In the fig., a 2.4m diameter cylinder plugs a rectangular hole in a tank that is 1.4m long. With what force is the cylinder pressed against the bottom of the tank due to the 2.7m depth of water?

.  In the fig., a 2.4m diameter cylinder plugs a rectangular hole in a tank that is 1.4m long. With what force is the cylinder pressed against the bottom of the tank due to the 2.7m depth of water?
In the fig., a 2.4m diameter cylinder plugs a rectangular hole in a tank that is 1.4m long. With what force is the cylinder pressed against the bottom of the tank due to the 2.7m depth of water?
Gate Force Problems

Problem Statement

In the figure, a 2.4 m diameter cylinder plugs a rectangular hole in a tank that is 1.4 m long. With what force is the cylinder pressed against the bottom of the tank due to the 2.7 m depth of water?

Solution

1. Analyze the Forces

Water is above the curved portion CDE, whereas it is below the curved portions AC and BE. For AC and BE, the imaginary weight of water vertically above them is considered, and the vertical force on these parts acts upwards.

2. Net Vertical Force

The net vertical force is given by:

\( F_V = (F_V)_{\text{CDE}} \text{ (down)} – (F_V)_{\text{AC}} \text{ (up)} – (F_V)_{\text{BE}} \text{ (up)} \)

This simplifies to:

\( F_V = \gamma \cdot \left[ \text{Volume above CDE} – \text{Volume above AC} – \text{Volume above BE} \right] \)

3. Volumes

The volumes are calculated as:

\( \text{Volume above CDE} = \text{Volume of rectangle CEMN} – \text{Volume of semicircle CDE} \)
\( \text{Volume above AC} = \text{Volume of rectangle CRPN} + \text{Volume of sector COA} – \text{Volume of triangle ROA} \)
\( \text{Volume above BE} = \text{Volume of rectangle SEMQ} + \text{Volume of sector BOE} – \text{Volume of triangle BOS} \)

4. Substitute Values

Substituting the dimensions and solving:

\( F_V = 9810 \times \Bigg[ (2.4 \times 2.1 \times 1.4 – \frac{\pi \times 1.2^2}{2} \times 1.4) \)
\( – \big( 0.16 \times 2.1 \times 1.4 + \frac{30}{360} \times \pi \times 1.2^2 \times 1.4 – \frac{1}{2} \times 1.04 \times 0.6 \times 1.4 \big) \)
\( – \big( 0.16 \times 2.1 \times 1.4 + \frac{30}{360} \times \pi \times 1.2^2 \times 1.4 – \frac{1}{2} \times 1.04 \times 0.6 \times 1.4 \big) \Bigg] \)
\( F_V = 27139 \, \text{N (down)} \)
Result:
  • The net vertical force pressing the cylinder against the bottom of the tank is \( F_V = 27139 \, \text{N} \).

Explanation

  • Curved Surfaces: The force on curved surfaces is calculated considering the imaginary weight of water above them and subtracting upward forces.
  • Volume Calculations: The volumes of the rectangle, semicircles, sectors, and triangles are used to determine the forces.
  • Net Force: The net vertical force is the result of combining downward and upward forces.

Physical Meaning

This problem demonstrates the hydrostatic forces acting on a cylindrical plug and their distribution on curved surfaces. Such calculations are crucial in designing tank closures and ensuring structural stability under fluid pressure.

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