To solve this problem involving a collar sliding on a smooth vertical bar with given masses, we need to analyze the forces acting on both masses \(m_A\) and \(m_B\). Let's assume there's a string or a rope connecting mass \(m_A\) and mass \(m_B\) over a pulley system.
### Given:
- Mass of collar \(A\) (\(m_A\)) = 20 kg
- Mass of object \(B\) (\(m_B\)) = 10 kg
### Objective:
- Determine the normal force on the collar \(A\).
### Assumptions:
- The bar is smooth, implying no friction between the collar and the bar.
- The system is in equilibrium or we are considering the forces right before any acceleration if it is not in equilibrium.
### Analysis:
#### Forces on Mass \(A\):
- Weight of \(A\) (\(W_A\)) = \(m_A \cdot g\) = 20 kg \(\cdot\) 9.8 m/s² = 196 N (downward)
- Tension in the string (\(T\)), acting upwards
#### Forces on Mass \(B\):
- Weight of \(B\) (\(W_B\)) = \(m_B \cdot g\) = 10 kg \(\cdot\) 9.8 m/s² = 98 N (downward)
- Tension in the string (\(T\)), acting upwards
#### Equations of Motion:
Since the system is connected and assuming it is in equilibrium (no acceleration), the tension \(T\) must be the same throughout the string.
For mass \(A\):
\[ T = m_A \cdot g - N \]
Where \(N\) is the normal force exerted by the bar on the collar \(A\).
For mass \(B\):
\[ T = m_B \cdot g \]
Since the tension \(T\) is the same in both equations, we can set them equal:
\[ m_B \cdot g = m_A \cdot g - N \]
Solving for \(N\):
\[ N = m_A \cdot g - m_B \cdot g \]
Substitute the given values:
\[ N = (20 \cdot 9.8) - (10 \cdot 9.8) \]
\[ N = 196 - 98 \]
\[ N = 98 \, \text{N} \]
### Conclusion:
The normal force \(N\) exerted by the smooth vertical bar on the collar \(A\) is 98 N.
