Torque

A mechanic loosening a stubborn wheel nut does not push harder — she reaches for a longer spanner. The force in her hand may be exactly the same twenty newtons either way, yet the longer handle turns the nut and the short one does not. Something about where and how a force is applied, not just how big it is, decides whether a body starts to spin. That something is torque, the rotational twin of force, and it is the single idea that explains why door handles sit far from the hinge, why a see-saw balances at particular distances rather than particular weights alone, and why tightening a bolt with your fingers near its head is so much harder than with a wrench. Torque is defined precisely as a cross product, $\vec\tau = \vec r\times\vec F$ , but the physics behind the symbols is intuitive: swing a force around a pivot and only the part of it that actually turns the body counts. A push straight along the spanner, toward or away from the nut, does nothing to turn it at all — every ounce of that force is wasted. A push at right angles to the spanner turns it hardest. Everything in between is governed by a single $\sin\theta$ factor, and mastering that factor is the difference between shrugging off a rotational-mechanics problem and getting tangled in it. This section builds torque from its vector definition through the everyday moment-arm picture, connects it to the rotational equation of motion $\tau = I\alpha$ that governs how fast a body spins up, and works through the classic spanner-and-nut calculations that CBSE boards and competitive exams both love to ask. By the end you should be able to look at any force applied to any lever, wrench, door or rod and immediately identify how much of it actually turns the object — and in which sense.

Key takeaways

Torque is the turning effect of a force, defined as the vector cross product $\vec\tau = \vec r\times\vec F$ , where $\vec r$ runs from the axis (or pivot) to the point where the force acts.
Magnitude: $\tau = rF\sin\theta$ , where $\theta$ is the angle between $\vec r$ and $\vec F$ — torque is maximum when the force is perpendicular to $\vec r$ and zero when it is along $\vec r$ .
Moment arm form: $\tau = Fd$ , with $d = r\sin\theta$ the perpendicular distance from the axis to the line of action of the force — the everyday way spanner and see-saw problems are actually solved.
Direction by the right-hand rule: $\vec\tau$ points along the rotation axis, perpendicular to the plane of $\vec r$ and $\vec F$ ; by convention anticlockwise torques (viewed from the chosen side) are taken positive.
Torque drives angular acceleration: $\tau = I\alpha$ is the rotational analogue of $F = ma$ , with moment of inertia $I$ playing the role of mass — a body's resistance to being spun up.
SI unit newton-metre (N·m) — numerically the same combination as the joule, but torque and work/energy are physically different (cross product vs dot product) and are never interchanged.
Key idea: torque tau equals r cross F, with magnitude rF sine theta or, more usably, force times the perpendicular moment arm Fd, and it is torque — not force alone — that produces angular acceleration through tau equals I alpha.

The core insight

Torque $\vec\tau = \vec r\times\vec F$ measures the turning effect of a force about an axis, with magnitude $\tau = rF\sin\theta = Fd$ (moment arm $d = r\sin\theta$ ) and direction along the axis by the right-hand rule, and it is this quantity — not force by itself — that drives angular acceleration through $\tau = I\alpha$ .

Where it fits in the syllabus

Torque (moment of a force) is a core, prescribed idea in the CBSE Class 11 chapter "System of Particles and Rotational Motion." The board expects the vector definition $\vec\tau = \vec r\times\vec F$ , the magnitude formula $\tau = rF\sin\theta$ , the moment-arm shortcut $\tau = Fd$ , and the link to angular acceleration via $\tau = I\alpha$ (moment of inertia $I$ is developed as its own topic elsewhere in this chapter, and used here only as the constant of proportionality). Torque also underlies the equilibrium-of-rigid-bodies and principle-of-moments applications later in the same chapter, so getting the sign convention and moment-arm logic right here pays off directly in those problems.

