Determine the tensions in the cords AB and BC for equilibrium of 30 kg block.[Figure: a 30 kg block hangs vertically from joint B. Cord BA goes up to the left to a support A and makes 30° with the horizontal at B; cord BC goes up to the right to a su…

Given: \(W = 30 \times 9.81 = 294.3\) N. At joint B three concurrent forces act: \(T_{AB}\) (30° above horizontal, to the left), \(T_{BC}\) (40° above horizontal, to the right) and W (down).

Lami's theorem. Angles between the forces: between \(T_{AB}\) and \(T_{BC}\) = 180° − 30° − 40° = 110°; between \(T_{BC}\) and W = 90° + 40° = 130°; between W and \(T_{AB}\) = 90° + 30° = 120°.

\(\dfrac{T_{AB}}{\sin 130^\circ} = \dfrac{T_{BC}}{\sin 120^\circ} = \dfrac{294.3}{\sin 110^\circ} = 313.19\)

\(T_{AB} = 313.19 \times 0.7660 = 239.92\ \text{N}\), \(T_{BC} = 313.19 \times 0.8660 = 271.23\ \text{N}\)

Check: \(\sum F_x = 271.23\cos40^\circ - 239.92\cos30^\circ = 207.77 - 207.77 = 0\); \(\sum F_y = 239.92\sin30^\circ + 271.23\sin40^\circ = 119.96 + 174.34 = 294.3\). ✔

Answer: \(T_{AB} = 239.9\ \text{N}\), \(T_{BC} = 271.2\ \text{N}\).