Refer to the fig. AB is the rod. The positions of the knife-edges K1 and k2, the centre of gravity of the rod, G and the suspended weight p are shown in fig.
Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross-section and homogeneous; AB = 70 cm, AG = 35 cm. AP = 30 cm. PG = 5 cm. AK1 = BK2 = 10 cm and K1 G = K2 G = 25 cm.
Also, w = weight of the rod = 4 kg. and w1 = suspended weight = 6kg; R1 and R2 are the normal reactions of the support at the knife-edges.
For translational equilibrium of the rod,
R1 +R2 - W1 - W =0..........(i)
Note W1 and w act vertically down and R1 and R2 act vertically ap.
For considering rotational equilibrium, we take moments of the forces.
A convenient rotational equilibrium, we take moments of the forces. A convenient point to take moments about, is G. The moments of K2 and w1 are anticlockwise (+ve) whereas the moment of K1 is clockwise (-ve).
For rotational equilibrium:
R1 (K1 G) + W1 (PG)- R2 (k2 G) = 0.........(ii)
It is given that W = 4 g N and W1 = 6g N.
Where g is acceleration due to the gravity. we take g = 9.8 ms-2.
R1 + R2 - 4gN - 6gN = 0
or, R1 + R2 = 10 g N.........(iii)
= 98 N.
Thus, the reactions of the support are 43.12 N at K1 and 54.88 N at k2.