Simple Machines: Levers, Pulleys and Gears

Beginner

How can one person lift a car with a jack, or pull out a nail with a claw hammer? With simple machines: levers, pulleys, ramps, wheels and axles, screws and gears.

Step 1. A machine changes the force you need. A crowbar lets your hand push gently on the long end while the short end heaves up a heavy rock.

Step 2. Nothing comes free. Physicists measure work as force × distance, and a machine cannot hand out more work than you put in. So if it shrinks the force, it stretches the distance: push with a quarter of the force and you must push four times as far.

Step 3. Pulleys show this best. In the animation, one rope runs up and down between a fixed upper block and a lower block carrying the crate. Four strands, numbered 1 to 4, hold the lower block, so they share the crate's weight and your hand pulls with only a quarter of it: compare the short red "your pull" bar with the purple weight bar. But to raise the crate 1 m, each strand must shorten by 1 m, so 4 m of rope passes through your hand: the gold bar grows four times as fast as the green one.

Step 4. A ramp is the same trick: pushing a box up a long, gentle slope takes less force than lifting it straight up, but over a longer path.

Check yourself: a smooth ramp 4 m long rises 1 m. What push moves a 400 N crate up it? Answer: 100 N. Four times the distance means a quarter of the force.

Real machines do slightly worse than this ideal: friction in pulleys and bearings turns part of your effort into heat, so you always pull a little harder.

Intermediate

What a machine changes, and what it cannot

A hand pushes gently on the long end of a crowbar, and the short end heaves up a heavy rock. A simple machine takes the force you apply, the effort $F_{\mathrm{in}}$, which moves through a distance $d_{\mathrm{in}}$, and acts on a load with a force $F_{\mathrm{out}}$ through a distance $d_{\mathrm{out}}$; for a steady lift, $F_{\mathrm{out}}$ is the load's weight $mg$. Two pure numbers describe it: the mechanical advantage, measured on the real machine, and the ideal one, set by the geometry alone.

• $\mathrm{MA}=\frac{F_{\mathrm{out}}}{F_{\mathrm{in}}}$
• $\mathrm{IMA}=\frac{d_{\mathrm{in}}}{d_{\mathrm{out}}}$

A machine has no energy source of its own. Without friction the work put in equals the work delivered, $W_{\mathrm{in}}=W_{\mathrm{out}}$, so $F_{\mathrm{in}}{d}_{\mathrm{in}}=F_{\mathrm{out}}{d}_{\mathrm{out}}$ and $\mathrm{MA}=\mathrm{IMA}$: force is multiplied exactly as much as distance is divided. With friction, part of the input becomes heat $Q$, so $W_{\mathrm{in}}=W_{\mathrm{out}}+Q$ and the efficiency
$\eta =\frac{W_{\mathrm{out}}}{W_{\mathrm{in}}}=\frac{\mathrm{MA}}{\mathrm{IMA}}$
is less than 1. Friction changes the forces but not the distances, so the ratio of the works is the ratio of $\mathrm{MA}$ to $\mathrm{IMA}$.

Pulleys: count the supporting strands

A pulley is a grooved wheel, a sheave, turning freely on an axle, so the tension $T$ in the rope is the same on both sides. A single fixed pulley therefore only turns the pull around, with $\mathrm{MA}=1$, although pulling down, helped by your own weight, is easier.

A block and tackle runs one rope up and down between a fixed upper block and a lower block that carries the load. If $n$ strands hold up the moving block, they share the load, $nT=mg$, so the pull on the free end is $F=T=mg/n$. The geometry agrees: when the lower block rises by $h$, each strand shortens by $h$, and the spare rope, $d=nh$, comes out through your hand. Count only the strands attached to the moving block. Friction raises the pull by the factor $1/\eta$:
$\text{Block and tackle}:d=nh,F=\frac{mg}{n\eta }$

Worked example: the hoist in the animation

The 40 kg crate weighs $mg=40\cdot 9.81=392.4\mathrm{N}$, and the four strands numbered in the novice view hold the lower block, so $n=4$ and, in the ideal machine of this layer,
$T=F=\frac{mg}{n}=\frac{392.4}{4}=98.1\mathrm{N}$

Every blue $T$ arrow on the strands has this length, and so does the red $F$ arrow at the hand. The crate rises $h=1.00\mathrm{m}$ at 0.25 m/s while the hand pulls $d=4\cdot 1.00=4.00\mathrm{m}$ of rope at 1 m/s, along the dashed gold line. The two works are
$W_{\mathrm{in}}=Fd=98.1\cdot 4.00=392.4\mathrm{J}$
$W_{\mathrm{out}}=mgh=392.4\cdot 1.00=392.4\mathrm{J}$

These are the final values on the two readouts, which climb together. The top line of the panel, $\mathrm{MA}=mg/F=d/h=4$, sums it up: a quarter of the force, four times the distance, the same work.

