Let’s analyze and solve the problem described in the uploaded image.

Problem Breakdown:

1. Given:

   • A ladder rests against a smooth vertical wall (no friction at the top).
   • The lower end of the ladder rests on a rough horizontal plane with a coefficient of static friction $\mu_s = \frac{1}{4}$.
   • The ladder is about to slide.
   • We need to calculate the angle of inclination $\theta$ of the ladder with the vertical.

2. Force Analysis:

   • At the bottom (horizontal surface):
     • There is a normal force $N$ and a friction force $f$, where $f \leq \mu_s N$.
   • At the top (smooth wall):
     • The wall exerts only a horizontal reaction force $R$.
   • The weight of the ladder $W$ acts vertically down from its center of gravity.

3. Conditions:

   • Since the ladder is on the verge of sliding, the frictional force $f$ is at its maximum:
     $f = \mu_s N = \frac{1}{4} N$
   • The ladder is in static equilibrium, so:
     • Sum of horizontal forces = 0
     • Sum of vertical forces = 0
     • Sum of moments about any point = 0

Solution:

Equilibrium Equations:

1. Horizontal Forces: $R = f = \frac{1}{4} N$

2. Vertical Forces: $N = W$

3. Moments about the Bottom Point (to eliminate $N$ and $f$): Taking moments about point $A$, we get: $R \cdot L \cdot \sin\theta = W \cdot \frac{L}{2} \cdot \cos\theta$ Simplify $L$ (the length of the ladder cancels out): $R \sin\theta = \frac{W}{2} \cos\theta$

   Substituting $R = \frac{1}{4} W$: $\frac{1}{4} W \sin\theta = \frac{W}{2} \cos\theta$

   Cancel $W$ from both sides: $\frac{1}{4} \sin\theta = \frac{1}{2} \cos\theta$

4. Simplify for $\tan\theta$: Divide through by $\cos\theta$: $\frac{\sin\theta}{\cos\theta} = 2$

   So: $\tan\theta = 2$

Final Answer:

The angle of inclination $\theta$ is: $\theta = \tan^{-1}(2)$

Thus, the correct option is: (a) $\tan^{-1}(2)$

Let me know if you need further clarification!

Related Questions:

1. What is the physical significance of the coefficient of friction in this problem?
2. How would the angle change if the coefficient of friction were higher?
3. What forces would act on the ladder if the wall were rough as well?
4. How do we calculate the reaction forces at the base and wall in general for such problems?
5. Can the ladder remain in equilibrium if the friction coefficient is less than $\frac{1}{4}$? Why or why not?

Tip:

Always write down all equilibrium equations clearly for such problems to avoid missing any force or moment.