The physical picture

A force can act on a rigid body at any point, and the same force applied at different points, or in different directions, has very different turning consequences. To capture this, fix a reference point or axis (typically the pivot the body rotates about) and let $\vec r$ be the position vector from that axis to the point where the force $\vec F$ acts. The torque of $\vec F$ about the axis is defined as the vector cross product:

\vec\tau = \vec r\times\vec F

where: $\vec r$ is the position vector from the axis to the point of application of the force, and $\vec F$ is the applied force. Because it is a cross product, $\vec\tau$ is perpendicular to the plane containing $\vec r$ and $\vec F$ , and its sense (which way along the axis it points) follows the right-hand rule — curl the fingers of the right hand from $\vec r$ toward $\vec F$ through the smaller angle, and the thumb gives the direction of $\vec\tau$ . Physically, this direction is the axis about which the force tends to spin the body, and its sign tells you whether that spin is anticlockwise or clockwise as seen from a chosen side. The magnitude of the torque follows directly from the definition of the cross product:

\tau = rF\sin\theta

where: $r = |\vec r|$ is the distance from the axis to the point of application, $F = |\vec F|$ is the magnitude of the force, and $\theta$ is the angle between $\vec r$ and $\vec F$ measured the short way. This single factor $\sin\theta$ explains almost every torque puzzle a student meets: when $\theta = 90°$ , $\sin\theta = 1$ and torque is maximum for that $r$ and $F$ ; when $\theta = 0°$ or $180°$ (force along or opposite to $\vec r$ ), $\sin\theta = 0$ and the torque vanishes completely, however large the force. A more physically intuitive route to the same magnitude uses the moment arm (or perpendicular distance) $d$ , the shortest distance from the axis to the line of action of the force — the infinite straight line along which $\vec F$ points, extended in both directions if needed. Since $d = r\sin\theta$ :

\tau = Fd

where: $F$ is the magnitude of the force and $d$ is the perpendicular distance from the axis to the force's line of action. This is the form used in almost every practical problem: you do not need to resolve $\vec r$ and $\vec F$ into components at all — just drop a perpendicular from the axis onto the line along which the force acts, measure that length, and multiply by the force. It is exactly why a longer spanner, or pushing a door far from its hinge, produces a larger turning effect for the same muscular force: the moment arm $d$ is larger.

Figure

A pivot point with a rod extending to a force applied at an angle, showing the position vector, the angle between the rod and the force, and the perpendicular moment arm dropped from the pivot to the line of action of the force

Torque plays the role in rotational motion that force plays in linear motion. Just as an unbalanced force produces linear acceleration through $F = ma$ , an unbalanced torque produces angular acceleration through $\tau = I\alpha$ , where $I$ is the body's moment of inertia — its rotational inertia, or resistance to being spun up, developed as its own topic elsewhere in this chapter. When several torques act on a rigid body — from an applied force, its weight, and reaction forces at a support — they combine as vectors along the rotation axis; for the coplanar problems typical at this level, this reduces to adding signed scalars about a chosen axis, with anticlockwise torques usually taken positive and clockwise negative. This scalar bookkeeping is exactly what the equilibrium and principle-of-moments topics later in this chapter build on. The SI unit of torque is the newton-metre (N·m). Numerically this is the same combination of units as the joule (the unit of work and energy), but the two quantities are not interchangeable: work is a scalar built from the dot product $W = \vec F\cdot\vec d$ (force along the direction of displacement), while torque is a vector-like quantity built from the cross product $\vec\tau = \vec r\times\vec F$ (force perpendicular to the position vector matters most). Examiners routinely test whether a student conflates the two just because the units look alike.

Derivation AFrom the cross-product definition to the everyday spanner formula

The goal of this derivation is to show, starting only from the vector definition $\vec\tau = \vec r\times\vec F$ , how the working formula $\tau = Fd$ used in every practical torque calculation follows directly from it, and why that formula is exactly equivalent to $\tau = rF\sin\theta$ — so that a student can move confidently between the "position vector and angle" picture used in the definition and the "perpendicular moment arm" picture used in spanner, door and see-saw problems, without ever losing track of which quantity is which or why the two routes must agree.

Step 1. Start from the definition. For a force $\vec F$ applied at a point with position vector $\vec r$ from the axis, the torque is $\vec\tau = \vec r\times\vec F$ . By the standard property of the vector cross product, the magnitude of $\vec\tau$ is $\tau = |\vec r||\vec F|\sin\theta = rF\sin\theta$ , where $\theta$ is the angle between $\vec r$ and $\vec F$ measured through the smaller value. This magnitude formula is exact and needs no approximation.

Step 2. Resolve the force into two perpendicular components relative to $\vec r$ : a component $F\cos\theta$ along the direction of $\vec r$ (call it the radial component) and a component $F\sin\theta$ perpendicular to $\vec r$ (the tangential component). Only the tangential component has any turning effect about the axis — the radial component points straight along the line joining the axis to the point of application and, however large, cannot rotate the body about that axis, since a force along $\vec r$ has no lever arm to act through.

