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Torque as a Cross Product

You’ll be able to

Only the perpendicular component turns things

Torque is the rotational analogue of force, and it depends on where and in what direction a force is applied, not merely on how large it is. The magnitude is τ = rF sin θ, where θ is the angle between the position vector from the pivot and the force. A force directed straight at the pivot has θ = 0 and produces no torque at all, no matter how strong — which is why pushing a door toward its hinges never opens it.

Torque as a vector product
τ = r × F |τ| = rF sin θ = r_⊥ F = r F_⊥
The three magnitude forms are equivalent: use the full angle, or the perpendicular distance from the pivot to the line of action (the moment arm), or the component of force perpendicular to r. Pick whichever the diagram makes easiest.

The moment arm

The moment arm r_⊥ is the perpendicular distance from the pivot to the line of action of the force — the infinite line along which the force points, extended in both directions if necessary. Sliding a force along its own line of action changes nothing about the torque, which is a useful simplification. In practice, drawing the line of action and dropping a perpendicular to it from the pivot is usually faster and less error-prone than hunting for the correct angle to put in sin θ.

Direction, and why torque is a vector

Torque is a vector because rotation has an axis and a sense. Its direction is given by the right-hand rule: point the fingers along r, curl them toward F, and the thumb gives the direction of τ, perpendicular to the plane containing r and F. In two-dimensional problems this reduces to a sign — counterclockwise positive, clockwise negative, by the usual convention. The vector nature is not decoration: it is what makes conservation of angular momentum a statement about a direction as well as a magnitude, and it is why a spinning gyroscope precesses instead of falling.

Worked example

A force of 50 N is applied at a point 0.30 m from a pivot, at 40° to the line joining the pivot to the point of application. Find the torque magnitude and identify the moment arm.

  1. 1.Use τ = rF sin θ with r = 0.30 m, F = 50 N and θ = 40°.
  2. 2.sin 40° ≈ 0.643.
  3. 3.τ = (0.30)(50)(0.643) ≈ 9.6 N·m.
  4. 4.Equivalently, the moment arm is r_⊥ = r sin θ = 0.30(0.643) ≈ 0.193 m, and τ = r_⊥F = 0.193(50) ≈ 9.6 N·m — the same result by the other route.
Answer: τ ≈ 9.6 N·m, with a moment arm of about 0.19 m; applying the same force perpendicular to r would give the maximum 15 N·m
Checkpoint

A force is applied to a wrench directly along the line toward the bolt at its center. The torque about the bolt is —

Tip

The angle in τ = rF sin θ is between r and F, not between the force and the horizontal or between the force and the object. Sketching the line of action and measuring the perpendicular distance to the pivot avoids the angle question altogether and is worth making a habit.

Checkpoint

A force in the plane of the page is applied to a disk that can rotate about an axis perpendicular to the page. The torque vector points —

Answer the 2 checkpoints as you read.

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