Problem Statement
A rectangular opening 2 m wide and 1 m deep in the vertical side of a tank is closed by a sluice gate of the same size. The gate can turn about the horizontal centroidal axis. Determine: (i) the total pressure on the sluice gate and (ii) the torque on the sluice gate. The head of water above the upper edge of the gate is 1.5 m.
Given Data
- Gate width, \( b = 2 \, \text{m} \)
- Gate depth, \( d = 1 \, \text{m} \)
- Pivot: Horizontal centroidal axis
- Head above upper edge = 1.5 m
- Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)
Solution
(i) Total Pressure (Force, \(F\))
First, calculate the area of the gate and the depth of its centroid.
Now, calculate the total force on the gate.
(ii) Torque on the Sluice Gate (\(T\))
The torque is created by the hydrostatic force \(F\) acting at the centre of pressure, which is below the pivot point (the centroid). The lever arm is the distance between these two points.
First, find the depth of the centre of pressure (\(h^*\)).
The lever arm is the distance between the centre of pressure and the centroid.
Finally, calculate the torque.
(i) Total Pressure (Force): \( F = 39.24 \, \text{kN} \).
(ii) Torque on the Sluice Gate: \( T \approx 1635 \, \text{N-m} \).
Explanation of Concepts
Total Pressure (Force): This is the total hydrostatic force exerted by the water on the gate. It is calculated by finding the pressure at the gate's geometric center (centroid) and multiplying it by the total area of the gate.
Torque and Centre of Pressure: The gate is pivoted at its centroid, but the total hydrostatic force acts at a slightly lower point called the centre of pressure. This is because the water pressure is greater at the bottom of the gate than at the top. The offset between the point of force application (centre of pressure) and the pivot point (centroid) creates a lever arm. The torque is the turning moment produced by the hydrostatic force acting over this lever arm (\( T = F \times \text{Lever Arm} \)).
