KAHIBARO
Table of Contents
From turning effect to vector quantity
Torque describes how a force tends to rotate an object about a chosen point or axis. In this chapter, the focus is the full vector definition of torque, which tells us not only how large the turning effect is, but also its direction in space.
The vector definition is
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
where $\vec{\tau}$ is the torque, $\vec{r}$ is the position vector from the pivot or reference point to the point where the force is applied, and $\vec{F}$ is the force.
This definition uses the cross product, so torque is a vector perpendicular to the plane formed by $\vec{r}$ and $\vec{F}$.
The vector definition of torque is
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
Torque depends on both the force and where that force is applied.
Meaning of the position vector
The vector $\vec{r}$ begins at the chosen origin, usually the pivot, axle, or point about which rotation is considered. It ends at the point where the force acts.
This is very important. A force by itself does not determine torque. The same force can produce different torques depending on where it is applied.
If the force is applied at position
$$
\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}
$$
and the force is
$$
\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}
$$
then the torque is found from their cross product.
Component form
Using the determinant form of the cross product,
$$
\vec{\tau} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
F_x & F_y & F_z
\end{vmatrix}
$$
So,
$$
\vec{\tau} =
(yF_z - zF_y)\hat{i}
-
(xF_z - zF_x)\hat{j}
+
(xF_y - yF_x)\hat{k}
$$
This gives the three components of torque directly.
In component form,
$$
\vec{\tau} =
(yF_z - zF_y)\hat{i}
+
(zF_x - xF_z)\hat{j}
+
(xF_y - yF_x)\hat{k}
$$
Be careful with signs when expanding the cross product.
Magnitude of torque
Because torque is a cross product, its magnitude is
$$
|\vec{\tau}| = rF\sin\theta
$$
where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.
This formula shows that only the part of the force perpendicular to $\vec{r}$ produces torque. If the force points directly along $\vec{r}$, then $\sin\theta = 0$, and the torque is zero.
If the force is perpendicular to $\vec{r}$, then $\sin\theta = 1$, and the torque has its maximum magnitude:
$$
|\vec{\tau}| = rF
$$
Direction of torque
The direction of $\vec{\tau}$ is determined by the right hand rule for the cross product.
Point the fingers of your right hand in the direction of $\vec{r}$. Curl them toward $\vec{F}$. Your thumb points in the direction of $\vec{\tau}$.
This means torque points along the axis of rotation, not along the direction of motion of the object.
For motion in the $xy$ plane, the torque often points along the $z$ axis. A counterclockwise tendency usually corresponds to positive $z$, and a clockwise tendency usually corresponds to negative $z$.
Direction rule for torque:
Use the right hand rule on
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
The order matters. In general,
$$
\vec{r} \times \vec{F} \neq \vec{F} \times \vec{r}
$$
In fact,
$$
\vec{F} \times \vec{r} = -\vec{\tau}
$$
Geometric interpretation
The magnitude $rF\sin\theta$ can also be understood geometrically. It is the product of the distance from the pivot and the perpendicular part of the force. This connects naturally to the idea of lever arm, but here the vector definition gives the complete three dimensional description.
The area of the parallelogram formed by $\vec{r}$ and $\vec{F}$ is also $rF\sin\theta$, so the torque magnitude equals that area.
A simple two dimensional example
Suppose a force acts in the $xy$ plane. Let
$$
\vec{r} = 2\hat{i} \text{ m}, \qquad \vec{F} = 3\hat{j} \text{ N}
$$
Then
$$
\vec{\tau} = \vec{r} \times \vec{F}
= (2\hat{i}) \times (3\hat{j})
= 6(\hat{i}\times\hat{j})
= 6\hat{k} \text{ N m}
$$
So the torque has magnitude $6\text{ N m}$ and points in the positive $z$ direction.
If instead the force were downward,
$$
\vec{F} = -3\hat{j} \text{ N}
$$
then
$$
\vec{\tau} = (2\hat{i}) \times (-3\hat{j}) = -6\hat{k} \text{ N m}
$$
The magnitude is the same, but the direction is opposite.
Example with zero torque
Let
$$
\vec{r} = 2\hat{i} \text{ m}, \qquad \vec{F} = 5\hat{i} \text{ N}
$$
Since the vectors are parallel, $\theta = 0$, so
$$
\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}
$$
Even though a force is present, it does not create any turning effect about the chosen point.
Three dimensional example
Consider
$$
\vec{r} = (1,2,0)\text{ m}, \qquad \vec{F} = (3,0,4)\text{ N}
$$
Then
$$
\vec{\tau} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 0 \\
3 & 0 & 4
\end{vmatrix}
$$
So
$$
\vec{\tau}
= (2\cdot 4 - 0\cdot 0)\hat{i}
-
(1\cdot 4 - 0\cdot 3)\hat{j}
+
(1\cdot 0 - 2\cdot 3)\hat{k}
$$
which gives
$$
\vec{\tau} = 8\hat{i} - 4\hat{j} - 6\hat{k}\text{ N m}
$$
This example shows that in three dimensions, torque can have components along several axes at once.
Comparison of situations
| Situation | Angle between $\vec{r}$ and $\vec{F}$ | Torque magnitude |
|---|---|---|
| Force parallel to $\vec{r}$ | $0^\circ$ | $0$ |
| Force anti parallel to $\vec{r}$ | $180^\circ$ | $0$ |
| Force perpendicular to $\vec{r}$ | $90^\circ$ | maximum, $rF$ |
| General angle | $\theta$ | $rF\sin\theta$ |
Choice of reference point
Torque always depends on the point about which it is calculated. If you change the origin, the vector $\vec{r}$ changes, so the torque usually changes too.
For the same force, one point may give zero torque while another gives a nonzero torque. Therefore, always identify the pivot or reference point before computing $\vec{\tau}$.
Torque must be calculated about a specified point.
Changing the reference point changes $\vec{r}$, and therefore may change $\vec{\tau}$.
Units of torque
From
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
the SI unit of torque is meter times newton:
$$
\text{N m}
$$
This has the same unit combination as energy, but torque and energy are different physical quantities. Torque is a vector, while energy is a scalar.
Visualizing vector torque
In this drawing, $\vec{r}$ and $\vec{F}$ lie in the plane of the page. The torque vector points perpendicular to that plane.
Common mistakes
A common mistake is using the full force magnitude even when the force is not perpendicular to $\vec{r}$. The correct magnitude uses $\sin\theta$.
Another common mistake is reversing the order of the cross product. Since cross products are not commutative, reversing the order changes the sign.
Students also sometimes forget that torque is about a point. Without stating the reference point, the calculation is incomplete.
Key idea
The vector definition of torque gives a complete description of rotational effect in three dimensions. It combines magnitude and direction in one expression:
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
Its magnitude is
$$
|\vec{\tau}| = rF\sin\theta
$$
and its direction is given by the right hand rule.
Essential results:
$$
\vec{\tau} = \vec{r} \times \vec{F}
$$
$$
|\vec{\tau}| = rF\sin\theta
$$
Torque is perpendicular to the plane containing $\vec{r}$ and $\vec{F}$, and its direction is found with the right hand rule.