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 distance x from A. The normal reaction on A is.. And on B is......

Options

A. `N_(A)=2w(1-x//d),N_(B)=wx//d`
B. `N_(A)=w(1-x//d),N_(B)=wx//d`
C. `N_(A)=2w(1-x//d),N_(B)=2wx//d`
D. `N_(A)=wx//d,N_(B)=w(1-(x)/(d))`

Answer

To solve the problem, we need to find the normal reactions at points A and B when a rod of weight \( W \) is supported by two parallel knife edges A and B, with the center of mass of the rod located at a distance \( x \) from A and the distance between A and B being \( d \).

### Step-by-Step Solution:

1. **Identify the Forces:**
- The rod has a weight \( W \) acting downwards at its center of mass.
- There are normal reactions \( N_A \) at point A and \( N_B \) at point B acting upwards.

2. **Set Up the Moment Equation About Point B:**
- To find the normal reaction at A (\( N_A \)), we can take moments about point B.
- The moment due to the weight of the rod about point B is \( W \times (d - x) \) (the distance from B to the center of mass).
- The moment due to the normal force at A is \( N_A \times d \) (the distance from A to B).
- Setting the sum of moments about point B to zero for equilibrium:
\[ N_A \cdot d - W \cdot (d - x) = 0 \]

3. **Solve for \( N_A \):**
- Rearranging the equation gives:
\[ N_A \cdot d = W \cdot (d - x) \]
- Dividing both sides by \( d \):
\[ N_A = \frac{W \cdot (d - x)}{d} \]
- This can also be expressed as:
\[ N_A = W \left(1 - \frac{x}{d}\right) \]

4. **Set Up the Moment Equation About Point A:**
- Now, to find the normal reaction at B (\( N_B \)), we take moments about point A.
- The moment due to the weight of the rod about point A is \( W \cdot x \) (the distance from A to the center of mass).
- The moment due to the normal force at B is \( N_B \cdot d \):
\[ W \cdot x - N_B \cdot d = 0 \]

5. **Solve for \( N_B \):**
- Rearranging the equation gives:
\[ N_B \cdot d = W \cdot x \]
- Dividing both sides by \( d \):
\[ N_B = \frac{W \cdot x}{d} \]

### Final Answers:
- The normal reaction at point A is:
\[ N_A = \frac{W \cdot (d - x)}{d} \]
- The normal reaction at point B is:
\[ N_B = \frac{W \cdot x}{d} \]