Reading the rig: arrows, bars and the ceiling

The red pull bar is a quarter of the purple weight bar, while the gold rope bar grows four times as fast as the green height bar. Think of each work as the area of a rectangle, force times distance: low and wide for the hand, tall and narrow for the crate, equal in area.

The arrows form a free-body diagram of the lower block: four $T$ arrows pull it up and the weight pulls it down, so $4T=mg$. The ceiling, however, carries more than the crate itself. The upper block is pulled down by all four strands and also by the free end, because you pull downward, so the ceiling must hold $5T=490.5\mathrm{N}$, which is $mg+F$.

Levers, gears and ramps

A lever is a rigid bar turning about a fixed point, the fulcrum. In a small turn each point moves in proportion to its distance from the fulcrum, so the arms $L_{\mathrm{in}}$ and $L_{\mathrm{out}}$, both measured from the fulcrum, give $\mathrm{IMA}=L_{\mathrm{in}}/L_{\mathrm{out}}$, and the torques balance:
$\text{Lever}:F_{\mathrm{in}}{L}_{\mathrm{in}}=F_{\mathrm{out}}{L}_{\mathrm{out}}$

What sits in the middle sets the class: the fulcrum in class 1 (crowbar, scissors), the load in class 2 (wheelbarrow, nutcracker; always $\mathrm{MA}>1$), the effort in class 3 (tweezers, your forearm; always $\mathrm{MA}<1$, trading force for speed and reach). A wheel and axle is a lever that keeps turning: a crank of radius $R$ on a drum of radius $r$ has $\mathrm{IMA}=R/r$.

Gears are toothed wheels whose teeth all have the same size, so the number of teeth $N$ is proportional to the radius:
$\text{Gears}:\frac{{\tau }_{\mathrm{out}}}{{\tau }_{\mathrm{in}}}=\frac{{\omega }_{\mathrm{in}}}{{\omega }_{\mathrm{out}}}=\frac{N_{\mathrm{out}}}{N_{\mathrm{in}}}$
Here $\tau$ is the torque, in N·m, and $\omega$ the rate of rotation. In the expert inset an 8-tooth pinion drives a 24-tooth gear: a third of the speed, three times the torque.

A smooth ramp of length $s$ rising $h$ needs a push along it of only $mgh/s$, so $\mathrm{IMA}=s/h$: a ramp 4 m long rising 1 m is the animation's tackle in another form.

Worked example: a geared hand winch

A building-site winch lifts a 150 kg pallet of tiles. Its 0.25 m crank turns a 10-tooth pinion, which drives a 40-tooth gear on the shaft of a drum of radius 0.050 m. The efficiency is 75 %. What force must you apply to the handle?

Crank and drum act as a wheel and axle with a factor $0.25/0.050=5$, and the gears add $40/10=4$. Ratios multiply along a chain of machines, so $\mathrm{IMA}=5\cdot 4=20$, and
$\mathrm{MA}=\eta \mathrm{IMA}=0.75\cdot 20=15$
$F_{\mathrm{in}}=\frac{mg}{\mathrm{MA}}=\frac{150\cdot 9.81}{15}=\frac{1471.5}{15}=98.1\mathrm{N}$

Check it with distances: one turn of the drum winds $2\pi \cdot 0.050=0.314\mathrm{m}$ of rope and takes four turns of the crank, $4\cdot 2\pi \cdot 0.25=6.28\mathrm{m}$ of hand travel, twenty times as far. A 1 m lift costs $98.1\cdot 20=1962\mathrm{J}$ for 1471.5 J of output; the other 490.5 J warm the gears and bearings.

