Question

The 12 m boom AB weighs 1 kN, the distance of the centre of gravity G being 6 m from A. For the position shown, determine the tension T in the cable and the reaction at B [Ref. Fig (a)].

Answer 1

The free body diagram of the boom is shown in Fig (b). 

ΣM_A = 0, gives 

T sin 15° × 12 – 2.5 × 12 cos 30° – 1 × 6 cos 30° = 0 

T = 10.0382 kN. 

∑ H = 0, gives 

H_A – T cos 15° = 0 

H_A = 9.6962 kN

∑V = 0, gives 

V_A = 1 + 2.5 + T sin 15° = 6.0981 kN

R_A = \(\sqrt{V_A^2+H_A^2}\)

R_A = 11.4544 kN.

α = tan^-1 \(\frac{6.0981}{9.6962}\)

= 32.17° as shown in Fig.(c).