Compute the horizontal and vertical components of the total force acting on a curved surface AB, which is in the form of a quadrant of a circle of radius 2 m. Take the width of the gate as unity.

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Pressure on a Curved Surface

Problem Statement

Compute the horizontal and vertical components of the total force acting on a curved surface AB, which is in the form of a quadrant of a circle of radius 2 m. Take the width of the gate as unity.

Given Data

  • Radius of quadrant, \( r = 2 \, \text{m} \)
  • Width of gate, \( w = 1 \, \text{m} \)
  • Depth of top edge C from surface = \( 1.5 \, \text{m} \)

Diagram

The curved gate (quadrant of a circle) submerged in water.

Diagram of the curved gate

Solution

1. Horizontal Component (\(F_x\))

The horizontal force \(F_x\) is the total pressure force on the vertical projection of the curved surface. The projection of AB is the vertical plane OB, which is a rectangle.

$$ \text{Projected Area, } A_{proj} = \text{height} \times \text{width} $$ $$ A_{proj} = 2 \, \text{m} \times 1 \, \text{m} $$ $$ A_{proj} = 2 \, \text{m}^2 $$
$$ \text{Depth of Centroid, } \bar{h} = (\text{Depth to O}) + \frac{\text{Height of OB}}{2} $$ $$ \bar{h} = 1.5 + \frac{2}{2} $$ $$ \bar{h} = 1.5 + 1 $$ $$ \bar{h} = 2.5 \, \text{m} $$
$$ F_x = \rho g A_{proj} \bar{h} $$ $$ F_x = 1000 \times 9.81 \times 2 \times 2.5 $$ $$ F_x = 49050 \, \text{N} $$

Point of Application for \(F_x\)

The depth of the centre of pressure for the horizontal force is \(h^*\).

$$ I_G = \frac{w d^3}{12} = \frac{1 \times 2^3}{12} $$ $$ I_G = \frac{8}{12} $$ $$ I_G = \frac{2}{3} \, \text{m}^4 $$
$$ h^* = \frac{I_G}{A_{proj} \bar{h}} + \bar{h} $$ $$ h^* = \frac{2/3}{2 \times 2.5} + 2.5 $$ $$ h^* = \frac{2/3}{5} + 2.5 $$ $$ h^* = \frac{2}{15} + 2.5 $$ $$ h^* = 0.1333 + 2.5 $$ $$ h^* = 2.633 \, \text{m} $$

2. Vertical Component (\(F_y\))

The vertical force \(F_y\) is equal to the weight of the volume of water vertically above the curved surface AB (the volume of DABOC).

$$ V = V_{DAOC} + V_{AOB} $$ $$ V = (\text{Area of rectangle DAOC} \times w) + (\text{Area of quadrant AOB} \times w) $$ $$ V = (1.5 \times 2 \times 1) + \left( \frac{\pi r^2}{4} \times 1 \right) $$ $$ V = 3 + \frac{\pi (2)^2}{4} $$ $$ V = 3 + \pi $$ $$ V \approx 3 + 3.14159 $$ $$ V \approx 6.14159 \, \text{m}^3 $$
$$ F_y = \rho g V $$ $$ F_y = 1000 \times 9.81 \times (3 + \pi) $$ $$ F_y = 9810 \times 6.14159 $$ $$ F_y \approx 60249.1 \, \text{N} $$
Final Results:

Horizontal Force: \( F_x = 49050 \, \text{N} \)

Depth of Horizontal Force: \( h^* \approx 2.633 \, \text{m} \)

Vertical Force: \( F_y \approx 60249 \, \text{N} \)

Explanation of Concepts

Horizontal Force (\(F_x\)): For any curved surface, the total horizontal force is found by projecting the surface onto a vertical plane. The problem then simplifies to finding the force on that flat, projected area, which can be solved using the standard formula \(F = \rho g A \bar{h}\).

Vertical Force (\(F_y\)): The vertical component of the force is equal to the weight of the fluid that is directly above the curved surface, extending up to the free surface. This is found by calculating the volume of that fluid and multiplying by its specific weight (\(\rho g\)).

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