# What is the tension in the rope?

Tension calculator: the tension in a rope holding a weight (T = m(g + a)), in two ropes at angles, in an Atwood machine, or pulling a block across a table, in newtons and pounds-force.

- Page: https://www.acalculator.org/physics/tension-calculator
- JSON spec: https://www.acalculator.org/physics/tension-calculator.json
- Version: cdf118644c5e

## Default answer

Example with the default inputs (Setup One rope, Mass (m) 154.323583529414 lb, Acceleration (a, m/s²) 0, Gravity (g, m/s²) 9.80665): T = 686.466 N (154.324 lbf).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| setup | Setup | How the rope holds or pulls the masses. |
| m | Mass (m) | The mass the rope or ropes hold. |
| m1 | Mass 1 (m₁) | Atwood: the mass on one side. Table: the block on the table. |
| m2 | Mass 2 (m₂) | The hanging mass on the other side of the pulley. |
| a | Acceleration (a, m/s²) | How fast the mass speeds up: positive up, negative down, 0 at rest or at a steady speed. |
| th1 | Angle of rope 1 (θ₁, degrees) | The angle of the first rope above level, more than 0° and up to 90°. |
| th2 | Angle of rope 2 (θ₂, degrees) | The angle of the second rope above level, more than 0° and up to 90°. |
| g | Gravity (g, m/s²) | The acceleration of gravity: 9.80665 m/s² is standard on Earth. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The tension (or both tensions) to 6 significant figures. |
| tension | Tension (N) | T in newtons (rope 1 with two ropes). |
| lbf | Tension (lbf) | T in pounds-force (rope 1 with two ropes). |
| tension2 | Tension in rope 2 (N) | T₂ in newtons. |
| lbf2 | Tension in rope 2 (lbf) | T₂ in pounds-force. |
| weight | Weight (N) | m × g of the held mass, or m₂ × g of the hanging mass. |
| accel | Acceleration (m/s²) | Atwood: m₂ moves down (negative: up). Table: the blocks speed up. |

## Method

One rope: T = m(g + a). Two ropes: T₁ = mg cos θ₂ ÷ sin(θ₁ + θ₂), T₂ = mg cos θ₁ ÷ sin(θ₁ + θ₂). Atwood: T = 2m₁m₂g ÷ (m₁ + m₂), a = (m₂ − m₁)g ÷ (m₁ + m₂). Table: T = m₁m₂g ÷ (m₁ + m₂), a = m₂g ÷ (m₁ + m₂).

## Assumptions

- Light ropes that do not stretch, over pulleys with no mass and no friction; the table has no friction.
- The masses are at rest or move in a straight line along the rope; for two ropes the weight hangs still.

## Worked examples

1. setup = hanging, m = 70, a = 0, g = 9.80665 gives tension = 686.4655, result = T = 686.466 N (154.324 lbf). Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (a rope holding a weight: T = w = mg; Example 5.13: a 70.0 kg tightrope walker, wire at 5.0° each side, T = mg ÷ (2 sin 5.0°) = 3,930 N), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).
2. setup = hanging, m = 75, a = 1.2, g = 9.8 gives tension = 825. Source: OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws (Example 6.1: a 15.0 kg traffic light on wires at 30° and 45°, T₁ = 108 N and T₂ = 132 N; Example 6.3: 75.0 kg at 1.20 m/s² up, 825 N; Example 6.4: a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂); Example 6.5: an Atwood machine of 2.00 kg and 4.00 kg, a = 3.27 m/s², T = 26.1 N), https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws (retrieved 2026-10-19), Example 6.3 (the same sum for a scale in an elevator).
3. setup = two, m = 15, th1 = 30, th2 = 45, g = 9.8 gives tension = 107.611469, tension2 = 131.796594. Source: OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws (Example 6.1: a 15.0 kg traffic light on wires at 30° and 45°, T₁ = 108 N and T₂ = 132 N; Example 6.3: 75.0 kg at 1.20 m/s² up, 825 N; Example 6.4: a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂); Example 6.5: an Atwood machine of 2.00 kg and 4.00 kg, a = 3.27 m/s², T = 26.1 N), https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws (retrieved 2026-10-19).
4. setup = two, m = 70, th1 = 5, th2 = 5, g = 9.8 gives tension = 3,935.483643, tension2 = 3,935.483643. Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (a rope holding a weight: T = w = mg; Example 5.13: a 70.0 kg tightrope walker, wire at 5.0° each side, T = mg ÷ (2 sin 5.0°) = 3,930 N), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).
5. setup = atwood, m1 = 2, m2 = 4, g = 9.8 gives tension = 26.133333, a = 3.266667. Source: OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws (Example 6.1: a 15.0 kg traffic light on wires at 30° and 45°, T₁ = 108 N and T₂ = 132 N; Example 6.3: 75.0 kg at 1.20 m/s² up, 825 N; Example 6.4: a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂); Example 6.5: an Atwood machine of 2.00 kg and 4.00 kg, a = 3.27 m/s², T = 26.1 N), https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws (retrieved 2026-10-19).
6. setup = table, m1 = 5, m2 = 3, g = 9.8 gives tension = 18.375, a = 3.675. Source: OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws (Example 6.1: a 15.0 kg traffic light on wires at 30° and 45°, T₁ = 108 N and T₂ = 132 N; Example 6.3: 75.0 kg at 1.20 m/s² up, 825 N; Example 6.4: a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂); Example 6.5: an Atwood machine of 2.00 kg and 4.00 kg, a = 3.27 m/s², T = 26.1 N), https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws (retrieved 2026-10-19), Example 6.4 (a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂)).

