Torque as a Vector and the Couple

Torque about a point

The last lesson measured torque about an axis. Often no axis is given. A spanner may be pushed at any angle, and a body in space turns about no fixed line. So we define torque about a point. Let O be any point, and let a force \(\vec F\) act at a point P. Let \(\vec r\) be the position vector of P from O. The torque of \(\vec F\) about O is the cross product of \(\vec r\) and \(\vec F\). \[ \vec\tau = \vec r \times \vec F \]

The magnitude \(rF\sin\theta\) is the same product as before, so nothing learnt about moment arms is lost. What is new is the direction. The vector \(\vec\tau\) is perpendicular to the plane that holds \(\vec r\) and \(\vec F\). The right-hand rule picks one of the two perpendicular directions. Curl the fingers of the right hand from \(\vec r\) towards \(\vec F\). The thumb then points along \(\vec\tau\).

For a drawing on a page this is simple. A force that tends to turn the body anticlockwise about O gives a torque pointing out of the page. A clockwise turning gives a torque pointing into the page. Out of the page is drawn as a dot in a ring, like the tip of an arrow coming at you. Into the page is drawn as a cross in a ring, like its tail.

The torque in components

Write \(\vec r = x\hat i + y\hat j + z\hat k\) and \(\vec F = F_x\hat i + F_y\hat j + F_z\hat k\). The cross product is then the determinant \[ \vec\tau = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ F_x & F_y & F_z \end{vmatrix} = (yF_z - zF_y)\,\hat i + (zF_x - xF_z)\,\hat j + (xF_y - yF_x)\,\hat k \] For a force in the xy plane, acting at a point of that plane, only the last term is left: \(\tau_z = xF_y - yF_x\). A positive \(\tau_z\) points out of the page and means anticlockwise turning.

In Fig 2.1 the two forces have the same size, \(\sqrt{29} \approx 5.39\,\text{N}\), and act at the same point. Yet their torques differ in size and in sign. The first force points steeply upward, to the left of the line OP as seen from O, and turns the body anticlockwise. The second points to the right of that line and turns it clockwise. The dashed line is the line of action, and the violet segment is the moment arm. It is \(11/5.39 \approx 2.04\,\text{m}\) in the first case and \(4/5.39 \approx 0.74\,\text{m}\) in the second.

About a point and about an axis

Torque about a point and torque about an axis are related, but they are not the same thing. Suppose a body is held so that it can turn only about a fixed axis through O, like a door on its hinges. Only the part of \(\vec\tau\) along that axis can turn it. The other parts try to twist the axis itself, and the bearings or hinges push back.

So the torque about an axis is the component of \(\vec\tau\) along the axis, \(\tau_{\text{axis}} = \vec\tau\cdot\hat n\), where \(\hat n\) is a unit vector along it. This component has the same value whichever point of the axis is used as O. For the torque of the last example, the component along the z axis is \(-5\,\text{N m}\), and the component along the x axis is \(+5\,\text{N m}\).

Let every force lie in one plane, with the axis perpendicular to it. Then this is exactly the torque \(rF\sin\theta\) of the last lesson, with its sign. Most problems of this chapter are of that kind. Then a torque is a signed number, and \(\tau_z = xF_y - yF_x\) gives the number with its sign.

The couple

Two equal and opposite forces that act along different lines form a couple. The two hands on a steering wheel apply a couple. So do the two fingers that turn a tap, and the fingers that turn a key. The net force of a couple is zero, so it gives the centre of mass no acceleration. Yet it turns the body.

Let \(\vec F\) act at \(\vec r_1\) and \(-\vec F\) act at \(\vec r_2\), both measured from some point O. The total torque about O is \[ \vec\tau = \vec r_1 \times \vec F + \vec r_2 \times (-\vec F) = (\vec r_1 - \vec r_2) \times \vec F \] The result contains only \(\vec r_1 - \vec r_2\), the position of one force relative to the other. The point O has vanished from the result. So the torque of a couple is the same about every point. Its size is \(Fd\), where \(d\) is the perpendicular distance between the two lines of action.

Put O at \(x = 0.1\,\text{m}\), between the lines. The upward force gives \(15 \times 0.1 = 1.5\,\text{N m}\), and the downward force gives \(15 \times 0.3 = 4.5\,\text{N m}\), both anticlockwise. Now put O at \(x = -0.6\,\text{m}\), outside both lines. The upward force gives \(+12.0\,\text{N m}\) and the downward force \(-6.0\,\text{N m}\). The share of each force changes, but the total is \(6.0\,\text{N m}\) every time.

A couple cannot be balanced by any single force, since a single force would leave a net force. Only another couple, of equal and opposite torque, can balance a couple. A wing nut held still against a turning spanner is an example. This is why a couple is treated as a thing in itself: a pure turning, with no push.

Now return to the bus driver on the ghat road. Her two hands apply a couple: equal and opposite forces on the two sides of the wheel. The net force on the wheel is zero, so the steering column is not pushed towards either hand. The torque of a couple is \(Fd\) about every point. So the turning depends on the gap between her hands, not on where she grips the rim.