Step 3. Because only $F\sin\theta$ (the perpendicular component of the force) contributes, the torque can equally be written as $\tau = r\,(F\sin\theta)$ — the full distance $r$ times the perpendicular component of the force. This is algebraically identical to $\tau = rF\sin\theta$ from Step 1, just grouped differently: instead of " $r$ times $F$ times $\sin\theta$ " read as one block, it is now " $r$ times the useful part of $F$ ."

Step 4. Regroup the same product the other way: $\tau = (r\sin\theta)\,F$ . Define $d = r\sin\theta$ — this is precisely the perpendicular distance from the axis to the line of action of $\vec F$ (drop a perpendicular from the axis onto the extended line along which $\vec F$ points; simple geometry of the right triangle formed by $\vec r$ , that perpendicular, and the line of action shows its length is exactly $r\sin\theta$ ). With this definition, $\tau = Fd$ .

Step 5. Confirm the two routes agree and see why the moment-arm form is more useful in practice. Both $\tau = r(F\sin\theta)$ and $\tau = (r\sin\theta)F$ are the same product $rF\sin\theta$ regrouped, so $\tau = Fd$ and $\tau = rF\sin\theta$ are exactly equivalent — never two different answers, only two ways of measuring the same turning effect. The moment-arm form is preferred in real problems because $d$ can often be read directly from a diagram (the perpendicular distance from a hinge to a line drawn along the force) without ever needing to identify $\vec r$ explicitly or measure the angle $\theta$ between it and $\vec F$ .

Worked examples

Board level - spanner turning a nut

A force of $20\ \text{N}$ is applied at the end of a spanner of length $0.3\ \text{m}$ , perpendicular to the spanner, to turn a nut. Find the torque about the nut.

Step 1. Identify $r$ , $F$ and $\theta$ . Here $r = 0.3\ \text{m}$ is the length of the spanner (the position vector from the nut to the point of application), $F = 20\ \text{N}$ , and since the force is applied perpendicular to the spanner, $\theta = 90°$ , so $\sin\theta = 1$ .

Step 2. Apply $\tau = rF\sin\theta$ . \tau = 0.3\times20\times1 = 6\ \text{N·m}

Step 3. Interpret the result. Because $\theta = 90°$ makes $\sin\theta$ as large as it can be, this is the maximum possible torque obtainable from this force and this spanner length — no other angle of application would give a larger turning effect. The torque about the nut is 6 N·m.

Competitive level - same spanner, force at an angle, plus a comparison

The same $20\ \text{N}$ force is now applied at the end of the same $0.3\ \text{m}$ spanner, but at an angle of $30°$ to the spanner instead of perpendicular to it. Find the torque, find the perpendicular moment arm directly, and compare the turning effect with the perpendicular case above.

Step 1. Apply $\tau = rF\sin\theta$ with $\theta = 30°$ . \tau = 0.3\times20\times\sin30° = 6\times0.5 = 3\ \text{N·m}

Step 2. Cross-check with the moment-arm route. The perpendicular distance from the nut to the line along which the force acts is $d = r\sin\theta = 0.3\times\sin30° = 0.3\times0.5 = 0.15\ \text{m}$ , so $\tau = Fd = 20\times0.15 = 3\ \text{N}\,\cdot\,\text{m}$ — exactly the same answer by an independent route, confirming the two formulas agree.

Step 3. Compare with the perpendicular case. The torque here, $3\ \text{N}\,\cdot\,\text{m}$ , is exactly half of the $6\ \text{N}\,\cdot\,\text{m}$ found when the same force was applied perpendicular to the spanner. The torque is 3 N·m, precisely half the maximum torque available from this force and spanner. This is why mechanics — and everyday tool use — favours applying force perpendicular to a lever arm: for a fixed force magnitude, the turning effect is maximised at $\theta = 90°$ and falls away as $\sin\theta$ for any other angle. Timing note: in a timed paper, always check first whether the force is already perpendicular to the arm — if so, skip straight to $\tau = rF$ and save the trigonometry. Only bring in $\sin\theta$ (or the equivalent moment-arm calculation $d = r\sin\theta$ ) when the angle is explicitly not $90°$ ; misreading a perpendicular case as needing $\sin\theta$ wastes precious minutes and invites arithmetic slips.

