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Section 4.6 Application: Torque

One of the most important physical applications of the cross product is torque, the rotational analog of force. A force makes an object accelerate; a torque makes it rotate. How effective a force is at producing rotation depends not only on its magnitude, but also on where it is applied and in what direction. Think of pushing a door: pushing hard near the hinge barely moves it, while a light push at the doorknob swings it open easily. Pushing at the doorknob but toward the hinge does nothing at all. The cross product captures all three effects at once.

Definition 4.18. Torque.

Suppose a force \(\vec F\) is applied at a point whose position vector relative to the axis of rotation is \(\vec r\text{.}\) The torque produced by \(\vec F\) is the vector
\begin{equation} \vec\tau = \vec r \times \vec F.\tag{4.7} \end{equation}
Everything we know about the cross product now translates into physics. By (4.1), the magnitude of the torque is
\begin{equation} |\vec\tau| = |\vec r||\vec F||\sin\theta|,\tag{4.8} \end{equation}
where \(\theta\) is the angle between \(\vec r\) and \(\vec F\text{,}\) and the direction of \(\vec\tau\) is given by the right-hand rule: \(\vec\tau\) points along the axis of rotation, and curling the fingers of the right hand from \(\vec r\) toward \(\vec F\) gives the sense of the rotation. Torque has units of meter-newtons (\(\text{m}\cdot\text{N}\)) in SI.
The quantity in (4.8) can be grouped in two useful ways:
\begin{align*} |\vec\tau| \amp= \underbrace{\left(|\vec r|\sin\theta\right)}_{r_\perp}|\vec F| = r_\perp |\vec F|,\\ |\vec\tau| \amp= |\vec r|\underbrace{\left(|\vec F|\sin\theta\right)}_{F_\perp} = |\vec r|\, F_\perp. \end{align*}
The first grouping says the torque is the force times the lever arm \(r_\perp\text{,}\) the perpendicular distance from the axis to the line along which the force acts. The second says it is the distance \(|\vec r|\) times \(F_\perp\text{,}\) the component of the force perpendicular to \(\vec r\text{;}\) the parallel component \(F_\parallel\) points straight at the axis and produces no rotation.

Example 4.20. Biceps torque.

The biceps muscle exerts a vertical force of \(|\vec F| = 700\) N on the lower arm, attached at a point \(5.0\) cm from the elbow joint (the axis of rotation). Calculate the torque about the elbow when
  1. the forearm is horizontal, and
  2. the forearm makes a \(30^\circ\) angle below the horizontal.
Figure 4.21. The torque \(\vec\tau = \vec r \times \vec F\) of the biceps force about the elbow. When the forearm drops \(30^\circ\) below the horizontal, the angle between \(\vec r\) and \(\vec F\) becomes \(120^\circ\text{,}\) the lever arm \(r_\perp = |\vec r|\sin\theta\) shortens, and the torque decreases from \(35\) to \(30\) \(\text{m}\cdot\text{N}\text{.}\)
described in detail following the image
Side-by-side diagrams of an arm. In (a), the forearm is horizontal and a 700 N biceps force acts 5.0 cm from the elbow axis. In (b), the forearm is angled 30 degrees below horizontal, making a 60 degree angle with the vertical, with lever arm R.
Figure 4.22. The biceps exerts a 700 N force on the forearm: (a) forearm horizontal, with the muscle attached 5.0 cm from the elbow axis; (b) forearm at 30° below horizontal, with lever arm \(R\text{.}\)
Solution.
Place the elbow at the origin and let the force act in the plane \(z = 0\text{,}\) with \(\vec F = \langle 0, 700, 0\rangle\) pointing straight up.
Part 1. The forearm is horizontal, so \(\vec r = \langle 0.050, 0, 0\rangle\) and the angle between \(\vec r\) and \(\vec F\) is \(\theta = 90^\circ\text{.}\) Then
\begin{align*} \vec\tau = \vec r \times \vec F \amp= \begin{vmatrix} \vec i \amp \vec j \amp \vec k \\ 0.050 \amp 0 \amp 0 \\ 0 \amp 700 \amp 0 \end{vmatrix} = \langle 0, 0, (0.050)(700)\rangle = \langle 0, 0, 35 \rangle, \end{align*}
so \(|\vec\tau| = 35\ \text{m}\cdot\text{N}\text{.}\) Equivalently, \(|\vec\tau| = |\vec r||\vec F|\sin 90^\circ = (0.050)(700)(1) = 35\ \text{m}\cdot\text{N}\text{.}\) The torque points in the direction \(+\vec k\) (out of the page), which by the right-hand rule corresponds to a counterclockwise rotation of the forearm about the elbow.
Part 2. Now the forearm points \(30^\circ\) below the horizontal, so
\begin{equation} \vec r = 0.050\,\langle \cos(-30^\circ), \sin(-30^\circ), 0\rangle = \langle 0.0433, -0.025, 0\rangle,\tag{4.9} \end{equation}
while \(\vec F = \langle 0, 700, 0\rangle\) is unchanged. The cross product gives
\begin{align*} \vec\tau = \vec r \times \vec F \amp= \begin{vmatrix} \vec i \amp \vec j \amp \vec k \\ 0.0433 \amp -0.025 \amp 0 \\ 0 \amp 700 \amp 0 \end{vmatrix} = \langle 0, 0, (0.0433)(700) - (-0.025)(0)\rangle\\ \amp= \langle 0, 0, 30.3 \rangle \approx \langle 0, 0, 30 \rangle. \end{align*}
So \(|\vec\tau| \approx 30\ \text{m}\cdot\text{N}\text{,}\) again counterclockwise. To check with (4.8): the vector \(\vec r\) sits at \(-30^\circ\) and \(\vec F\) at \(90^\circ\text{,}\) so the angle between them is \(\theta = 120^\circ\text{,}\) and
\begin{equation} |\vec\tau| = |\vec r||\vec F|\sin 120^\circ = (0.050)(700)\left(\tfrac{\sqrt 3}{2}\right) \approx 30\ \text{m}\cdot\text{N}.\tag{4.10} \end{equation}
Note that \(\sin 120^\circ = \sin 60^\circ\text{,}\) so this agrees with the lever-arm point of view: the lever arm has shortened to \(r_\perp = (0.050)\sin 60^\circ \approx 0.043\) m. The arm exerts less torque at this angle than when the forearm is horizontal — weight machines at gyms are often designed to account for exactly this variation.
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