A rectangular plate 1.25 m × 2.4 m is immersed in a liquid of relative density 0.85 with its 1.25 m side horizontal and just at the water surface. If the plane of the plate makes an angle of 60° with the horizontal, the pressure force on one side of the plate is :
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2012
Detailed Explanation
The total hydrostatic thrust on an inclined submerged plane surface equals the product of the liquid's specific weight, the surface area, and the vertical depth of the centroid from the free liquid surface.
1. Identify the given parameters:
- Plate dimensions: width \(W = 1.25\text{ m}\), inclined length \(L = 2.4\text{ m}\)
- Area of plate: \(A = W \times L = 1.25\text{ m} \times 2.4\text{ m} = 3.0\text{ m}^2\)
- Liquid relative density: \(S = 0.85 \;\implies\; \rho = 850\text{ kg/m}^3\)
- Inclination angle with horizontal: \(\theta = 60^\circ\)
- Top edge (\(1.25\text{ m}\)) lies horizontally at the free surface.
2. Calculate the vertical depth to the centroid (\(\bar{h}\)):
\[ \bar{h} = \frac{L}{2} \sin\theta = \frac{2.4\text{ m}}{2} \sin 60^\circ = 1.2 \times \frac{\sqrt{3}}{2} = 1.2 \times 0.866025 \approx 1.03923\text{ m} \]
3. Calculate the total hydrostatic thrust (\(P\)):
\[ P = \rho g A \bar{h} \]
\[ P = 850\text{ kg/m}^3 \times 9.81\text{ m/s}^2 \times 3.0\text{ m}^2 \times 1.03923\text{ m} \]
\[ P = 8338.5 \times 3.11769 \approx 25,996.8\text{ N} \approx 26.0\text{ kN} \]
Why “26.0 kN” Is Correct
The centroid of the inclined plate is located at a vertical depth of \(1.2 \sin 60^\circ = 1.0392\text{ m}\). Multiplying the liquid's specific weight (\(850 \times 9.81 = 8338.5\text{ N/m}^3\)), the area (\(3.0\text{ m}^2\)), and the centroid depth gives \(25,997\text{ N}\), which rounds to \(26.0\text{ kN}\).
Why the Other Options Are Wrong
- A. 15.0 kN: 15.0 kN results from omitting sin 60° or using an incorrect area calculation.
- B. 30.0 kN: 30.0 kN results from using water (SG = 1.0) and taking centroid depth as 1.0 m or using g = 10 m/s².
- C. 30.6 kN: 30.6 kN results from calculating total force using horizontal projection without inclination: 850 × 9.81 × 3.0 × 1.2 = 30.0 kN.
Key Point
Inclined plane thrust P = ρ·g·A·h̄ where h̄ = (L/2)·sin θ from the free surface.
Key Concepts for Students
- Hydrostatic force on an inclined surface depends solely on the vertical depth h̄ of its centroid.
- Inclination angle θ affects the centroid depth via h̄ = ȳ sin θ, where ȳ is distance along the inclined plane.
- Centre of pressure depth along the plane is y* = ȳ + I_G / (A ȳ), corresponding to vertical depth h* = y* sin θ.
Practice more: this is question 27 in Hydrostatic Pressure MCQs – Part 2. Try the whole set and check every answer instantly.