Tension Force Calculator – Ropes, Slings, Pulleys and Rigging Angles
Tensionis the pulling force carried along a rope, cable, string, chain or sling. Unlike weight, it has no formula of its own: it is whatever value makes Newton's laws close for the bodies the rope connects. That is why a single expression cannot answer every tension problem, and why this tension force calculator works as a small system solver with one mode per rigging arrangement — from a mass hanging on a string to a multi-leg sling under a crane hook.
When tension equals weight, and when it does not
A stationary mass on one vertical rope is the only case where T = m·g is the whole story. A 12 kg load gives T = 12 × 9.81 = 117.72 N. Every other configuration bends that answer in one of three ways. Angle divides the useful vertical pull by sin θ, so a rope leaning away from vertical must carry more tension to hold the same load. Acceleration adds or subtracts m·a, which is why a rope in a lift rising at 2.5 m/s² carries T = m(g + a) — about 25 % more than the weight — and why a rope in free fall carries nothing at all. Rotation adds the centripetal requirement m·v²/r, so a string whirled in a vertical loop is tightest at the bottom and slackest at the top.
Two ropes, and the counter-intuitive 30° result
Hang a 20 kg load from two ropes, each 30° above horizontal, and each rope carries T = W / (2 sin θ) = 196.2 / 1 = 196.2 N — the entire weight of the load, in each rope. Adding a second rope bought nothing, because the shallow angle exactly doubled the load factor. When the two angles differ, the tensions split unevenly: T_A = W·cos β / sin(α+β) and T_B = W·cos α / sin(α+β). A 50 kg load on ropes at 30° and 60° gives 245.25 N and 424.79 N — the steeper rope always takes the larger share, which is the opposite of most people's intuition.
Sling angles and the load factor
In rigging, angles are measured from horizontal, and the amplification is the load factor 1 / sin θ: 1.000 at 90°, 1.155 at 60°, 1.414 at 45°, 2.000 at 30°, 3.864 at 15° and 11.474 at 5°. Closing a two-leg bridle from 60° to 30° raises each leg tension by 73 % without changing the load by a gram. A 500 kg load on two legs at 60° puts 2 831.90 N in each leg; at 30° the same load puts 4 905 N in each — enough to break through the 4 000 N working load limit of a 20 kN sling on a 5:1 design factor.
Dynamics: pulleys, inclines and lifts
An ideal pulley is massless and frictionless, so it takes no torque to turn and the tension is uniform along the whole string. In an Atwood machine the two masses share one tension value, T = 2·m₁·m₂·g / (m₁ + m₂), which always lies strictly between the two weights — 8 kg against 3 kg gives 42.81 N, bracketed by 29.43 N and 78.48 N. Add a slope and friction and the solver must first decide whether the system moves at all, comparing the driving force m₂·g against m₁·g·sin θ plus the maximum friction μ·m₁·g·cos θ before writing any equation.
Rotational modes
A conical pendulum splits its tension in two: the vertical component holds the weight, giving T = m·g / cos θ, while the horizontal component supplies the centripetal force that keeps the bob turning. In vertical circular motion the tension is T = m·v²/r + m·g·cos φ, measured from the bottom of the loop. Below v = √(g·r) at the top the string goes slack — a string cannot push, so the honest answer is zero tension and an object that has left the circular path, not a negative number.
Working load limits and safety factors
For lifting work the tension is only half the question. The working load limit is the sling's minimum breaking load divided by a design factor, typically 5:1 for general lifting and 7:1 where people are suspended. This calculator reports the utilisation as a percentage of the WLL and the realised safety factor MBL / T, and it tells you the shallowest sling angle that still passes. Every mode is also solved twice by independent routes and the residual is shown, so a silent sin/cos swap cannot hide behind a plausible-looking answer.