The tank in the fig. contains oil (sp gr = 0.8) and water as shown. Find the resultant force on side ABC and its point of application. ABC is 1.2m wide.

Gate Force Problems

Problem Statement

The tank in the figure contains oil (specific gravity = 0.8) and water as shown. Find the resultant force on side ABC and its point of application. ABC is 1.2 m wide.

Solution

1. Specific Weight of Oil

The specific weight of oil is:

\( \gamma_{\text{oil}} = 0.8 \times 9810 = 7848 \, \text{N/m}^3 \)

2. Area of Sections

The area of section AB (oil side) is:

\( A_1 = 1.2 \times 3 = 3.6 \, \text{m}^2 \)

The area of section BC (water side) is:

\( A_2 = 1.2 \times 1.8 = 2.16 \, \text{m}^2 \)

3. Location of Center of Gravity (CG)

The location of CG for section AB is:

\( y_{1, \text{CG}} = \frac{3}{2} = 1.5 \, \text{m} \)

4. Force on Section AB

The hydrostatic force on section AB is:

\( F_{AB} = \gamma_{\text{oil}} \cdot A_1 \cdot y_{1, \text{CG}} \)

\( F_{AB} = 7848 \cdot 3.6 \cdot 1.5 = 42379 \, \text{N} \)

The point of application of \( F_{AB} \) is:

\( h = \frac{2}{3} \cdot 3 = 2 \, \text{m below A} \)

5. Force on Section BC

Water acts on BC, and the equivalent depth of water for 3 m of oil is:

\( h_{\text{equiv}} = \frac{\gamma_{\text{oil}} \cdot h_{\text{oil}}}{\gamma} = \frac{7848 \cdot 3}{9810} = 2.4 \, \text{m} \)

The location of CG for the imaginary water surface (IWS) is:

\( y_{2, \text{CG}} = 2.4 + \frac{1.8}{2} = 3.3 \, \text{m} \)

The hydrostatic force on section BC is:

\( F_{BC} = \gamma \cdot A_2 \cdot y_{2, \text{CG}} \)

\( F_{BC} = 9810 \cdot 2.16 \cdot 3.3 = 69925 \, \text{N} \)

The point of application of \( F_{BC} \) from A is:

\( h = y_{2, \text{CG}} + \frac{I_G}{A_2 \cdot y_{2, \text{CG}}} \)

\( h = 3.3 + \frac{\frac{1}{12} \cdot 1.2 \cdot 1.8^3}{2.16 \cdot 3.3} = 3.38 \, \text{m} \)

Adding the depth from oil:

\( h = 3.38 + 0.6 = 3.98 \, \text{m from A} \)

6. Total Force on Side ABC

The total force is:

\( F = F_{AB} + F_{BC} = 42379 + 69925 = 112304 \, \text{N} = 112.304 \, \text{kN} \)

Taking moments about A:

\( 112304 \cdot y = 42379 \cdot 2 + 69925 \cdot 3.98 \)

\( y = 3.23 \, \text{m} \)

The resultant force acts at 3.23 m below A.

Results:

Total Force on Side ABC: \( F = 112.304 \, \text{kN} \)
Point of Application: \( y = 3.23 \, \text{m below A} \)

Explanation

Oil and Water: The specific weights and depths of oil and water are used to compute the hydrostatic forces.
CG Location: The CG locations are determined based on the dimensions and the depth of the sections.
Force Calculation: Hydrostatic forces are computed separately for oil and water sections, and their points of application are calculated.
Total Force: The total force is the sum of forces on sections AB and BC, with moments taken about point A to find the line of action.

Physical Meaning

This problem illustrates the calculation of hydrostatic forces on a submerged surface composed of two different fluids. The principles are critical in designing and analyzing tanks and fluid systems subjected to multiple fluid layers.