$A$ metal bar $70 \; cm$ long and $4.00 \; kg$ in mass is supported on two knife-edges placed $10 \; cm$ from each end. $A$ $6.00 \; kg$ load is suspended at $30 \; cm$ from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross-section and homogeneous.)

(N/A) Let the rod be $AB$. The positions of the knife-edges are $K_1$ and $K_2$. The centre of gravity of the rod is at $G$,and the suspended load is at $P$.
The weight of the rod $W = mg = 4.00 \times 9.8 = 39.2 \; N$ acts at its centre of gravity $G$. Since the rod is uniform and homogeneous,$G$ is at the centre of the rod.
Given: $AB = 70 \; cm$,so $AG = 35 \; cm$.
The load $W_1 = 6.00 \times 9.8 = 58.8 \; N$ is suspended at $P$,where $AP = 30 \; cm$.
Thus,$PG = AG - AP = 35 \; cm - 30 \; cm = 5 \; cm$.
The knife-edges are at $AK_1 = 10 \; cm$ and $BK_2 = 10 \; cm$.
Therefore,$K_1G = AG - AK_1 = 35 \; cm - 10 \; cm = 25 \; cm$.
And $K_2G = BG - BK_2 = 35 \; cm - 10 \; cm = 25 \; cm$.
For translational equilibrium:
$R_1 + R_2 = W + W_1 = (4.00 + 6.00) \times 9.8 = 98.0 \; N$.
For rotational equilibrium,taking moments about $G$:
$R_1(K_1G) - W_1(PG) - R_2(K_2G) = 0$
$R_1(0.25) - 58.8(0.05) - R_2(0.25) = 0$
$0.25(R_1 - R_2) = 2.94$
$R_1 - R_2 = 11.76 \; N$.
Solving the two equations:
$R_1 + R_2 = 98.0$
$R_1 - R_2 = 11.76$
Adding them: $2R_1 = 109.76 \implies R_1 = 54.88 \; N$.
Subtracting them: $2R_2 = 86.24 \implies R_2 = 43.12 \; N$.
Thus,the reactions are $R_1 = 54.88 \; N$ and $R_2 = 43.12 \; N$.