A circular plate of 3 m diameter is under water with its plane making an angle of 30° with the water surface. If the top edge of the plate is 1 m below the water surface, find the force on one side of the plate and its location.

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Inclined Circular Plate Problem

Problem Statement

A circular plate of 3 m diameter is under water with its plane making an angle of 30° with the water surface. If the top edge of the plate is 1 m below the water surface, find the force on one side of the plate and its location.

Given Data

  • Diameter of plate, \( D = 3 \, \text{m} \)
  • Angle with free surface, \( \theta = 30^\circ \)
  • Depth of top edge = 1 m
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Diagram of the Plate

Diagram of an inclined circular plate submerged in water

Solution

(i) Force on the Plate (\(F\))

First, calculate the area of the circular plate.

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

Next, find the vertical depth of the plate's centroid (\(\bar{h}\)). The centroid is at a radius (\(D/2\)) distance along the incline from the upper edge.

$$ \bar{h} = (\text{Depth of top edge}) + \left( \frac{D}{2} \sin\theta \right) $$ $$ \bar{h} = 1 + \left( \frac{3}{2} \sin 30^\circ \right) $$ $$ \bar{h} = 1 + (1.5 \times 0.5) $$ $$ \bar{h} = 1 + 0.75 $$ $$ \bar{h} = 1.75 \, \text{m} $$

Now, calculate the total force on the plate.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 7.0686 \times 1.75 $$ $$ F \approx 121354 \, \text{N} $$

(ii) Location 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 \(h^*\).

$$ h^* = \frac{3.976 \times (\sin 30^\circ)^2}{7.0686 \times 1.75} + 1.75 $$ $$ h^* = \frac{3.976 \times (0.5)^2}{12.37} + 1.75 $$ $$ h^* = \frac{3.976 \times 0.25}{12.37} + 1.75 $$ $$ h^* = \frac{0.994}{12.37} + 1.75 $$ $$ h^* \approx 0.0803 + 1.75 $$ $$ h^* \approx 1.8303 \, \text{m} $$
Final Results:

Force on the Plate: \( F \approx 121.35 \, \text{kN} \).

Location of Centre of Pressure: \( h^* \approx 1.83 \, \text{m} \) (vertical depth from free surface).

Explanation of Concepts

Inclined Surfaces: When a surface is inclined, we must use trigonometry to find the true vertical depth of its centroid. The force calculation (\(F = \rho g A \bar{h}\)) always uses this vertical depth (\(\bar{h}\)), not the distance measured along the incline.

Centre of Pressure on an Inclined Plane: The formula for the centre of pressure must also account for the inclination. The term \(I_G \sin^2 \theta\) in the numerator 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. It is always deeper than the centroid's vertical depth.

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A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are 2 m long and the tank contains a liquid of specific gravity 1.15. The centre of the aperture is 1.5 m below the free surface. Calculate the thrust exerted on the plate by the liquid and the position of its centre of pressure.

A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are 2 m long and the tank contains a liquid of specific gravity 1.15. The centre of the aperture is 1.5 m below the free surface. Calculate the thrust exerted on the plate by the liquid and the position of its centre of pressure.

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