In short
Tension is the pulling force transmitted through a string, rope, or chain. In a massless (ideal) string, the tension is the same at every point along its length. In a rope with mass, tension varies — for a vertical rope hanging under its own weight, the tension at a distance x from the bottom is T(x) = \frac{M}{L}xg, greatest at the top and zero at the free bottom end. Unlike rigid rods, flexible strings can only pull, never push.
A clothesline stretches between two hooks on the wall of an Indian balcony. You hang a wet bedsheet in the middle. The line sags, the hooks groan, and if the sheet is heavy enough, the line snaps. Something inside that line was pulling — pulling so hard it broke. That pulling force has a name: tension.
Tension is not something the string "has" like mass or colour. It is something the string does. Every tiny segment of the string pulls on the segments next to it, passing the force along like a chain of people in a tug-of-war, each gripping the next person's hand. Cut the string at any point, and the two halves fly apart — proof that each half was pulling on the other.
This is different from a rigid rod. Push one end of a bamboo stick, and the other end pushes the object it touches. A stick can push or pull. But try pushing a rope — it buckles, goes slack, does nothing. A flexible string can only transmit a pull. That single constraint — strings can only pull, never push — shapes every problem in this chapter.
What tension actually is
Pick up a bucket of water from a well using a rope. The bucket hangs from one end of the rope, your hand holds the other. Gravity pulls the bucket down. Your hand pulls up. The rope transmits your pull all the way down to the bucket. The force with which any cross-section of the rope pulls on the section below it — that is the tension at that cross-section.
Notice the key idea: tension is a contact force that acts in both directions at every point. If you make an imaginary cut anywhere along the rope, the piece above pulls up on the piece below, and the piece below pulls down on the piece above. These are Newton's third law pairs — equal in magnitude, opposite in direction.
The ideal string: massless and inextensible
Most problems in your textbook use an "ideal" string — one that is massless and does not stretch. These two assumptions simplify the physics enormously, and they are reasonable for a thin nylon cord holding a few kilograms.
Assumptions: The string is massless (m_\text{string} = 0) and inextensible (its length does not change). There is no friction or air resistance.
Why tension is the same everywhere in a massless string
Take a small segment of the string. It has mass \Delta m (which is zero for a massless string). Call the tension on the left end T_1 and on the right end T_2. Apply Newton's second law to this segment:
Why: the net force on the segment equals its mass times its acceleration. T_2 pulls one way, T_1 pulls the other.
Since \Delta m = 0 for a massless string:
Why: if the mass of the segment is zero, the net force on it must be zero — regardless of how large the acceleration is. This means both ends of the segment carry the same tension.
This holds for every segment. So the tension is the same at every point along a massless string. This is the fundamental property that makes ideal-string problems tractable: you have one unknown T, not a different tension at every point.
What "inextensible" means
An inextensible string does not stretch. This means every point on the string has the same acceleration as every other point — and the same acceleration as whatever is attached to the ends. If block A accelerates at 3 m/s² to the right, block B (connected by an inextensible string) also accelerates at 3 m/s² to the right. The string is a rigid constraint on motion.
A rope with mass: tension varies along the length
Real ropes have mass. A thick jute rope used to pull water from a village well, a climbing rope, the long rope in a school tug-of-war — these are not massless. And when the rope has mass, the tension is no longer the same everywhere.
Deriving the tension in a hanging rope
Take a uniform rope of total mass M and length L, hanging vertically from a fixed support. The rope is in equilibrium — nothing is accelerating. You want to find the tension T(x) at a distance x from the bottom end.
Step 1. Identify the segment. Consider the portion of the rope from the bottom end (free) up to a height x.
The rope is uniform, so its linear mass density (mass per unit length) is:
Why: the rope's mass is distributed evenly. A piece of length x has mass \lambda x = \frac{M}{L}x.
Step 2. Draw the FBD of this segment. Two forces act on it:
- Tension T(x) at the top of the segment, pulling upward — this is the force the rest of the rope (above the cut) exerts on the segment.
- Weight of the segment, \frac{M}{L}xg, pulling downward.
There is no force at the bottom because the bottom end is free (nothing attached).
Step 3. Apply Newton's second law. The segment is in equilibrium (a = 0), so the net force is zero:
Why: equilibrium means no acceleration. The upward tension must exactly balance the downward weight of all the rope below the cut.
Step 4. Solve for T(x).
Why: the tension at height x equals the weight of the rope hanging below that point. This makes physical sense — the rope at height x must support everything beneath it.
