A rectangular plane surface of width 2 m and height 3 m is placed vertically in water. What will be the location of center of pressure of the surface when its upper edge is horizontal and lies 2.5 m below the free surface of water?

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This question was previously asked in

SJVN ET Mechanical 2019 Official Paper

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Answer 1

Concept:

The depth of center of pressure for a vertically submerged plane surface is given by:

\( h_{cp} = \bar{h} + \frac{I_G}{A \cdot \bar{h}} \)

Given:

Width, \( b = 2~\text{m} \)
Height, \( h = 3~\text{m} \)
Top edge depth = 2.5 m → Centroid depth \( \bar{h} = 2.5 + \frac{3}{2} = 4~\text{m} \)

Calculation:

\( A = 2 \cdot 3 = 6~\text{m}^2 \)

\( I_G = \frac{b h^3}{12} = \frac{2 \cdot 27}{12} = 4.5~\text{m}^4 \)

\( h_{cp} = 4 + \frac{4.5}{6 \cdot 4} = 4 + 0.1875 = \boxed{4.1875~\text{m}} \)