A 40 kg block hangs from joint B. Cord BA goes up to the left to a fixed support A, making 35° with the horizontal at B, and cord BC goes up to the right to a fixed support C, making 50° with the horizontal at B. Determine the tensions in cords AB an…

Answer: \(T_{AB}=253.2\) N, \(T_{BC}=322.7\) N.

Weight \(W=40\times9.81=392.4\) N. FBD of joint B: \(T_{AB}\) up-left at 35°, \(T_{BC}\) up-right at 50°, W downward.

\(\sum F_x=0:\ -T_{AB}\cos35^\circ+T_{BC}\cos50^\circ=0\Rightarrow T_{AB}=T_{BC}\dfrac{\cos50^\circ}{\cos35^\circ}=0.7847\,T_{BC}\)

\(\sum F_y=0:\ T_{AB}\sin35^\circ+T_{BC}\sin50^\circ=392.4\)

\(T_{BC}(0.7847\times0.5736+0.7660)=392.4\Rightarrow T_{BC}(1.2161)=392.4\Rightarrow T_{BC}=322.7\) N

\(T_{AB}=0.7847\times322.7=253.2\) N.

Check by Lami's theorem: angles between the forces: \(T_{AB}\)–\(T_{BC}\): 95°; \(T_{BC}\)–W: 140°; W–\(T_{AB}\): 125°.

\(\dfrac{T_{AB}}{\sin140^\circ}=\dfrac{T_{BC}}{\sin125^\circ}=\dfrac{392.4}{\sin95^\circ}=393.9\Rightarrow T_{AB}=253.2\) N, \(T_{BC}=322.7\) N ✓