This result tells you three things immediately:
- At the bottom (x = 0): T(0) = 0. The free end carries no load. Nothing hangs below it.
- At the top (x = L): T(L) = Mg. The support point carries the full weight of the rope.
- In between: tension increases linearly from bottom to top. The higher you go, the more rope hangs below, the greater the tension.
This is why a long, heavy rope used in tug-of-war is most likely to snap near the top — near where it is tied or held. That is where the tension is greatest.
Explore the tension profile yourself
The formula says tension climbs linearly from zero at the free bottom to Mg at the support. Drag the point below to move along the rope's height and watch the tension — and the mass it supports — update in real time.
Rigid rods versus flexible strings
You have two objects connected by something. Is it a string or a rod? The physics changes.
| Property | Flexible string | Rigid rod |
|---|---|---|
| Can pull? | Yes | Yes |
| Can push? | No — goes slack | Yes |
| Tension sign | Always \geq 0 | Can be positive (tension) or negative (compression) |
| Bends under load? | Yes — takes the shape that makes every segment in equilibrium | No — maintains its shape |
| Constraint | Length is fixed (inextensible), but direction is free | Both length and direction are fixed |
Think of it this way: tie a cricket ball to a string and whirl it in a circle. The string pulls the ball inward. Now try pushing the ball outward with the string — you cannot. The string goes limp. Replace the string with a thin bamboo stick, and you can push the ball away. A rod transmits both tension and compression; a string transmits only tension.
This constraint matters in problems. If you solve for the tension in a string and get a negative number, the physical meaning is that the string should be pushing — but strings cannot push. So a negative tension means the string has gone slack, and your model (with the string taut) no longer applies.
Worked examples
Example 1: Two blocks connected by a light string
A 4 kg block and a 6 kg block sit on a smooth (frictionless) horizontal surface, connected by a light (massless) string. A horizontal force of 50 N is applied to the 6 kg block. Find the acceleration of the system and the tension in the string.
Step 1. Find the acceleration of the whole system.
Treat both blocks plus the string as a single system. The only external horizontal force is the 50 N pull. The total mass is 4 + 6 = 10 kg.
Why: the string is massless and inextensible, so both blocks accelerate together at the same rate. The internal tension forces cancel when you consider the whole system — only external forces matter for the system's acceleration.
Step 2. Find the tension in the string.
Now isolate the 4 kg block alone. The only horizontal force on it is the tension T, pulling it to the right.
Why: the 4 kg block has no other horizontal force. The string must supply exactly the force needed to give this block an acceleration of 5 m/s².
Step 3. Verify using the 6 kg block.
For the 6 kg block, the applied force is 50 N to the right, and the tension is T to the left (the 4 kg block pulls back on the 6 kg block via the string).
Why: both approaches give T = 20 N. The 6 kg block equation is a consistency check — Newton's second law applied to either block must give the same tension, because the string is massless.
Result: The system accelerates at a = 5 m/s², and the tension in the string is T = 20 N.
What this shows: In a connected-body problem on a frictionless surface, the system method gives you the acceleration quickly, and isolating one block gives you the tension. The tension is always less than the applied force — the string only needs to accelerate the block behind it, not the whole system.
Example 2: A heavy rope hanging vertically
A rope of length 10 m and mass 5 kg hangs vertically from a ceiling. Find the tension in the rope at the top, at the middle (5 m from the bottom), and at the bottom.
Step 1. Write the tension formula.
For a uniform rope of mass M = 5 kg and length L = 10 m, the tension at height x from the bottom is:
Why: the tension at height x supports the weight of the rope below it. The linear mass density is M/L = 0.5 kg/m, so the mass below x is 0.5x kg, and its weight is 0.5 \times 9.8 \times x N.
Step 2. Tension at the bottom (x = 0).
Why: the bottom end is free. There is no rope below it, so there is nothing to support. The tension vanishes.
Step 3. Tension at the middle (x = 5 m).
Why: the midpoint supports half the rope's mass. Half of 5 kg is 2.5 kg, and 2.5 \times 9.8 = 24.5 N. This is exactly half the total weight.
Step 4. Tension at the top (x = 10 m).
Why: the top supports the entire rope. The full weight is Mg = 5 \times 9.8 = 49 N. This is the maximum tension anywhere in the rope.
Result: The tension is 0 N at the bottom, 24.5 N at the middle, and 49 N at the top. The tension increases linearly from bottom to top.
