Tension, Pulleys & Connected Systems

A crate sits on a table. A rope runs from it, over a little wheel at the table's edge, and down to a bucket hanging in the air. Let go, and the bucket falls, the crate slides, and the two of them somehow agree on a speed. They are not touching. The only thing they share is a rope.

That rope is doing something worth understanding. It carries a force from one object to another, it changes the direction of that force when it goes round the wheel, and it forces the two objects to move in lockstep. Get those three ideas straight and a whole family of problems (Atwood machines, blocks on inclines, trains of blocks, block and tackle, monkeys climbing ropes) becomes the same problem wearing different clothes.

In the friction lesson you drew a free body diagram for one block and solved it. Nothing about that changes here. What is new is that you now draw two diagrams, get two equations, and need one more relation (the constraint) to close the system.

What a rope can and cannot do

A rope pulls. It never pushes. That is the whole of it, and it is the first thing to check when a sign comes out wrong: if your algebra hands you a negative tension, the rope has gone slack and your model is broken, not merely off by a minus sign.

The magnitude of that pull is the tension $T$, measured in newtons. Pick any point along the rope and cut it there in your imagination. The tension at that point is the force each half exerts on the other. Both halves feel the same size of pull, in opposite directions: that is Newton's third law applied to a cut.

Now ask whether $T$ is the same at every point. Take a short segment of rope of mass $\delta m$, with tension $T_{\text{above}}$ pulling up on it and $T_{\text{below}}$ pulling down, in a system accelerating upward at $a$. Newton gives

$T_{\text{above}} - T_{\text{below}} - \delta m\, g = \delta m\, a.$

So the tensions differ by $\delta m\,(g+a)$. Set $\delta m = 0$ (a massless or "light" rope) and the difference vanishes. The tension is the same everywhere along a single massless rope. That is the entire content of the word "light" in a problem statement, and it is why you are allowed to write one symbol $T$ instead of a different one at every point.

An ideal pulley carries this round a corner. It has no mass and no bearing friction, so it needs no net torque to spin, so the rope does not have to pull harder on one side than the other. A pulley changes the direction of the tension and nothing else.

The Atwood machine

Two masses, $m_1$ and $m_2$, hang from the ends of a light rope over an ideal pulley. Say $m_1 > m_2$, so $m_1$ goes down and $m_2$ goes up.

Draw each block on its own. On $m_1$: weight $m_1 g$ down, tension $T$ up. On $m_2$: weight $m_2 g$ down, the same $T$ up. Now the constraint. The rope cannot stretch, so whatever distance $m_1$ falls, $m_2$ rises: same distance, same speed, same magnitude of acceleration $a$.

The trick that saves sign errors is to choose "positive" separately for each block, in the direction you expect it to move. Down is positive for $m_1$, up is positive for $m_2$. Then both blocks have the same positive $a$:

$m_1 g - T = m_1 a, \qquad T - m_2 g = m_2 a.$

Two equations, two unknowns. Add them, and $T$ cancels:

$(m_1 - m_2) g = (m_1 + m_2) a \quad\Longrightarrow\quad a = \frac{m_1 - m_2}{m_1 + m_2}\, g.$

Put that back into either equation to get the tension:

$T = m_1(g - a) = \frac{2 m_1 m_2}{m_1 + m_2}\, g.$

Read what those say. The numerator $m_1 - m_2$ is the unbalanced weight; the denominator $m_1 + m_2$ is all the mass that has to be moved. Equal masses give $a = 0$ and $T = m g$, a stalemate. Let $m_2 \to 0$ and you get $a \to g$ and $T \to 0$: free fall, with a rope that has forgotten it exists.

That tension result is worth staring at. $T$ lies strictly between $m_2 g$ and $m_1 g$. It cannot equal $m_1 g$, because then $m_1$ would not accelerate. A very common wrong answer is "the tension is the weight of the lighter block". It is not, unless nothing is moving.

Worked example 1: the crate and the bucket

A $4.0\;\mathrm{kg}$ block on a frictionless table is joined by a light string over a pulley at the table edge to a $2.0\;\mathrm{kg}$ block hanging freely. Find the acceleration and the tension. Use $g = 9.8\;\mathrm{m/s^2}$.

Start on blank paper with two free body diagrams, not with an equation.

The table block, call it $M = 4.0\;\mathrm{kg}$, feels its weight down, the normal force up, and $T$ horizontally toward the pulley. Vertically it is not accelerating, so $N = Mg$ and the vertical direction tells you nothing else. Horizontally, taking "toward the pulley" as positive:

$T = M a.$

The hanging block, $m = 2.0\;\mathrm{kg}$, feels $mg$ down and $T$ up. Taking down as positive:

$mg - T = m a.$

Same $a$, because the string is inextensible: the block slides exactly as fast as the bucket drops. Substitute the first into the second:

$mg - Ma = ma \quad\Longrightarrow\quad a = \frac{m}{M+m}\,g = \frac{2.0}{6.0}(9.8) = 3.27\;\mathrm{m/s^2}.$

Then $T = Ma = (4.0)(3.27) = 13.1\;\mathrm{N}$.

