A rod can turn freely about one of its ends which is fixed. At the other end a horizontal force equal to half the weight of the rod is acting. In the position of equilibrium, the rod is inclined to the vertical at an angle

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None

Answer

To solve the problem of the rod in equilibrium, we will analyze the forces and torques acting on the rod.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Rod:**
- Let the weight of the rod be \( W \).
- A horizontal force \( F \) is acting at the free end of the rod, where \( F = \frac{1}{2} W \).

2. **Determine the Position of the Center of Mass:**
- The center of mass of the rod is located at its midpoint, which is at a distance of \( \frac{L}{2} \) from the pivot point (the fixed end).

3. **Set Up the Torque Equation:**
- The torque due to the weight of the rod (acting downwards at the center of mass) is given by:
\[ \tau_{weight} = W \cdot \left(\frac{L}{2} \cdot \cos(\theta)\right) \]
- The torque due to the horizontal force \( F \) is given by:
\[ \tau_{force} = F \cdot (L \cdot \sin(\theta)) \]
- Since \( F = \frac{1}{2} W \), we can substitute this into the torque equation:
\[ \tau_{force} = \left(\frac{1}{2} W\right) \cdot (L \cdot \sin(\theta)) \]

4. **Equate the Torques for Equilibrium:**
- For the rod to be in equilibrium, the clockwise torque must equal the anticlockwise torque:
\[ W \cdot \left(\frac{L}{2} \cdot \cos(\theta)\right) = \left(\frac{1}{2} W\right) \cdot (L \cdot \sin(\theta)) \]

5. **Simplify the Equation:**
- Cancel \( W \) and \( L \) from both sides (assuming \( W \) and \( L \) are non-zero):
\[ \frac{1}{2} \cos(\theta) = \frac{1}{2} \sin(\theta) \]
- This simplifies to:
\[ \cos(\theta) = \sin(\theta) \]

6. **Solve for the Angle \( \theta \):**
- The equation \( \cos(\theta) = \sin(\theta) \) implies that:
\[ \tan(\theta) = 1 \]
- Therefore, the angle \( \theta \) is:
\[ \theta = 45^\circ \]

### Final Answer:
The rod is inclined to the vertical at an angle of \( 45^\circ \).