Tension Calculator

Find the tension in a rope holding a hanging mass or elevator, joining two masses over a pulley, or supporting a weight from two angled ropes.

🪢 Tension Calculator
Hanging mass
kg
Upward acceleration0 at rest or constant speed, positive up, negative down
m/s²
Wherepicking a body fills g, which you can still edit
Gravitational acceleration g
m/s²
Rope breaking strength (optional)blank to skip the safety check
N
Setup
Massesm2 is the hanging mass that pulls
m1 (kg)
m2 (kg)
Table friction coefficientbetween the block m1 and the table
Hanging mass
kg
Angle of each rope above the horizontalboth ropes at the same angle, 90 is vertical
°
Span between the supports (optional)gives the sag and the rope length
m
Rope breaking strength (optional)blank to skip the safety check
N
Rope tension T
—
Weight m g
—
Tension compared with weight
—
Motion
—
Acceleration in g
—
Safety factor
—
Largest safe upward acceleration
—

Force diagram, to scale: the blue arrow is the rope tension pulling up and the red arrow is the weight pulling down.

Rope tension T
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Acceleration
—
What moves
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Weight of m2
—
Tension compared with m2 g
—
Table friction on m1
—
Normal force on m1
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Setup diagram: the masses are labelled m1 and m2 and the rope runs over the pulley. The a value is the acceleration of the system.

Tension in each rope
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Tension compared with weight
—
Vertical part of each tension
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Horizontal part of each tension
—
Weight m g
—
Sag of the load
—
Length of each rope
—
Safety factor
—
Smallest safe angle
—

Force diagram: the two blue arrows are the rope tensions along the ropes and the red arrow is the weight. The arrow lengths are fixed for readability, the numbers are in the labels.

🪢 What is Tension?

Tension is the pulling force that a rope, string, cable, chain or rod carries when it is stretched tight between two points. It acts along the length of the rope and pulls equally on whatever is attached to each end. A rope can pull but it cannot push, so a rope that goes slack has no tension. Tension is not a fixed property of the rope, it is whatever force is needed for the situation: the rope tension rises and falls to satisfy Newton's laws for the objects it is attached to. To find it, you draw the forces on the object, write Newton's second law along the rope and solve for T. This calculator does that for three of the most common textbook setups and shows a diagram and a chart for each one.

In the hanging mass or elevator mode, a load hangs from a vertical rope. If the load is at rest or moves at a constant speed the tension equals the weight, m g. If it accelerates upward the rope must supply the extra force m a, so T = m (g + a), and if it accelerates downward the tension drops, falling to zero in free fall. That is why a lift feels heavier as it starts to rise and lighter as it starts to descend. An optional breaking strength gives a safety factor and the largest upward acceleration the rope can take. In the two masses over a pulley mode you can study an Atwood machine, where two masses hang on either side of a pulley, or a block on a table pulled over the edge by a hanging mass, with friction. The tension in each case is less than the weight of the heavier hanging mass, because that mass is accelerating downward. For the table case, see the friction calculator and the normal force calculator for the forces on the block.

In the two angled ropes mode a weight hangs from the middle of two ropes that slope upward at the same angle, like a sign on two cables or a hammock. Only the vertical parts of the tensions hold the weight, so 2 T sin(theta) = m g and T = m g / (2 sin(theta)). The flatter the ropes, the larger the tension. At 30 degrees each rope carries exactly the weight, and at 5 degrees almost six times as much. The horizontal parts cancel each other but pull the two supports toward each other, which is why anchors must be strong. You can add the span between the supports to get the sag of the load and the length of each rope, and a breaking strength to get the smallest safe angle. The sum of the forces on the hanging object is the topic of the net force calculator, and the underlying acceleration law is on the Newton's second law calculator. Ropes on slopes are covered by the inclined plane calculator.

