Question

A mental bar 70 cm long and 4 kg in mass supported on two knife edge placed 10 cm. from each end. A 6 kg weight is suspended at 40 cm. from end. A 6 kg weight is suspended at 40 cm. From one end. Find the reaction at the knife-edges. (Assume the bar to be of uniform cross- section and homogeneous).

Answer 1

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.