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 W_{1}, 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 + W_{1}

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, W_{1} = 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 W_{1} are producing counterclockwise torques 4W and 2.4W_{1}, respectively. Here, 4 m and 2.4 m are the distances of W and W_{1} from the pivot point A, respectively.

8R_{A} = 4W + 2.4W_{1}. Put the values of W and W_{1 }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.

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