The models are idealised. The rope is massless and does not stretch, the pulley is frictionless and massless, the angled ropes are straight and symmetric, and the loads are static values without shock. Real ropes sag under their own weight, stretch, wear and fail first at knots, and a falling load can briefly produce several times the static tension. Use the results for physics problems and rough checks, and rely on rated working loads, safety standards and a qualified engineer for anything that holds a person or a valuable load.

📐 Formula

Hanging: T = m (g + a)    Atwood: T = 2 m1 m2 g ÷ (m1 + m2)    Angled ropes: T = m g ÷ (2 sin θ)
T = the rope tension in newtons (N). m = the mass in kg. g = 9.80665 m/s² on Earth (standard gravity), 1.62 on the Moon and 3.721 on Mars, or any value you type.
a = the upward acceleration of the load in m/s². It is 0 at rest or at constant speed, positive upward and negative downward. A rope cannot pull a load downward faster than g.
Atwood machine: a = (m2 − m1) g ÷ (m1 + m2), with the heavier mass going down. The same tension T acts on both masses.
Block on a table: a = (m2 − μ m1) g ÷ (m1 + m2) and T = m1 (a + μ g) when m2 > μ m1. Otherwise nothing moves and T = m2 g.
Angled ropes: θ is the angle of each rope above the horizontal. The horizontal part of each tension is T cos θ, the vertical part is half the weight, the sag is (span / 2) tan θ and each rope has length (span / 2) / cos θ.
Safety factor: breaking strength ÷ tension. The largest upward acceleration before the rope breaks is Tbreak / m − g, and the smallest safe rope angle is asin(m g / (2 Tbreak)).
Example: a 70 kg load accelerating upward at 1.5 m/s² gives T = 70 × (9.80665 + 1.5) = 791.47 N.

📖 How to Use This Calculator

Steps

1
Choose the setup. Use Hanging mass or elevator for a vertical rope, Two masses over a pulley for an Atwood or table system, or Two angled ropes for a weight hung from two ropes.
2
Enter the masses and the setup values. Give the mass and the vertical acceleration, or the two masses and the table friction, or the mass and the angle of the ropes above the horizontal.
3
Add a breaking strength if you want a check. Enter the rope's breaking strength in newtons to see the safety factor, the largest acceleration or the smallest angle that is still within it.
4
Read the tension, the diagram and the chart. The results list the tension and the related forces, a diagram is drawn from your numbers and a chart shows how the tension changes with acceleration, with m2 or with the angle.

💡 Example Calculations

Example 1 — A Mass Hanging at Rest

A 10 kg mass hangs motionless from a rope on Earth

1
With a = 0 the tension equals the weight: T = m (g + a) = 10 × (9.80665 + 0).
2
T = 98.066 N, which is exactly 1.000 times the weight.
Rope tension = 98.066 N
Try this example →

Example 2 — An Elevator Accelerating Upward

A 70 kg load accelerates upward at 1.5 m/s² on a rope rated at 1500 N

1
T = 70 × (9.80665 + 1.5) = 791.47 N, while the weight is 686.47 N.
2
The tension is 791.47 / 686.47 = 1.153 times the weight.
3
Safety factor = 1500 / 791.47 = 1.895. The largest upward acceleration is 1500 / 70 − 9.80665 = 11.622 m/s².
Rope tension = 791.47 N, safety factor = 1.895
Try this example →

Example 3 — An Elevator Accelerating Downward

An 80 kg load accelerates downward at 2 m/s²

1
Downward is negative, so a = −2 and T = 80 × (9.80665 − 2) = 624.53 N.
2
The weight is 80 × 9.80665 = 784.53 N, so the tension is 0.7961 times the weight and the load feels lighter.
Rope tension = 624.53 N
Try this example →

Example 4 — An Atwood Machine

Masses of 3 kg and 5 kg hang over an ideal pulley

1
a = (5 − 3) × 9.80665 / (3 + 5) = 2.452 m/s², with the 5 kg mass going down.
2
T = 2 × 3 × 5 × 9.80665 / 8 = 36.775 N.
3
The weights are 29.420 N and 49.033 N, so T lies between them, at 0.7500 times the weight of the 5 kg mass.
Tension = 36.775 N, a = 2.452 m/s²
Try this example →

