Problem Statement
Determine the total pressure and centre of pressure on an isosceles triangular plate of base 4 m and altitude 4 m when it is immersed vertically in an oil of specific gravity 0.9. The base of the plate coincides with the free surface of the oil.
Given Data
- Base of plate, \(b = 4 \, \text{m}\)
- Altitude of plate, \(h_{plate} = 4 \, \text{m}\)
- Area, \(A = \frac{1}{2} b \times h_{plate} = \frac{1}{2} \times 4 \times 4 = 8.0 \, \text{m}^2\)
- Specific gravity of oil, \(S = 0.9\)
- Density of oil, \(\rho = 0.9 \times 1000 = 900 \, \text{kg/m}^3\)
Diagram
Visual representation of the vertically immersed triangular plate.
Solution
(i) Total Pressure (Force)
First, determine the depth of the centre of gravity (C.G.) of the plate from the free surface. For a triangle, the C.G. is located at one-third of the altitude from the base.
The total pressure (force) is given by the formula:
Substitute the known values:
(ii) Position of Centre of Pressure
The depth of the centre of pressure (\(h^*\)) from the free surface is given by:
Where \(I_G\) is the moment of inertia of the triangle about its C.G.:
Substitute the values to find \(h^*\):
Total Pressure (Force): \( F \approx 94.18 \, \text{kN} \)
Centre of Pressure: \( h^* = 2.0 \, \text{m} \) below the free surface
Explanation of Concepts
Centre of Gravity (\(\bar{h}\)): This is the geometric centre of the object, the average location of all the points of the plate. For a triangle, it is one-third of the way from the base to the apex.
Total Pressure (Force): This is the net hydrostatic force acting on the plate. It is calculated by multiplying the pressure at the plate's centre of gravity by the total area of the plate.
Centre of Pressure (\(h^*\)): This is the point on the surface where the total hydrostatic force effectively acts. Since fluid pressure increases with depth, the lower parts of the plate experience more force than the upper parts. This shifts the centre of pressure to a point deeper than the centre of gravity.
Physical Meaning
The results indicate that the total force on the plate is approximately 94.18 kN. The centre of pressure is located at a depth of 2.0 m, while the centre of gravity is at 1.333 m.
This difference is crucial. The force from the oil is not distributed evenly; it is zero at the surface (the base of the triangle) and increases linearly to a maximum at the plate's tip (vertex). The resultant force, therefore, acts at a point halfway down the plate's height (2.0 m). This demonstrates that for a triangular surface with its base at the liquid's surface, the centre of pressure is always located at \(h_{plate}/2\), which is significantly lower than its geometric centre.
