A gate having a quadrant shape of radius 2 m is shown in the figure. Find the resultant force due to water per metre length of the gate. Also, find the angle at which this resultant force will act.

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Quadrant Gate Fluid Pressure Problem

Problem Statement

A gate having a quadrant shape of radius 2 m is shown in the figure. Find the resultant force due to water per metre length of the gate. Also, find the angle at which this resultant force will act.

Given Data

  • Radius of Gate, \( R = 2 \, \text{m} \)
  • Width of Gate, \( b = 1 \, \text{m} \) (per metre length)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram

The quadrant gate AB submerged in water.

Diagram of the quadrant gate

Solution

To find the resultant force on a curved surface, we calculate its horizontal and vertical components separately.

1. Horizontal Force (\(F_x\))

The horizontal force is the total pressure on the projected area of the curved surface onto a vertical plane. The projected area is a rectangle (BO) of height 2 m and width 1 m.

$$ A_{proj} = \text{Radius} \times \text{Width} = 2 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^2 $$

The depth of the centroid of this projected area from the free surface is:

$$ \bar{h} = \frac{2}{2} = 1 \, \text{m} $$

The horizontal force is calculated as:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ F_x = 1000 \times 9.81 \times 2 \times 1 $$ $$ F_x = 19620 \, \text{N} $$

2. Vertical Force (\(F_y\))

The vertical force is equal to the weight of the (real or imaginary) fluid column above the curved surface. In this case, it's the weight of the water in the area AOB.

The volume of water above the gate per unit width is:

$$ \text{Volume (V)} = \text{Area of Quadrant AOB} \times \text{Width} $$ $$ V = \left( \frac{\pi R^2}{4} \right) \times 1 $$ $$ V = \frac{\pi (2)^2}{4} \times 1 = \pi \, \text{m}^3 $$

The vertical force is the weight of this volume:

$$ F_y = \rho g V $$ $$ F_y = 1000 \times 9.81 \times \pi $$ $$ F_y \approx 30819 \, \text{N} $$

3. Resultant Force (\(F\))

The resultant force is the vector sum of the horizontal and vertical components.

$$ F = \sqrt{F_x^2 + F_y^2} $$ $$ F = \sqrt{(19620)^2 + (30819)^2} $$ $$ F = \sqrt{384944400 + 949810761} $$ $$ F = \sqrt{1334755161} $$ $$ F \approx 36534.4 \, \text{N} $$

4. Angle of Action (\(\theta\))

The angle the resultant force makes with the horizontal is found using trigonometry.

$$ \tan \theta = \frac{F_y}{F_x} $$ $$ \tan \theta = \frac{30819}{19620} \approx 1.5708 $$ $$ \theta = \tan^{-1}(1.5708) $$ $$ \theta \approx 57.52^\circ \text{ or } 57^\circ 31' $$
Final Result:

The resultant force is \( F \approx 36534.4 \, \text{N} \).

The angle of action with the horizontal is \( \theta \approx 57^\circ 31' \).

Explanation of Concepts

Horizontal Force on Curved Surface: The horizontal component of the hydrostatic force on any curved surface is always equal to the force on the vertical projection of that surface. This allows us to simplify the problem by analyzing a simple flat plate.

Vertical Force on Curved Surface: The vertical component is equal to the weight of the fluid volume directly above the curved surface, extending up to the free surface. This force acts through the centroid (center of gravity) of that fluid volume.

Resultant Force: Since the horizontal and vertical forces are perpendicular, the total resultant force can be found using the Pythagorean theorem, and its direction can be determined with basic trigonometry.

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