Example 5 — A Block on a Table

A 4 kg block on a table with mu = 0.2 is pulled by a 2 kg mass hanging over the edge

1
The hanging weight 19.613 N exceeds the friction limit 0.2 × 4 × 9.80665 = 7.845 N, so the system moves.
2
a = (2 − 0.2 × 4) × 9.80665 / (4 + 2) = 1.961 m/s².
3
T = 4 × (1.961 + 0.2 × 9.80665) = 15.691 N, which is less than the 19.613 N hanging weight.
Tension = 15.691 N, a = 1.961 m/s²
Try this example →

Example 6 — A Sign on Two Ropes

A 20 kg sign hangs from two ropes at 30° above the horizontal, with supports 6 m apart

1
Weight = 20 × 9.80665 = 196.133 N. T = 196.133 / (2 sin 30°) = 196.13 N in each rope.
2
Each rope pulls up with 98.066 N and sideways with 196.133 × cos 30° = 169.86 N.
3
Sag = 3 × tan 30° = 1.732 m and each rope is 3 / cos 30° = 3.464 m long.
Tension in each rope = 196.13 N
Try this example →

Example 7 — Ropes That Are Too Flat

A 50 kg load hangs from two ropes at 5°, each rated at 2000 N

1
T = 50 × 9.80665 / (2 sin 5°) = 2,812.97 N, which is 5.737 times the weight.
2
Safety factor = 2000 / 2812.97 = 0.7110, below 1, so the ropes break.
3
The smallest safe angle is asin(490.33 / 4000) = 7.041°.
Tension = 2,812.97 N, safe only above 7.041°
Try this example →

❓ Frequently Asked Questions

What is tension in physics?+
Tension is the pulling force transmitted along a rope, string, cable or rod when it is pulled tight at its ends. It acts along the rope and pulls equally on the objects at both ends. A 10 kg mass hanging at rest on Earth puts 98.07 N of tension in its rope.
How do you calculate the tension in a rope holding a hanging mass?+
Use T = m (g + a), with up positive. For a mass at rest or at constant speed a = 0, so T = m g. A 10 kg mass gives 10 x 9.80665 = 98.07 N. The tension equals the weight only when the load is not accelerating.
What is the tension in an elevator cable when it accelerates?+
T = m (g + a), where a is positive when the elevator accelerates upward. A 70 kg load accelerating up at 1.5 m/s squared needs 70 x (9.80665 + 1.5) = 791.47 N, which is 1.153 times its weight. Accelerating down at 2 m/s squared an 80 kg load needs only 624.53 N.
How do you find the tension in an Atwood machine?+
For masses m1 and m2 over an ideal pulley, T = 2 m1 m2 g / (m1 + m2) and a = (m2 - m1) g / (m1 + m2). For 3 kg and 5 kg the acceleration is 2.452 m/s squared and the tension is 36.775 N, between the two weights of 29.42 N and 49.03 N.
How do you find the tension for a block on a table pulled by a hanging mass?+
With a friction coefficient mu, a = (m2 - mu m1) g / (m1 + m2) and T = m1 (a + mu g). A 4 kg block with mu = 0.2 pulled by a 2 kg hanging mass moves at 1.961 m/s squared with a tension of 15.69 N, less than the 19.61 N weight of the hanging mass.
How do you calculate the tension in two ropes at an angle?+
For two symmetric ropes at an angle theta above the horizontal, the vertical parts must hold the weight: 2 T sin(theta) = m g, so T = m g / (2 sin theta). A 20 kg load on ropes at 30 degrees puts 196.13 N in each rope, equal to the weight.
Why does a flatter rope have more tension?+
Because only the vertical part of the tension holds the weight, and a flatter rope has a smaller vertical part. To supply the same vertical force the rope must pull harder, T = m g / (2 sin theta), so the tension grows without limit as the angle approaches zero.
Can the tension in a rope be larger than the weight it holds?+
Yes. It is larger than the weight when the load accelerates upward and when two ropes hold it at less than 30 degrees from the horizontal. A 50 kg load on ropes at 5 degrees puts 2812.97 N in each rope, 5.7 times its weight.
Is the tension the same on both sides of a pulley?+
Yes, for an ideal massless rope over a frictionless massless pulley. That is why the two masses share one tension T. A real pulley with mass and friction makes the two sides differ, which this calculator does not model.
What is a safety factor for a rope?+
It is the breaking strength divided by the working tension. A rope rated at 1500 N carrying 791.47 N has a safety factor of 1.895. Real applications require much larger factors, because shocks, wear and knots reduce the usable strength.
What happens to the tension in free fall?+
If the load accelerates downward at exactly g the rope tension is zero, because T = m (g + a) with a = -g. A rope cannot push, so a load cannot be pulled down faster than g by the rope. The calculator reports a message for accelerations below -g.

