = \frac{2}{3} $$

(i) Total Force Exerted by Water (\(F\))

Next, calculate the area of the circular plate.

$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (3)^2 $$ $$ A \approx 7.0686 \, \text{m}^2 $$

Now, calculate the total force on the plate.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 7.0686 \times 2 $$ $$ F \approx 138686 \, \text{N} $$

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

The vertical depth of the centre of pressure for an inclined plane is given by:

$$ h^* = \frac{I_G \sin^2 \theta}{A \bar{h}} + \bar{h} $$

First, calculate the moment of inertia (\(I_G\)) for the circular plate.

$$ I_G = \frac{\pi D^4}{64} $$ $$ I_G = \frac{\pi \times (3)^4}{64} $$ $$ I_G = \frac{81\pi}{64} \approx 3.976 \, \text{m}^4 $$

Now, substitute all values into the formula for \(h^*\).

$$ h^* = \frac{3.976 \times (2/3)^2}{7.0686 \times 2} + 2 $$ $$ h^* = \frac{3.976 \times (4/9)}{14.1372} + 2 $$ $$ h^* = \frac{1.767}{14.1372} + 2 $$ $$ h^* \approx 0.125 + 2 $$ $$ h^* = 2.125 \, \text{m} $$
Final Results:

(i) Total Force Exerted: \( F \approx 138.69 \, \text{kN} \).

(ii) Position of Centre of Pressure: \( h^* = 2.125 \, \text{m} \) (vertical depth from free surface).

Explanation of Concepts

Geometric Setup: From the least and greatest depths, we can determine both the depth of the geometric center (centroid, \(\bar{h}\)) and the angle of inclination (\(\theta\)) of the plate. The centroid's depth is the average of the top and bottom depths, and the sine of the angle is the ratio of the vertical rise to the length of the plate (its diameter).

Total Force: The total hydrostatic force is calculated based on the pressure at the centroid's vertical depth (\(\bar{h}\)). This gives the average pressure acting over the entire area of the plate.

Centre of Pressure on an Inclined Plane: The formula for the centre of pressure (\(h^*\)) must account for the inclination. The term \(I_G \sin^2 \theta\) correctly projects the moment of inertia's effect onto the vertical axis. The result, \(h^*\), gives the true vertical depth of the centre of pressure from the free surface, which is always deeper than the centroid's vertical depth.

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