Torque

Push a heavy door right at its hinges. Put your full weight behind it. The door barely moves. Now walk to the far edge and push at the handle with one hand. The door swings open.

The force was the same both times. The door was the same. The hinges were the same. But the effect was completely different. What changed was where you pushed — and that single change multiplied the turning effect eightfold.

This is torque. A force alone does not tell you whether something will rotate. You need three pieces of information: how hard you push, how far from the pivot you push, and at what angle. Torque bundles all three into a single quantity that predicts the rotation.

What torque measures

Think of torque as the rotational analogue of force. Force causes translation — it makes things speed up, slow down, or change direction along a straight line. Torque causes rotation — it makes things spin faster, spin slower, or start spinning in the first place. Just as Newton's second law says \vec{F} = m\vec{a} for straight-line motion, there is a rotational version: \tau = I\alpha, where I is the moment of inertia and \alpha is the angular acceleration. But that law is the subject of the next article. Here, you focus on understanding \tau itself.

Start with what you already know from the door. The turning effect depends on three things:

1. The force F — push harder, and the door swings faster.
2. The distance from the pivot r — push farther from the hinges, and the same force produces a bigger turning effect. This is why door handles are always mounted at the far edge, never near the hinges.
3. The angle \theta between the force and the lever arm — push perpendicular to the door, and you get the maximum turning effect. Push along the door (toward the hinges), and the door does not rotate at all, no matter how hard you push.

The force applied 10 cm from the hinge produces a torque of 40 \times 0.10 = 4 N·m. The same force applied 80 cm from the hinge produces 40 \times 0.80 = 32 N·m. That factor of eight came entirely from choosing where to push. A mechanic tightening a bolt on a scooter knows this instinctively — reach for a longer spanner when the bolt won't budge.

The definition — \vec{\tau} = \vec{r} \times \vec{F}

Reading the definition. Start with \vec{r} — it points from the pivot to the exact spot where the force acts. Then \vec{F} — the force itself. The cross product combines them into a vector that captures both the magnitude of the turning effect and the axis about which the rotation happens.

The cross product is the right tool here because it naturally encodes two things at once: the magnitude (the sine of the angle means only the perpendicular component of force contributes) and the direction (perpendicular to the plane, telling you the axis of rotation). If the force and the lever arm both lie in the plane of this page, the torque vector points either out of the page or into the page.

Why sin θ? At \theta = 90° (force perpendicular to the lever arm), \sin\theta = 1 and the torque is at its maximum — every bit of force goes into turning. At \theta = 0° or 180° (force along the lever arm), \sin\theta = 0 and the torque vanishes — pushing directly toward or away from the pivot compresses or stretches the lever but does not rotate it.

Units. Although N·m has the same dimensions as the joule (kg·m²/s²), the newton-metre is kept as a separate name to distinguish torque from energy. Torque is a turning effect; energy is stored capacity for work. They are different physical concepts that happen to share dimensions.

Three ways to compute the same torque

The formula \tau = rF\sin\theta can be grouped in two other useful ways. All three give the same number — choose whichever matches the information in your problem.

Form 1 — use the angle directly:

Multiply the distance, the force, and the sine of the angle between them. This is the definition.

Form 2 — the moment arm:

Group the sine with r:

Here r_\perp = r\sin\theta is the moment arm (or lever arm) — the perpendicular distance from the pivot O to the line of action of the force.

Why does r\sin\theta equal the perpendicular distance? Drop a perpendicular from O to the infinite line along which the force acts (the "line of action"). In the right triangle formed by O, the point of application P, and the foot of the perpendicular, the hypotenuse is r and the opposite side is the perpendicular distance. Since the angle at P between OP and the line of action is the supplement of \theta (or \theta itself, depending on orientation), the opposite side is r\sin\theta.

Form 3 — the perpendicular force component:

Group the sine with F:

Here F_\perp = F\sin\theta is the component of force perpendicular to the position vector \vec{r}.

Why does only the perpendicular component matter? Decompose \vec{F} into a radial component F_\parallel = F\cos\theta (along \vec{r}) and a tangential component F_\perp = F\sin\theta (perpendicular to \vec{r}). The radial component pulls directly toward or away from the pivot — it loads the hinge but does not turn the door. Only the tangential component creates rotation.

Three forms, one torque:

Each form highlights a different piece of the geometry. The moment-arm form (r_\perp F) is often the fastest in practice, because the perpendicular distance from the pivot to the force's line of action is usually easy to read off a diagram. The force-component form (rF_\perp) is useful when you already know the perpendicular component of force (for example, when forces are resolved into components along and across a beam). The full form (rF\sin\theta) is what you use when you know all three quantities directly.

Explore how angle affects torque

Fix r = 1 m and F = 10 N. The only variable left is the angle \theta between the lever arm and the force. Drag the red point below to see how torque depends on this angle.

Notice the symmetry: \theta = 30° and \theta = 150° give the same torque magnitude. This is because \sin 30° = \sin 150° = 0.5. Physically, a force tilted 30° toward the lever and a force tilted 30° away from it (on the other side) are equally effective at producing rotation — though they rotate the body in opposite directions.

Sign convention — which way is positive?

