Problem Statement
Find the magnitude and direction of the resultant pressure force on a curved face of a dam which is shaped according to the relation \( y = \frac{x^2}{6} \). The height of water retained by the dam is 12m. Assume unit width of the dam.
Solution
Equation of the Dam
Given equation of the dam:
\( y = \frac{x^2}{6} \)Solving for \( x \):
\( x = \sqrt{6y} \)Area of Region OAB
Consider an element of thickness \( dy \) and length \( x \) at a distance \( y \) from the base.
Area of the element: \( x dy \)
\( \text{Area}_{OAB} = \int_0^{12} \sqrt{6y} dy \)Computing the integral:
\( = \sqrt{6} \times \frac{2}{3} \left[ y^{3/2} \right]_0^{12} = 67.882 \, \text{m}^2 \)Horizontal Force \( F_x \)
Using \( F_x = \gamma A \bar{y} \):
\( F_x = 9810 \times (12 \times 1) \times 6 = 706320 \, N \)Vertical Force \( F_y \)
Weight of water vertically above the dam:
\( F_y = \gamma \times \text{Area}_{OAB} \times L \)Substituting values:
\( F_y = 9810 \times 67.882 \times 1 = 665922 \, N \)5. Resultant Force \( F_R \)
Using \( F_R = \sqrt{F_x^2 + F_y^2} \):
\( F_R = \sqrt{(706320)^2 + (665922)^2} = 970742 \, N = 970.742 \, \text{kN} \)Direction of resultant force:
\( \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) = \tan^{-1} \left( \frac{665922}{706320} \right) = 43.310^\circ \)Explanation
This problem is solved using fundamental concepts of hydrostatics and integration:
- Hydrostatic Pressure: The force exerted by a fluid in equilibrium due to its weight.
- Integration: Used to determine the area under the curved dam face, which helps calculate the vertical force.
- Resultant Force Calculation: The total force acting on the dam is found by combining horizontal and vertical forces using vector addition.
- Inclination of Force: The angle of the resultant force is derived using trigonometric relations.