Pulleys: how a rope and a few wheels lift a heavy load

Everything near the Earth’s surface is pulled downwards by gravity. The size of that pull, the object’s weight, is its mass multiplied by a number $g$ that tells us how hard the Earth pulls on each kilogram: $g \approx 9.8\,\text{N/kg}$, so each kilogram is pulled down with about $9.8$ newtons. We write the weight as $mg$, which means $m \times g$. For our 4 kg load, $mg = 4\,\text{kg} \times 9.8\,\text{N/kg} = 39.2\,\text{N}$.

In this lesson we always lift slowly and steadily, at a constant speed. As the Newton’s laws page explains, an object moving at constant speed in a straight line, or not moving at all, has forces on it that balance: the upward forces add up to exactly the downward ones. So to lift the load steadily by hand we pull up with exactly $39.2\,\text{N}$, no more and no less. (Pulling harder for a moment starts the load moving; after that, a balanced pull keeps it moving at that speed.) The question of this page is whether a rope and pulleys can let us lift the same load with a smaller pull.

When a rope is pulled tight, every part of it pulls on its neighbours. Hold one end of a tight rope and you feel it pulling your hand along the rope; the load at the other end feels the same kind of pull. The size of this pull is called the tension in the rope, and we will write it $T$. A rope can only pull, never push, and it always pulls along its own length.

The tension is the same all along one tight rope. Why? Look at a short piece of the rope. The rope on its left pulls it one way and the rope on its right pulls it the other way. Our rope is very light, so this piece has almost no mass. If the two pulls were different, the difference would be an unbalanced force on almost no mass, and the piece would speed up enormously, which a rope in a steady lift does not do. So the two pulls on every piece are equal, and the tension is the same from one end of the rope to the other.

A pulley lets a rope bend round a corner without changing this. The wheel turns freely on its axle, so the rope does not have to slide over anything. If the tension on one side of the wheel were bigger than on the other, it would turn the wheel faster and faster; a light wheel with no friction turning steadily must have equal pulls on its two sides. So when the rope leaves the wheel it still carries the same tension $T$, now in a new direction. Our model makes two simplifications, a light rope and a frictionless wheel; the last chapter shows what changes with real ones.

Start with the simplest arrangement. One pulley hangs from the beam on a hook and does not move, so it is called a fixed pulley. The rope is tied to the load, goes up and over the wheel, and comes down to a red grip in our hand. Between the rope and the grip is a spring balance, so we can read the tension we are pulling with.

Pull the rope with the slider below and watch three things: the reading on the balance, which way the load moves compared with your hand, and the two rulers. The blue ruler shows how far the load has risen; the orange one shows how far your hand has pulled. The dark marks painted on the rope every 20 cm show the rope itself sliding round the wheel.

The load hangs on one strand of rope, and that strand pulls it up with the tension $T$. For a steady lift this one upward force must balance the weight, so $T = mg$: for 4 kg, $T = 39.2\,\text{N}$, exactly what we needed by hand. The rope is one piece, so when the hand pulls 30 cm of rope down, 30 cm of rope comes up on the other side and the load rises 30 cm. A fixed pulley saves no force and no distance. What it changes is the direction: we pull down, the load goes up, and pulling down is easier because we can use our own weight.

Before we add more pulleys, we need a reliable way to find the force, because soon the rope will wind back and forth and it will be hard to see which parts are doing what. Here is the method. Draw a box round the part that moves upwards with the load, and look only at the forces that cross the edge of the box. Gravity pulls the load inside the box down with $mg$. A rope can cross the edge too, and every strand of rope that the box cuts pulls on what is inside, upwards, with the rope’s tension $T$.

For a steady lift the forces on everything inside the box balance. If the box cuts $n$ strands, the upward forces add up to $nT$, and they must equal the downward weight: $nT = mg$. Dividing both sides by $n$ gives the tension, $T = mg / n$. Forces that act between two things both inside the box, such as the moving block pulling on the load, are left out: one pulls the other up and the other pulls it down just as hard, so together they add nothing.

Try it on the fixed pulley. Switch on the box and the arrows: the box holds only the load, it cuts one strand, so $n = 1$ and $T = mg / 1 = mg$, which is what we found. The orange arrow at the top of the box is the tension in that strand, and the blue arrow is the weight; the two are the same length.

Now tie the rope to the beam instead. It comes down, passes under a wheel, and goes back up to the hand. This wheel is not fixed to anything: it hangs in the loop of the rope, and the load hangs from a ring under it, so when the load rises the wheel rises too. It is called a movable pulley.

Draw the box round the movable pulley and the load, since they move up together. Its top edge now cuts two strands of rope: the one that comes down from the beam and the one that goes up to the hand. Both belong to the same rope, so both pull up with the same tension $T$. With $n = 2$, $2T = mg$ and $T = mg / 2$ (half the weight). The beam is pulling up on one strand and our hand on the other, and the beam takes half the weight.

Check this in the picture: the balance should read half the weight. Notice that the hand now pulls upwards; we will fix that later.

Pull 20 cm of rope and look at the rulers: the load has risen only 10 cm. Why only half? Think about the rope’s length, which never changes. When the movable pulley and load rise by a height $h$, the strand from the beam down to the wheel becomes $h$ shorter, because its top stays tied to the beam while its bottom rises. The strand from the wheel up to the hand also becomes $h$ shorter at its bottom end. That is $h + h = 2h$ of rope no longer needed in the two strands, and since the rope cannot disappear, all of it must have come out at the hand. So the hand pulls $s = 2h$.

