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A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in (Figure 1). It is found that θ1 = 29 ∘ when the center rope is adjusted to be perfectly horizontal. What is the tension in the rope 3?

To find the tension in the rope 3 (\(T_3\)) in this system, we can approach this problem by balancing the forces acting on the system. Here's a step-by-step explanation:

### Given:
- Mass of the cat: \( m_1 = 2.0 \, \text{kg} \)
- Mass of the dog: \( m_3 = 4.0 \, \text{kg} \)
- Angle \( \theta_1 = 29^\circ \)
- The center rope (rope 2) is perfectly horizontal, meaning there is no vertical force component in that section.

We'll work through the following concepts:
1. The gravitational force acting on the cat and dog are:
   - \( F_{\text{cat}} = m_1 g = 2.0 \times 9.8 = 19.6 \, \text{N} \)
   - \( F_{\text{dog}} = m_3 g = 4.0 \times 9.8 = 39.2 \, \text{N} \)

Since the system is in equilibrium, the sum of forces in both horizontal and vertical directions must be zero.

### For the vertical forces:
- The tension in rope 1 (\(T_1\)) supports the weight of the cat, and the tension in rope 3 (\(T_3\)) supports the weight of the dog.
- Vertically, we have:
  \[
  T_1 \sin(\theta_1) = F_{\text{cat}} = 19.6 \, \text{N}
  \]
  \[
  T_3 \sin(\theta_3) = F_{\text{dog}} = 39.2 \, \text{N}
  \]

Now, the horizontal force components must balance because the center rope (rope 2) is horizontal:
\[
T_1 \cos(\theta_1) = T_3 \cos(\theta_3)
\]
This gives us the relationship between \(T_1\) and \(T_3\). From the first two equations, we can solve for \(T_1\) and \(T_3\).

### Step 1: Solve for \(T_1\)
From the vertical force balance for the cat:
\[
T_1 = \frac{F_{\text{cat}}}{\sin(\theta_1)} = \frac{19.6}{\sin(29^\circ)}
\]
Let me calculate that.

### Step 2: Solve for \(T_3\)
Once we have \(T_1\), we can use the horizontal force balance to solve for \(T_3\). 
\[
T_3 = \frac{T_1 \cos(\theta_1)}{\cos(\theta_3)}
\]
However, we don't have \( \theta_3 \), so we need to assume or calculate it based on the symmetry or additional information.

Let me calculate \(T_1\) first.The tension in rope 1 (\(T_1\)) is approximately \(40.43 \, \text{N}\).

### Step 2: Solve for \(T_3\)
Now, assuming symmetry (i.e., \( \theta_3 = \theta_1 = 29^\circ \)), we can solve for \(T_3\) using the horizontal force balance:
\[
T_3 = \frac{T_1 \cos(\theta_1)}{\cos(\theta_3)}
\]
Since \( \theta_3 = \theta_1 \), the equation simplifies to \( T_3 = T_1 \).

Thus, \(T_3 = 40.43 \, \text{N}\).

Would you like further details or explanations?

Here are five related questions to expand your understanding:
1. How would the tension change if the angles \( \theta_1 \) and \( \theta_3 \) were different?
2. What would happen to the tension in the ropes if the mass of the cat increased?
3. How can we determine the angle \( \theta_3 \) if it's not symmetric?
4. How would the tensions be affected if the center rope was not horizontal?
5. Can we solve this problem using vector components?

**Tip**: When dealing with systems in equilibrium, always start by balancing both vertical and horizontal forces to ensure all forces are accounted for.

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable. It is found that θ1 = 29° when the center rope is adjusted to be perfectly horizontal. What is the tension in the rope 3?

Calculate the tension in rope 3 for a system with a 2.0 kg steel cat and a 4.0 kg steel dog. Explore equilibrium forces, tension, and trigonometric relationships. Suitable for students in high school physics, Grades 11-12.

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