Since the beam is in equilibrium, we apply the two conditions of equilibrium.

∑F = 0 and ∑τ = 0

Now consider the forces on the beam. There are two normal reactions from the two supports, R_A and R_B, acting upward, weight of the beam W acting downward and weight of the suspended load, say W₁, which is also acting down. The first condition of the equilibrium implies that the upward forces must be equal to the downward force.

R_A + R_B = W + W₁

Now mass of the girder = 450 kg ⇒ Weight of the girder, W = 450 × 9.8 = 4410 N

Similarly, mass of the load = 220 kg ⇒ Weight of the load, W₁ = 220 × 9.8 = 2156 N

Upward forces, R_A and R_B are not known and to be determined. However, for equilibrium,

R_A + R_B = 4410 + 2156 = 6566 N           …          (A)

Now apply the second condition of equilibrium, considering point B to be the pivot. R_A is producing clockwise torque = 8 × R_A and W and W₁ are producing counterclockwise torques 4W and 2.4W₁, respectively. Here, 4 m and 2.4 m are the distances of W and W₁ from the pivot point A, respectively.

8R_A = 4W + 2.4W₁. Put the values of W and W₁ from the above equations,

8R_A = 4 × 4410 + 2.4 × 2156 ⇒ 8R_A = 17640 + 5174.4 = 22814.4 N

⇒ R_A = 22814.4/8 ⇒ R_A = 2851.8 N         …          (X)

Now put this value of R_A in equation (A),

2851.8 + R_B = 6566 ⇒ R_B = 6566 – 2851.8 = 3714.2 N          …           (Y)

Equations (X) and (Y) give the respective reactions of the supports.