Torque

Figure B.7: Application of torque $\tau$ about the origin given by a tangential force $f_t$ on a lever arm of length $R$ .

When twisting things, the rotational force we apply about the center is called a torque (or moment, or moment of force). Informally, we think of the torque as the tangential applied force $f_t$ times the moment arm (length of the lever arm) $R$

$$\tau \isdefs f_tR \protect$$ (B.26)

as depicted in Fig.B.7. The moment arm is the distance from the applied force to the point being twisted. For example, in the case of a wrench turning a bolt, $f_t$ is the force applied at the end of the wrench by one's hand, orthogonal to the wrench, while the moment arm $R$ is the length of the wrench. Doubling the length of the wrench doubles the torque. This is an example of leverage. When $R$ is increased, a given twisting angle $\theta$ is spread out over a larger arc length $R\theta$ , thereby reducing the tangential force $f_t$ required to assert a given torque $\tau$ .

For more general applied forces $\underline{f}\in\mathbb{R}^3$ , we may compute the tangential component $\underline {f}_t$ by projecting $\underline {f}$ onto the tangent direction. More precisely, the torque $\tau$ about the origin $\underline {0}$ applied at a point $\underline{x}\in\mathbb{R}^3$ may be defined by

$$\underline{\tau}\isdefs \underline{x}\times \underline{f} \protect$$ (B.27)

where $\underline {f}$ is the applied force (at $\underline {x}$ ) and $\times$ denotes the cross product, introduced above in §B.4.12.

Note that the torque vector $\underline{\tau}$ is orthogonal to both the lever arm and the tangential-force direction. It thus points in the direction of the angular velocity vector (along the axis of rotation).

The torque magnitude is

$$\tau \isdefs \vert\vert\,\tau\,\vert\vert \eqsp \vert\vert\,\underline{x}\,\vert\vert \cdot \vert\vert\,\underline{f}\,\vert\vert \cdot\sin(\theta),$$

where $\theta$ denotes the angle from $\underline {x}$ to $\underline {f}$ . We can interpret $\vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$ as the length of the projection of $\underline {f}$ onto the tangent direction (the line orthogonal to $\underline {x}$ in the direction of the force), so that we can write

$$\tau \eqsp \vert\vert\,\underline{x}\,\vert\vert \cdot \vert\vert\,\underline{f}_t\,\vert\vert$$

where $\vert\vert\,\underline{f}_t\,\vert\vert = \vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$ , thus getting back to Eq.(B.26).