Tension is the pull inside a rope, cable or string — and the answer changes completely with the setup: the same 10 kg mass loads a rope with 98 newtons standing still, more in a rising lift, and over 113 newtons split across two angled ropes. This tension calculator handles the four canonical arrangements — single hanging mass, accelerating mass, two symmetric ropes, and the Atwood machine — so the geometry you actually have is the one that gets calculated.
The Hanging Mass: T = m(g + a)
Start with the still case the others modify: a mass hanging motionless from a rope pulls with its full weight, T = mg — 10 kg × 9.81 = 98.1 N. Acceleration edits the answer in the direction intuition suggests once you think in forces: a lift accelerating upward at a makes the rope supply both gravity’s defeat and the acceleration, T = m(g + a); accelerating downward (a negative) lightens the rope; and in free fall the mass and rope drop together and tension vanishes — the weightlessness of the falling lift, delivered by one bracket. This is also why crane and rigging ratings carry dynamic factors: a load that is merely heavy standing still becomes much heavier the moment it is jerked upward.
Two Ropes: The Angle Tax
Split a load between two symmetric ropes angled θ from vertical and equilibrium demands their vertical components sum to the weight: 2T cos θ = mg, so T = mg ÷ (2 cos θ). The cosine is the tax collector. Vertical ropes (θ = 0) split the load evenly — 49.1 N each for our 10 kg. At θ = 30° each rope already carries 56.6 N — more than its “share.” At 60° each carries the entire weight, 98.1 N, and the geometry gains nothing; beyond that the tension explodes toward infinity as θ → 90°. Every rigger, climber and tent designer lives inside this curve: anchor angles are kept small because the mathematics of wide angles is punitive, and the innocent-looking horizontal tightrope is the most loaded rope in the circus.
The Atwood Machine: One Rope, One Tension
Two masses over an ideal (massless, frictionless) pulley share a single rope tension, found by writing Newton’s second law for each mass and eliminating the acceleration: T = 2 m₁ m₂ g ÷ (m₁ + m₂), with system acceleration a = (m₁ − m₂)g ÷ (m₁ + m₂). For 10 kg against 6 kg: T = 2 × 10 × 6 × 9.81 ÷ 16 ≈ 73.6 N — less than the heavy weight (98.1 N, which is why it accelerates down) and more than the light one (58.9 N, which is why it accelerates up). George Atwood built the device in 1784 to slow gravity down to measurable speeds; it remains the cleanest two-body problem in mechanics, and the template for lift counterweights everywhere.
What “Tension” Assumes Here
The calculations idealise the rope as massless and inextensible, which is why tension is uniform along it (a massive rope carries its own weight too, and tension then grows toward the support — the reason long cables are tapered in honest engineering). Real ropes also stretch, knots halve strengths, and pulleys add friction that splits the “one tension” into two. Treat these answers as the physics core — exact for the ideal system, and the baseline from which a rigger’s safety factors (commonly 5:1 or 10:1 on rated loads) are then applied with the humility the idealisations demand.
Tension in the Wild
Elevator cables sized for m(g + a) with a at the emergency-brake deceleration, not the cruise. Guitar strings, where tension sets pitch (T in the wave-speed formula √(T/μ)) and a semitone is a measurable newton difference. Climbing protection, where a fall factor converts metres of drop into kilonewtons at the anchor through rope stretch dynamics far beyond the static formulas here. Bridge stays and tent guys, living in the two-rope angle tax. And the humble tug-of-war, where the rope’s tension equals whichever team’s pull is smaller plus acceleration corrections — a result that surprises every first-year class it meets.
Common Mistakes
Entering weight (newtons) where the formula wants mass (kilograms) — the calculator’s g multiplication is not optional bookkeeping. Splitting a two-rope load by simply halving the weight, ignoring the cosine. Giving the Atwood machine two different tensions because the masses differ (in the ideal machine the rope enforces one). Forgetting the sign of acceleration in a descending lift. And measuring the rope angle from the horizontal while the formula reads it from the vertical — cos and sin swap, and so does the answer; this page’s angle is explicitly from vertical, marked on the input.
Related Tools
The forces bench continues in the Physics Calculators hub — the Centrifugal Force Calculator takes force into circular motion, and the Gear Ratio Calculator trades force against speed in machines.
Frequently Asked Questions
How do you calculate tension in a rope?
It depends on the setup. A hanging mass: T = m(g + a) — 98.1 N for 10 kg at rest. Two symmetric ropes at angle θ from vertical: T = mg/(2 cos θ) each. An Atwood machine: T = 2m₁m₂g/(m₁ + m₂).
Why do angled ropes carry more tension?
Only the vertical components of the two tensions hold the weight, and angling the ropes shrinks those components by cos θ. At 60 degrees from vertical each rope carries the full weight; near 90 degrees the required tension approaches infinity.
What is the tension in an Atwood machine?
The same throughout the rope (ideal pulley): T = 2m₁m₂g/(m₁+m₂). For 10 kg and 6 kg it is about 73.6 N — between the two weights, which is why the heavy side falls and the light side rises.
Does tension change in an accelerating lift?
Yes: T = m(g + a) upward. Accelerating up raises the tension above the weight; accelerating down lowers it; in free fall (a = −g) tension is zero — the rope and mass fall together.
Is tension the same throughout a rope?
For a massless rope over a frictionless pulley, yes. A real heavy rope carries increasing tension toward its support because each point also holds the rope below it — the idealisation this calculator states openly.
What units does tension use?
Newtons in SI (1 N = 1 kg·m/s²). To feel a value in pounds-force, divide by 4.448 — the 98.1 N from a 10 kg mass is about 22 lbf.