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

Answer: (B)

$\frac{W(d- x)}{d}$

Answer 2

Given situation is shown in figure.

$N_1 =$ Normal reaction on A
$N_2 =$ Normal reaction on B
$W =$ Weight of the rod
In vertical equilibrium,
$N_1 + N_2 = W \, \, \, \, \, ......(i)$
Torque balance about centre of mass of the rod,
$N_1 x = N_2 (d - x)$
Putting value of $N_2$ from equation (i)
$N_1 x = (W - N_1) (d - x)$
$\Rightarrow N_1 x = Wd - Wx - N_1 d + N_1 x$
$\Rightarrow N_1 d = W(d - x)$
$\therefore N_1 = \frac{W(d - x)}{d}$