Pulley systems

Every crane on a construction site, every gym cable machine, and every well-bucket rope shares one humble ancestor: the pulley, a grooved wheel that a rope runs over. Some pulleys are lazy tricks — they change only the direction you pull, not the effort needed. Others genuinely multiply your force, letting a person lift a car engine or hoist a sail that would otherwise be impossible to move by hand. The ICSE Class 10 syllabus asks you to know exactly three standard arrangements — the single fixed pulley, the single movable pulley, and the block and tackle — and to reason about each using the same three numbers every machine is judged by: mechanical advantage, velocity ratio, and efficiency. Get these three ideas straight, know which pulley does what, and pulley problems become fast, mechanical arithmetic rather than a source of confusion.

The core insight

A pulley system is just a machine whose mechanical advantage and velocity ratio depend entirely on how many rope segments actually support the moving load.

The core idea

Pulleys sit inside the "Machines" section of the ICSE unit "Force, Work, Power and Energy," alongside levers. Every machine, pulleys included, is described with the same four quantities:

MA = \frac{L}{E}, \qquad VR = \frac{\text{distance moved by effort}}{\text{distance moved by load}}, \qquad \eta = \frac{MA}{VR} \times 100\%

where: $L$ is the load (the weight being lifted, in newtons), $E$ is the effort (the force you apply, in newtons), $MA$ is mechanical advantage, $VR$ is velocity ratio, and $\eta$ is efficiency. The syllabus wants you to know these ideal values for three arrangements, not derive them from scratch.

Single fixed pulley. The wheel is anchored to a fixed support such as a beam; it does not move. The rope runs over it, load on one end, effort on the other. Pulling the effort down by a distance $d$ raises the load by the same $d$ — one rope segment does all the work. For the ideal case, $MA = 1$ and $VR = 1$ . It multiplies nothing; its only job is to change the direction of your pull, which is why wells and flagpoles use it — pulling down is far more convenient than lifting straight up.

Single movable pulley. Here the wheel is attached to the load itself and rises with it, while the rope passes over a separate fixed support with the effort applied at the free end. Two rope segments now share the load between them, so the effort needed is only half the load — but the effort end must travel twice as far as the load rises. Ideally, $MA = 2$ and $VR = 2$ .

Block and tackle. Two blocks of pulleys — an upper fixed block and a lower movable block — are threaded by one continuous rope. This is the arrangement behind real cranes, hoists, and gym machines, because it can be built to give large mechanical advantage. At ICSE level the rule is stated, not derived:

MA = \text{number of rope segments supporting the movable (lower) block}

and for an ideal block and tackle $VR$ equals that same count, so $MA = VR$ and efficiency is $100\%$ — only in the idealised, frictionless case. Real pulleys have friction at every wheel, so the actual $MA$ falls below the ideal value and efficiency drops below $100\%$ .

Figure

Three pulley arrangements side by side showing a single fixed pulley with one supporting segment, a single movable pulley with two supporting segments, and a block and tackle with four supporting segments under the load

Derivation AWhere the single movable pulley's MA = 2 comes from

The goal is to show, using only the free-body idea of tension being the same throughout a single ideal rope and the requirement that the movable pulley's forces balance, why an ideal single movable pulley supports twice the load for a given effort, so that a rope segment carrying tension $E$ can lift a load $L = 2E$ — the origin of the value $MA = 2$ that the syllabus asks you to simply know and apply.

Step 1. Picture the movable pulley hanging from two vertical segments of the same continuous rope, with the load $L$ hooked to its axle. Because it is one ideal (massless, frictionless) rope threaded around one ideal (frictionless, weightless) wheel, the tension $T$ is identical in both segments — there is no way for the rope to "know" which side is which.

Step 2. The movable pulley and its load are in equilibrium (it is lifted slowly, so acceleration is effectively zero), so the total upward pull from the rope must balance the downward weight of the load. Both segments pull upward on the pulley, so the total upward force is $T + T = 2T$ .

Step 3. Balancing forces gives $2T = L$ , so $T = L/2$ . But the tension in the rope is exactly the effort you apply at the free end, so $E = T = L/2$ .

