A rope holding a weight seems too simple to need a theory. The theory arrives when the rope is not vertical, when there are two ropes, when the weight sits on a slope, or when the rope runs over a pulley to a second weight — and in each case the answer comes from the same move: isolate the body, draw every force on it, and require the forces to balance (or to produce exactly the acceleration the body has). Tension is a force, measured in newtons, and it is the same at both ends of a massless rope. Everything else in this guide is geometry applied to that one fact.

What tension is

Tension is a pulling force transmitted through a flexible connector — a rope, cable, string, chain or belt. It pulls away from the body at each end of the connector, along the connector’s line, and for an ideal (massless, inextensible) rope it has the same magnitude at both ends and everywhere along its length. A rope can only pull, never push: a rope in compression goes slack, and a slack rope carries no tension at all. Real ropes have mass and stretch, and the tension then varies along the length and the rope extends under load, but for the problems in this guide — and for most engineering estimates — the ideal rope is the model, and its error is smaller than the rounding in the inputs.

The unit is the newton (N): 1 N accelerates 1 kg at 1 m/s². Weight is a force too, and the weight of a mass m is mg, with g = 9.81 m/s² at the Earth’s surface: a 10 kg mass weighs 98.1 N, a 20 kg mass 196.2 N. Tension problems are weight problems with geometry attached, and the geometry is where the answers stop being obvious.

One rope, one hanging mass: the base case

A mass m hangs at rest from a vertical rope. Two forces act on the mass: weight mg downward and tension T upward. At rest the forces balance, so T = mg. A 10 kg mass gives T = 10 × 9.81 = 98.1 N. If the mass accelerates upward at a, the tension must supply both the weight and the acceleration: T = m(g + a). Accelerating the same 10 kg mass upward at 2 m/s² needs T = 10 × 11.81 = 118.1 N. Accelerating it downward at 2 m/s² needs T = 10 × 7.81 = 78.1 N — and in free fall (a = g downward) the tension is zero, which is why a falling elevator’s cable goes slack in the idealised case and why astronauts in orbit feel weightless: everything accelerates together, and no tension is needed to hold anything up.

The sign convention is the only discipline the case needs: choose upward as positive, write the sum of forces as ma, and let the algebra carry the signs. Guessing the direction of the tension and then checking it against the equation is how most errors enter; writing the equation first is how they are avoided.

Two ropes at angles: the geometry that multiplies tension

Hang the 10 kg mass from two ropes, each making 30° with the horizontal, symmetric about the vertical. Each rope contributes a vertical component T·sin(30°) = 0.5T, and the two together must balance the weight: 2 × 0.5T = 98.1, so T = 98.1 N per rope — the same as the single vertical rope, because at 30° the two ropes’ vertical components add to exactly the weight. Lower the angle to 10° and the arithmetic changes sharply: each rope contributes T·sin(10°) = 0.1736T, so 2 × 0.1736T = 98.1 gives T = 282.4 N per rope, nearly three times the weight. At 5°, it is 563 N per rope. As the ropes approach horizontal, the tension approaches infinity — which is why a washing line, a hammock, or a cable strung nearly level sags: the sag is the rope finding the angle at which the tension is finite, and a rope pulled truly straight between two fixed points under a central load would need infinite tension to stay straight.

The horizontal components cancel in the symmetric case — each rope pulls outward with T·cos(θ), equal and opposite — and that cancellation is a check on the setup: if the angles are unequal, the horizontal components do not cancel, the mass hangs off-centre, and the two tensions differ. The general case is two equations (horizontal balance, vertical balance) for two unknowns, and it solves the same way: resolve each tension into components, sum each direction to zero, and solve. The tension calculation on this site carries the component arithmetic for the common configurations, so the geometry can be checked against a number rather than trusted from a diagram.

