Find the magnitude and direction of the resultant force due to water acting on a roller gate of cylindrical form of 4.0 m diameter, when the gate is placed on the dam in such a way that water is just going to spill. Take the length of the gate as 8 m.

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Cylindrical Roller Gate Fluid Pressure Problem

Problem Statement

Find the magnitude and direction of the resultant force due to water acting on a roller gate of cylindrical form of 4.0 m diameter, when the gate is placed on the dam in such a way that water is just going to spill. Take the length of the gate as 8 m.

Given Data

  • Diameter of Gate, \( D = 4.0 \, \text{m} \)
  • Radius of Gate, \( R = D/2 = 2.0 \, \text{m} \)
  • Length of Gate, \( l = 8 \, \text{m} \)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram of Roller Gate

The cylindrical roller gate ACB submerged in water, with water at the spill level.

Diagram of the cylindrical roller gate

Solution

To find the resultant force on the curved gate, 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 (AOB) is a rectangle with height equal to the diameter and width equal to the gate length.

$$ A_{proj} = \text{Diameter} \times \text{Length} = 4 \, \text{m} \times 8 \, \text{m} = 32 \, \text{m}^2 $$

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

$$ \bar{h} = \frac{\text{Diameter}}{2} = \frac{4}{2} = 2 \, \text{m} $$

The horizontal force is calculated as:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ F_x = 1000 \times 9.81 \times 32 \times 2 $$ $$ F_x = 627840 \, \text{N} $$

2. Vertical Force (\(F_y\))

The vertical force is equal to the weight of the water supported by the curved surface ACB. This volume is a semi-cylinder.

The volume of water supported by the gate per unit length is:

$$ \text{Volume (V)} = (\text{Area of semi-circle ACB}) \times \text{Length} $$ $$ V = \left( \frac{\pi R^2}{2} \right) \times l $$ $$ V = \frac{\pi (2)^2}{2} \times 8 = 16\pi \approx 50.265 \, \text{m}^3 $$

The vertical force is the weight of this volume:

$$ F_y = \rho g V $$ $$ F_y = 1000 \times 9.81 \times 16\pi $$ $$ F_y \approx 493104 \, \text{N} $$

3. Resultant Force (\(F\))

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

$$ F = \sqrt{F_x^2 + F_y^2} $$ $$ F = \sqrt{(627840)^2 + (493104)^2} $$ $$ F = \sqrt{3.94 \times 10^{11} + 2.43 \times 10^{11}} $$ $$ F = \sqrt{6.37 \times 10^{11}} $$ $$ F \approx 798328 \, \text{N} $$

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

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

$$ \tan \theta = \frac{F_y}{F_x} $$ $$ \tan \theta = \frac{493104}{627840} \approx 0.7853 $$ $$ \theta = \tan^{-1}(0.7853) $$ $$ \theta \approx 38.14^\circ $$
Final Result:

The resultant force is \( F \approx 798328 \, \text{N} \) or \( 798.33 \, \text{kN} \).

The angle of action with the horizontal is \( \theta \approx 38.14^\circ \).

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 (or supported by) 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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