A semicircular plate with radius 9 m is submerged vertically in water so that the top is 1 m above the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s? for the acceleration due to gravity. Recall that the mass density of water is 1000 kg/m.) N
A semicircular plate with radius 9 m is submerged vertically in water so that the top is 1 m above the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s? for the acceleration due to gravity. Recall that the mass density of water is 1000 kg/m.) N
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![### Hydrostatic Force on a Submerged Semicircular Plate
**Problem Statement:**
A semicircular plate with a radius of 9 meters is submerged vertically in water so that the top is 1 meter above the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s² for the acceleration due to gravity. Recall that the mass density of water is 1000 kg/m³.)
**Integral Expression:**
\[
2 \rho g \int_{0}^{9 - 1} (\sqrt{9^2 - x^2}) x \, dx
\]
where:
- \(\rho = 1000 \) kg/m³ (mass density of water)
- \(g = 9.8 \) m/s² (acceleration due to gravity)
- \(x\) ranges from the top to the bottom of the submerged plate
**Diagram Description:**
The diagram illustrates a semicircular plate fully submerged in water with a portion extending 1 meter above the water surface. The depth is denoted \(x\) and the horizontal span is \( \sqrt{9^2 - x^2} \), representing the distance across the plate at a depth \(x\).
### Analytical Approach:
1. **Hydrostatic Pressure Calculation:**
\[
P = \rho g h
\]
where \(h\) is the depth below the water surface.
2. **Force Calculation:**
The differential force \(dF\) on a thin strip at depth \(x\):
\[
dF = P \cdot dA
\]
where \(dA\) is the area of the strip.
\[
dA = 2 (\sqrt{9^2 - x^2}) \cdot dx
\]
Hence:
\[
dF = \rho g x \cdot 2 (\sqrt{9^2 - x^2}) \cdot dx
\]
3. **Integral Setup:**
\[
F = 2 \rho g \int_{0}^{8} (\sqrt{9^2 - x^2}) x \, dx
\]
**Evaluate the Integral:**
The solution involves integrating the function over the given limits and then multiplying by](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ecf1484-f95f-4119-9858-43c8382b92c0%2Fc5683634-dd33-4981-9c34-0df2d089b7ed%2Fjnefqpy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hydrostatic Force on a Submerged Semicircular Plate
**Problem Statement:**
A semicircular plate with a radius of 9 meters is submerged vertically in water so that the top is 1 meter above the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s² for the acceleration due to gravity. Recall that the mass density of water is 1000 kg/m³.)
**Integral Expression:**
\[
2 \rho g \int_{0}^{9 - 1} (\sqrt{9^2 - x^2}) x \, dx
\]
where:
- \(\rho = 1000 \) kg/m³ (mass density of water)
- \(g = 9.8 \) m/s² (acceleration due to gravity)
- \(x\) ranges from the top to the bottom of the submerged plate
**Diagram Description:**
The diagram illustrates a semicircular plate fully submerged in water with a portion extending 1 meter above the water surface. The depth is denoted \(x\) and the horizontal span is \( \sqrt{9^2 - x^2} \), representing the distance across the plate at a depth \(x\).
### Analytical Approach:
1. **Hydrostatic Pressure Calculation:**
\[
P = \rho g h
\]
where \(h\) is the depth below the water surface.
2. **Force Calculation:**
The differential force \(dF\) on a thin strip at depth \(x\):
\[
dF = P \cdot dA
\]
where \(dA\) is the area of the strip.
\[
dA = 2 (\sqrt{9^2 - x^2}) \cdot dx
\]
Hence:
\[
dF = \rho g x \cdot 2 (\sqrt{9^2 - x^2}) \cdot dx
\]
3. **Integral Setup:**
\[
F = 2 \rho g \int_{0}^{8} (\sqrt{9^2 - x^2}) x \, dx
\]
**Evaluate the Integral:**
The solution involves integrating the function over the given limits and then multiplying by
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