Common mistakes

• Counting every length of rope. The fifth vertical length in the animation runs from the upper block to your hand and holds nothing up: $n=4$, and dividing by 5 would give 78.5 N.
• Believing that a machine saves work. It saves force, never work: the real hoist of the expert view takes 490.5 J to deliver 392.4 J.
• Measuring lever arms from the load. A 1.5 m crowbar on a fulcrum 0.10 m from its tip has arms of 1.40 m and 0.10 m, so $\mathrm{MA}=14$, not 15.
• Taking the losses as a share of the output. At $\eta =0.80$ the hoist needs $392.4/0.80=490.5\mathrm{J}$: the heat, 98.1 J, is a fifth of the input, not 20 % of the output.
• Calling a machine with $\mathrm{MA}\le 1$ useless. A fixed pulley changes the direction of the pull, and tweezers trade force for fine movement.

Where you meet it

On a bicycle a 48-tooth chainring driving a 16-tooth sprocket turns the wheel three times per pedal turn: gears used the other way round, for speed. Your forearm is a class 3 lever: the biceps pulls about 4 cm from the elbow and the hand is 35 cm away, so holding a 50 N weight takes about 440 N of muscle force.

The hydraulic press of Fluid Pressure and Pascal's Principle is a machine whose advantage is a ratio of piston areas; transformers in Motors, Generators and Transformers trade voltage for current as gears trade torque for speed; and efficiency appears again, with a stricter limit, in First Law of Thermodynamics and Heat Engines.

Advanced

Why every ideal machine obeys the same rule

Levers, pulleys, gears and ramps look nothing alike, yet they all trade force for distance in the same proportion. This section shows why, with an argument that uses no forces at all, only the fact that perpetual motion is impossible.

Call a machine reversible if, without friction, it runs backwards as readily as forwards: whatever it does one way, it can undo the other way. Now take two reversible machines. Each is driven by lowering the same 100 N weight by 1 m, and each uses that to lift the same load. The first lifts it by $x_1$, the second by $x_2$, and suppose that $x_2>x_1$.

Run the second machine forwards: the weight goes down 1 m and the load goes up $x_2$. Then run the first machine backwards: it lowers the load by $x_1$ and, in doing so, raises the 100 N weight back up by 1 m. The weight is where it started and both machines are as they were, yet the load hangs $x_2-x_1$ higher than before. Repeat the cycle and you raise the load for nothing, as often as you like.

Such perpetual motion is never observed, so the assumption was wrong: $x_2\le x_1$. The argument works just as well with the two machines swapped, which gives $x_1\le x_2$. Hence $x_1=x_2$: all reversible machines lift a given load equally high.

So one reversible machine is enough to find the rule for all of them, and a balanced lever is the simplest. Tip it slightly, and the weight lowered on one side times the height it falls equals the weight raised on the other side times the height it rises. A force times the distance moved along it is the work $W$, so for every reversible machine the input work equals the output work, $W_{\mathrm{in}}=W_{\mathrm{out}}$.

A machine with friction is not reversible, and the same argument limits it. If it lifted the load higher than a reversible machine, running the reversible one backwards would again give something for nothing. So $W_{\mathrm{out}}\le W_{\mathrm{in}}$, and the efficiency $\eta =W_{\mathrm{out}}/W_{\mathrm{in}}$ is at most 1. Sadi Carnot used the same reasoning in 1824 to show that no heat engine beats a reversible one working between the same two temperatures.

Virtual work: the advantage from geometry

Legend

$F_1$	Effort pushing down on the end of the long arm: the input force.	$\mathrm{N}$
$F_2$	Load pressing down on the end of the short arm: the output force.	$\mathrm{N}$
$L_1$	Length of the effort arm, from the fulcrum to the effort.	$\mathrm{m}$
$L_2$	Length of the load arm, from the fulcrum to the load.	$\mathrm{m}$
$\delta \theta$	Tiny angle through which the lever is tipped, drawn large.	$\mathrm{rad}$
$\delta x_1$	Distance the effort end moves down: $L_1\delta \theta$.	$\mathrm{m}$
$\delta x_2$	Distance the load end moves up: $L_2\delta \theta$.	$\mathrm{m}$

The rule $W_{\mathrm{in}}=W_{\mathrm{out}}$ turns a question about a machine's forces into a question about its geometry: how far each end moves. Figure 1 shows this for a lever in balance. The effort $F_1$ pushes down on the end of the long arm $L_1$, and the load $F_2$ presses down on the end of the short arm $L_2$; the subscripts 1 and 2 stand for "in" and "out".

