A rod of weight `w` is supported by two parallel knife edges `A` and `B` and is in equilibrium in a horizontal position. The knives are at a distance `d` from each other. The centre of mass of the rod is at a distance `x` from `A`.

Options

A. `(Wd)/(x)`
B. `(W(d-x))/(x)`
C. `(W(d-x))/(d)`
D. `(Wx)/(d)`

Answer

To solve the problem of a rod supported by two parallel knife edges A and B, we will analyze the forces and torques acting on the rod in equilibrium.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Weight of the rod, \( W \)
- Distance between the knife edges, \( d \)
- Distance from knife edge A to the center of mass of the rod, \( x \)

2. **Draw a Free Body Diagram:**
- Draw a horizontal line representing the rod.
- Mark points A and B where the knife edges support the rod.
- Indicate the weight \( W \) acting downwards at the center of mass of the rod (which is at distance \( x \) from A).
- The distance from B to the center of mass is \( d - x \).

3. **Set Up the Equilibrium Conditions:**
- For vertical equilibrium, the sum of the upward forces must equal the downward forces:
\[ N_1 + N_2 = W \quad \text{(Equation 1)} \]
where \( N_1 \) is the normal reaction at A and \( N_2 \) is the normal reaction at B.

4. **Torque Equilibrium:**
- Choose the center of mass of the rod as the pivot point to analyze torques. The torque due to \( N_1 \) and \( N_2 \) must balance out.
- The torque due to \( N_1 \) (acting at a distance \( x \) from the center of mass) is:
\[ \tau_{N_1} = N_1 \cdot x \]
- The torque due to \( N_2 \) (acting at a distance \( d - x \) from the center of mass) is:
\[ \tau_{N_2} = N_2 \cdot (d - x) \]
- Setting the torques equal gives:
\[ N_1 \cdot x = N_2 \cdot (d - x) \quad \text{(Equation 2)} \]

5. **Substituting \( N_2 \) from Equation 1 into Equation 2:**
- From Equation 1, express \( N_2 \):
\[ N_2 = W - N_1 \]
- Substitute this into Equation 2:
\[ N_1 \cdot x = (W - N_1) \cdot (d - x) \]

6. **Expanding and Rearranging:**
- Expanding the equation:
\[ N_1 \cdot x = W(d - x) - N_1(d - x) \]
- Rearranging gives:
\[ N_1 \cdot x + N_1(d - x) = W(d - x) \]
- Factor out \( N_1 \):
\[ N_1 \cdot (x + d - x) = W(d - x) \]
- Thus:
\[ N_1 \cdot d = W(d - x) \]

7. **Solving for \( N_1 \):**
- Finally, we can solve for \( N_1 \):
\[ N_1 = \frac{W(d - x)}{d} \]

### Final Result:
The normal reaction at knife edge A is:
\[ N_1 = \frac{W(d - x)}{d} \]