The cable at 45° from vertical carries more tension than the 60° one because:

It is closer to vertical, so it supports more of the load

2068 Shrawan · 5 marks

Forces in cables AB and AC

Short answer

Apply equilibrium at joint B. T_AB sin45° = T_AC sin60°, and vertical equilibrium gives T_AB = 89.66 N, T_AC = 73.21 N. Alternatively, using Lami’s theorem yields the same tensions.

Determine the forces in cables AB and AC as shown in figure below. [Figure: cables AB and BC fixed to the ceiling at A and C, meeting at B at angles 45° and 60° to the vertical, with a 100 N load hanging from B.]

Reading of the figure: cable AB (from the support on the left) makes 45° with the vertical, the other cable (from the support on the right, joined at B) makes 60° with the vertical, and a load of 100 N hangs from B. The right-hand cable is the one called AC in the question (the supports are marked A and C), so I call its tension T_AC.

FBD of joint B

Forces: T_AB up-left (45° from vertical), T_AC up-right (60° from vertical), 100 N down.

ΣFx = 0: T_AC sin 60° − T_AB sin 45° = 0 …(1) ΣFy = 0: T_AB cos 45° + T_AC cos 60° − 100 = 0 …(2)

From (1): T_AC = T_AB × (0.7071/0.8660) = 0.8165 T_AB

Substituting in (2): 0.7071 T_AB + 0.5 × 0.8165 T_AB = 100 1.1154 T_AB = 100 T_AB = 89.66 N T_AC = 0.8165 × 89.66 = 73.21 N

(Check by Lami's theorem. The angle between T_AC and the load is 120°, between T_AB and the load is 135°, and between the two cables is 105°. So T_AB / sin 120° = T_AC / sin 135° = 100 / sin 105° = 103.53, which gives T_AB = 89.66 N and T_AC = 73.21 N.)

Answer: T_AB = 89.66 N, T_AC (right cable) = 73.21 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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