Gate AB in the fig. is 1.25m wide and hinged at A. Gage G reads -12.5KN/m2, while oil (sp gr = 0.75) is in the right tank. What horizontal force must be applied at B for equilibrium of gate AB?

Gate Force Problems

Problem Statement

Gate AB in the figure is 1.25 m wide and hinged at A. Gauge G reads \(-12.5 \, \text{kN/m}^2\), while oil (specific gravity = 0.75) is in the right tank. What horizontal force must be applied at B for the equilibrium of gate AB?

Solution

1. Specific Weight of Oil

The specific weight of oil is:

\( \gamma_{\text{oil}} = 0.75 \times 9810 = 7357.5 \, \text{N/m}^3 \)

2. Calculate Area and Location of CG for the Right Side

The area of the gate’s right side is:

\( A = 1.25 \times 1.8 = 2.25 \, \text{m}^2 \)

The location of the CG for the right side is:

\( \bar{y}_1 = \frac{1.8}{2} = 0.9 \, \text{m} \)

3. Force on AB at the Right Side

The force on AB at the right side is:

\( F_{\text{oil}} = \gamma_{\text{oil}} \cdot A \cdot \bar{y}_1 \)

\( F_{\text{oil}} = 7357.5 \times 2.25 \times 0.9 = 14899 \, \text{N} \)

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

\( \text{Point of application} = \frac{2}{3} \times 1.8 = 1.2 \, \text{m from A} \)

4. Force on the Left Side

For the left side, convert the negative pressure due to air to an equivalent head in water:

\( \text{Equivalent depth of water} = \frac{P}{\gamma} = \frac{-12.5}{9.81} = -1.27 \, \text{m} \)

This negative pressure head is equivalent to having 1.27 m less water above A. The location of the CG from the imaginary water surface (IWS) is:

\( \bar{y}_2 = 2.33 + \frac{1.8}{2} = 3.23 \, \text{m} \)

The force due to water is:

\( F_{\text{water}} = \gamma \cdot A \cdot \bar{y}_2 \)

\( F_{\text{water}} = 9810 \times 2.25 \times 3.23 = 71294 \, \text{N} \)

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

\( \bar{h} = \bar{y}_2 + \frac{I_G}{A \cdot \bar{y}_2} \)

\( \bar{h} = 3.23 + \frac{\frac{1}{12} \times 1.25 \times 1.8^3}{2.25 \times 3.23} = 3.31 \, \text{m} \)

From the IWS, the application point is:

\( 3.31 – 2.33 = 0.98 \, \text{m from A} \)

5. Horizontal Force at B

Taking moments about A:

\( P \cdot 1.8 + 14899 \cdot 1.2 – 71294 \cdot 0.98 = 0 \)

\( P = 28883 \, \text{N} = 28.883 \, \text{kN} \)

Results:

Horizontal Force at B: \( P = 28.883 \, \text{kN} \)

Explanation

Right Side Force: The hydrostatic force due to oil is calculated based on its specific weight and the gate’s dimensions.
Left Side Force: Negative gauge pressure is converted to an equivalent water head, and the corresponding force is calculated.
Equilibrium: Moments about A ensure that the gate remains balanced, allowing the horizontal force at B to be determined.

Physical Meaning

This problem illustrates the interplay of hydrostatic forces on a hinged gate due to fluids of different densities and gauge pressures. It highlights the importance of force equilibrium for structural stability.