Physics · Ch 6 — System of Particles and Rotational Motion

Vector (Cross) Product and the Moment of a Force -- Torque

6.5

Vector (Cross) Product and the Moment of a Force -- Torque

Multiplying two vectors together can be done in two genuinely different ways. The scalar (dot) product $\vec A \cdot \vec B = AB\cos\theta$ produces an ordinary number and measures how much one

vector runs along the direction of the other. The vector (cross) product, needed for everything in

this chapter, instead produces a NEW VECTOR, defined for two vectors $\vec A$ and $\vec B$ separated by

angle $\theta$ as

\vec C = \vec A \times \vec B, \qquad |\vec C| = AB\sin\theta

with $\vec C$ directed perpendicular to the plane containing both $\vec A$ and $\vec B$ , its exact sense

(one of the two possible perpendicular directions) fixed by the right-hand rule: curl the fingers of

the right hand from $\vec A$ towards $\vec B$ through the smaller of the two angles between them, and the

extended thumb points along $\vec C$ . Unlike ordinary multiplication, the cross product is not commutative: reversing the order flips the sign, $\vec B \times \vec A = -\vec A \times \vec B$ , because

reversing the order of the two vectors in the right-hand rule reverses which way the thumb points.

Torque (also called the moment of a force) is defined using exactly this cross product. If a force

$\vec F$ acts at a point whose position vector, measured from some chosen reference point $O$ , is $\vec r$ , the torque of that force about $O$ is

\vec\tau = \vec r \times \vec F, \qquad |\vec\tau| = rF\sin\theta

where $\theta$ is the angle between $\vec r$ and $\vec F$ . Torque is the rotational analogue of force: just

as a force is what is needed to change a body's state of translational motion, torque is what is needed to

change a body's state of rotational motion about the chosen point or axis. Two features of the formula

$\tau = rF\sin\theta$ are worth noting directly: torque is zero whenever the force acts exactly along

the line joining $O$ to the point of application ( $\theta = 0$ or $180^\circ$ , since $\sin\theta = 0$ then)

-- a push or pull directed straight at (or away from) the pivot produces no turning effect at all, however

large; and torque is largest, for a given force magnitude and given $r$ , when the force is applied …

Figure 1Torque as the vector (cross) product of position and force

What this figure shows. A point $O$ chosen as the origin, with a dashed position vector $\vec r$ drawn from $O$ to a point $P$ where a force $\vec F$ acts, the force arrow drawn at $P$ making some angle $\theta$ (marked) with $\vec r$ , neither along nor perpendicular to it in general. A dashed parallelogram is sketched spanning $\vec r$ and $\vec F$ , with its enclosed area shaded, labelled as equal to $|\vec\tau| = rF\sin\theta$ . A separate small right-hand-rule inset shows the right hand's fingers curling from $\vec r$ toward $\vec F$ through the angle $\theta$ , with the thumb extended along the resulting torque vector $\vec\tau = \vec r \times \vec F$ , drawn perpendicu …

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