Answer :
To solve this problem, we need to use principles from fluid mechanics.
Part (i): Total Pressure on the Plate
The total pressure exerted by the fluid on the plate is equal to the weight of the column of fluid directly above it.
Determine the area of the plate: The area [tex]A[/tex] of the plate is given by its length and width.
[tex]A = 3 \text{ m} \times 1 \text{ m} = 3 \text{ m}^2[/tex]Calculate the pressure exerted by water: The pressure exerted at a depth [tex]h[/tex] is given by [tex]P = \rho g h[/tex], where:
- [tex]\rho[/tex] is the density of water (approximately [tex]1000 \text{ kg/m}^3[/tex]),
- [tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \text{ m/s}^2[/tex]), and
- [tex]h[/tex] is the depth to the center of the plate.
Since the plate is 1 meter below the surface, the average depth [tex]h[/tex] of the center is at:
[tex]h = 1 + \frac{1.5}{2} = 2.5 \text{ m}[/tex]
[tex]P = 1000 \times 9.81 \times 2.5 = 24525 \text{ N/m}^2[/tex]Calculate the total force (total pressure) on the plate using [tex]F = P \times A[/tex].
[tex]F = 24525 \text{ N/m}^2 \times 3 \text{ m}^2 = 73575 \text{ N}[/tex]
Part (ii): Position of the Centre of Pressure
The center of pressure is the point where the total force acts on the plate.
- Calculate the center of pressure:
The vertical distance to the center of pressure [tex]h_{cp}[/tex] from the surface is given by:
[tex]h_{cp} = h + \frac{I_g}{A \cdot h}[/tex]
where [tex]I_g[/tex] is the second moment of area (or the moment of inertia) about the horizontal axis through the surface:
[tex]I_g = \frac{b \times h^3}{12} = \frac{1 \times 3^3}{12} = \frac{27}{12} = 2.25 \text{ m}^4[/tex]
Substituting for [tex]h_{cp}[/tex]:
[tex]h_{cp} = 2.5 + \frac{2.25}{3 \times 2.5}[/tex]
[tex]h_{cp} = 2.5 + 0.3 = 2.8 \text{ m}[/tex]
Hence, the center of pressure is 2.8 meters below the surface of the water.