## FAQ

### How do you calculate tension in a rope?

For a rope holding a weight still, the tension equals the weight: T = mg. A 70 kg load gives T = 70 × 9.80665 = 686.5 N. If the load speeds up, T = m(g + a): lifting 75.0 kg at 1.20 m/s² takes 75 × (9.8 + 1.2) = 825 N.

### How do you find the tension in two ropes at angles?

Balance the forces both ways: T₁ cos θ₁ = T₂ cos θ₂ and T₁ sin θ₁ + T₂ sin θ₂ = mg. That gives T₁ = mg cos θ₂ ÷ sin(θ₁ + θ₂) and T₂ = mg cos θ₁ ÷ sin(θ₁ + θ₂). A 15.0 kg traffic light on wires at 30° and 45° pulls 108 N and 132 N.

### Why is the tension in a tightrope so large?

With both sides at a small angle θ, T = mg ÷ (2 sin θ), and sin θ is small. A 70.0 kg walker on a wire sagging 5.0° each side makes T = 686 ÷ (2 × sin 5°) = 686 ÷ 0.174311 = 3,935 N, almost six times the weight.

### How do you calculate tension in an Atwood machine?

Two masses on one rope over a pulley: a = (m₂ − m₁)g ÷ (m₁ + m₂) and T = 2m₁m₂g ÷ (m₁ + m₂). With 2.00 kg and 4.00 kg, a = 3.27 m/s² and T = 26.1 N, between the two weights (19.6 N and 39.2 N).

### What is the tension when a hanging block pulls another across a table?

With no friction, a = m₂g ÷ (m₁ + m₂) and T = m₁m₂g ÷ (m₁ + m₂), less than the hanging weight. A 3.00 kg block pulling a 5.00 kg block gives a = 3.675 m/s² and T = 18.375 N.

### Can tension push?

No. A rope can only pull, so tension is never negative. If a load speeds downward faster than gravity, the rope goes slack, and the page gives no answer.

## Sources

- OpenStax, University Physics Volume 1, §5.6 Common Forces: tension in a rope holding a weight, T = w = mg; Example 5.13 (a 70.0 kg tightrope walker on a wire at 5.0° each side: T = mg ÷ (2 sin 5.0°) = 3,930 N, almost six times the 686 N weight), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces
- OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws: Example 6.1 (a 15.0 kg traffic light on wires at 30° and 45°: T₁ = 108 N, T₂ = 132 N); Example 6.3 (75.0 kg at 1.20 m/s² up: 825 N); Example 6.4 (a block on a frictionless table and a hanging block: a = m₂g ÷ (m₁ + m₂), T = m₁m₂g ÷ (m₁ + m₂)); Example 6.5 (an Atwood machine of 2.00 kg and 4.00 kg: a = 3.27 m/s², T = 26.1 N), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws
- NIST, CODATA 2022: standard acceleration of gravity g_n = 9.80665 m/s² (exact), retrieved 2026-10-19. https://physics.nist.gov/cgi-bin/cuu/Value?gn
- NIST Special Publication 811 (2008), Appendix B.8: 1 lb = 0.45359237 kg, 1 oz = 0.028349523125 kg, 1 lbf = 4.4482216152605 N, retrieved 2026-10-19. https://www.nist.gov/pml/special-publication-811