Sanity check: $mg = 19.6\;\mathrm{N}$, and the tension came out less than that, as it must for the bucket to fall at all.

There is a shortcut hiding here. Adding the two equations killed $T$ and left $mg = (M+m)a$: the driving force is the hanging weight, and the mass being moved is everything on the rope. You can go straight to that if you only want $a$. But you can only do it when nothing internal is asked for: the moment a question says "find the tension", you need the individual equations back.

If that table had friction with $\mu_k$, the table block's equation becomes $T - \mu_k M g = M a$ and everything else is unchanged. Try it with $M = 5.0\;\mathrm{kg}$, $\mu_k = 0.20$, $m = 3.0\;\mathrm{kg}$: adding the two equations gives $mg - \mu_k M g = (M+m) a$, so $a = (29.4 - 9.8)/8.0 = 2.45\;\mathrm{m/s^2}$.

Writing the constraint properly

For two blocks on one straight rope, the constraint is obvious: same speed, same acceleration. For anything more tangled, guessing is how people lose points. The reliable method is to write the total rope length as a sum of segment lengths, note that it is constant, and differentiate twice.

Take a movable pulley: a load hangs from a pulley that is itself held up by two parallel segments of one rope. One end of the rope is anchored to the ceiling, the rope goes down round the movable pulley, up over a fixed pulley, and out to your hand.

Measure downward from the ceiling. Let $y$ be the depth of the movable pulley and $x$ be the depth of the free end. The rope length is

$L = y + y + x + (\text{constants for the bits round the fixed pulley}) = 2y + x + \text{const}.$

$L$ never changes, so $2\dot y + \dot x = 0$ and $2\ddot y + \ddot x = 0$. In words: the free end moves twice as far, twice as fast, and with twice the acceleration of the load.

$a_{\text{end}} = 2\, a_{\text{load}}.$

Meanwhile the force side goes the other way. Two rope segments pull up on the movable pulley, each with tension $T$, so the upward force on the load assembly is $2T$. You pull with $T$ and the load feels $2T$. Twice the force over half the distance, which is exactly what has to happen if energy is not to appear from nowhere.

Worked example 2: a movable pulley with a counterweight

An $8.0\;\mathrm{kg}$ block hangs from one end of a rope. The rope runs up over a fixed pulley, down around a movable pulley carrying a $5.0\;\mathrm{kg}$ load, and its other end is tied to the ceiling. Find the acceleration of the $5.0\;\mathrm{kg}$ load.

Two idealizations do the heavy lifting. One rope means one tension $T$ throughout, including the two segments holding the movable pulley. And the $8.0\;\mathrm{kg}$ block is the free end, so its acceleration is twice the load's. Call the load's acceleration $a$ upward; then the block's acceleration is $2a$ downward.

Block, taking down as positive:

$m_1 g - T = m_1 (2a), \qquad (8.0)(9.8) - T = 16.0\,a.$

Movable pulley plus load, taking up as positive, with $2T$ up and $m_2 g$ down:

$2T - m_2 g = m_2 a, \qquad 2T - (5.0)(9.8) = 5.0\,a.$

From the first, $T = 78.4 - 16.0 a$. Substitute:

$2(78.4 - 16.0a) - 49.0 = 5.0a$ $156.8 - 32.0a - 49.0 = 5.0a$ $107.8 = 37.0 a \quad\Longrightarrow\quad a = 2.91\;\mathrm{m/s^2}.$

The load accelerates up at $2.91\;\mathrm{m/s^2}$ and the block falls at $5.83\;\mathrm{m/s^2}$. Check the plausibility: $5.0\;\mathrm{kg}$ hanging on two segments behaves like a much lighter load, so the $8.0\;\mathrm{kg}$ block wins easily, and it should. Also check the block's number is below $g$. It is.

Trains of blocks, and where to cut

Blocks in a line joined by strings need no pulleys, but the same two ideas. Three blocks of $1.0$, $2.0$ and $3.0\;\mathrm{kg}$ sit connected in that order on a frictionless floor, and an $18\;\mathrm{N}$ force pulls the $1.0\;\mathrm{kg}$ block. Find the tension between the $2.0$ and the $3.0\;\mathrm{kg}$ blocks.

Everything moves together, so treat the train as one object of $6.0\;\mathrm{kg}$:

$a = \frac{18}{6.0} = 3.0\;\mathrm{m/s^2}.$

Now cut the train at the string you care about, and look at just one side of the cut. The clean side here is the $3.0\;\mathrm{kg}$ block on its own: the only horizontal force on it is that tension, so

$T = (3.0)(3.0) = 9.0\;\mathrm{N}.$

Choosing which side of the cut to take is the whole skill. Take the side with fewer forces on it. Had you taken the $1.0 + 2.0$ side instead you would write $18 - T = (3.0)(3.0)$ and get the same $9.0\;\mathrm{N}$, with more chances to slip.

This also explains why a string nearer the back of a train carries less tension: it only has to accelerate the blocks behind it.

