Question

acceleration of blocks connected by a frictionless pulley on an inclined plane angle greater than 20 degrees

Answer 1

To analyze the acceleration of blocks connected by a frictionless pulley on an inclined plane with an angle greater than 20 degrees, we can follow these steps:

### System Description

Assume we have two blocks:
- Block \( m_1 \) is on an inclined plane with angle \( \theta \) (greater than 20 degrees).
- Block \( m_2 \) is hanging vertically and connected to \( m_1 \) via a pulley.

### Forces Acting on Each Block

1. **For Block on the Incline (\( m_1 \))**:
- The gravitational force acting down the incline: \( F_{\text{gravity, incline}} = m_1 g \sin(\theta) \)
- The tension force in the string: \( T \)

Using Newton's second law, the equation of motion for block \( m_1 \) is:
\[
m_1 a = T - m_1 g \sin(\theta) \tag{1}
\]

2. **For Hanging Block (\( m_2 \))**:
- The gravitational force acting down: \( F_{\text{gravity, hanging}} = m_2 g \)
- The tension force in the string: \( T \)

The equation of motion for block \( m_2 \) is:
\[
m_2 a = m_2 g - T \tag{2}
\]

### Combine the Equations

We have two equations (1) and (2). We can solve for \( T \) in both equations:

From equation (1):
\[
T = m_1 a + m_1 g \sin(\theta) \tag{3}
\]

From equation (2):
\[
T = m_2 g - m_2 a \tag{4}
\]

### Set Equations Equal

Setting equations (3) and (4) equal to each other gives:
\[
m_1 a + m_1 g \sin(\theta) = m_2 g - m_2 a
\]

Now, combining like terms:
\[
m_1 a + m_2 a = m_2 g - m_1 g \sin(\theta)
\]
\[
(m_1 + m_2) a = m_2 g - m_1 g \sin(\theta)
\]

### Solving for Acceleration

Now, we can solve for acceleration \( a \):
\[
a = \frac{m_2 g - m_1 g \sin(\theta)}{m_1 + m_2}
\]

### Conclusion

Thus, the acceleration of the blocks connected by the frictionless pulley with respect to the setup on the inclined plane is given by the formula:
\[
a = \frac{m_2 g - m_1 g \sin(\theta)}{m_1 + m_2}
\]

### Notes

- Ensure the angle \( \theta \) is in radians if you're using a calculator that requires it.
- The acceleration will be positive if \( m_2 \) is large enough to accelerate the system downward, overcoming the component of gravity acting on \( m_1 \). If \( m_1 g \sin(\theta) > m_2 g \), the blocks will not accelerate in the direction assumed.
- Adjustments for any additional forces (like friction) would change the analysis.