The marks on the rope show this happening. Watch the mark nearest the beam: it hardly moves, because that strand is tied there. A mark on the hand strand moves up twice as fast as the wheel. The wheel turns because its left side rests on still rope while its centre rises.

Half the force and twice the distance: is this a coincidence? Lift our 4 kg load by 1 m: the hand pulls up with 39.2 N over 1 m, so it does about 39 J of work ($1\,\text{J} = 1\,\text{N} \times 1\,\text{m}$). A steady lift needs an upward force $mg$ over the height $h$, so the load stores $mgh$ as gravitational energy. The work and energy page covers this; if you have not read it, the numbers here are enough to follow the check. Our hand does work $Fs$, and the load gains $mgh$. For the movable pulley, $F = T = mg/2$ and $s = 2h$, so the hand’s work is $\tfrac{mg}{2} \times 2h = mgh$, exactly the energy the load gains.

So a pulley never gives us energy for free. Energy is conserved: whatever the load gains, our hand has to supply. A pulley only lets us trade: a smaller force over a longer distance. The readout below works out both sides in joules ($1\,\text{J} = 1\,\text{N} \times 1\,\text{m}$, so the centimetres are first changed into metres) for whatever you have pulled.

Pulling upwards is awkward, so add a fixed pulley to the beam beside the movable one. The rope leaves the movable pulley, goes up and over the fixed one, and comes down to the hand. Now we pull downwards again.

Has the force changed? Draw the box round what moves up: the movable pulley and the load. Its top edge cuts the strand from the beam and the strand going up to the fixed pulley, so $n = 2$ again and $T = mg / 2$. The strand that comes down to our hand is outside the box: it pulls on our hand, not on the moving part, so it does not count. The fixed pulley changed the direction of the pull and nothing else, just as it did in chapter 3.

To save more force, the rope must hold up the moving part with more strands. Here two movable pulleys hang side by side and are joined by a bar, the spreader, which carries the load. Two fixed pulleys hang from the beam. The rope is tied to the beam and then follows four steps: (1) down and under the first movable pulley; (2) up and over the first fixed pulley; (3) down and under the second movable pulley; (4) up and over the second fixed pulley, then down to the hand. This arrangement is called a block and tackle.

Draw the box round the two movable pulleys, the bar and the load. Its top edge cuts four strands, so $4T = mg$ and $T = mg/4$. For 4 kg, $T = 39.2\,\text{N} \div 4 = 9.8\,\text{N}$, a quarter of the weight. The fifth strand, down to the hand, is outside the box again.

The distance follows the same argument as before. When the moving part rises $h$, each of the four strands holding it becomes $h$ shorter, which frees $4h$ of rope, and all of it comes out at the hand: $s = 4h$. Pull 40 cm and the load rises 10 cm. Then the energy check works again: $Fs = \tfrac{mg}{4} \times 4h = mgh$.

Every arrangement on this page follows the same two rules. If the box round the moving part cuts $n$ strands, then $T = mg / n$ and $s = nh$. More strands mean a smaller force and a longer pull, and their product, the work, stays $mgh$.

Most mistakes with pulleys come from counting strands that do not hold up the moving part. Three habits help.

Real pulleys need a little more force than $mg / n$, for two reasons. The movable blocks and the bar have weight of their own, and they rise with the load, so the box contains their weight too. And the wheels rub on their axles and the rope must bend round them, so some of our work turns into heat instead of lifting. Engineers describe the second effect by an efficiency $\eta$ (the Greek letter eta), the fraction of the hand’s work that is not turned into heat by friction: $\eta = 0.85$ means 85%. Out of 100 J of work by the hand, 85 J lifts the load and 15 J becomes heat.

Put both effects into the numbers. With a 0.5 kg block, four strands and $\eta = 0.85$, the mass to lift is $4 + 0.5 = 4.5\,\text{kg}$, whose weight is $4.5 \times 9.8 = 44.1\,\text{N}$. The hand must supply $\dfrac{44.1}{4 \times 0.85} \approx 13.0\,\text{N}$, not $9.8\,\text{N}$. In letters, the box now holds a mass $m + m_{\text{block}}$, and only the fraction $\eta$ of the hand’s work $Fs$ goes into lifting it. Lifting it by $h$ needs $(m + m_{\text{block}})\,gh$, and $s = nh$, so $\eta F nh = (m + m_{\text{block}})\,gh$. Dividing both sides by $\eta nh$ gives $F = \dfrac{(m + m_{\text{block}})\,g}{n\,\eta}$. Try it with the controls: with a 0.5 kg block, four strands and $\eta = 0.85$, lifting 4 kg takes $\dfrac{4.5 \times 9.8}{4 \times 0.85} \approx 13.0\,\text{N}$ instead of $9.8\,\text{N}$. Adding more and more pulleys helps less and less, because each one adds weight and friction.

This is why a crane’s hook block has several sheaves but not dozens, and why sailors choose the number of strands to suit the job. The ideas from this page still decide the design: count the strands that hold the moving part, and remember that every newton saved costs extra rope.