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This problem involves determining the **tension in the cable** and the **horizontal force on the horse** when lifting a hay bale. Let's analyze this step by step.

### Given Information:
- **Mass of horse** \( m_A = 700 \, \text{kg} \)
- **Mass of hay bale** \( m_B = 50 \, \text{kg} \)
- **Length of cable** \( r = 8 \, \text{m} \)
- **Angle of cable** \( \theta = 60^\circ \)
- Horse's velocity \( v_A = 2 \, \text{m/s} \) (to the right)
- Horse's acceleration \( a_A = 0.5 \, \text{m/s}^2 \) (to the left)

### Unknowns:
1. **Tension in the cable** \( T \)
2. **Horizontal force between the ground and the horse’s feet** \( F_h \)

### Assumptions:
- Neglect the mass of the pulley.
- The system is frictionless (no pulley friction).
  
### Approach:
1. **For Part (a) - Tension in the cable**: 
   We need to use Newton's second law applied to the hay bale and the geometry of the problem. The tension can be broken into components: 
   \[
   T_x = T \cos(\theta), \quad T_y = T \sin(\theta)
   \]
   Considering the forces acting on the hay bale, we can balance the forces in the vertical direction (gravity acting on the bale and the vertical component of tension).

2. **For Part (b) - Horizontal force on the horse**:
   The horizontal force on the horse can be computed using Newton's second law applied to the horse. This involves finding the force due to the tension in the horizontal direction and combining it with the net force acting on the horse due to its acceleration.

Let me now proceed with solving these parts.

### Part (a) - Tension in the Cable:
The hay bale \( B \) experiences a vertical force due to gravity:
\[
F_g = m_B g = 50 \times 9.81 = 490.5 \, \text{N}
\]
To balance this force, the vertical component of the tension must counteract the weight:
\[
T_y = T \sin(\theta) = 490.5 \, \text{N}
\]
Given \( \theta = 60^\circ \), we can solve for \( T \):
\[
T = \frac{490.5}{\sin(60^\circ)}
\]

### Part (b) - Horizontal Force on the Horse:
The horizontal force \( F_h \) is composed of two parts:
- The force from the horizontal component of the tension \( T_x \), and
- The net force required to accelerate the horse \( F_{net} = m_A a_A \).

Let me calculate these now.### Results:
1. **Tension in the cable** \( T \) is approximately **566.38 N**.
2. **Horizontal force between the ground and the horse's feet** \( F_h \) is approximately **633.19 N**.

Would you like a detailed breakdown of the steps or have any further questions?

Here are some related questions for further practice:
1. How would the tension change if the angle \( \theta \) increased to \( 75^\circ \)?
2. What would be the horizontal force if the horse's acceleration doubled?
3. How would the system behave if the hay bale's mass were halved?
4. How would friction at the pulley affect the tension in the cable?
5. Can you derive the relationship between the horizontal and vertical components of tension for different angles?

**Tip**: Always ensure that you decompose forces into their horizontal and vertical components when working with inclined systems like this.

A 700-kg horse A lifts a 50-kg hay bale B using a pulley system. The problem involves determining the tension in the cable and the horizontal force between the ground and the horse's feet. At a specific instant, the system conditions include: r = 8 m, θ = 60°, the velocity of the horse is 2 m/s to the right, and the acceleration is 0.5 m/s² to the left.

In this problem, a 700-kg horse lifts a 50-kg hay bale using a pulley system. This detailed solution explains how to calculate the tension in the cable and the horizontal force on the horse's feet using Newton's Second Law and trigonometric concepts.

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