What this shows: In a heavy rope hanging under its own weight, every point carries a different tension. The farther up you go, the more rope hangs below, and the greater the tension. The graph of T versus x is a straight line — a direct consequence of the rope being uniform. If the rope were thicker at the bottom, the line would curve.
Common confusions
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"Tension is a property of the string." Not exactly. Tension is a force — it has magnitude and direction, and it acts on specific objects. The string is the medium through which tension is transmitted, but the tension is not stored in the string the way mass is. Cut the string, and the tension vanishes instantly.
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"A heavy rope has the same tension everywhere." No — that is only true for a massless string. In a rope with mass, the tension varies. For a vertical rope in equilibrium, the variation is linear: T(x) = (M/L)xg. In accelerating systems, the variation can be different.
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"Tension always equals the weight of what hangs from the string." Only in equilibrium. If you accelerate a bucket upward using a rope, the tension exceeds the bucket's weight — the extra force is what produces the upward acceleration. If the bucket is in free fall (both you and the bucket falling together), the tension is zero even though the bucket still has weight.
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"The string can push if you push hard enough." No. A flexible string, by definition, cannot transmit a compressive force. If the calculation gives a negative tension, the string has gone slack. You need to change your model (perhaps the objects are no longer connected).
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"Massless strings don't exist, so why study them?" Because the approximation is excellent for most problems. A thin nylon thread weighing 2 g connecting two 5 kg blocks contributes 0.04% of the total mass — ignoring it changes the answer by less than the rounding error. The massless-string model is not a lie; it is a good approximation stated honestly.
If you are comfortable with tension in ideal strings and the formula for tension in a hanging rope, you have the tools for most textbook problems. What follows is for readers who want the general case — tension in an accelerating massive rope — and the formal connection to constraints.
Tension in a massive rope that is accelerating
Suppose the same rope of mass M and length L hangs vertically, but instead of being in equilibrium, the entire rope accelerates upward with acceleration a. (Imagine pulling the top end upward with a force greater than Mg.)
Take the same segment from the bottom to height x. The mass of this segment is still (M/L)x. But now the net force on it is not zero — it has an upward acceleration a.
Why: every segment needs a net upward force to accelerate upward. The tension must now support both the weight and provide the force for acceleration. Replace g with (g + a) — this is the effective gravitational field in an accelerating frame.
At the top (x = L): T(L) = M(g + a). This is the force your hand must exert to accelerate the entire rope upward.
At the bottom (x = 0): T(0) = 0, just as before — the free end still carries nothing.
The tension still varies linearly, but the slope is steeper. The faster you accelerate the rope, the greater the tension at every point.
Why constraints matter: degrees of freedom
A massless, inextensible string connecting two blocks imposes a constraint: both blocks must have the same acceleration (in magnitude, along the string direction). This reduces the number of unknowns in the problem. Without the string, two blocks on a surface have two independent accelerations. With the string, they share one acceleration, and you gain a new unknown (the tension) — but you also gain a new equation (Newton's second law for the second block). The problem remains solvable.
For a massive string, the constraint is more subtle. The string itself is a continuous body with infinitely many points. The tension varies continuously, and you need calculus — integrating Newton's second law over the string's length — to find the tension function T(x). The linear result T(x) = (M/L)x(g + a) is the simplest case (uniform rope, uniform acceleration). A non-uniform rope (thicker at one end) gives a nonlinear T(x).
When the string goes slack
Consider a block hanging from a string attached to the ceiling of a lift. If the lift accelerates downward at exactly g (free fall), the block and the ceiling fall together — there is no relative motion between them, and the tension in the string drops to zero. The string goes slack.
More generally, the string goes slack whenever the calculated tension would be negative. In problems with swinging objects (a ball on a string whirled in a vertical circle), the tension is T = \frac{mv^2}{r} - mg\cos\theta at angle \theta from the top. When v is too small, this expression goes negative, meaning the string cannot maintain the circular path — the ball falls inward, and the string goes limp. The critical speed at the top of the circle, where T = 0 and the string is barely taut, is v_{\text{min}} = \sqrt{gR}.
Where this leads next
- Pulleys and String Constraints — how pulleys redirect tension and create acceleration constraints between connected objects.
- Applications: Connected Bodies and Systems — Atwood machines, blocks on inclines, and multi-body problems using the techniques from this article.
- Free Body Diagrams — the systematic method for identifying all forces on a body, which every tension problem requires.
- Newton's Second Law — the foundation for every calculation in this article: net force equals mass times acceleration.
- Friction — in real-world problems, surfaces are not frictionless, and friction adds another force to the free body diagram alongside tension.