1. Equilibrium Equations

• Horizontal Forces: \( N_{wall} = f_{floor} \)
• Vertical Forces: \( N_{floor} = mg \)
• Torque about the bottom end: \[ N_{wall} (L \sin \theta) = mg (\frac{L}{2} \cos \theta) \] \[ N_{wall} = \frac{mg}{2} \cot \theta \]

2. Effect of Expansion

When the rod is heated, it expands, so length \( L \) increases. Assuming the friction at the floor holds the bottom end stationary (since it is below the limiting value), the expansion causes the top end to slide up the frictionless wall.

This means the angle \( \theta \) increases.

3. Change in Forces

We found that both the wall contact force and floor friction are proportional to \( \cot \theta \):

\[ N_{wall} = f_{floor} = \frac{mg}{2} \cot \theta \]

As \( \theta \) increases (rod becomes steeper), \( \cot \theta \) decreases.

Therefore, both \( N_{wall} \) and \( f_{floor} \) decrease.