Q: The12m boom AB weight 1 KN, the distance of center of gravity G is 6 m from point A. For the position given, determine the tension T in cable and reaction at B.
Sol.: The free body diagram of boom is shown in the figure given to us
MA = 0
T sin 15º X 12 - 2.5 X 12 cos 30º - 1 X 6 cos 30º = 0
T = 10.0382 KN .......ANS
Reaction at point B = (2.52 + 102 + 10 X 2.5 X cos 75º)1/2
RB = 10.61 KN .......ANS