The same push, further out, and why that is a different quantity

A bolt that will not move under a hand on a short spanner moves under the same hand on a long one. Nothing about the hand has changed. The force is the same force, delivered by the same arm at the same effort, and the bolt can tell the difference.

That is a problem for a description of mechanics built entirely on forces. A free-body diagram records the forces acting on an object and their directions, and from those it predicts how the object accelerates. Two diagrams with identical arrows predict identical motion. But the two spanners give identical arrows — one force of a hundred and twenty newtons, pointing the same way — and produce different outcomes.

The missing information is where the force is applied. A force has a magnitude and a direction, and for translation that is all that matters — the centre of mass of an object accelerates according to the vector sum of the forces on it and nothing else. For rotation there is a third piece of information, the line the force acts along, and the whole of this rung is the construction that extracts a single number from it.

The quantity, built rather than quoted #

The number is called the torque, and the most common way of writing it,

$$\tau = rF\sin\theta,$$

hides what it is by presenting it as an arithmetic recipe. The construction underneath is a piece of geometry with no arithmetic in it at all.

Extend the force’s line of action — the infinite straight line through the point of application, running along the direction of the force, in both directions. Then drop a perpendicular from the pivot to that line. The length of that perpendicular is the moment arm, and the torque is the force multiplied by it.

Two things follow immediately and neither is obvious from the formula.

The point of application does not matter, only the line. Sliding the hand along the spanner changes $r$, and sliding it along the line of action does not change the torque at all, because the perpendicular from the pivot to a line is a property of the line. A rope pulling on a capstan exerts the same torque whether it is gripped near the drum or three metres back along the rope.

A force through the pivot does nothing. If the line of action passes through the pivot, the perpendicular has zero length, and the torque is zero however large the force. Pushing a door directly at its hinge, however hard, will not close it. This is why a free-body diagram of a rotating object can carry an enormous force at the axle and simply ignore it in the rotational bookkeeping: the axle’s reaction always passes through the axle.

Why the sine, and where it comes from #

The angle enters through the perpendicular and not through any property of rotation. When the force is at right angles to the spanner, the line of action is already perpendicular to the shaft, so the moment arm is the full length $r$. Tilt the force by an angle $\theta$ from the shaft and the perpendicular distance shrinks to $r\sin\theta$, by the same right-angled triangle that appears whenever a length is projected onto a direction.

The projection that produces the sine is the same one that resolves a vector into two perpendicular components — familiar from velocities, applied here to a distance. The moment arm is the component of the position vector perpendicular to the force, and the trigonometry is identical to the velocity case while the physics is not: one is a decomposition of a motion and the other is a measure of leverage.

There is a second, equivalent way to read the same triangle, and it is worth having both because different problems make one or the other obvious. Instead of shrinking the distance, resolve the force into a component along the spanner and one across it. The along-the-shaft component pulls the spanner out of the bolt or pushes it in; it exerts no turn. The across-the-shaft component, of size $F\sin\theta$, acts at the full distance $r$.

Shrink the distance and keep the force, or keep the distance and shrink the force: the product is the same, and the two readings are the same triangle read along different sides. The inclined plane uses the identical manoeuvre on the weight of a block, and the habit of choosing which of the two to resolve is most of what makes one setup of a problem easier than another.

The flatness near the right angle #

The torque is largest when the force is perpendicular to the spanner, which is unsurprising, and it is insensitive near that maximum, which is more useful.

At $90°$ the sine is $1$. At $75°$ it is $0.966$, and at $60°$ it is $0.866$. So a pull fifteen degrees off square loses three per cent of its effect, and one thirty degrees off loses thirteen. A mechanic working in a confined space, unable to get a clean pull, is not losing much — and knowing that is worth more than the exact number, because it says the geometry does not have to be right, only roughly right.

That flatness has exactly the structure of the flat maximum in projectile range, and for the same reason: any smooth function is flat at its own maximum, so a quantity being optimised is always forgiving of small errors near the optimum. Both are stationary points, and the practical consequence — the setting does not need to be exact — is one of the more transferable facts in mechanics.