Worked example 3: two inclines and a pulley

A $4.0\;\mathrm{kg}$ block sits on a frictionless $30^\circ$ incline. Over a pulley at the apex, a rope connects it to a $5.0\;\mathrm{kg}$ block on a frictionless $45^\circ$ incline on the other side. Find the acceleration.

Nothing conceptual is new, but the geometry is. From the friction lesson, the component of weight along a slope of angle $\theta$ is $mg\sin\theta$, and the component pressing into the surface is $mg\cos\theta$. Only the along-slope component matters here, because the ropes and motion are along the slopes.

Guess a direction: the $5.0\;\mathrm{kg}$ block on the steeper side slides down. Take positive as "down the $45^\circ$ slope, and therefore up the $30^\circ$ slope", which is one consistent direction round the rope.

$m_2 g\sin 45^\circ - T = m_2 a, \qquad T - m_1 g \sin 30^\circ = m_1 a.$

Add them:

$g(m_2 \sin 45^\circ - m_1 \sin 30^\circ) = (m_1 + m_2)\,a.$

Now the numbers. $\sin 45^\circ = 0.7071$, so $m_2 \sin 45^\circ = 3.536$. $\sin 30^\circ = 0.5$, so $m_1 \sin 30^\circ = 2.00$. The difference is $1.536\;\mathrm{kg}$, and the total mass is $9.0\;\mathrm{kg}$:

$a = \frac{(9.8)(1.536)}{9.0} = \frac{15.05}{9.0} = 1.67\;\mathrm{m/s^2}.$

Positive, so the guessed direction was right. Had it come out negative, the magnitude would still be correct and you would just report the motion as the other way. That is the practical reason to guess rather than agonise.

When the frame moves, and when the rope has mass

Two more situations break the simplest picture, and both appear in the problems.

An Atwood machine inside an accelerating elevator. If the elevator accelerates upward at $A$, then relative to the elevator everything behaves as if gravity were stronger: $g_{\text{eff}} = g + A$. Every formula you derived carries over with $g$ replaced by $g_{\text{eff}}$. For $2.0$ and $3.0\;\mathrm{kg}$ in an elevator accelerating up at $2.0\;\mathrm{m/s^2}$:

$T = \frac{2 m_1 m_2}{m_1+m_2}\,g_{\text{eff}} = \frac{2(2.0)(3.0)}{5.0}(11.8) = (2.4)(11.8) = 28.3\;\mathrm{N}.$

A rope with mass. Then the tension is not uniform, and you go back to the segment argument from the start of this lesson. The tension at any cut must support everything below the cut and accelerate it. A $7.0\;\mathrm{kg}$ block hangs from a uniform rope of mass $3.0\;\mathrm{kg}$, all accelerating upward at $2.0\;\mathrm{m/s^2}$. At a point one quarter of the way up the rope from the block, what is below is the block plus a quarter of the rope:

$m_{\text{below}} = 7.0 + 0.25(3.0) = 7.75\;\mathrm{kg},$ $T = m_{\text{below}}(g + a) = (7.75)(11.8) = 91.5\;\mathrm{N}.$

The same cut-and-look-below rule handles the monkey problems. A monkey climbing a rope is not a passenger: it pulls on the rope with its hands, and the rope pulls back. Write $T - mg = m a_{\text{monkey}}$ where $a_{\text{monkey}}$ is measured relative to the ground. If a problem gives the monkey's acceleration relative to the rope, convert first: the rope's own acceleration on the monkey's side is minus the counterweight's, so $a_{\text{monkey}} = a_{\text{rel}} - a_{\text{block}}$ when the block rises.

How to recognise this technique

Reach for connected-system methods when a problem statement contains any of these signals.

A rope, string, cord or cable joining two things that both move. The giveaway is that you cannot solve either object alone: each free body diagram has an unknown $T$ in it, so you need both diagrams plus a constraint.

The words light, massless, ideal or frictionless pulley. These are permission slips. They mean one tension for the whole rope, and they are stated precisely because without them the problem would be much harder.

A question that asks for a tension specifically. That tells you the add-the-equations shortcut is not enough on its own; you will need to go back into one block's equation. And it tells you which block to cut at: the one with the fewest other forces on it.

Anything hanging from something else that hangs, or any pulley that is itself free to move. Both mean a constraint that is not simply $a_1 = a_2$. Write the rope length as a sum, differentiate twice, and trust the result over your intuition.

A pulley axle question. The rope pulls down on the axle with $T$ on each side, so the total is $2T$: for the $5.0$ and $3.0\;\mathrm{kg}$ Atwood machine, $2(36.75) = 73.5\;\mathrm{N}$, which is less than the combined weight of $78.4\;\mathrm{N}$ because part of the system is accelerating downward.

Your routine, every time: draw a separate free body diagram for each object, choose a positive direction for each one in the direction you expect it to move, write $F = ma$ for each, write the length constraint, then solve. Do not skip the constraint and do not merge objects together until you have checked that the question does not ask about the force between them.

Next you will meet a connector that is not rigid at all, one whose pull depends on how far you stretch it: springs and Hooke's law.