Ametal bar 70cm long and 4.00 kg in mass supported on two knief -edge placed 10 cm from each end. a 6.00 kg weight is suspended at 30 cm from one end . find the reaction at the knife-edge ?

First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.

To determine F1: (reference is F4, so F4=0)

∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)

To determine F4: (reference is F1, so F1=0)

∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge) 

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way