Problem Statement
Determine the total pressure and centre of pressure on an isosceles triangular plate of base 5 m and altitude 5 m when the plate is immersed vertically in an oil of sp. gr. 0.8. The base of the plate is 1 m below the free surface of the oil.
Given Data
- Base of plate, \( b = 5 \, \text{m} \)
- Altitude of plate, \( d = 5 \, \text{m} \)
- Specific Gravity of oil, \( S.G. = 0.8 \)
- Density of oil, \( \rho = 0.8 \times 1000 = 800 \, \text{kg/m}^3 \)
- Depth of base below free surface = 1 m
Solution
(i) Total Pressure (Force, \(F\))
First, we calculate the area of the plate and the depth of its centroid (\(\bar{h}\)).
The centroid of a triangle is \(d/3\) from its base. Since the plate is inverted, the centroid is below the base.
Now, we can calculate the total force.
(ii) Position of Centre of Pressure (\(h^*\))
The depth of the centre of pressure is found using the parallel axis theorem.
First, we need the moment of inertia about the centroid (\(I_G\)) for a triangle.
Now, we substitute all values into the formula for \(h^*\).
Total Pressure (Force): \( F = 261.6 \, \text{kN} \).
Depth of Centre of Pressure: \( h^* \approx 3.188 \, \text{m} \) below the free surface.
Explanation of Concepts
Centroid of an Inverted Triangle: The problem describes an inverted triangle where the base is closer to the surface than the apex. The centroid is the geometric center, which for a triangle is one-third of the altitude from the base. We add this distance to the depth of the base to find the centroid's total depth from the free surface.
Centre of Pressure: The total hydrostatic force, or "total pressure," acts at the centre of pressure, not the centroid. Because pressure increases with depth, this point is always vertically below the centroid for a submerged plane. The distance between them depends on the object's shape (through its moment of inertia, \(I_G\)) and its depth.
