4. A trough is filled with a liquid of density 875 kg/m³. The ends of the trough are equilateral triangles with sides 14 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s² for the acceleration due to gravity.) Show all your steps clearly.
4. A trough is filled with a liquid of density 875 kg/m³. The ends of the trough are equilateral triangles with sides 14 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s² for the acceleration due to gravity.) Show all your steps clearly.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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![**Hydrostatics Problem: Hydrostatic Force on a Trough End**
**Problem Statement:**
A trough is filled with a liquid of density 875 kg/m³. The ends of the trough are equilateral triangles with sides 14 m long and a vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s² for the acceleration due to gravity.) Show all your steps clearly.
**Solution Steps:**
1. **Identify the Shape and Dimensions:**
- The end of the trough is an equilateral triangle with side length 14 m.
- The height \( h \) of an equilateral triangle can be found using the formula:
\[
h = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 14 \text{ m}
\]
2. **Calculate the Height of the Triangle:**
\[
h = \frac{\sqrt{3}}{2} \times 14 = 7\sqrt{3} \text{ m}
\]
3. **Hydrostatic Pressure:**
- Hydrostatic pressure at a depth \( y \) is given by:
\[
P(y) = \rho \times g \times y
\]
where \( \rho = 875 \text{ kg/m}^3 \) and \( g = 9.8 \text{ m/s}^2 \).
4. **Hydrostatic Force Calculation:**
- Force is the integral of pressure over the area:
\[
F = \int \rho g y \cdot \text{width(y)} \cdot dy
\]
- At height \( y \) from the bottom, the width of the triangle can be found using similar triangles:
\[
\text{width} = \frac{14}{h} \times y = \frac{14}{7\sqrt{3}} \times y = \frac{2y}{\sqrt{3}}
\]
- Therefore, the force is:
\[
F = \int_0^{7\sqrt{3}} 875 \cdot 9.8 \cdot y \cdot \frac{2y}{\sqrt{3}} \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F916b3e43-3e70-4159-b0c4-23957999f35d%2F501b3527-62fb-42ab-a006-74cebb17ff7e%2F47ky62_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Hydrostatics Problem: Hydrostatic Force on a Trough End**
**Problem Statement:**
A trough is filled with a liquid of density 875 kg/m³. The ends of the trough are equilateral triangles with sides 14 m long and a vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s² for the acceleration due to gravity.) Show all your steps clearly.
**Solution Steps:**
1. **Identify the Shape and Dimensions:**
- The end of the trough is an equilateral triangle with side length 14 m.
- The height \( h \) of an equilateral triangle can be found using the formula:
\[
h = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 14 \text{ m}
\]
2. **Calculate the Height of the Triangle:**
\[
h = \frac{\sqrt{3}}{2} \times 14 = 7\sqrt{3} \text{ m}
\]
3. **Hydrostatic Pressure:**
- Hydrostatic pressure at a depth \( y \) is given by:
\[
P(y) = \rho \times g \times y
\]
where \( \rho = 875 \text{ kg/m}^3 \) and \( g = 9.8 \text{ m/s}^2 \).
4. **Hydrostatic Force Calculation:**
- Force is the integral of pressure over the area:
\[
F = \int \rho g y \cdot \text{width(y)} \cdot dy
\]
- At height \( y \) from the bottom, the width of the triangle can be found using similar triangles:
\[
\text{width} = \frac{14}{h} \times y = \frac{14}{7\sqrt{3}} \times y = \frac{2y}{\sqrt{3}}
\]
- Therefore, the force is:
\[
F = \int_0^{7\sqrt{3}} 875 \cdot 9.8 \cdot y \cdot \frac{2y}{\sqrt{3}} \,
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