the rod $AB$, the positions of the knife edges $K _{1}$ and $K _{2},$ the centre of gravity of the rod at $G$ and the suspended load at $P$.

Note the weight of the rod $W$ acts at its centre of gravity $G$. The rod is uniform in cross section and homogeneous; hence $G$ is at the centre of the rod; $AB =70 cm . AG =35 cm , AP$

$=30 cm , PG =5 cm , AK _{1}= BK _{2}=10 cm$ and $K _{1} G$

$= K _{2} G =25 cm .$ Also, $W=$ weight of the rod $=$

$4.00 kg$ and $W_{1}=$ suspended load $=6.00 kg$ $R_{1}$ and $R_{2}$ are the normal reactions of the support at the knife edges. For translational equilibrium of the rod. $R_{1}+R_{2}-W_{1}-W=0$

Note $W_{1}$ and $W$ act vertically down and $R_{1}$ and $R_{2}$ act vertically up.

For considering rotational equilibrium, we take moments of the forces. A conventent point to take moments about is $G$. The moments of $R _{2}$ and $W _{1}$ are anticlockwise $(+ve)$, whereas the moment of $R _{1}$ is clockwise $(-ve)$. For rotational equilibrium, $-R_{1}\left( K _{1} G \right)+W_{1}( PG )+R_{2}\left( K _{2} G \right)=0$

It is given that $W=4.00 g { N }$ and $W_{1}=6.00 g$

N. where $g=$ acceleration due to gravity. We take $g=9.8 m / s ^{2}$

With numerical values inserted,

$R_{1}+R_{2}-4.00 g-6.00 g=0$

or $R_{1}+R_{2}=10.00 g N$

$=98.00 N$

$-0.25 R_{1}+0.05 W_{1}+0.25 R_{2}=0$

or $R_{1}-R_{2}=1.2 g N =11.76 N$

$R_{1}=54.88 N$

$R_{2}=43.12 N$

Thus the reactions of the support are about $55 N$ at $K _{1}$ and $43 N$ at $K _{2}$