A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a

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 $x$ from $A$. The normal reaction on $A$ is\

Professor PC Thomas Classes · 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,\ $$ N_{1}+N_{2}=W 	ag{i} $$ Torque balance about centre of mass of the rod, $$ N_{1} x=N_{2}(d-x) $$ Putting value of $N_{2}$ from equation (i)\ $N_{1} x=\left(W-N_{1}ight)(d-x) \Rightarrow N_{1} x=W d-W x-N_{1} d+N_{1} x$\ $\Rightarrow \quad N_{1} d=W(d-x) ; 	herefore \quad N_{1}=\frac{W(d-x)}{d}$\