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)

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)
$N_{1}x=(W-N_1)(d-x)$
$\Rightarrow N_{1}x=Wd-Wx-N_{1}d+N_{1}x$
$\Rightarrow N_{1}d=W(d-x)$
$\therefore N_1=\frac{W(d-x)}{d}$