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To solve this problem, we need to analyze the balance of forces across the entire hanger structure. The weights on each side must equalize to maintain the overall balance, starting from the individual parts and working up.

### Analyzing the Hanger from Left to Right

1. **Left Side of the Hanger (13 kg)**
   - The left side has a total labeled weight of 13 kg, divided across three main branches of black boxes.
   - Let \( x \) be the weight of each black box in this section.
   - We’ll calculate the balance of weights across the connections by setting up equations based on equal torques.

2. **Right Side of the Hanger (5 kg) and Watermelon**
   - On the right side, we have a total labeled weight of 5 kg plus the watermelon with an unknown weight \( W \).
   - The goal is to find \( W \) so that both sides of the structure stay balanced.

I'll calculate each part systematically and give a step-by-step explanation of how to deduce the unknowns here. Let's proceed with these calculations.

Give me a moment to work through this problem.### Step-by-Step Solution:

#### Step 1: Set up Equations for Each Part of the Structure
The left and right sides of the hanger are balanced, which means:

\[
\text{Left Side Total Weight} = \text{Right Side Total Weight}
\]

Let’s denote:
- \( x \) as the weight of each black box.
- \( W \) as the weight of the watermelon.

#### Step 2: Breakdown of the Left Side (Total Weight = 13 kg)

1. The 13 kg label represents the total weight exerted on the left side.
2. This 13 kg weight is split across three sections of black boxes in a balanced configuration.

Based on the diagram, analyze each black box group to determine how they contribute to the total weight on the left side.

#### Step 3: Breakdown of the Right Side (Total Weight = 5 kg + Watermelon Weight \( W \))

On the right side:
- A 5 kg labeled weight and an additional unknown weight, \( W \) (the watermelon).
  
#### Step 4: Solve the Balance Equation

Since the structure is balanced overall, we equate both sides:

\[
13 \text{ kg (Left Side)} = 5 \text{ kg} + W \text{ (Right Side)}
\]

This simplifies to:

\[
W = 13 - 5 = 8 \text{ kg}
\]

### Final Answer:
The weight of the watermelon \( W \) is **8 kg**.

Would you like a more detailed breakdown of each black box's individual contributions, or do you have any questions about this solution?

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### Related Questions for Further Practice:

1. How would the balance change if the weight of each black box on the left side was different?
2. If the watermelon was placed at a different point, how would that affect the balance?
3. Can we solve for the weight of each black box on the left side independently?
4. What if the right side had additional weights? How would you adjust the solution?
5. How does the length of each hanger bar affect the overall balance in such problems?

### Tip:
In balance problems, always start by breaking down each side individually and then set up a balance equation based on equal total weights or torques.

The hanger shown in the diagram is balanced. The weight of each white box is marked. The weight of each black box is unknown. The black boxes may or may not weigh the same. Find the weight of the watermelon. Assume that all of the parts of the hanger itself (both the horizontal bars and vertical strings) weigh nothing.

Explore how to determine the weight of a watermelon in a balanced hanger system with unknown weights. This problem involves algebraic concepts, equilibrium, and solving linear equations, suitable for students in Grades 6-8.

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