Now imagine tipping the lever through a tiny angle $\delta \theta$, drawn large in the figure so that it can be seen. A point at distance $L$ from the fulcrum moves along an arc of length $L\delta \theta$, and for a tiny angle that arc is a short step straight up or down. Look at the two arrows at the ends of the lever: the effort end goes down by $\delta x_1=L_1\delta \theta$ and the load end goes up by $\delta x_2=L_2\delta \theta$. Their ratio, $\delta x_1/\delta x_2=L_1/L_2$, is 2 in the picture and depends on the lengths of the arms alone.

Now apply the work rule. Because the lever is in balance, the movement can be made as slow as we like, so it gives the lever no kinetic energy; because the lever is reversible, none of the work turns into heat. All the work done by the effort therefore goes into raising the load:
$F_{\mathrm{in}}\delta x_{\mathrm{in}}=F_{\mathrm{out}}\delta x_{\mathrm{out}}$

The $\delta x$ are virtual displacements: imagined rather than performed, and so small that the forces do not change while they happen. Only their ratio matters, and the constraints of the machine fix it, here the rigid lever turning about its fulcrum. For the lever the balance reads $F_1\delta x_1=F_2\delta x_2$. Put in the two displacements and $\delta \theta$ cancels, leaving the familiar torque balance $F_1{L}_1=F_2{L}_2$.

This is the principle of virtual work, and it gives a machine's advantage without a single free-body diagram. Two ratios describe any machine: the mechanical advantage $\mathrm{MA}=F_{\mathrm{out}}/F_{\mathrm{in}}$, the factor by which it multiplies force, and the ideal mechanical advantage $\mathrm{IMA}=\delta x_{\mathrm{in}}/\delta x_{\mathrm{out}}$, the ratio of the distances moved. Virtual work says that for a reversible machine the two are equal; for the lever both are $L_1/L_2$. With friction, $\eta =W_{\mathrm{out}}/W_{\mathrm{in}}=\mathrm{MA}/\mathrm{IMA}$.

The tackle: a rope of fixed length

Legend

$y$	Distance from the upper block to the lower one: the length of each supporting strand.	$\mathrm{m}$
$\ell$	Length of rope from the upper block down to the hand.	$\mathrm{m}$
$n$	Number of strands holding up the lower block, here 4.
$\delta h$	Height through which the load is raised.	$\mathrm{m}$
$n\delta h$	Distance the hand comes down: $n$ times $\delta h$.	$\mathrm{m}$

A block and tackle, the hoist of the animation, keeps its geometry in the rope. The rope does not stretch, and that single fact fixes how far the hand must pull. In Figure 2 the lower block, which carries the crate, hangs a distance $y$ below the upper block on $n=4$ strands, each of length $y$, and $\ell$ is the length of rope from the upper block down to the hand.

The length of rope in the $n$ strands plus the free end is the whole rope, so it stays the same:
$ny+\ell =\mathrm{const},\delta \ell =-n\delta y$

Now raise the load by $\delta h$, so that the lower block comes up and $\delta y=-\delta h$; the dashed boxes show where the crate and the hand were before. Each of the $n$ strands is now shorter by $\delta h$. The rope they give up, $n\delta h$ in all, goes to the free end, $\delta \ell =n\delta h$, so the hand comes down by $n\delta h$, the long arrow on the right.

Virtual work then gives the pull. The hand's force $F$ acts over $n\delta h$ while the weight $mg$ rises by $\delta h$, so $Fn\delta h=mg\delta h$ and $F=mg/n$: the $\mathrm{IMA}$ is $n$, the number of strands holding up the lower block. Dividing both distances by the time taken gives the speeds, $v_{\mathrm{in}}=n{v}_{\mathrm{out}}$: in the animation, 1 m/s at the hand and 0.25 m/s at the crate.

Gears and a slope by the same rule

The same reasoning handles machines that turn or slide. Two gears in mesh have teeth that touch and move together, so their rims share one speed: ${\omega }_{\mathrm{in}}{r}_{\mathrm{in}}={\omega }_{\mathrm{out}}{r}_{\mathrm{out}}$, where $\omega$ is the angular speed and $r$ the radius. The teeth have the same spacing on both gears, so each radius is proportional to its number of teeth, and the speeds are in the inverse ratio of the tooth counts.