The mirror image is the other end. At $10°$ off the shaft the sine is $0.17$, so five-sixths of the effort is being spent pulling the spanner off the bolt. That is where rounded bolt heads come from, and it is a geometric failure rather than a strength one.

The lever, and what it is really trading #

A lever is the same construction with two forces on it, and the ancient statement — that a small force far out balances a large force close in — is the statement that two torques of opposite sense are equal.

$$F_1 r_1 = F_2 r_2.$$

The trade is usually described as force for distance, and the second half of that is the part worth being careful about. If the lever turns through a small angle $\phi$, the far end sweeps an arc of length $r_1\phi$ and the near end sweeps $r_2\phi$. The work done at the input is $F_1 r_1 \phi$; the work delivered at the output is $F_2 r_2 \phi$. Those are equal, because the torques are.

So the lever multiplies force and divides displacement, and conserves their product exactly. It is not a source of anything. What it provides is a conversion, and the fact that the conversion is lossless in the ideal case is a restatement of energy conservation rather than an independent fact about levers — which is why every simple machine, the pulley and the screw and the gear train included, obeys the same accounting.

The word for the ratio is mechanical advantage, and it equals $r_1/r_2$: a pure number, read off the geometry, with no physics in it. That is a recurring shape in this subject and it turns up again in how a shape decides a falloff exponent — a quantity that looks like it should depend on the forces involved and turns out to depend only on the arrangement.

The pendulum, seen as a torque #

The clearest case of a torque that has already appeared on this site is one that was written down as a force.

A pendulum is a torque problem in disguise. Resolve the weight along and across the rod — as the small-angle rung does — and the across-the-rod component, $mg\sin\theta$, is exactly the component that survives the moment-arm construction. The along-the-rod component points at the pivot and cannot turn anything, which is the same sentence the moment arm says with a sine in it.

The restoring torque about the pivot is $\tau = -mgL\sin\theta$: the weight $mg$ acting at distance $L$, reduced by the sine of the angle from the rod. The minus sign says it turns the bob back towards the bottom.

Nothing about that is new physics. What is new is that it explains why the pendulum’s along-the-rod component was discarded without comment on that rung: it passes through the pivot, so its moment arm is zero. The rod’s tension does the same. Of all the forces on the bob, exactly one has a line of action missing the pivot, and it is the only one that appears in the equation of motion.

That is the practical value of the torque construction. It is a filter. In a problem with a fixed axis, every force whose line passes through the axis can be deleted from the rotational equation before any arithmetic begins — and in most real problems that is most of the forces.

Which pivot, and the answer that is not “the obvious one” #

The torque of a given force is not a property of the force. It is a property of the force and a chosen point, and changing the point changes the number.

This is often taught as a warning and is better taught as a tool, because the choice is free. For a body in equilibrium the sum of the torques is zero about every point — not merely about the real hinge — which turns one equation into an unlimited supply of them, and the art is picking the point that makes the unknowns vanish.

The standard move: take moments about the point where an unknown force acts. Its line of action passes through the chosen point, its moment arm is zero, and it drops out of the equation. A ladder against a wall has an unknown normal force at the wall and an unknown friction force at the floor; taking moments about the foot removes both floor unknowns in one line.

The translational half of the same problem runs alongside and neither contains the other. Forces summed as vectors give the acceleration of the centre of mass; the same forces, each multiplied by its own moment arm about a chosen point, give the angular acceleration. A block on a slope needs both sums to be answered completely, and the two use the same list of forces to compute different things.

For a body that is not in equilibrium the freedom is narrower and worth stating precisely, because it is the commonest place to go wrong. The rotational equation of motion holds about a fixed axis, or about the centre of mass, and not in general about an arbitrary moving point. The centre of mass is privileged for rotation in the same way it is privileged in a collision, and for a related reason: it is the point about which the internal forces of a body contribute nothing.

Where the model stops #

The construction above treats the spanner as a rigid line and the bolt as a point, and both idealisations have a range.

Rigidity. A real spanner bends. A long enough one bends enough that the force is no longer applied where the drawing says, and the extra length stops buying what the geometry promises. Torque wrenches are calibrated against this: the reading is taken from the deflection of a beam, which means the instrument’s accuracy depends on the elasticity it is trying to ignore.

