Question
What is the force required to hold a 2.5-meter-long, 0.5-meter-wide rectangular gate submerged vertically in water 1.8 meters above the bottom of the gate?
Answers
GPT-4o mini
To find the force required to hold a submerged rectangular gate in place, we can calculate the hydrostatic force acting on the gate due to the water pressure.
### Data Given:
- Width of gate, \( w = 0.5 \, \text{m} \)
- Height of gate, \( h = 2.5 \, \text{m} \)
- Depth of water above the bottom of the gate, \( d = 1.8 \, \text{m} \)
### Step 1: Determine the Position of the Gate
The top of the gate is \( d - h = 1.8 \, \text{m} - 2.5 \, \text{m} = -0.7 \, \text{m} \). This indicates that the top of the gate is actually above the water level, and the bottom of the gate is 1.8 m below the water surface.
### Step 2: Calculate the Depths
- Depth at the top of the gate, \( d_{top} = 1.8 \, \text{m} - 2.5 \, \text{m} = -0.7 \, \text{m} \) (as stated, it is above water)
- Depth at the bottom of the gate, \( d_{bottom} = 1.8 \, \text{m} \)
### Step 3: Find the Average Depth
Since the gate extends from \( -0.7 \, \text{m} \) to \( 1.8 \, \text{m} \):
\[
\text{Average depth} = \frac{d_{top} + d_{bottom}}{2} = \frac{(-0.7) + 1.8}{2} = \frac{1.1}{2} = 0.55 \, \text{m}
\]
### Step 4: Calculate Hydrostatic Pressure
The hydrostatic pressure \( P \) is given by:
\[
P = \rho g h
\]
Where:
- \( \rho \) (density of water) \( \approx 1000 \, \text{kg/m}^3 \)
- \( g \) (acceleration due to gravity) \( \approx 9.81 \, \text{m/s}^2 \)
- \( h \) is the average depth of the gate \( = 0.55 \, \text{m} \)
Calculating the pressure:
\[
P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.55 \, \text{m} \approx 5395.5 \, \text{Pa}
\]
### Step 5: Calculate the Area of the Gate
Area \( A \) of the gate is given by:
\[
A = \text{Width} \times \text{Height} = 0.5 \, \text{m} \times 2.5 \, \text{m} = 1.25 \, \text{m}^2
\]
### Step 6: Calculate the Total Force
The total hydrostatic force \( F \) acting on the gate can be expressed as:
\[
F = P \times A
\]
Substituting in the values:
\[
F = 5395.5 \, \text{Pa} \times 1.25 \, \text{m}^2 = 6744.375 \, \text{N}
\]
### Conclusion
The force required to hold the 2.5-meter-long, 0.5-meter-wide rectangular gate submerged vertically in water held at a depth of 1.8 meters above the bottom (with the calculated average pressure) is approximately:
\[
\boxed{6744.38 \, \text{N}}
\]
### Data Given:
- Width of gate, \( w = 0.5 \, \text{m} \)
- Height of gate, \( h = 2.5 \, \text{m} \)
- Depth of water above the bottom of the gate, \( d = 1.8 \, \text{m} \)
### Step 1: Determine the Position of the Gate
The top of the gate is \( d - h = 1.8 \, \text{m} - 2.5 \, \text{m} = -0.7 \, \text{m} \). This indicates that the top of the gate is actually above the water level, and the bottom of the gate is 1.8 m below the water surface.
### Step 2: Calculate the Depths
- Depth at the top of the gate, \( d_{top} = 1.8 \, \text{m} - 2.5 \, \text{m} = -0.7 \, \text{m} \) (as stated, it is above water)
- Depth at the bottom of the gate, \( d_{bottom} = 1.8 \, \text{m} \)
### Step 3: Find the Average Depth
Since the gate extends from \( -0.7 \, \text{m} \) to \( 1.8 \, \text{m} \):
\[
\text{Average depth} = \frac{d_{top} + d_{bottom}}{2} = \frac{(-0.7) + 1.8}{2} = \frac{1.1}{2} = 0.55 \, \text{m}
\]
### Step 4: Calculate Hydrostatic Pressure
The hydrostatic pressure \( P \) is given by:
\[
P = \rho g h
\]
Where:
- \( \rho \) (density of water) \( \approx 1000 \, \text{kg/m}^3 \)
- \( g \) (acceleration due to gravity) \( \approx 9.81 \, \text{m/s}^2 \)
- \( h \) is the average depth of the gate \( = 0.55 \, \text{m} \)
Calculating the pressure:
\[
P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.55 \, \text{m} \approx 5395.5 \, \text{Pa}
\]
### Step 5: Calculate the Area of the Gate
Area \( A \) of the gate is given by:
\[
A = \text{Width} \times \text{Height} = 0.5 \, \text{m} \times 2.5 \, \text{m} = 1.25 \, \text{m}^2
\]
### Step 6: Calculate the Total Force
The total hydrostatic force \( F \) acting on the gate can be expressed as:
\[
F = P \times A
\]
Substituting in the values:
\[
F = 5395.5 \, \text{Pa} \times 1.25 \, \text{m}^2 = 6744.375 \, \text{N}
\]
### Conclusion
The force required to hold the 2.5-meter-long, 0.5-meter-wide rectangular gate submerged vertically in water held at a depth of 1.8 meters above the bottom (with the calculated average pressure) is approximately:
\[
\boxed{6744.38 \, \text{N}}
\]