A mass on an incline, held by a rope

A 20 kg block rests on a frictionless incline of 30°, held by a rope parallel to the slope. Gravity acts vertically, but the slope constrains the motion to the plane, so the useful directions are along the slope and perpendicular to it. The component of weight along the slope is mg·sin(30°) = 196.2 × 0.5 = 98.1 N, pulling the block down the slope; the component perpendicular to the slope, mg·cos(30°) = 169.9 N, is balanced by the normal force from the surface and never reaches the rope. At rest, the tension equals the along-slope component: T = 98.1 N — half the weight, because the slope carries the other half through the normal force.

Add friction and the rope’s load falls further: static friction acts up the slope (opposing the tendency to slide down) with a maximum of μₛ times the normal force, so the tension needed at rest is the along-slope component minus whatever friction supplies, down to zero if friction alone can hold the block. Tilt the rope away from parallel and the tension splits again — only its along-slope component opposes the slide, and its perpendicular component reduces the normal force and therefore the friction. The incline case is the same equilibrium move as the hanging case, with the coordinate axes rotated to fit the surface.

Pulleys and two bodies: one tension, two masses

A massless, frictionless pulley changes a rope’s direction without changing its tension. Two masses, m₁ and m₂, hang from the two ends of one rope over such a pulley (an Atwood machine). The same tension T acts on both masses, upward on each; gravity acts downward on each. If m₂ is the heavier, the system accelerates with m₂ falling and m₁ rising, and the two equations — T − m₁g = m₁a and m₂g − T = m₂a — solve together to a = (m₂ − m₁)g / (m₁ + m₂) and T = 2m₁m₂g / (m₁ + m₂). With m₁ = 5 kg and m₂ = 10 kg: a = (5 × 9.81) / 15 = 3.27 m/s², and T = (2 × 5 × 10 × 9.81) / 15 = 65.4 N — between the two weights (49.05 N and 98.1 N), as it must be, since the lighter mass is pulled up and the heavier mass is held back.

Real pulleys have mass and friction, and the tension then differs on the two sides by the amount needed to accelerate the pulley itself — the difference is the torque that spins it. Belt drives, cranes and elevators are the same physics at scale: the tension in the loaded side and the slack side differ by the load being moved, and the machinery is sized on the larger tension. For problem-solving, the ideal pulley is the starting model and the correction is a second step, not a different subject.

Frequently asked questions

What is tension in physics?

Tension is the pulling force transmitted through a rope, cable, string or chain. It pulls away from the body at each end of the connector, along the connector’s line, and for an ideal massless rope it has the same magnitude at both ends. It is measured in newtons. A rope can only pull, never push — a rope in compression goes slack and carries no tension.

How do you calculate tension in a rope holding a weight?

For a mass at rest on a vertical rope, tension equals weight: T = mg, with g = 9.81 m/s². A 10 kg mass gives 98.1 N. If the mass accelerates upward at a, T = m(g + a); accelerating downward, T = m(g − a). For ropes at angles, resolve each tension into vertical and horizontal components and balance each direction separately.

Why is tension higher when a rope is nearly horizontal?

Because only the vertical component of the tension opposes gravity, and that component shrinks as the angle to the horizontal shrinks. Two ropes at 10° holding 10 kg carry about 282 N each — nearly three times the weight — and the tension grows without limit as the angle approaches zero. A rope pulled truly straight under a central load would need infinite tension, which is why loaded ropes always sag.

Is tension the same on both sides of a pulley?

For an ideal massless, frictionless pulley, yes — the pulley changes the rope’s direction without changing its tension, and the same tension acts on both sides. A real pulley with mass or friction carries different tensions on its two sides, the difference being what accelerates the pulley itself.

What is the difference between tension and weight?

Weight is the gravitational force on a mass, mg, acting downward. Tension is the pulling force a connector exerts, acting along the connector. In the simplest case — a mass hanging at rest from a vertical rope — the two are equal in magnitude, but in general they are different forces, in different directions, related by the equilibrium (or acceleration) equations, not by identity.

Does tension change along a rope?

For an ideal massless rope, no — tension is uniform along its length. For a real rope with mass, tension varies along the length, increasing toward the support by the weight of the rope below each point. A heavy cable holding a load carries noticeably more tension at the top than at the bottom, and that variation is part of sizing real cables and chains.