7.137 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
7.137 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Problem 7.137
A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
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### Explanation
This problem involves analyzing a catenary or suspended cable scenario, where the cable is subject to a constant weight per unit length and two fixed endpoints. The objective is to calculate the smallest sag allowed given the tension constraint.
#### Problem Breakdown
- **Cable Weight**: 2 pounds per foot.
- **Distance Between Points**: 160 feet.
- **Maximum Tension**: 400 pounds.
To solve the problem, use principles from statics and the equation for the catenary curve, which relates the sag to the tension and the weight per unit length.
### Solution Approach
1. **Identify Variables**:
- \( w = 2 \, \text{lb/ft} \) (weight per unit length)
- \( L = 160 \, \text{ft} \) (span length)
- \( T_{\text{max}} = 400 \, \text{lb} \) (maximum tension)
2. **Equations**:
- Use the catenary formula and statics principles to relate the sag, tension, and weight, typically given by:
\[
T = \sqrt{ \left( \frac{wL^2}{8d} \right)^2 + (wL)^2 }
\]
- Where \( d \) is the sag.
3. **Rearrange and Solve**:
- Solve for the sag \( d \) ensuring \( T \leq T_{\text{max}} \).
4. **Check Solution**:
- Confirm all values satisfy the physical constraints and principles.
This problem requires a comprehension of engineering mechanics, particularly the static equilibrium of structures. Completing the solution involves algebraic manipulation and possibly numerical methods if a direct solution isn't straightforward.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa86b2d75-dc89-44ee-816d-e64b811f53cb%2F887f34b6-fb85-49b6-b39b-18468ae8530d%2Frjrwrsh_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 7.137
A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
---
### Explanation
This problem involves analyzing a catenary or suspended cable scenario, where the cable is subject to a constant weight per unit length and two fixed endpoints. The objective is to calculate the smallest sag allowed given the tension constraint.
#### Problem Breakdown
- **Cable Weight**: 2 pounds per foot.
- **Distance Between Points**: 160 feet.
- **Maximum Tension**: 400 pounds.
To solve the problem, use principles from statics and the equation for the catenary curve, which relates the sag to the tension and the weight per unit length.
### Solution Approach
1. **Identify Variables**:
- \( w = 2 \, \text{lb/ft} \) (weight per unit length)
- \( L = 160 \, \text{ft} \) (span length)
- \( T_{\text{max}} = 400 \, \text{lb} \) (maximum tension)
2. **Equations**:
- Use the catenary formula and statics principles to relate the sag, tension, and weight, typically given by:
\[
T = \sqrt{ \left( \frac{wL^2}{8d} \right)^2 + (wL)^2 }
\]
- Where \( d \) is the sag.
3. **Rearrange and Solve**:
- Solve for the sag \( d \) ensuring \( T \leq T_{\text{max}} \).
4. **Check Solution**:
- Confirm all values satisfy the physical constraints and principles.
This problem requires a comprehension of engineering mechanics, particularly the static equilibrium of structures. Completing the solution involves algebraic manipulation and possibly numerical methods if a direct solution isn't straightforward.
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