A rectangular gate 6 m x 2 m is hinged at its base and inclined at 60° to the horizontal as shown. To keep the gate in a stable position, a counter weight of 29430 N is attached at the upper end of the gate. Find the depth of water at which the gate begins to fall. Neglect the weight of the gate and also friction at the hinge and pulley.

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Hinged Inclined Gate Problem

Problem Statement

A rectangular gate 6 m x 2 m is hinged at its base and inclined at 60° to the horizontal as shown. To keep the gate in a stable position, a counter weight of 29430 N is attached at the upper end of the gate. Find the depth of water at which the gate begins to fall. Neglect the weight of the gate and also friction at the hinge and pulley.

Given Data

  • Gate length, \( L = 6 \, \text{m} \)
  • Gate width, \( b = 2 \, \text{m} \)
  • Angle with horizontal, \( \theta = 60^\circ \)
  • Counter weight, \( W = 29430 \, \text{N} \)

Diagram of the Gate

Diagram of a hinged inclined gate with a counterweight

Solution

Let \(h\) be the depth of water above the hinge when the gate is about to fall. The gate will fall when the moment caused by the hydrostatic force about the hinge equals the moment caused by the counterweight.

1. Moment due to Counterweight (\(M_W\))

The counterweight acts at the full length of the gate from the hinge.

$$ M_W = W \times L $$ $$ M_W = 29430 \times 6 $$ $$ M_W = 176580 \, \text{N-m} $$

2. Hydrostatic Force (\(F\)) and Moment (\(M_F\))

The hydrostatic force depends on the depth \(h\). Let \(L_{sub}\) be the length of the submerged part of the gate.

$$ L_{sub} = \frac{h}{\sin \theta} = \frac{h}{\sin 60^\circ} $$

The submerged area \(A_{sub}\) and the depth of its centroid \(\bar{h}\) are:

$$ A_{sub} = b \times L_{sub} = 2 \times \frac{h}{\sin 60^\circ} $$ $$ \bar{h} = \frac{h}{2} $$

The hydrostatic force \(F\) is:

$$ F = \rho g A_{sub} \bar{h} $$ $$ F = 1000 \times 9.81 \times \left( \frac{2h}{\sin 60^\circ} \right) \times \frac{h}{2} $$ $$ F = \frac{9810 h^2}{\sin 60^\circ} $$

The force \(F\) acts at the centre of pressure. For a submerged rectangle hinged at the bottom, the lever arm from the hinge is \(L_{sub}/3\).

$$ \text{Lever Arm} = \frac{L_{sub}}{3} = \frac{h}{3 \sin 60^\circ} $$

The moment due to the hydrostatic force is:

$$ M_F = F \times \text{Lever Arm} $$ $$ M_F = \left( \frac{9810 h^2}{\sin 60^\circ} \right) \times \left( \frac{h}{3 \sin 60^\circ} \right) $$ $$ M_F = \frac{9810 h^3}{3 \sin^2 60^\circ} $$

3. Equating Moments to find \(h\)

The gate begins to fall when \(M_F = M_W\).

$$ \frac{9810 h^3}{3 \sin^2 60^\circ} = 176580 $$ $$ \frac{9810 h^3}{3 \times (0.866)^2} = 176580 $$ $$ \frac{9810 h^3}{3 \times 0.75} = 176580 $$ $$ \frac{9810 h^3}{2.25} = 176580 $$ $$ 4360 h^3 = 176580 $$ $$ h^3 = \frac{176580}{4360} $$ $$ h^3 = 40.5 $$ $$ h = \sqrt[3]{40.5} \approx 3.434 \, \text{m} $$
Final Result:

The gate begins to fall when the depth of water is \( h \approx 3.434 \, \text{m} \).

Explanation of Concepts

Principle of Moments: The core of this problem is balancing moments about a pivot point (the hinge). A moment is a turning force, calculated as Force × Perpendicular Distance (Lever Arm). The gate is in equilibrium as long as the clockwise moment from the counterweight is greater than or equal to the counter-clockwise moment from the water pressure.

Hydrostatic Moment: The total hydrostatic force (\(F\)) acts at a single point called the centre of pressure. For a rectangular surface submerged from the top edge and hinged at the bottom, this point is located at one-third of the submerged length up from the hinge. This distance (\(L_{sub}/3\)) serves as the lever arm for the hydrostatic force.

Condition for Failure: The gate "begins to fall" at the exact moment the counter-clockwise moment from the water pressure just equals the clockwise moment from the counterweight. By setting these two moments equal to each other, we can create an equation where the only unknown is the water depth (\(h\)), allowing us to solve for the critical depth.

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