A rectangular sluice gate is situated on the vertical wall of a lock. The vertical side of the sluice is 6 m in length and depth of centroid of the area is 8 m below the water surface. Prove that the depth of the centre of pressure is given by 8.375 m.

Rectangular Sluice Gate Problem

Problem Statement

Given Data

Shape: Rectangular
Depth of gate, $ d = 6 \, \text{m} $
Depth of centroid, $ \bar{h} = 8 \, \text{m} $
Let the width of the gate be $ b $.

Proof

The depth of the centre of pressure ($h^*$) for a vertically submerged surface is given by the formula:

$$ h^* = \frac{I_G}{A\bar{h}} + \bar{h} $$

First, we define the area ($A$) and the moment of inertia about the centroid ($I_G$) for the rectangular gate.

$$ A = b \times d = b \times 6 = 6b \, \text{m}^2 $$ $$ I_G = \frac{bd^3}{12} = \frac{b(6)^3}{12} = \frac{216b}{12} = 18b \, \text{m}^4 $$

Now, we substitute these values, along with the given depth of the centroid ($\bar{h} = 8$ m), into the formula for the centre of pressure.

$$ h^* = \frac{18b}{(6b)(8)} + 8 $$

The width of the gate, $b$, cancels out from the numerator and denominator:

$$ h^* = \frac{18}{48} + 8 $$ $$ h^* = 0.375 + 8 $$ $$ h^* = 8.375 \, \text{m} $$

Conclusion

Thus, it is proven that the depth of the centre of pressure on the sluice gate is 8.375 m based on the provided dimensions and depth.

Explanation of Concepts

Centre of Pressure vs. Centroid: The centroid is the geometric center of an object's area. The centre of pressure is the point where the total hydrostatic force acts. Because fluid pressure increases with depth, this point is always located below the centroid for a submerged vertical surface.

Formula Breakdown: The term $ \frac{I_G}{A\bar{h}} $ represents the vertical distance between the centroid and the centre of pressure. The calculation shows this distance is 0.375 m for this specific gate. Adding this to the centroid's depth ($ \bar{h} = 8 $ m) gives the final location of the centre of pressure.