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}$