A rectangular plane surface that lies in a vertical plane in water is 2 m wide and 3 m deep. Find the Total pressure if the upper edge is horizontal and 2 m below the free water surface.
206.01 kN
The question asks us to find the total pressure force exerted by water on a vertical rectangular plane surface. We are given the dimensions of the rectangle and its position relative to the free water surface.
The total pressure force on a submerged plane surface is the resultant force due to the varying pressure of the fluid over the surface area. For a plane surface, this total force (F) is given by the formula:
\[F = \rho g \bar{h} A\]
Where:
The area of the rectangular surface is the product of its width and depth.
\[A = \text{Width} \times \text{Depth} = 2 \, m \times 3 \, m = 6 \, m^2\]
For a rectangle, the centroid is located at the geometric center. The distance of the centroid from the upper edge is half of the depth.
Distance of centroid from upper edge = \(\frac{\text{Depth}}{2} = \frac{3 \, m}{2} = 1.5 \, m\)
The upper edge is located \(2 \, m\) below the free surface. Therefore, the depth of the centroid from the free surface (\(\bar{h}\)) is the sum of the depth of the upper edge and the distance of the centroid from the upper edge.
\[\bar{h} = \text{Depth of upper edge} + \text{Distance of centroid from upper edge}\]
\[\bar{h} = 2 \, m + 1.5 \, m = 3.5 \, m\]
Now, substitute the values of \(\rho\), \(g\), \(\bar{h}\), and \(A\) into the formula \(F = \rho g \bar{h} A\).
\[F = (1000 \, kg/m^3) \times (9.81 \, m/s^2) \times (3.5 \, m) \times (6 \, m^2)\]
\[F = 1000 \times 9.81 \times 3.5 \times 6 \, N\]
\[F = 9810 \times 21 \, N\]
\[F = 206010 \, N\]
To express the force in kiloNewtons (kN), divide by 1000.
\[F = \frac{206010}{1000} \, kN = 206.01 \, kN\]
The total pressure force acting on the vertical rectangular plane surface is \(206.01 \, kN\).
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