A horizontal clothesline is tied between 2 poles, 12 meters apart.
When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters.

What is the magnitude of the tension on the ends of the clothesline?

To solve for the magnitude of the tension at the ends of the clothesline, we can use the principles of static equilibrium and trigonometry.

### Given:
- Distance between the two poles: \( L = 12 \, \text{m} \)
- Sag (vertical displacement at the center): \( h = 4 \, \text{m} \)
- Mass of the object: \( m = 5 \, \text{kg} \)
- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)

### Approach:

1. **Forces acting on the system:**
   - The weight of the object acts vertically downward at the center of the clothesline: \( W = mg \).
   - The tension in the clothesline at the ends can be resolved into horizontal (\( T_x \)) and vertical (\( T_y \)) components.

2. **Symmetry:**
   - Since the object is at the midpoint, the tension forces at both ends will be equal.
   - The horizontal components of the tension forces on both sides are equal and opposite, so they cancel each other out.
   - The vertical components of the tension forces on both sides must sum to balance the weight of the object.

3. **Geometry of the problem:**
   - The horizontal distance from the center to either pole: \( \frac{L}{2} = 6 \, \text{m} \).
   - The vertical distance (sag): \( h = 4 \, \text{m} \).
   - The length of the segment from the center to either pole: \( \sqrt{\left(\frac{L}{2}\right)^2 + h^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \, \text{m} \).

4. **Trigonometric relationship:**
   - The angle \( \theta \) formed by the tension with the horizontal can be found using:
     \[
     \sin \theta = \frac{h}{\text{length of segment}} = \frac{4}{\sqrt{52}}
     \]

5. **Vertical force equilibrium:**
   - The vertical component of the tension must equal half the weight:
     \[
     T_y = T \sin \theta = \frac{mg}{2}
     \]
     Substituting values:
     \[
     T \sin \theta = \frac{5 \times 9.8}{2} = 24.5 \, \text{N}
     \]

6. **Solve for the total tension \( T \):**
   \[
   T = \frac{24.5}{\sin \theta} = \frac{24.5}{\frac{4}{\sqrt{52}}}
   \]
   Simplifying:
   \[
   T = \frac{24.5 \times \sqrt{52}}{4} \approx \frac{24.5 \times 7.2111}{4} \approx \frac{176.67195}{4} \approx 44.17 \, \text{N}
   \]

So, the magnitude of the tension on the ends of the clothesline is approximately **44.17 N**.

Would you like more details on any part of the solution, or do you have further questions?

### Related Questions:
1. How would the tension change if the sag were reduced to 2 meters?
2. What would happen to the tension if the mass were doubled?
3. How does the tension vary with the length of the clothesline if the sag remains the same?
4. If the mass were placed at a point other than the midpoint, how would the tensions in the two segments compare?
5. Can you derive a general formula for the tension in terms of \( L \), \( h \), and \( m \)?

### Tip:
Understanding how to resolve forces into their components is crucial for solving problems in static equilibrium.

Learn how to calculate the tension on the ends of a clothesline when a 5kg mass causes a 4m sag. This problem uses principles of static equilibrium and trigonometry to determine the tension forces. Suitable for advanced high school students and beyond.

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