The Overloaded Sling
During a stores lift, a crane hoists a crate of mass 240 kg using a two-leg sling. Each leg makes a 30° angle with the vertical, and the legs are symmetrical. The crane's safe working load for this configuration is 3.0 kN, and the sling wire is rated at 2.0 kN per leg. A cadet notices the crate swaying and that one leg appears tighter than the other. Take g = 9.81 m/s². The cadet is asked to check the lift before it is raised further and to explain the result to a deckhand who believes the load is shared as "half each".
1. Calculate the tension in each leg for the symmetrical case and compare it with the two limits.
Read the worked answer
Weight: W = mg = 240 × 9.81 = 2,354 N ≈ 2.35 kN.
For two symmetrical legs at angle θ to the vertical, vertical equilibrium gives 2T cos θ = W, so T = W ÷ (2 cos 30°).
cos 30° = 0.866, so 2 × 0.866 = 1.732.
T = 2,354 ÷ 1.732 ≈ 1,359 N ≈ 1.36 kN per leg.
Checks:
- Per-leg wire rating 2.0 kN: 1.36 < 2.0 → OK, margin ≈ 0.64 kN.
- Crane SWL 3.0 kN: total vertical load is 2.35 kN < 3.0 kN → OK.
So the symmetrical lift is within both limits, but the per-leg margin is only about 32%.
2. The deckhand says each leg carries half the weight, i.e. 1.18 kN. Show why this is wrong and what the true relationship is.
Read the worked answer
Half the weight is 2,354 ÷ 2 = 1,177 N ≈ 1.18 kN. The correct figure is 1.36 kN - about 15% higher.
The deckhand's error is to forget that each leg pulls along its own axis, and only the vertical component of that pull holds the crate up. The horizontal components of the two legs cancel each other, doing no lifting work at all.
The relationship: T = W ÷ (2 cos θ). As the legs splay further from vertical, cos θ falls and tension rises. At 30° the factor is 1 ÷ (2 × 0.866) = 0.577 × W, not 0.5 × W.
Practical point: the wider the sling angle from vertical, the greater the tension for the same load. This is why slings are kept as near vertical as practical, and why "half each" is only true when the legs are vertical (θ = 0).
3. The cadet observes one leg tighter than the other. Explain the consequence for the tighter leg and what should be done before lifting further.
Read the worked answer
If the load is not shared equally, the tighter leg carries more than 1.36 kN while the slack leg carries less. The total vertical support must still equal 2.35 kN, so the extra comes entirely onto one wire.
Consequence: the tighter leg approaches its 2.0 kN rating. If, for example, it took 70% of the vertical load, its tension would be roughly (0.70 × 2,354) ÷ 0.866 ≈ 1.90 kN - only about 5% below the wire rating, with no allowance for shock loading as the crate sways.
Action before lifting further: lower the load to a stable rest, stop the lift, re-rig so the legs are equal in length and the hook is centred over the crate's centre of gravity, and check the shackles and wire for damage. Swaying adds dynamic load on top of the static figure, so the margin is smaller than the arithmetic suggests. Do not continue the lift on the assumption that the numbers "just pass".