Common traps

Using the full distance $r$ instead of the correct perpendicular moment arm $d = r\sin\theta$ — torque depends on the component of $\vec r$ perpendicular to $\vec F$ (equivalently, the perpendicular distance from the axis to the line of action of the force), not on $r$ alone.
Forgetting that torque is exactly zero when the force is applied along the line joining the point of application to the axis ( $\theta = 0°$ or $180°$ ), no matter how large that force is — a common trick in board and competitive questions alike.
Treating torque and work/energy as interchangeable just because both carry the unit N·m — they are physically different quantities: torque comes from a cross product and is associated with rotation about an axis, while work/energy comes from a dot product and is a plain scalar.
Getting the right-hand rule wrong when several forces act at once, especially mixing up clockwise and anticlockwise sign conventions when torques must be added as signed scalars about a fixed axis.
Assuming a bigger force always means a bigger torque — a small force applied with a large moment arm, perpendicular to the arm, can produce a larger torque than a big force applied close to the axis or nearly along the arm.

You should be able to…

State the vector definition $\vec\tau = \vec r\times\vec F$ and compute its magnitude using either $\tau = rF\sin\theta$ or the moment-arm form $\tau = Fd$ .
Identify the perpendicular moment arm $d$ for a given force and pivot directly from a diagram, even when the force is drawn at an angle rather than perpendicular to the lever.
Explain, using a concrete example, why torque depends on the point of application and the direction of a force and not merely on its magnitude.
Recognise instantly when a force produces zero torque about a given axis ( $\theta = 0°$ or $180°$ ) without doing any calculation.
Distinguish torque from work/energy despite the shared N·m = J unit, and explain the physical reason (cross product versus dot product) the two are never the same quantity.
Relate torque to angular acceleration through $\tau = I\alpha$ , and explain qualitatively how a larger moment of inertia changes the angular acceleration produced by a fixed torque.

Ask the tutor: stuck on finding the moment arm when a force is drawn at an odd angle to the lever, or want more spanner-and-torque problems worked through step by step? Ask the AI tutor to walk you through it.

Why does torque depend on the angle theta and not just on the force and the distance from the axis?

Because only the part of the force that actually pushes the object around the axis can turn it — the part that pushes straight toward or away from the axis has nothing to turn against and is wasted. The factor $\sin\theta$ is exactly the fraction of the force that acts perpendicular to the position vector $\vec r$ , which is the only part that contributes to rotation. When $\theta = 90°$ the entire force is perpendicular to $\vec r$ and contributes fully, giving the maximum torque $\tau = rF$ ; when $\theta = 0°$ the force points straight along $\vec r$ and contributes nothing, giving zero torque however large $F$ is.

How do I find the moment arm d when the force is not drawn perpendicular to the lever in the diagram?

Extend the line along which the force acts (its line of action) as far as needed in both directions, treating it as an infinite straight line rather than just the short arrow drawn in the figure. Then drop a perpendicular from the axis to that extended line — the length of that perpendicular segment is $d$ . Algebraically, if you already know $r$ and $\theta$ , you can skip the drawing and simply compute $d = r\sin\theta$ , then use $\tau = Fd$ . Both routes always give the same answer, since they are the same product regrouped, as shown in the derivation above.

Is torque a vector or is it fine to just use plus and minus signs?

Formally torque is a vector, $\vec\tau = \vec r\times\vec F$ , pointing along the rotation axis with a direction fixed by the right-hand rule. In most CBSE Class 11 problems, however, all the forces and pivots lie in one flat plane, so every torque vector points along the same single axis (in or out of the page). In that common special case it is perfectly correct, and much faster, to drop the vector notation and treat torques as signed scalars — positive for anticlockwise, negative for clockwise (or vice versa, as long as you are consistent) — and simply add them up. Reach for the full cross product only when forces act in genuinely different planes, which is rare at this level.

Why doesn't a bigger moment of inertia mean a bigger torque?

Torque and moment of inertia are two separate ingredients that only get linked together through the equation $\tau = I\alpha$ , which is the rotational analogue of $F = ma$ . The torque $\tau$ is produced entirely by the applied force and where it acts — it does not know or care what the body's moment of inertia is. The moment of inertia $I$ (developed as its own topic elsewhere in this chapter) instead describes how much angular acceleration $\alpha$ a given torque manages to produce: a larger $I$ means the same torque produces a smaller $\alpha$ , exactly as a larger mass means the same force produces a smaller linear acceleration. So a bigger $I$ does not create more torque; it makes the body harder to spin up with whatever torque is applied.