A gate supporting water is shown in the figure. Find the height ‘h’ of the water so that the gate tips about the hinge. Take the width of the gate as unity (1 m).

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Tipping Gate Fluid Pressure Problem

Problem Statement

A gate supporting water is shown in the figure. Find the height 'h' of the water so that the gate tips about the hinge. Take the width of the gate as unity (1 m).

Given Data

  • Inclination of Gate, \( \theta = 60^\circ \)
  • Hinge position: 3 m vertically from the bottom of the gate.
  • Width of gate: \( 1 \, \text{m} \)

Diagram

The inclined gate hinged at point B.

Diagram of the inclined gate hinged at B

Solution

The gate will begin to tip about the hinge B when the resultant hydrostatic force acts exactly at point B. This means the centre of pressure must be at B.

1. Depth of Centre of Pressure (\(h^*\))

From the diagram, the vertical depth of the hinge B is 3 m below the water surface. Therefore, the depth of the centre of pressure, \(h^*\), must be:

$$ h^* = h - 3 $$

2. Calculating \(h^*\) using the Formula

First, we find the gate's geometric properties in terms of water depth \(h\).

$$ \text{Length of submerged gate, } AC = \frac{h}{\sin 60^\circ} $$ $$ AC = \frac{h}{\sqrt{3}/2} $$ $$ AC = \frac{2h}{\sqrt{3}} $$
$$ \text{Area, } A = AC \times \text{width} $$ $$ A = \frac{2h}{\sqrt{3}} \times 1 $$ $$ A = \frac{2h}{\sqrt{3}} \, \text{m}^2 $$
$$ \text{Depth of Centroid, } \bar{h} = \frac{h}{2} $$
$$ \text{Moment of Inertia, } I_G = \frac{b(AC)^3}{12} $$ $$ I_G = \frac{1 \times (\frac{2h}{\sqrt{3}})^3}{12} $$ $$ I_G = \frac{8h^3}{12 \times 3\sqrt{3}} $$ $$ I_G = \frac{2h^3}{9\sqrt{3}} $$

Now, we use the standard formula for the centre of pressure.

$$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{(\frac{2h^3}{9\sqrt{3}}) (\sin 60^\circ)^2}{(\frac{2h}{\sqrt{3}})(\frac{h}{2})} + \frac{h}{2} $$ $$ h^* = \frac{(\frac{2h^3}{9\sqrt{3}}) (\frac{\sqrt{3}}{2})^2}{\frac{h^2}{\sqrt{3}}} + \frac{h}{2} $$ $$ h^* = \frac{(\frac{2h^3}{9\sqrt{3}}) (\frac{3}{4})}{\frac{h^2}{\sqrt{3}}} + \frac{h}{2} $$ $$ h^* = \frac{\frac{h^3}{6\sqrt{3}}}{\frac{h^2}{\sqrt{3}}} + \frac{h}{2} $$ $$ h^* = \frac{h^3}{6\sqrt{3}} \times \frac{\sqrt{3}}{h^2} + \frac{h}{2} $$ $$ h^* = \frac{h}{6} + \frac{h}{2} $$ $$ h^* = \frac{h + 3h}{6} = \frac{4h}{6} = \frac{2h}{3} $$

3. Equating and Solving for \(h\)

Now we set the two expressions for \(h^*\) equal to each other to find the required water depth \(h\).

$$ h - 3 = \frac{2h}{3} $$ $$ 3(h - 3) = 2h $$ $$ 3h - 9 = 2h $$ $$ 3h - 2h = 9 $$ $$ h = 9 \, \text{m} $$
Final Result:

The height of the water must be \( h = 9 \, \text{m} \) for the gate to begin to tip.

Explanation of Concepts

Tipping Condition: A hinged gate will tip when the turning moment from the hydrostatic force becomes large enough. The limiting condition, where it just begins to tip, occurs when the total resultant hydrostatic force acts precisely at the hinge point. Any higher water level would cause the force to act below the hinge, creating a turning moment that opens the gate.

Centre of Pressure as Hinge: By setting the physical location of the hinge as the centre of pressure, we establish the critical condition. We derive two separate expressions for the depth of the centre of pressure (\(h^*\)): one based on the geometry of the setup (\(h-3\)) and one from the standard fluid mechanics formula. Equating these two allows us to solve for the unknown water height (\(h\)) that satisfies this tipping condition.

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