A hanging 3.16-kg block begins to accelerate a 3.85-kg block on a horizontal surface. But a rope connected to
the wall at 30.0 degrees (to the horizontal) keeps the block at equilibrium. What is the minimum tension in the
angled rope to keep the block from accelerating once it has stopped? The coefficient of static friction between
the surface and the 3.85-kg block is 0.239

To find the minimum tension in the angled rope that keeps the 3.85-kg block at equilibrium, we can follow these steps:

1. **Calculate the gravitational force on the hanging block (m₁ = 3.16 kg)**:
   \[
   F_{g1} = m_1 \cdot g = 3.16 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 31.00 \, \text{N}
   \]

2. **Calculate the normal force acting on the 3.85-kg block (m₂)**:
   - The weight of the block is:
   \[
   F_{g2} = m_2 \cdot g = 3.85 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 37.84 \, \text{N}
   \]

3. **Determine the maximum static friction force (F_friction)**:
   - The static friction force is given by:
   \[
   F_{\text{friction}} = \mu_s \cdot F_{\text{normal}} = 0.239 \cdot F_{g2} = 0.239 \cdot 37.84 \, \text{N} = 9.04 \, \text{N}
   \]

4. **Resolve the tension (T) in the rope into its components**:
   - The horizontal component of tension:
   \[
   T_x = T \cdot \cos(30^\circ)
   \]
   - The vertical component of tension:
   \[
   T_y = T \cdot \sin(30^\circ)
   \]

5. **Set up the equilibrium conditions**:
   - **Vertical forces**: The tension's vertical component must balance the weight of the hanging block:
   \[
   T_y = F_{g1} = 31.00 \, \text{N} \quad \Rightarrow \quad T \cdot \sin(30^\circ) = 31.00 \, \text{N}
   \]
   \[
   T \cdot 0.5 = 31.00 \quad \Rightarrow \quad T = \frac{31.00}{0.5} = 62.00 \, \text{N}
   \]

6. **Horizontal forces**:
   - The horizontal component of tension must equal the maximum static friction force:
   \[
   T_x = F_{\text{friction}} \quad \Rightarrow \quad T \cdot \cos(30^\circ) = 9.04 \, \text{N}
   \]
   \[
   T \cdot \left(\frac{\sqrt{3}}{2}\right) = 9.04 \quad \Rightarrow \quad T = \frac{9.04}{\frac{\sqrt{3}}{2}} \approx 10.43 \, \text{N}
   \]

7. **Finding the final tension**:
   To maintain equilibrium, we need to ensure that the tension satisfies both conditions. The greater tension required is from balancing the weight of the hanging block.

Thus, the **minimum tension** in the angled rope to keep the block from accelerating once it has stopped is approximately **62.00 N**.

### Summary
- The minimum tension required in the angled rope is **62.00 N**.

Would you like any further details or explanations? Here are some related questions you might consider:

1. How does changing the angle of the rope affect the tension needed?
2. What would happen if the coefficient of static friction increased?
3. How would the system behave if the block started to move?
4. What are the effects of using a pulley in this system?
5. How would the scenario change if the mass of the hanging block were doubled? 

**Tip**: Always check the equilibrium conditions in both the vertical and horizontal directions when analyzing forces in a system.

A hanging 3.16-kg block begins to accelerate a 3.85-kg block on a horizontal surface. But a rope connected to the wall at 30.0 degrees (to the horizontal) keeps the block at equilibrium. What is the minimum tension in the angled rope to keep the block from accelerating once it has stopped? The coefficient of static friction between the surface and the 3.85-kg block is 0.239.

Discover how to calculate the minimum tension in an angled rope required to maintain equilibrium for a 3.85-kg block under the influence of a 3.16-kg hanging block. This solution involves static friction and force equilibrium principles, ideal for high school physics students.

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