A rectangular gate, 3m wide and 2m deep and hinged at its upper horizontal surface sits in the
wall of a tank making 30o
to the vertical as shown. If the gate hinge can just support 300kNm,
determine the depth of water above the hinge for the gate to just open. What is the maximum
pressure on the gate? The correct answers are (3.94m, 55.7kPa). I just need help reaching those answers

This problem involves a rectangular gate submerged in water, and the objective is to determine the depth of water above the hinge required for the gate to open and the maximum pressure on the gate. We'll use principles from hydrostatics, including force due to water pressure and the moment generated by this force.

Let's work through this step by step:

### 1. **General Setup**:

- Gate dimensions: \( 3 \, \text{m} \) wide and \( 2 \, \text{m} \) deep.
- The gate is hinged at the top and is inclined at an angle of \( 30^\circ \) to the vertical.
- The gate hinge can support a moment of \( 300 \, \text{kNm} \).
  
Let \( h \) be the depth of water above the hinge.

### 2. **Hydrostatic Force on the Gate**:

The pressure at any point in a fluid increases linearly with depth, so the pressure at depth \( h \) is given by:
\[
p = \rho g h
\]
where:
- \( \rho \) is the density of water (\( 1000 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the depth of the water at a given point.

The total hydrostatic force \( F \) on the gate is the result of integrating the pressure over the area of the gate. The depth to the centroid of the gate is:
\[
\bar{h} = h + \frac{2}{2} = h + 1 \, \text{m}
\]
The force due to the water pressure on the gate is:
\[
F = \rho g \bar{h} A
\]
where \( A \) is the area of the gate. The area is:
\[
A = 3 \, \text{m} \times 2 \, \text{m} = 6 \, \text{m}^2
\]

Thus:
\[
F = \rho g \bar{h} A = 1000 \times 9.81 \times (h + 1) \times 6
\]

### 3. **Moment Calculation**:

The moment generated by the force about the hinge is the hydrostatic force multiplied by the distance from the hinge to the centroid of the pressure distribution. The location of the center of pressure is slightly below the centroid at a distance:
\[
\frac{I}{\bar{h}A} = \frac{d^2 \sin(30^\circ)}{12\bar{h}} 
\]

Now

A rectangular gate, 3m wide and 2m deep and hinged at its upper horizontal surface sits in the wall of a tank making 30 degrees to the vertical as shown. If the gate hinge can just support 300kNm, determine the depth of water above the hinge for the gate to just open. What is the maximum pressure on the gate? The correct answers are (3.94m, 55.7kPa). I just need help reaching those answers.

Solve a hydrostatics problem involving a rectangular gate, 3m wide and 2m deep, inclined at 30 degrees to the vertical. Learn how to calculate the depth of water for the gate to just open and find the maximum pressure. This solution applies hydrostatics and moment principles with detailed formulas for water pressure and force.

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