What is tension in physics?

Tension is the pulling force transmitted along a rope, string, cable or rod when it is pulled tight at its ends. It acts along the rope and pulls equally on the objects at both ends. A 10 kg mass hanging at rest on Earth puts 98.07 N of tension in its rope.

How do you calculate the tension in a rope holding a hanging mass?

Use T = m (g + a), with up positive. For a mass at rest or at constant speed a = 0, so T = m g. A 10 kg mass gives 10 x 9.80665 = 98.07 N. The tension equals the weight only when the load is not accelerating.

What is the tension in an elevator cable when it accelerates?

T = m (g + a), where a is positive when the elevator accelerates upward. A 70 kg load accelerating up at 1.5 m/s squared needs 70 x (9.80665 + 1.5) = 791.47 N, which is 1.153 times its weight. Accelerating down at 2 m/s squared an 80 kg load needs only 624.53 N.

How do you find the tension in an Atwood machine?

For masses m1 and m2 over an ideal pulley, T = 2 m1 m2 g / (m1 + m2) and a = (m2 - m1) g / (m1 + m2). For 3 kg and 5 kg the acceleration is 2.452 m/s squared and the tension is 36.775 N, between the two weights of 29.42 N and 49.03 N.

How do you find the tension for a block on a table pulled by a hanging mass?

With a friction coefficient mu, a = (m2 - mu m1) g / (m1 + m2) and T = m1 (a + mu g). A 4 kg block with mu = 0.2 pulled by a 2 kg hanging mass moves at 1.961 m/s squared with a tension of 15.69 N, less than the 19.61 N weight of the hanging mass.

How do you calculate the tension in two ropes at an angle?

For two symmetric ropes at an angle theta above the horizontal, the vertical parts must hold the weight: 2 T sin(theta) = m g, so T = m g / (2 sin theta). A 20 kg load on ropes at 30 degrees puts 196.13 N in each rope, equal to the weight.

Why does a flatter rope have more tension?

Because only the vertical part of the tension holds the weight, and a flatter rope has a smaller vertical part. To supply the same vertical force the rope must pull harder, T = m g / (2 sin theta), so the tension grows without limit as the angle approaches zero.

Can the tension in a rope be larger than the weight it holds?

Yes. It is larger than the weight when the load accelerates upward and when two ropes hold it at less than 30 degrees from the horizontal. A 50 kg load on ropes at 5 degrees puts 2812.97 N in each rope, 5.7 times its weight.

Is the tension the same on both sides of a pulley?

Yes, for an ideal massless rope over a frictionless massless pulley. That is why the two masses share one tension T. A real pulley with mass and friction makes the two sides differ, which this calculator does not model.

What is a safety factor for a rope?

It is the breaking strength divided by the working tension. A rope rated at 1500 N carrying 791.47 N has a safety factor of 1.895. Real applications require much larger factors, because shocks, wear and knots reduce the usable strength.