Consider the pulley system is used to keep blocks W_1 and W_2 in equilibrium, as shown in Figure P1. Block W_1 has a weight of {21} kN. The support at point C experiences a load of {85} kN. Determine the tension force at point S. Provide your answer in kN to at least 4 decimal places. [1 mark]

To solve this problem, we need to analyze the pulley system's equilibrium to determine the tension force at point S. 

### Given:
- Weight of block \( W_1 = 21 \) kN.
- Load on support at point \( C = 85 \) kN.
- The system is in equilibrium.

### Analyzing the Pulley System:

1. **Understanding the Load Distribution:**
   - The load at point \( C \) includes the tension in all the ropes connected to it.
   - In an ideal pulley system, the tension is the same throughout a single continuous rope.
   
2. **Number of Rope Segments:**
   - From the figure, there are multiple rope segments supporting the blocks through the pulleys. Typically, the tension in the ropes that directly connect to a weight will have a multiplying effect depending on how many segments support the weight.

3. **Relating Weights and Tensions:**
   - \( W_1 \) is directly supported by multiple segments of the rope.
   - Let’s denote the tension force in each rope segment as \( T \).

4. **Equilibrium at Point C:**
   - The total load at point \( C \) is the sum of tensions due to block \( W_1 \) and any additional segments contributing to the load.

5. **Load Distribution Calculation:**
   - Assume there are \( n \) segments of rope supporting \( W_1 \). Then the tension \( T \) in each segment is given by:
   \[
   nT = W_1
   \]
   - The force at point \( C \) is contributed by each segment of rope. For \( n \) segments, we have:
   \[
   \text{Load at } C = nT
   \]
   - Given the load at point \( C \) is 85 kN:
   \[
   85 = nT
   \]

6. **Find \( T \) and \( n \):**
   - We already have:
   \[
   W_1 = 21 \text{ kN} = nT
   \]
   \[
   nT = 85 \text{ kN}
   \]
   - Dividing the two equations to eliminate \( n \):
   \[
   \frac{21}{85} = \frac{nT}{nT} = \frac{nT}{nT} = \frac{T}{T} \implies n = \frac{85}{21}
   \]

7. **Calculating Tension \( T \):**
   \[
   T = \frac{W_1}{n} = \frac{21}{\frac{85}{21}}
   \]
   - Simplifying the above expression for \( T \):
   \[
   T = \frac{21 \times 21}{85} = \frac{441}{85}
   \]

8. **Final Tension Force Calculation:**
   \[
   T = 5.1882 \text{ kN}
   \]

### Answer:
The tension force at point \( S \) is **5.1882 kN**. 

Would you like more details or have any questions?

Here are five related questions to explore this topic further:

1. How would the tension change if the weight of \( W_1 \) was different?
2. What would be the effect on the tension if more pulleys were added to the system?
3. How can we calculate the force distribution in a more complex pulley system?
4. What happens to the equilibrium if one of the ropes is cut or breaks?
5. How do friction and real-world factors affect the ideal calculations in a pulley system?

**Tip:** When analyzing pulley systems, carefully count the number of rope segments supporting each weight, as this directly influences the tension and load calculations.

Learn how to calculate the tension force at point S in a pulley system with blocks W_1 and W_2 in equilibrium. This problem involves analyzing forces in equilibrium and calculating tensions in a mechanical system. Suitable for university-level physics or engineering students.

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