Question

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 a from A. The normal reaction on A is

• A. $ \frac{W(d- x)}{x}$
• B. $ \frac{W(d- x)}{d}$
• C. $ \frac{Wx}{d}$
• D. $ \frac{Wd}{x}$

Answer 1

The Correct Option is B

To find the normal reaction on point A, we need to understand the equilibrium condition of the rod supported by two knife edges.

1. The rod is in equilibrium under the action of its weight W and the normal reactions at points A and B. Let the normal reaction at A be R_A, and at B be R_B.
2. According to the problem, the distance between A and B is d, and the center of mass is at a distance a from A.
3. For equilibrium, the sum of the vertical forces should be zero. Therefore, we have:

R_A + R_B = W

4. For rotational equilibrium, taking moments about point B, we apply the principle that the sum of moments about any point should be zero:

R_A \cdot d = W \cdot (d - a)

5. Solving the above equation for R_A, we get:

R_A = \frac{W(d - a)}{d}

6. Thus, the normal reaction on point A is \frac{W(d - a)}{d}, which matches the given correct answer option.

This reasoning correctly identifies the moments and balances the forces, illustrating how the normal reaction is distributed based on the center of mass location.