Question

A metal rod of weight 'W' is supported by two parallel knife-edges A and B. The rod is in equilibrium in horizontal position. The distance 'between two knife-edges is 'r'. The centre of mass of the rod is at a distance 'x' from A. The normal reaction on A is

• A. $\frac{W \cdot r}{x}$
• B. $\frac{W \cdot x}{r}$
• C. $\frac{W \cdot (r-x)}{x}$
• D. $\frac{W \cdot (r-x)}{r}$

Answer 1

Answer: (D)

$\frac{W \cdot (r-x)}{r}$

Answer 2

Let the knife-edges A and B be at a distance 0 and r respectively with the centre of mass at distance x from A.

Vertical Equilibrium:

$N_A + N_B = W$

Taking Moment about B:

$N_A \cdot r = W \cdot (r – x)$

$\Rightarrow N_A = \frac{W \cdot (r – x)}{r}$

Thus, the normal reaction on A is $\frac{W \cdot (r – x)}{r}$.