Question

A massless rigid rod is in equilibrium as shown in figure. Find the normal

reaction at knife -edge 1

Answer 1

Let the normal reaction at knife edge $\sf{K_1}$ be $\sf{R_1}$ which is downwards.

The net rotational effect of the rod is zero at any point in the rod since it is in equilibrium. We consider net rotational effect about the knife edge $\sf{K_2.}$

According to $\sf{K_2,\ R_1}$ makes a rotational effect on the rod which is in anticlockwise direction (if we consider rod as shown in the fig.).

Similarly, according to $\sf{K_2,\ F}$ makes a rotational effect on the rod which is in clockwise direction.

Or, as in other words, $\sf{R_2}$ is acting at a distance $\sf{a}$ towards left from $\sf{K_2}$ and $\sf{F}$ is acting at a distance $\sf{b}$ towards right from $\sf{K_2.}$

Hence,

$\longrightarrow\sf{R_1(-a)+Fb=0}$

$\longrightarrow\sf{-R_1\,a+Fb=0}$

$\longrightarrow\sf{R_1\,a=Fb}$

$\longrightarrow\sf{\underline{\underline{R_1=F\left(\dfrac{b}{a}\right)}}}$

Hence $\sf{(D)\ F\left(\dfrac{b}{a}\right)}$ is the answer.