Step 4. Rearranging, $L = 2E$ , so $MA = L/E = 2$ for the ideal case. Because the load rises by however much both supporting segments shorten together, pulling the free end down by $2d$ raises the load by only $d$ , which is exactly $VR = 2$ — matching $MA = VR$ , as expected when there is no friction. This is the reasoning behind the block-and-tackle rule too: every extra rope segment supporting the movable block adds one more $T$ to the balance, so $MA$ simply equals the segment count.

Worked example

A single movable pulley lifts a load of $400\ \text{N}$ , and the load must rise by $1.5\ \text{m}$ .

Step 1. Find the ideal effort. For an ideal single movable pulley, $MA = 2$ , so $E = L/MA = 400/2 = 200\ \text{N}$ .

Step 2. Find the distance the effort end must move. $VR = 2$ , so the effort travels twice the load's distance: $2 \times 1.5 = 3\ \text{m}$ .

Step 3. Interpret. An effort of $200\ \text{N}$ raises the $400\ \text{N}$ load, but only because the rope is pulled through $3\ \text{m}$ to lift the load $1.5\ \text{m}$ — the force saved is paid for in extra distance.

Competitive stretch

Now suppose this same pulley is not ideal: friction means the actual effort needed is $250\ \text{N}$ rather than $200\ \text{N}$ , while $VR$ stays at $2$ (fixed by geometry, not friction).

Step 1. Find the actual $MA$ . $MA_{actual} = L/E_{actual} = 400/250 = 1.6$ .

Step 2. Find the efficiency. $\eta = \dfrac{MA}{VR} \times 100\% = \dfrac{1.6}{2} \times 100\% = 80\%$ .

Final answer: the real pulley needs $250\ \text{N}$ of effort, giving an actual $MA$ of $1.6$ (below the ideal $2$ ) and an efficiency of $80\%$ , since friction always keeps the real $MA$ below the ideal value.

Common traps

Thinking a single fixed pulley multiplies force — it does not; ideal $MA = 1$ , and its whole benefit is changing the direction of the effort.
Forgetting that a single movable pulley needs the effort to move twice as far as the load, even though the effort itself is only half the load.
Counting the wrong block when applying the "supporting strings" rule — always count segments holding up the movable (lower) block, not the fixed one.
Assuming real systems reach their ideal $MA$ and $VR$ — friction always lowers the actual $MA$ below the ideal, so efficiency is below $100\%$ ; only in the ideal case does $MA = VR$ .

Ask the tutor: Want to try a trickier block-and-tackle diagram or a few more MA, VR and efficiency problems worked step by step?

Why does a single fixed pulley feel easier to use even though it does not reduce the effort needed?

It is easier for a practical reason, not a force reason: pulling down uses your body weight and lets you brace your feet on the ground, which is far more comfortable and controlled than lifting a rope straight up over your head. The physics gives no force advantage at all — the ideal $MA$ stays exactly $1$ . What you gain is convenience of direction, which is precisely why the syllabus stresses that its "usefulness" is not about mechanical advantage.

How do I quickly count the supporting strings in a block and tackle diagram?

Trace the rope from where the effort is applied and count every segment that touches the lower, movable block — not the upper fixed block. Segments that only wrap around the fixed block, or the free end you pull, do not count. A neat trick is to mark each point where the rope crosses the movable block's pulleys with a tick; the number of ticks is the ideal $MA$ , and it also equals $VR$ for the ideal case.

If MA equals VR for an ideal pulley system, why do real systems have MA less than VR?

Because $VR$ is a purely geometric number — it depends only on how the rope is threaded and how far the effort must travel, so friction cannot change it. $MA$ , however, depends on the actual effort you must apply, and friction at every pulley wheel means you always need more effort than the frictionless ideal predicts. Since $MA = L/E_{actual}$ and $E_{actual}$ is larger than the ideal effort, $MA$ comes out smaller than $VR$ , and $\eta = MA/VR \times 100\%$ falls below $100\%$ .

Pulley systems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads