Divide the given figure into three parts
I-rectangle
II-rectangle
III-triangle
We have to find the centroid of the given structure.
S.No A(ft2) x(ft) xA(ft3)
I 3.6 Wa 1.8 6.48Wa
II 2.7 Wa 3.6+1/2(5.4) 17.01Wa
III 1.35 Wa 3.6+1/3(5.4) 7.29wa
=>ΣA = 7.65 Wa
ΣxA = 30.78Wa
=> Centroid, X = 30.78 Wa/7.65 Wa = 4.0235 ft
Now, draw the free body diagram of the given structure
Find the summation of moments about the load 7.65Wa
ΣM = 0
=> 6(4.0235-1.5) - Rr(4.0235-3.6) - 4.5(6-4.0235) - 1(7.2-4.0235) = 0
=> Rr = 7.25 kips
Resolve the forces along y axis
ΣY = 0
=> Rr + 7.65Wa - 6 - 4.5 - 1 = 0
=> Wa = 556 lb/ft