For a slope, take a block of mass $M$ on a smooth incline of length $s$ and height $h$, tied by a rope over a pulley at the top to a hanging mass $M^{\prime }$. As $M^{\prime }$ drops by $s$, it pulls the block the whole length of the slope and so raises it by $h$. The work balance $M^{\prime }gs=Mgh$ says the two balance when $M^{\prime }s=Mh$: for a 4 m slope rising 1 m, a quarter of $M$ is enough. Simon Stevin reached such results in 1586 another way: a closed loop of chain hung over a triangular block never starts turning by itself, which is again the impossibility of perpetual motion.

Losses, sheave by sheave

Legend

$F$	Pull of the hand on the free end of the rope.	$\mathrm{N}$
$T_4$	Tension in the strand next to the hand: ${\eta }_{s}F$.	$\mathrm{N}$
$T_3$	Tension in the second strand from the hand: ${\eta }_s^{2}F$.	$\mathrm{N}$
$T_2$	Tension in the third strand from the hand: ${\eta }_s^{3}F$.	$\mathrm{N}$
$T_1$	Tension in the fourth strand, tied to the upper block: ${\eta }_s^{4}F$.	$\mathrm{N}$
${\eta }_s$	Fraction of the incoming tension that one sheave passes on; below 1 because of friction.
$mg$	Weight of the load, held by the four strands together.	$\mathrm{N}$

So far every machine was reversible. Real pulleys have friction in their bearings, and the tensions in the rope show what it costs and where the cost appears. Figure 3 lays the rope of the animation's tackle out straight and draws every tension to one scale: the hand's pull $F$ on top, then the four strands, from $T_4$ next to the hand down to $T_1$, the strand tied to the upper block.

The model is simple. Each sheave, the grooved wheel of a pulley, passes on only a fraction ${\eta }_s$ of the tension it receives, a little less than 1, and the work lost warms its bearing. So each arrow in the figure is ${\eta }_s$ times the one above it, as the ${\eta }_s$ labels between them show.

Follow the rope back from the hand. It passes over one sheave before the first strand, so that strand carries $T_4={\eta }_{s}F$. One more sheave gives $T_3={\eta }_s^{2}F$, and so on to the anchored strand with $T_1={\eta }_s^{n}F$. The strands together hold up the load, which the bar at the bottom shows by laying the four arrows end to end against $mg$. Their sum is a geometric series:
$mg=F\left({\eta }_s+{\eta }_s^2+\dots +{\eta }_s^n\right)=F\frac{{\eta }_s(1-{\eta }_s^n)}{1-{\eta }_s}$

Now bring in the efficiency $\eta$ of the whole tackle. Its $\mathrm{IMA}$ is still $n$, so $\mathrm{MA}=n\eta$ and $F=mg/(n\eta )$. Put that into the sum and the load cancels:
$n\eta =\frac{{\eta }_s(1-{\eta }_s^n)}{1-{\eta }_s}$

For the animation, $\eta =0.80$ and $n=4$ require ${\eta }_s+{\eta }_s^2+{\eta }_s^3+{\eta }_s^4=3.2$, so ${\eta }_s\approx 0.913$. The 40 kg crate weighs 392.4 N, so the pull is $F=392.4/3.2=122.6\mathrm{N}$. The strand tensions $T_1$ to $T_4$ are 85.1, 93.2, 102.2 and 111.9 N, and they add up to exactly 392.4 N: in the figure the blue arrows lengthen towards the hand, and the red $F$ is the longest of all. The actual $\mathrm{MA}$ is $n\eta =3.2$ while the $\mathrm{IMA}$ is still 4, because friction changes the forces but not the geometry.

