Torque measures the rotational effect of a force about a chosen pivot or axis. In vector form, torque is defined by
the cross product
\[
\vec{\tau}=\vec r\times\vec F.
\]
Here, \(\vec r\) is the position vector from the pivot to the point where the force is applied, and \(\vec F\) is the
force vector.
1. Position vector from the pivot
If the pivot point is
\[
\vec r_{\mathrm{pivot}}=(x_p,y_p,z_p)
\]
and the force is applied at
\[
\vec r_{\mathrm{point}}=(x,y,z),
\]
then the relative position vector is
\[
\vec r=\vec r_{\mathrm{point}}-\vec r_{\mathrm{pivot}}.
\]
This relative vector is the moment arm vector used in the torque cross product.
2. Cross product formula
Let
\[
\vec r=(r_x,r_y,r_z),
\qquad
\vec F=(F_x,F_y,F_z).
\]
Then
\[
\vec{\tau}=\vec r\times\vec F.
\]
Component by component,
\[
\tau_x=r_yF_z-r_zF_y,
\]
\[
\tau_y=r_zF_x-r_xF_z,
\]
\[
\tau_z=r_xF_y-r_yF_x.
\]
3. Torque in 2D
In many introductory problems, the position and force are in the \(x\)-\(y\) plane. Then \(r_z=0\) and \(F_z=0\), so
only the \(z\)-component of torque remains:
\[
\tau_z=r_xF_y-r_yF_x.
\]
If \(\tau_z>0\), the torque points out of the page in the \(+\hat{k}\) direction. If \(\tau_z<0\), the torque points
into the page in the \(-\hat{k}\) direction.
4. Magnitude of torque
The torque magnitude can be found from the vector components:
\[
|\vec{\tau}|=\sqrt{\tau_x^2+\tau_y^2+\tau_z^2}.
\]
It can also be written as
\[
|\vec{\tau}|=rF\sin\phi,
\]
where \(\phi\) is the angle between \(\vec r\) and \(\vec F\).
The perpendicular moment arm is
\[
d_\perp=r\sin\phi.
\]
Therefore,
\[
|\vec{\tau}|=Fd_\perp.
\]
5. Direction and the right-hand rule
The direction of torque is perpendicular to the plane formed by \(\vec r\) and \(\vec F\). Use the right-hand rule:
- Point your fingers along \(\vec r\).
- Curl your fingers toward \(\vec F\).
- Your thumb points in the direction of \(\vec \tau\).
The order of the cross product matters:
\[
\vec r\times\vec F=-(\vec F\times\vec r).
\]
6. Multiple forces
If several forces act on the same body, compute the torque from each force and add the torque vectors:
\[
\vec{\tau}_{\mathrm{net}}=\sum_i\vec{\tau}_i.
\]
For each force,
\[
\vec{\tau}_i=\vec r_i\times\vec F_i.
\]
Static rotational equilibrium requires
\[
\vec{\tau}_{\mathrm{net}}=\vec 0.
\]
7. Units
Position is measured in meters and force is measured in newtons, so torque has units
\[
\mathrm{m}\cdot\mathrm{N}=\mathrm{N\,m}.
\]
Torque and energy can both have dimensions of newton-meters, but they are not the same kind of quantity. Torque is
a vector related to rotation; energy is a scalar.
8. Worked example
A force of magnitude \(25\ \mathrm{N}\) is applied at the point
\[
(0,4)\ \mathrm{m}
\]
relative to the pivot. To obtain a positive \(+\hat{k}\) torque with this point, take the force direction as
\(150^\circ\) from the positive \(x\)-axis:
\[
\vec F=(25\cos150^\circ,\;25\sin150^\circ,\;0).
\]
\[
\vec F\approx(-21.65,\;12.5,\;0)\ \mathrm{N}.
\]
The position vector is
\[
\vec r=(0,4,0)\ \mathrm{m}.
\]
The \(z\)-component is
\[
\tau_z=r_xF_y-r_yF_x.
\]
\[
\tau_z=(0)(12.5)-(4)(-21.65).
\]
\[
\tau_z\approx86.6\ \mathrm{N\,m}.
\]
Therefore,
\[
\vec{\tau}=(0,0,86.6)\ \mathrm{N\,m}.
\]
Since \(\tau_z>0\), the torque points out of the page in the \(+\hat{k}\) direction.
Key idea: only the component of force perpendicular to the position vector creates torque.
