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In most mechanics problems the string is "light", so the tension is one number, T, all the way along. Give the rope mass and that stops being true. Heavy-rope questions test whether you understand where the constant-tension rule comes from. This post gives one method, the imaginary cut, and applies it to four standard set-ups.
Why is tension the same everywhere in a light string?
The NCERT Class 11 chapter on laws of motion says it directly: "It is customary to use a constant tension T throughout the string. This assumption is true for a string of negligible mass." The reason is Newton's second law. The net force on any small piece of string equals its mass times its acceleration. If the mass is zero, the pulls on its two ends must be equal. Once the rope has mass, they can differ, and the difference accelerates or supports that piece. The JEE (Main) 2026 syllabus lists Newton's laws under Unit 3; check it for your exam year.
What is the cut-the-rope method?
Imagine cutting the rope at the point where you want the tension.
Choose the piece on one side of the cut, whichever has fewer forces on it.
The tension at the cut is the force that held that piece. Apply F = ma to the piece alone.
For a uniform rope of mass M and length L, a piece of length x has mass Mx/L.
A 2025 teaching paper, Teaching of tension force: a massive rope in equilibrium, studies a heavy rope fixed at both ends. It works from the forces on each small rope segment to the tension at every point, and reports that first-year engineering students who used it deduced the general rules and solved different problems.
Worked example 1: a rope hanging from a ceiling
A uniform rope of mass 2 kg and length 4 m hangs from a hook. Take g = 10 m/s². Find the tension 1 m below the hook.
Cut 1 m below the hook. Choose the lower piece: 3 m long, so its mass is 2 × 3/4 = 1.5 kg.
It hangs at rest, so the tension holds up its weight: T = 1.5 × 10 = 15 N.
In general, at height y above the free bottom end, T = (M/L)gy. It is zero at the bottom, Mg/2 at the middle and Mg = 20 N at the hook. The graph of T against y is a straight line.
Now hang a 5 kg block from the bottom. The piece below the cut now includes the block, so T = (5 + 1.5) × 10 = 65 N, and the tension at the hook is (5 + 2) × 10 = 70 N.
Worked example 2: a rope pulled along a smooth floor
A 3 kg rope, 6 m long, lies straight on a smooth floor. A force of 18 N pulls one end along the rope's length.
The whole rope accelerates at a = 18/3 = 6 m/s².
Cut 2 m from the trailing end. The piece behind the cut has mass 3 × 2/6 = 1 kg.
The only horizontal force on it is the tension at the cut, so T = 1 × 6 = 6 N.
In general, at distance x from the trailing end, T = Fx/L. It rises from zero at the free end to the full 18 N at the end being pulled, and is 9 N at the middle.
Worked example 3: a rope pulling a block
A 4 kg block on a smooth floor is pulled by a 1 kg rope, with 25 N applied to the rope's front end.
System acceleration: 25/(4 + 1) = 5 m/s².
Tension where the rope meets the block: the block alone needs 4 × 5 = 20 N.
Tension at the rope's midpoint: the piece behind the cut is the block plus half the rope, 4.5 kg, so T = 4.5 × 5 = 22.5 N.
The rope uses 5 N of the applied force to accelerate itself; a massless rope would pass all 25 N to the block.
Worked example 4: lifting a rope with acceleration
The rope from Example 1 is pulled up by its top end, accelerating at 2 m/s². For the piece of mass m′ below a cut, T = m′(g + a). At the top, T = 2 × 12 = 24 N; at the middle, 1 × 12 = 12 N.
How does this connect to waves on a string?
Wave speed on a string depends on the tension: OpenStax University Physics gives v = √(T/μ), where μ is the mass per length. On a hanging rope, T = μgy, so v = √(gy): a pulse speeds up as it climbs from the bottom. Adding up dy/√(gy) from the bottom to the top gives the travel time 2√(L/g). For the 4 m rope that is 2 × √0.4 ≈ 1.26 s.
Common mistakes and how to avoid them
Using one T for a rope with mass. Always ask: tension at which point?
Including the wrong piece. The tension at a cut supports or accelerates everything on the chosen side, including any block attached.
Mixing up which end is zero. The free end of a hanging or dragged rope has zero tension, not the end being held.
A practice routine
Day 1: Sketch T against position for each worked example. All four should be straight lines.
Day 2: Three hanging-rope problems with a block attached at the bottom.
Day 3: Two problems where a heavy rope passes over a smooth pulley, using the cut method on each side.
Day 4: Find the pulse travel time for ropes of 1 m and 9 m. Check the ratio is 1:3.
When the idea itself is the sticking point, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise tension and Newton's second law on their own. For a closely related energy problem, see our post on the work needed to pull a hanging chain onto a table.
Tonight, find the tension 3 m below the hook in Worked example 1 and check it is 5 N.
