Determine the position of center of pressure (from the surface) for a rectangular lamina of height h vertically inside the water.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29-01-2018 (Morning Shift)
Detailed Explanation
The centre of pressure for a vertical rectangular lamina submerged in water depends on its depth of submersion. When the upper edge of the lamina coincides with the free water surface, the pressure distribution is triangular.
1. Case 1: Upper edge of rectangular lamina at the free water surface:
- Centroid depth: \(\bar{h} = \frac{h}{2}\)
- Area: \(A = b \times h\)
- Centroidal moment of inertia: \(I_G = \frac{b h^3}{12}\)
2. Calculate centre of pressure \(h^*\):
\[ h^* = \bar{h} + \frac{I_G}{A \bar{h}} = \frac{h}{2} + \frac{\frac{b h^3}{12}}{(b h) \left(\frac{h}{2}\right)} = \frac{h}{2} + \frac{h}{6} = \frac{4h}{6} = \frac{2h}{3} \]
3. Case 2 (Alternative reading in some commission exams): If the top edge is submerged to depth \(h\) (so the plate lies between depth \(h\) and \(2h\)):
\[ \bar{h} = h + \frac{h}{2} = \frac{3h}{2} \;\implies\; h^* = \frac{3h}{2} + \frac{\frac{b h^3}{12}}{(b h)\left(\frac{3h}{2}\right)} = \frac{3h}{2} + \frac{h}{18} = \frac{14h}{9} \]
Due to this wording ambiguity, SSC JE cancelled this question or accepted \(2h/3\) representing the standard case where the plate begins at the surface.
Why “2h/3” Is Correct
The triangular pressure prism on a vertical rectangle starting at the water surface has its centroid at two-thirds of the total height from the apex (surface), giving \(h^* = 2h/3\).
Why the Other Options Are Wrong
- A. h/3: h/3 is the distance from the bottom edge to the centre of pressure (h − 2h/3 = h/3).
- B. h/2: h/2 is the depth to the geometric centroid (h̄), not the centre of pressure.
- D. 3h/2: 3h/2 exceeds the height h of the plate itself and represents centroid depth if submerged to depth h.
Key Point
Vertical rectangular surface with top edge at free surface: h̄ = h/2 and h* = 2h/3.
Key Concepts for Students
- The pressure prism is a triangular wedge; its centroid is at 2/3 of its depth from the top.
- When submerged deeper, the pressure prism becomes trapezoidal and h* moves closer to the centroid h̄.
- Centre of pressure can never lie above the centroid for any vertical surface.
Practice more: this is question 18 in Hydrostatic Pressure MCQs – Part 2. Try the whole set and check every answer instantly.