Find out the reactions at hinge:
A board ABCD is held in position as shown in Figure by a cable BE and a hinge at A. If the weight of the board is 5 kN, find out the reactions at hinge A and the tension T in the cable.
Assume the components of the reaction at A be HA and VA as illustrated in Figure. The board is at rest under the action of four forces HA, VA, T and W.
Taking moments of all of forces about A, we obtain
∑ M A = 0
∴ - (T sin 30o) × 1.6 + W × 0.8 = 0
∴ - 0.8T + 5 × 0.8 = 0
∴ T = 4 /0.8 = 5 kN
And, ∑ Fx = 0
∴ H A - T cos 30o = 0
∴ H A = T cos 30o
(On putting T = 5 kN) ∴ H A = 5 × 0.866 = 4.33 kN
Also, ∑ Fy = 0
∴ VA + T sin 30o - W = 0
∴ V A = W - T sin 30o
= 5 - 5 × 0.5
= 2.5 kN
= 5 kN
θ = tan -1 ( VA / H A)
= tan -1 (2.50 /4.33)
Reaction RA has a magnitude of 5 kN and is inclined at 30o to the horizontal.
[Note : You can attempt the same problem considering the board to be subjected to 3 forces only viz., RA, T and W. As three of the forces keep the body at rest, these should be concurrent. Find the point of concurrence and get the values of unknowns RA and T, using
∑ Fx = 0 and ∑ Fy = 0
knowing the direction of RA, you might also use Lami's theorem either.]