A rectangular gate 5 m x 2 m is hinged at its base and inclined at 60° to the horizontal. To keep the gate in a stable position, a counterweight of 5000 kgf 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 friction.

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Force on a Hinged Sluice Gate

Problem Statement

An inclined rectangular sluice gate AB, 1.2 m by 5 m size, is installed to control the discharge of water. The end A is hinged. Determine the force normal to the gate applied at B to open it.

Given Data

  • Gate Dimensions (Length x Width): \( 1.2 \, \text{m} \times 5.0 \, \text{m} \)
  • Inclination Angle, \( \theta = 45^\circ \)
  • Depth to Hinge A: \( 5.0 \, \text{m} - 1.2 \sin 45^\circ \)
  • Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

Diagram

Illustration of the hinged rectangular sluice gate.

Diagram of the inclined sluice gate

Solution

To find the force \(P\) required to open the gate, we need to balance the moments about the hinge A. The hydrostatic force \(F\) will create a closing moment, which must be overcome by the opening moment from force \(P\).

1. Total Hydrostatic Force (\(F\))

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

$$ \text{Area, } A = 1.2 \, \text{m} \times 5.0 \, \text{m} = 6.0 \, \text{m}^2 $$ $$ \bar{h} = (5.0 - 1.2 \sin 45^\circ) + \frac{1.2}{2} \sin 45^\circ $$ $$ \bar{h} = 5.0 - 0.6 \sin 45^\circ $$ $$ \bar{h} = 5.0 - 0.6 \times \frac{1}{\sqrt{2}} \approx 4.576 \, \text{m} $$

Now, calculate the total hydrostatic force \(F\):

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 6.0 \times 4.576 $$ $$ F \approx 269343 \, \text{N} $$

2. Centre of Pressure and Moment Arm

Next, find the location where the force \(F\) acts. This is the centre of pressure, at a vertical depth \(h^*\).

$$ I_G = \frac{\text{width} \times \text{length}^3}{12} $$ $$ I_G = \frac{5.0 \times (1.2)^3}{12} = 0.72 \, \text{m}^4 $$ $$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{0.72 \times (\sin 45^\circ)^2}{6.0 \times 4.576} + 4.576 $$ $$ h^* = \frac{0.72 \times 0.5}{27.456} + 4.576 $$ $$ h^* \approx 0.0131 + 4.576 $$ $$ h^* \approx 4.589 \, \text{m} $$

To find the moment arm, we determine the positions along the incline of the gate.

$$ \text{Distance from surface to Centroid, } y_{cg} = \frac{\bar{h}}{\sin 45^\circ} $$ $$ y_{cg} = \frac{4.576}{1/\sqrt{2}} $$ $$ y_{cg} \approx 6.471 \, \text{m} $$ $$ \text{Distance from CG to CP along incline, } y_{dist} = \frac{I_G}{A y_{cg}} $$ $$ y_{dist} = \frac{0.72}{6 \times 6.471} $$ $$ y_{dist} \approx 0.0185 \, \text{m} $$ $$ \text{Moment Arm about hinge A, } AH = (\text{dist. A to CG}) + y_{dist} $$ $$ AH = \frac{1.2}{2} + 0.0185 $$ $$ AH = 0.6 + 0.0185 $$ $$ AH = 0.6185 \, \text{m} $$

3. Moment Balance

Now, balance the moments about the hinge A to solve for \(P\).

$$ P \times (\text{Length of Gate}) = F \times (AH) $$ $$ P \times 1.2 = 269343 \times 0.6185 $$ $$ P = \frac{269343 \times 0.6185}{1.2} $$ $$ P \approx 138800 \, \text{N} $$
Final Result:

The force normal to the gate required at B is approximately \( P \approx 138.8 \, \text{kN} \).

Explanation of Concepts

Moment Balance About a Hinge: To determine the force required to open the gate, we analyze the turning effects (moments) around the hinge point A. The water exerts a hydrostatic force (\(F\)) which acts at the center of pressure, creating a moment that tries to keep the gate shut. The applied force (\(P\)) creates an opposing moment to open it. The gate will begin to open when the moment from force \(P\) just exceeds the moment from the hydrostatic force.

Moment Arm: The effectiveness of a force in causing rotation depends on its moment arm—the perpendicular distance from the pivot (hinge) to the line of action of the force. Here, the moment arm for the hydrostatic force is the distance along the gate from the hinge (A) to the center of pressure (H).

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