In two-dimensional problems (and most introductory physics problems are two-dimensional), the torque vector points either out of the page or into the page. Rather than specifying a 3D direction each time, you assign a sign:

• Positive torque (+τ): tends to produce counterclockwise (CCW) rotation
• Negative torque (−τ): tends to produce clockwise (CW) rotation

This is a convention, not a law of physics. Some textbooks reverse it. The NCERT and most JEE resources use counterclockwise as positive, and that is what padho-wiki uses throughout.

In three dimensions, the sign convention is replaced by the right-hand rule: curl the fingers of your right hand from \vec{r} toward \vec{F}, and your thumb points in the direction of \vec{\tau}. If that direction is out of the page, the torque is positive (CCW). If into the page, negative (CW). The sign convention and the right-hand rule always agree.

Torque about a point and torque about an axis

There is a subtle but important distinction. The formula \vec{\tau} = \vec{r} \times \vec{F} gives the torque about a point — a full vector in 3D space. But many real objects can only rotate about a fixed axis: a door about its vertical hinges, a wheel about its axle, a see-saw about its central pivot. In those cases, what matters is the torque about the axis — only the component of \vec{\tau} along that axis contributes to the rotation.

A door can only rotate about a vertical axis through its hinges. If you push the door horizontally and perpendicular to its surface, the torque vector points vertically (along the axis), and it contributes fully to swinging the door open. But if you push vertically downward on the door's edge, the torque vector is horizontal — perpendicular to the hinge axis. This torque tries to rip the door off its hinges, but it does not rotate it about the hinge axis. Only the component of torque along the axis of rotation matters.

For two-dimensional problems where all forces lie in one plane and the rotation axis is perpendicular to that plane, this distinction collapses: the torque about the point is the torque about the axis. This is why you can ignore the distinction in most introductory problems.

The couple — pure rotation with zero net force

A couple is a pair of forces that are equal in magnitude, opposite in direction, and applied at different points on the same body. The net force is zero — a couple cannot translate the body. But the net torque is not zero — a couple produces pure rotation.

Consider two forces: \vec{F} at point A and -\vec{F} at point B, where A and B are separated by a perpendicular distance d between the lines of action.

Deriving the couple's torque:

Compute the net torque about an arbitrary point O:

Why: each force contributes a torque about O. The position vectors \vec{r}_A and \vec{r}_B point from O to the points where the forces are applied.

Why: factor out \vec{F} from the cross product (the cross product distributes over vector addition). The result depends only on the difference \vec{r}_A - \vec{r}_B = \vec{BA} — the vector from B to A.

Why: the position of O has cancelled out. The torque of a couple does not depend on where you choose the pivot — it is the same about every point. This is a remarkable property: a couple has no preferred centre of rotation.

When the forces are perpendicular to the line joining A and B:

Why: |\vec{BA}| = d, the angle between \vec{BA} and \vec{F} is 90°, and \sin 90° = 1.

You encounter couples constantly. Turning a bicycle handlebar — your right hand pushes forward, your left hand pulls backward. Opening a jar of pickle (achaar) — your fingers grip opposite edges and twist in opposite directions. Steering a scooter. Turning a water tap. In each case, two hands apply equal and opposite forces, and the object rotates without translating.

Worked examples

Common confusions

• "Torque and force are the same thing." They are not. Force causes translation (\vec{F} = m\vec{a}); torque causes rotation (\vec{\tau} = I\vec{\alpha}). A body can have zero net force and nonzero net torque — that is a couple. It can have nonzero net force and zero net torque — a single force applied at the pivot. The two quantities describe different aspects of how a force acts on a body.

• "A bigger force always means bigger torque." Not necessarily. A 100 N force applied directly at the pivot (r = 0) produces zero torque. A 10 N force applied 2 m from the pivot produces 20 N·m. Distance and angle matter as much as force magnitude. This is the whole point of a lever.

• "Torque depends on the choice of pivot." For a single force, yes — changing the pivot changes \vec{r}, which changes \vec{\tau}. This is why you must always specify the point about which you are computing torque. But for a couple, the torque is the same about every point — the pivot cancels out in the derivation.

• "The SI unit of torque is the joule." It is not, even though N·m and J share the same dimensions (kg·m²/s²). The newton-metre is deliberately kept as a separate name. Torque is a turning effect; energy is a scalar quantity measuring capacity for work. A torque of 20 N·m does not mean the body has 20 J of energy — it means a force is producing a turning effect of that magnitude. Torque can do work (W = \tau\,\Delta\theta), but torque itself is not work.

• "If the force is along the lever arm, the force has no effect." The force has no rotational effect — no torque. But it still acts on the body: it compresses or tensions the lever. The radial component F_\parallel = F\cos\theta loads the pivot (the hinge carries that force) but produces zero turning. Only the perpendicular component F_\perp = F\sin\theta creates torque.

Where this leads next

• Newton's Second Law for Rotation — the full equation \tau = I\alpha, connecting torque to moment of inertia and angular acceleration.
• Angular Momentum — the rotational analogue of linear momentum, and the quantity whose rate of change equals torque.
• Conservation of Angular Momentum — what happens when the net torque is zero: spinning skaters speed up, gyroscopes resist tilting, and planetary orbits obey Kepler's second law.
• Moment of Inertia — the rotational analogue of mass, determining how much angular acceleration a given torque produces.
• Equilibrium of Rigid Bodies — applying \sum \vec{F} = 0 and \sum \vec{\tau} = 0 together to solve for unknown forces in structures and machines.