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To solve this problem, let’s analyze the forces and use equilibrium conditions.

### Problem Breakdown
1. We have a uniform rod \( AB \) of weight \( W \), inclined at \( 45^\circ \) to the horizontal.
2. The rod rests with one end \( A \) against a smooth vertical wall (no friction at point \( A \)), and the other end \( B \) on a rough horizontal ground with a coefficient of static friction \( \mu = \frac{3}{4} \).
3. A horizontal force \( F \) is applied at point \( B \) to make the rod move towards the wall.

### Given Data
- Weight of the rod \( = W \).
- Coefficient of static friction \( \mu = \frac{3}{4} \).
- Angle \( \theta = 45^\circ \).

### Step 1: Identify Forces Acting on the Rod
1. **Weight \( W \)** acts vertically downward at the center of gravity of the rod (midpoint).
2. **Normal Force \( N \)** acts upward at point \( B \) from the ground.
3. **Frictional Force \( f \)** at point \( B \), which is \( f = \mu N = \frac{3}{4} N \), acting horizontally.
4. **Horizontal Force \( F \)** applied at point \( B \).
5. **Reaction Force \( R \)** at point \( A \), which is horizontal because the wall is smooth.

### Step 2: Set Up Equilibrium Equations
Since the rod is in static equilibrium, we have the following conditions:

1. **Sum of Vertical Forces = 0:**
   \[
   N = W
   \]

2. **Sum of Horizontal Forces = 0:**
   \[
   F = R + f
   \]
   Substituting \( f = \frac{3}{4}N \):
   \[
   F = R + \frac{3}{4}W
   \]

3. **Taking Moments about Point \( B \) to eliminate \( R \):**
   - Length of the rod \( AB \) = \( L \).
   - The weight \( W \) acts at the midpoint, so its perpendicular distance from \( B \) is \( \frac{L}{2} \sin 45^\circ = \frac{L}{2} \cdot \frac{\sqrt{2}}{2} = \frac{L \sqrt{2}}{4} \).
   - The horizontal reaction \( R \) at \( A \) has a perpendicular distance \( L \cos 45^\circ = L \cdot \frac{\sqrt{2}}{2} = \frac{L \sqrt{2}}{2} \).

   Now, summing moments about point \( B \):
   \[
   R \cdot \frac{L \sqrt{2}}{2} = W \cdot \frac{L \sqrt{2}}{4}
   \]
   Simplifying this:
   \[
   R = \frac{W}{2}
   \]

### Step 3: Substitute \( R \) in the Horizontal Force Equation
Using \( F = R + \frac{3}{4}W \):
\[
F = \frac{W}{2} + \frac{3}{4}W = \frac{5}{4}W
\]

### Answer
The required horizontal force \( F \) is:
\[
\boxed{\frac{5}{4}W}
\]

So, the correct answer is **(b) \( \frac{5}{4} W \)**.

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Would you like further clarification or have any questions? Here are some related questions for further exploration:

1. How would the answer change if the angle of inclination were different?
2. What if the coefficient of friction at point \( B \) were \( 1 \)?
3. How would the horizontal force change if the wall at \( A \) were not smooth?
4. Can you determine the normal reaction if the angle were different?
5. What if the force was applied at a point other than \( B \)?

**Tip:** In equilibrium problems, taking moments about a point can often simplify calculations by eliminating unknowns directly acting through that point.

AB is a uniform rod of weight (W) rests with its ends A against a smooth vertical wall and its end B on a rough horizontal ground, the coefficient of static friction between them is 3/4. If a horizontal force acted on the rod at the point B to make it about to move towards the wall when the rod inclined to the horizontal by an angle of measure 45°, then the magnitude of the horizontal force = __________.

Find the horizontal force needed to keep a uniform rod in equilibrium on an inclined plane at a 45° angle with a friction coefficient of 3/4. This solution uses equilibrium of forces and torque principles, covering statics and friction concepts suitable for advanced high school students.

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