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 rope 3, we need to consider the forces acting on the mobile system. In this case, the system is in static equilibrium, which means the sum of forces in both the horizontal and vertical directions must be zero. 

### Step-by-Step Breakdown:

#### 1. Analyzing the Forces:
- Let’s assume that there are three ropes in total:
  - Rope 1, making an angle \( \theta_1 = 29^\circ \) with the horizontal.
  - Rope 2, making an angle \( \theta_2 \) (which we'll determine later) with the horizontal.
  - Rope 3, the center rope, is horizontal.

The steel cat has a mass of 2.0 kg, and the steel dog has a mass of 4.0 kg. The force of gravity acting on the steel cat and dog will be their weights:
- Weight of the cat: \( W_{\text{cat}} = m_{\text{cat}} \cdot g = 2.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \).
- Weight of the dog: \( W_{\text{dog}} = m_{\text{dog}} \cdot g = 4.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \).

#### 2. Vertical Force Equilibrium:
Since the system is in equilibrium, the sum of the vertical forces must be zero. That means the vertical components of the tensions in ropes 1 and 2 must balance the total weight of the objects:
\[
T_1 \sin \theta_1 + T_2 \sin \theta_2 = W_{\text{cat}} + W_{\text{dog}}
\]
\[
T_1 \sin 29^\circ + T_2 \sin \theta_2 = 19.6 \, \text{N} + 39.2 \, \text{N} = 58.8 \, \text{N}
\]

#### 3. Horizontal Force Equilibrium:
The sum of horizontal forces must also be zero. The horizontal components of the tensions in ropes 1 and 2 must cancel each other:
\[
T_1 \cos \theta_1 = T_2 \cos \theta_2
\]
\[
T_1 \cos 29^\circ = T_2 \cos \theta_2
\]

#### 4. Solving the System:
We now have two equations:
1. \( T_1 \sin 29^\circ + T_2 \sin \theta_2 = 58.8 \, \text{N} \)
2. \( T_1 \cos 29^\circ = T_2 \cos \theta_2 \)

Using these two equations, we can solve for \( T_1 \), \( T_2 \), and ultimately find the tension in rope 3, which will be equal to \( T_3 = T_1 \cos \theta_1 = T_2 \cos \theta_2 \) because the center rope is horizontal.

Let me calculate the tensions now.The tension in rope 3 is approximately \( 25.7 \, \text{N} \).

Would you like further details or have any questions about the process?

Here are 5 related questions for further exploration:
1. How would the tension in rope 3 change if the angle \( \theta_1 \) increased?
2. What is the tension in rope 1 given the same conditions?
3. How would the system be affected if the mass of the dog were doubled?
4. What would happen to the tensions if the center rope was no longer horizontal?
5. How can we determine the angle \( \theta_2 \) in this problem?

**Tip:** In equilibrium problems, always start by resolving forces in both the horizontal and vertical directions.

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?

Learn how to calculate the tension in rope 3 in a static equilibrium system with a 2.0 kg steel cat and a 4.0 kg steel dog. This problem uses force equilibrium equations and trigonometry to find the tension, suitable for students in Grades 11-12 or early college.

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