Tension Calculator
Find the tension in a rope, cable, or string for five common scenarios: a single vertical rope, one or two angled support ropes, an Atwood machine (two masses over a pulley), or a horizontal pull chain. Enter the inputs below and get the tension force instantly, with a full worked solution.
Formula
Worked example
A 5 kg mass at rest: T = 5 x 9.807 = 49.03 N. The same mass accelerating upward at 2 m/s²: T = 5 x (9.807 + 2) = 59.03 N. Two symmetric ropes each at 30 deg from vertical holding 20 kg: T per rope = (20 x 9.807) / (2 x cos 30) = 196.1 / 1.732 = 113.2 N. Atwood machine with 8 kg and 3 kg: T = (2 x 8 x 3 x 9.807) / (8 + 3) = 42.9 N, a = (8 - 3) / 11 x 9.807 = 4.46 m/s².
What tension is and where it comes from
Tension is the pulling force transmitted along a rope, cable, string, or chain when it is pulled taut by forces acting from opposite ends. It always acts along the line of the rope and, for an ideal massless rope, has the same magnitude at every cross-section. A key physical constraint is that a rope can only pull, never push: it supports a load by pulling up on it while the load pulls down. This is why tension is always positive or zero in a real system. When a single mass hangs from a fixed support at rest, the rope tension exactly equals the weight, T = mg. Once the mass accelerates vertically, Newton's second law adds or subtracts from that weight: T = m(g + a), where a is positive upward.
Angled ropes and how the angle amplifies tension
When a rope is not perfectly vertical it carries both a vertical component (supporting the weight) and a horizontal component (preventing sideways movement). Only the vertical component balances gravity, so the rope must carry extra force to provide that same vertical component at an angle. A single angled rope gives T = W / cos(theta), where theta is measured from the vertical. At 0 deg the rope is straight down and T = W; at 60 deg from vertical the rope carries twice the weight. For two symmetric ropes each at angle theta: T = W / (2 cos theta). For two asymmetric ropes at angles alpha and beta from vertical, each tension is found by resolving force components and solving the simultaneous equations: T1 = W sin(beta) / sin(alpha + beta) and T2 = W sin(alpha) / sin(alpha + beta). The rope at the shallower angle always carries more tension because its vertical component per unit of total force is larger.
Atwood machine and pulley systems
An Atwood machine consists of two masses connected by a rope over a frictionless, massless pulley. Because the rope is continuous, the same tension acts on both sides. Writing Newton's second law for each mass and solving simultaneously gives T = 2m1m2g / (m1 + m2) and a system acceleration of a = (m1 - m2)g / (m1 + m2). When the masses are equal, a = 0 and T = mg (the weight of one mass), which is a useful static calibration check. The heavier mass accelerates downward and the lighter one upward, but both feel the same rope tension pulling them toward the pulley. In real pulley systems, rope mass, pulley inertia, and bearing friction all modify these results, but the ideal formulas are excellent starting points for design.
Horizontal pull chains and why tension decreases along the chain
When a force F is applied to a chain of objects on a frictionless horizontal surface, the whole system accelerates at a = F / (m1 + m2 + ...). The rope between any two consecutive objects only needs to pull the trailing objects, not the whole chain. So the tension in the rope between object A and object B equals the trailing mass times the common acceleration. This means tension is highest at the front of the chain (where the force is applied) and decreases toward the rear. This is why towing chains can fail at the coupling even when the vehicle engine is far within its rated output: the coupling carries almost the full applied force.
Choosing a rope or cable and safety factors
Rope and cable manufacturers publish a minimum breaking load (MBL) and a working load limit (WLL), which is the MBL divided by a safety factor. For general lifting the safety factor is typically 5 to 7; for life-safety applications (climbing, rescue) it can reach 10 to 15. The tension calculated here is the actual force the rope must carry. If that force exceeds the WLL, the rope is undersized. Dynamic events such as shock loading during a sudden jerk, impacts, or resonance can create peak forces several times the calculated static or steady-state tension, which is one reason generous safety factors are essential. The reverse-solve mode in this calculator lets you start from a rated tension and work backwards to find the maximum safe mass for a given acceleration.
Tension vs. angle for a 10 kg mass (T = mg / cos theta)
| Angle from vertical (deg) | cos(theta) | Tension (N) | Factor vs. vertical |
|---|---|---|---|
| 0 | 1 | 98.1 | 1.00x |
| 15 | 0.966 | 101.6 | 1.04x |
| 30 | 0.866 | 113.3 | 1.15x |
| 45 | 0.707 | 138.7 | 1.41x |
| 60 | 0.5 | 196.1 | 2.00x |
| 75 | 0.259 | 378.9 | 3.86x |
| 85 | 0.087 | 1126 | 11.5x |
A single angled rope supporting 10 kg (weight = 98.07 N). As the rope tilts, tension rises sharply.
Frequently asked questions
How do you calculate tension in a vertical rope?
For a single mass hanging from a vertical rope, use T = m(g + a). With zero acceleration this simplifies to T = mg, the weight. If the mass accelerates upward at a m/s² the tension increases; if it accelerates downward the tension decreases. At a downward acceleration equal to g (free fall) the tension is zero and the rope goes slack.
How does rope angle affect tension?
Tilting a rope away from vertical forces it to carry a horizontal component as well as a vertical one. Only the vertical component supports the weight, so the rope must be more taut overall. For a single rope at angle theta from vertical, T = mg / cos(theta). At 60 deg from vertical the tension is twice the weight; at 80 deg it is nearly six times the weight. This is why suspension bridge cables that sag even slightly create enormous forces at the towers.
What is an Atwood machine and how does the tension formula work?
An Atwood machine is two masses connected by a rope over a frictionless pulley. The heavier mass falls and the lighter one rises at the same acceleration. Because the rope is continuous and the pulley ideal, the tension is the same throughout: T = 2m1m2g / (m1 + m2). If the masses are equal the tension is mg and the system does not accelerate. The more unequal the masses, the lower the tension and the faster the acceleration.
Why is tension lower in a trailing object than in the applied force?
In a pull-chain (objects linked by ropes on a frictionless surface), the applied force accelerates the whole system. The rope between any two objects only has to pull the trailing objects, so its tension equals the trailing mass times the common acceleration. The rope at the very back of the chain carries the least tension; the rope closest to the applied force carries the most.
How do I find the maximum mass a rated rope can lift?
Use the reverse mode and enter the rope's working load limit (not the breaking strength). The calculator divides the rated tension by (g + a) to give the maximum mass. If the load will be hoisted (upward acceleration a > 0), the maximum mass is lower than the static case. Always apply a safety factor, especially for dynamic loads or life-safety applications.
What units does this tension calculator use?
Primary inputs can be entered in kilograms or pounds using the unit toggle next to each mass field. Tension results are shown in newtons (SI), kilogram-force (kgf), and pound-force (lbf). Angles are in degrees and acceleration in m/s². 1 kgf = 9.807 N = 2.205 lbf.