Find the magnitude and direction of the resultant water pressure acting on a curved face of a dam which is shaped according to the relation y =x2/9 . The height of the water retained by the dam is 10 m. Consider the width of the dam as unity (1 m).

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Curved Dam Fluid Pressure Problem

Problem Statement

Find the magnitude and direction of the resultant water pressure acting on a curved face of a dam which is shaped according to the relation \( y = \frac{x^2}{9} \). The height of the water retained by the dam is 10 m. Consider the width of the dam as unity (1 m).

Given Data

  • Equation of the curve, \( y = \frac{x^2}{9} \implies x = 3\sqrt{y} \)
  • Height of water, \( h_{total} = 10 \, \text{m} \)
  • Width of dam, \( b = 1 \, \text{m} \)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram of Curved Dam

Diagram of the curved dam face

Solution

1. Horizontal Force Component (\(F_x\))

The horizontal force is the pressure on the vertically projected area (BC).

$$ A_{proj} = \text{Height} \times \text{Width} $$ $$ = 10 \, \text{m} \times 1 \, \text{m} = 10 \, \text{m}^2 $$

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

$$ \bar{h} = \frac{\text{Height}}{2} $$ $$ = \frac{10}{2} = 5 \, \text{m} $$

The horizontal force is calculated as:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ = 1000 \times 9.81 \times 10 \times 5 $$ $$ = 490500 \, \text{N} $$

2. Vertical Force Component (\(F_y\))

The vertical force is equal to the weight of the water supported by the curved surface AB (the volume of water in portion ABC).

We find the area of ABC by integrating the curve's equation from y=0 to y=10.

$$ \text{Area (ABC)} = \int_{0}^{10} x \,dy $$ $$ = \int_{0}^{10} 3\sqrt{y} \,dy = 3 \int_{0}^{10} y^{1/2} \,dy $$ $$ = 3 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{10} = 2 \left[ y^{3/2} \right]_{0}^{10} $$ $$ = 2 [10^{3/2} - 0] = 2 \times 31.622 $$ $$ \approx 63.245 \, \text{m}^2 $$

The vertical force is the weight of this volume of water:

$$ F_y = \rho g \times \text{Area (ABC)} \times \text{Width} $$ $$ = 1000 \times 9.81 \times 63.245 \times 1 $$ $$ \approx 620439 \, \text{N} $$

3. Resultant Force (\(F\)) and Direction (\(\theta\))

The magnitude of the resultant force is found using the Pythagorean theorem:

$$ F = \sqrt{F_x^2 + F_y^2} $$ $$ = \sqrt{(490500)^2 + (620439)^2} $$ $$ \approx 790907 \, \text{N} $$

The direction of the resultant force relative to the horizontal is:

$$ \tan \theta = \frac{F_y}{F_x} $$ $$ = \frac{620439}{490500} \approx 1.265 $$ $$ \theta = \tan^{-1}(1.265) \approx 51.67^\circ \text{ or } 51^\circ 40' $$
Final Result:

The resultant force is \( F \approx 790907 \, \text{N} \) or \( 790.91 \, \text{kN} \).

The direction of the resultant force is \( \theta \approx 51^\circ 40' \) with the horizontal.

Explanation of Concepts

Force on a Parabolic Surface: For surfaces defined by a mathematical equation (like a parabola), we cannot use simple geometric formulas for area. Instead, we use integral calculus to find the exact area under the curve. This area is then used to calculate the volume of the supported fluid, which in turn gives us the vertical component of the hydrostatic force.

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