Reading the energy graph

Legend

$t$	Time since the start.	$\mathrm{s}$
$W$	Work done so far.	$\mathrm{J}$
$W_1$	Input work: the hand's pull times the length of rope pulled in, $Fd$.	$\mathrm{J}$
$W_2$	Output work: the potential energy gained by the load, $mgh$.	$\mathrm{J}$
$Fd$	Input work at the end of the lift.	$\mathrm{J}$
$mgh$	Output work at the end of the lift.	$\mathrm{J}$
$t_1$	Instant the crate starts to rise.	$\mathrm{s}$
$t_2$	Instant the crate stops rising.	$\mathrm{s}$
$\Delta t$	A short stretch of time.	$\mathrm{s}$
$\Delta W_1$	Input work done during $\Delta t$.	$\mathrm{J}$
$\Delta W_2$	Output work done during $\Delta t$.	$\mathrm{J}$
$Q$	Heat made by friction in the sheaves: the final gap between the two works.	$\mathrm{J}$

The work balance can also be watched as it happens. Figure 4 plots the two works against time during the lift, the subscripts 1 and 2 again standing for "in" and "out". The input work is $W_1=W_{\mathrm{in}}(t)=Fd(t)$, where $d$ is the length of rope the hand has pulled in, and the output work is $W_2=W_{\mathrm{out}}(t)=mgh(t)$, where $h$ is the height the crate has risen.

Read the graph from left to right. Nothing happens before $t_1=0.4$ s. From then on the crate rises at a constant speed, so both works grow along straight lines. At $t_2=4.4$ s the lift ends: the hand has pulled in 4 m of rope and the crate is 1 m higher. The lines level off at $Fd=490.5\mathrm{J}$ and $mgh=392.4\mathrm{J}$.

Each slope is a power, the rate of doing work. Look at the two small triangles: over the same stretch $\Delta t$ the input line rises by $\Delta W_1$ and the output line by $\Delta W_2$. The hand pulls 122.6 N at 1 m/s, an input power of 122.6 W, while the crate gains potential energy at $392.4\cdot 0.25=98.1\mathrm{W}$. The ratio of the slopes is 0.80 at every instant, $\Delta W_2/\Delta W_1=\eta$, so the efficiency is also the ratio of output power to input power.

The shaded gap between the lines is the work lost to friction so far. It widens at $122.6-98.1=24.5\mathrm{W}$ and ends at $Q=490.5-392.4=98.1\mathrm{J}$, the bracket at the right of the figure and the value on the animation's heat readout. After $t_2$ both lines are flat, because a machine that has stopped does no work, however hard it is held.

More strands, gear trains and self-locking

The loss model answers three practical questions: whether more strands always help, how losses add up along a chain of gears, and why some machines hold their load by themselves.

More strands do not help without limit. As $n$ grows, ${\eta }_s^n\to 0$, so the actual $\mathrm{MA}={\eta }_s(1-{\eta }_s^n)/(1-{\eta }_s)$ levels off at ${\eta }_s/(1-{\eta }_s)\approx 10.5$. Going from 4 to 8 strands raises it only from 3.2 to 5.4, while $\eta$ falls from 0.80 to 0.68. Figure 3 shows the reason: each extra strand lies further from the hand and carries less tension than the one before.

Ideal gears pass on the power unchanged, ${\tau }_{\mathrm{in}}{\omega }_{\mathrm{in}}={\tau }_{\mathrm{out}}{\omega }_{\mathrm{out}}$, where $\tau$ is the torque: a gear that turns three times more slowly turns with three times the torque. In the animation's inset an 8-tooth pinion drives a 24-tooth gear; the pinion turns 270° while the gear turns 90°, and ${\tau }_2=3{\tau }_1$, with 1 for the pinion and 2 for the gear. Along a gear train the stage ratios multiply, two 1 : 3 stages giving 1 : 9, and so do the efficiencies. A good stage loses only a few per cent, so two stages at 0.98 still pass on 96 %.

A screw jack or a worm gear must hold its load when you let go; such a machine is called self-locking. To see when that happens, suppose friction costs the same heat $Q$ whichever way the machine runs. Lifting then needs $W_{\mathrm{in}}=W_{\mathrm{out}}+Q$. If the load tried to drive the machine backwards, the descending load would supply only $W_{\mathrm{out}}$ but would still have to pay $Q$; it cannot, and so stays put, if $W_{\mathrm{out}}\le Q$. In terms of the lifting efficiency, that is $\eta =W_{\mathrm{out}}/(W_{\mathrm{out}}+Q)\le 0.5$.

