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Tensions in tugboat ropes pulling vessel

Short answer

Using equilibrium, T_AB sin30° = T_BC sin40° and T_AB cos30° + T_BC cos40° = 150 kN. Solving gives T_AB = 102.6 kN and T_BC = 79.8 kN.

A commercial vessel is being pulled into harbour for unloading by two tugboats as shown in figure knowing the vessel requires 150 kN along its axis to move it steadily. Compute the tensions in rope AB and BC when α = 40°. [Figure: vessel B pulled by ropes to tugboats A (above) and C (below); one rope at 30° to the vessel's axis, the other at α; resultant 150 kN along the axis.]

Reading of the figure: the vessel B needs a pull of 150 kN along its axis. Rope BA (to tugboat A) makes 30° with the axis, and rope BC (to tugboat C) makes α = 40° with the axis, on the other side. The two rope tensions have resultant 150 kN along the axis.

Let T_AB and T_BC be the tensions, x along the axis.

ΣFy = 0 (perpendicular to the axis, the components must cancel): T_AB sin 30° = T_BC sin 40° T_BC = T_AB × (0.5/0.6428) = 0.7779 T_AB …(1)

ΣFx = 150 kN (along the axis): T_AB cos 30° + T_BC cos 40° = 150 …(2)

Put (1) in (2): T_AB (0.8660 + 0.7779 × 0.7660) = 150 T_AB (0.8660 + 0.5959) = 150 T_AB = 150/1.4619 = 102.6 kN T_BC = 0.7779 × 102.6 = 79.8 kN

(Check with the sine rule: 150/sin 110° = 159.6; T_AB = 159.6 × sin 40° = 102.6 kN; T_BC = 159.6 × sin 30° = 79.8 kN.)

Answer: T_AB = 102.6 kN, T_BC = 79.8 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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