Static Equilibrium and Slipping Condition of a Leaning Ladder

To determine the correct conditions for static equilibrium when the ladder is on the verge of slipping, we analyze the forces acting on the uniform ladder of mass $m$ and length $L$.

1. Analysis of Forces and Equilibrium Conditions

The forces acting on the ladder are:

• Weight of the ladder, $mg$, acting vertically downwards at its center of mass ($\frac{L}{2}$ from either end).
• Normal reaction force from the floor, $N_2$, acting vertically upwards.
• Friction force from the floor, $f_2$, acting horizontally towards the wall.
• Normal reaction force from the wall, $N_1$, acting horizontally away from the wall.
• Friction force from the wall, $f_1$, acting vertically upwards.

Equilibrium Equations:

1. Horizontal Force Balance ($\sum F_x = 0$): $N_1 = f_2$

2. Vertical Force Balance ($\sum F_y = 0$): $N_2 + f_1 = mg$

3. Torque Balance about the Bottom Contact Point ($\sum \tau_{\text{bottom}} = 0$): $N_1 (L \sin\theta) + f_1 (L \cos\theta) - mg \left(\frac{L}{2} \cos\theta\right) = 0$ Dividing the entire torque equation by $L \cos\theta$: $N_1 \tan\theta + f_1 = \frac{mg}{2} \quad \text{--- (Equation 1)}$

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2. Evaluation of Options

Case 1: When $\mu_1 \neq 0$ and $\mu_2 \neq 0$ (Checking Option C)

When the ladder is on the verge of slipping, the frictional forces reach their maximum static limit: $f_1 = \mu_1 N_1 \quad \text{and} \quad f_2 = \mu_2 N_2$

From the horizontal equilibrium condition: $N_1 = f_2 = \mu_2 N_2$

Substituting $N_1$ into the vertical equilibrium equation: $N_2 + f_1 = mg \implies N_2 + \mu_1 N_1 = mg$ $N_2 + \mu_1 (\mu_2 N_2) = mg$ $N_2 (1 + \mu_1 \mu_2) = mg \implies N_2 = \frac{mg}{1 + \mu_1 \mu_2}$

Thus, Option C is correct.

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Case 2: When $\mu_1 = 0$ and $\mu_2 \neq 0$ (Checking Options A and D)

If the wall is smooth ($\mu_1 = 0$), there is no friction force at the wall, i.e., $f_1 = 0$.

Substituting $f_1 = 0$ into Equation 1: $N_1 \tan\theta + 0 = \frac{mg}{2} \implies N_1 \tan\theta = \frac{mg}{2}$

Also, from vertical equilibrium: $N_2 + 0 = mg \implies N_2 = mg$

Thus:

• Option D correctly gives $N_1 \tan\theta = \frac{mg}{2}$.
• Option A incorrectly states $N_2 \tan\theta = \frac{mg}{2}$.

Therefore, Option D is correct, while Option A is incorrect.

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Case 3: When $\mu_1 \neq 0$ and $\mu_2 = 0$ (Checking Option B)

If the floor is smooth ($\mu_2 = 0$), then $f_2 = 0$, which implies $N_1 = 0$. If $N_1 = 0$, static equilibrium cannot be maintained because $N_1 \tan\theta + f_1 = \frac{mg}{2}$ would require $f_1 = \frac{mg}{2}$, but maximum possible friction at the wall is $f_{1,\max} = \mu_1 N_1 = 0$. Thus, the ladder cannot remain in equilibrium without floor friction.

Therefore, Option B is incorrect.

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Conclusion

The correct options are C and D.