The point pivot. A bolt head is not a point and neither is an axle. The reaction is distributed over a contact area, and the “torque about the axis” is a resultant of a continuous distribution of small forces. Where the contact is large — a shaft in a plain bearing — friction at the contact produces a torque of its own that opposes the motion and does not appear anywhere in the geometry above.

Statics only, so far. Everything on this rung is about the turning effect of a force. Nothing here says how fast the object will actually turn. That question needs a second quantity — the resistance to being turned, which depends on how the mass is arranged rather than on how much of it there is — and that is the next rung.

The plane. Torque here has been a number with a sign, which works when everything happens in one plane. In three dimensions it is a vector, defined as $\mathbf{r}\times\mathbf{F}$, and it points along the axis about which the force tends to turn things. The plane case is the component of that vector perpendicular to the page, and every statement above survives the generalisation — but the intuition that a torque “points” somewhere, at right angles to both the force and the arm, does not come free with the two-dimensional picture and is where gyroscopes stop making sense.

The units, and a piece of pedantry that is not pedantry #

A torque is a force times a distance, so its unit is the newton metre. So is the joule. They are dimensionally identical and they are never interchangeable, and the reason is a good check on whether the construction has been understood.

Work is force times distance along the force. Torque is force times distance perpendicular to it. The two quantities use the two different projections of the same pair of vectors — one takes the component that survives when the vectors are parallel, the other takes what survives when they are square — and combining them is meaningless in the same way that adding an area to a length is.

The convention that keeps them apart is to write torque in newton metres and never in joules, which is a naming rule standing in for a geometric distinction. When a torque is multiplied by an angle in radians the result genuinely is energy, and the radian, being a ratio of lengths, is silently dimensionless. That silence is where the confusion enters, and it is worth noticing rather than resolving.

The proof that came two thousand years early #

The law of the lever is older than the concept of force by roughly two millennia, and the way Archimedes established it is worth recovering, because it does not use one.

His argument is a symmetry argument. Take a weightless beam with equal weights at equal distances from a central support: it balances, because there is nothing to distinguish the two sides — an appeal of exactly the kind the three falloff exponents rest on, where a symmetry claim does the work that a calculation would otherwise have to. Then observe that a weight may be replaced by two half-weights placed symmetrically about its position without disturbing anything, since each half is balanced about the point it replaced.

Repeated, those two moves generate the whole law. A weight of four at distance one is replaced by four unit weights spread symmetrically about that point; a weight of one at distance four is left alone; and the resulting arrangement is symmetric about the pivot, so it balances. Nothing has been assumed except that symmetric arrangements do not fall over.

The modern derivation runs the other way — define torque, sum it, set the sum to zero — and gets the same answer with less insight into why the product of force and distance is the thing that matters rather than their sum or their ratio. Archimedes’ construction shows that the product is what survives repeated subdivision, which is close to a proof that no other combination could work.

Two habits from that argument recur throughout this site. The first is that a symmetry can determine an answer without any equation, which is Gauss’s law in a sentence. The second is that replacing a body by an equivalent distribution of smaller bodies, chosen to make a symmetry visible, is a general technique — the same one that turns a rigid object into a collection of point masses on the next rung, and that turns a charged plate into a row of point charges whenever a field has to be traced.

Where the ladder goes next #

This anchor has a long way to run. The immediate next rung is the moment of inertia — what decides the angular acceleration a given torque produces, and why a hoop and a disc of the same mass behave differently. After it: angular momentum, and why it is conserved when no external torque acts; the parallel-axis theorem, which relates the resistance about any axis to the resistance about the centre of mass; the gyroscope, where the vector nature of the torque produces motion at right angles to the push and every intuition built here fails; rolling as a constraint rather than a force law; the couple, a pair of equal opposite forces with no resultant and a torque about every point; and static equilibrium of structures, where taking moments about a well-chosen point is the entire technique.

The idea to carry forward is smaller than any of those. A force is not the whole description of a push. Where the push acts — the line, not the point — carries information that the arrow does not, and the perpendicular from the pivot to that line is the way to get it out.