A car jack puts numbers on this. With a 0.20 m handle, one turn takes the hand round a circle of $2\pi \cdot 0.20=1.26\mathrm{m}$, while a screw advancing 2.0 mm per turn raises the car 0.0020 m, so $\mathrm{IMA}=2\pi \cdot 0.20/0.0020\approx 630$. At $\eta =0.30$ the actual $\mathrm{MA}$ is about 190, and about 21 N at the handle lifts 4000 N. Since 0.30 is below one half, the car stays up when you let go.

Worked problem: lowering the crate

Run the animation's tackle backwards: lower the 40 kg crate by 1 m at a steady speed, with ${\eta }_s=0.9127$ at every sheave. How hard must the hand hold, and where does the energy go?

First, which way does friction act? The rope now runs from the hand into the tackle, and friction at each sheave still opposes its motion. So at each sheave the tension on the side the rope is moving towards must be the larger one: it is $1/{\eta }_s$ times the tension on the hand's side. The strand next to the hand carries $F/{\eta }_s$, the next $F/{\eta }_s^2$, and so on up to $F/{\eta }_s^4$ in the anchored strand. Together they still hold the load:
$mg=F\left(\frac{1}{{\eta }_s}+\frac{1}{{\eta }_s^2}+\frac{1}{{\eta }_s^3}+\frac{1}{{\eta }_s^4}\right)=5.052F$
$F=\frac{392.4}{5.052}=77.67\mathrm{N}$

These tensions are the four values of the lift in reverse order, from 111.9 N in the anchored strand down to 85.1 N next to the hand. The holding force is ${\eta }_s^5$ times the lifting force, about 63 % of it: friction now helps the hand instead of fighting it.

Now the energy. The sinking crate releases $mgh=392.4\mathrm{J}$. The hand lets out 4 m of rope while resisting with 77.67 N, so it absorbs $77.67\cdot 4.00=310.7\mathrm{J}$, and the sheaves turn the remaining $392.4-310.7=81.7\mathrm{J}$ into heat.

If the bearings grip about as well at rest as in motion, any pull between 77.7 N and 122.6 N holds the crate still: below that range it sinks, above it it rises. Yet the holding force never reaches zero. With a lifting efficiency of 0.80, well above one half, the tackle is not self-locking, and if you let go the crate runs down.

Limits of validity

Every result above rests on idealisations: a massless rope that does not stretch, weightless sheaves, rigid levers, and slow, steady motion. Each can be relaxed, and the work balance shows what changes.

A 2.0 kg lower block is lifted with the crate, so it adds 19.6 N to the load and raises the ideal pull from 98.1 N to 103.0 N. Accelerating the crate upward at 0.5 m/s² needs a net upward force, $nT=m(g+a)$, or 103.1 N per strand, and the extra work becomes kinetic energy rather than height. Real rope also stretches, and bearings have a loss that does not grow with the load, so a lightly loaded tackle does worse than a constant ${\eta }_s$ predicts.

The work balance itself is the conservation of energy, and it has never been seen to fail. What changes at tiny scales is the smooth running.

A kinesin motor protein shows this. It walks along a filament inside a cell in 8 nm steps and stalls against a load of about 6 pN. It therefore does at most $(6\times 10^{-12}\mathrm{N})(8\times 10^{-9}\mathrm{m})\approx 4.8\times 10^{-20}\mathrm{J}$ of work per step, only about 11 times the thermal energy $kT\approx 4.3\times 10^{-21}\mathrm{J}$ at body temperature, where $k$ is Boltzmann's constant and $T$ here the absolute temperature. Each step spends one ATP molecule, worth about twice that work, and random thermal kicks are a large part of the motion. At this scale the balance holds only on average, and the Second Law decides which way the machine can run.

Where this leads

Virtual work is a shortcut to the torque balances of Static Equilibrium and Levers, and Rotational Dynamics and Torque supplies the torque and the power $P=\tau \omega$ that a gear pair passes on. The hydraulic press of Fluid Pressure and Pascal's Principle has an ideal advantage equal to a ratio of areas: 100 for pistons of 2 cm and 20 cm diameter. Transformers, in Motors, Generators and Transformers, multiply voltage by their turns ratio and divide current by it, passing on power as gears do. Rope stretch belongs to Elasticity: Hooke's Law, Stress and Strain. The reversible-machine argument returns in First Law of Thermodynamics and Heat Engines and The Second Law of Thermodynamics and Entropy, where it caps the efficiency of every heat engine.