The vector product and the torque of a force

Turning needs more than a force

Push a door at its hinge and nothing happens. Push it at the handle and it swings easily. What turns a body is not just the size of a force, but where and which way it acts. That turning effect is the torque.

The vector product

To describe it we need a new product of two vectors, the vector product (or cross product):

It is a vector at right angles to both $\vec{A}$ and $\vec{B}$. Curl the fingers of your right hand from $\vec{A}$ to $\vec{B}$: your thumb points along $\vec{A} \times \vec{B}$.

The order matters: $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$. For the unit vectors, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$, and any vector crossed with itself gives zero.

Torque

The torque of a force $\vec{F}$ acting at $\vec{r}$ from a point O is

in newton metres. The length $r\sin\theta$ is the moment arm, the perpendicular distance from O to the line of the force. So torque is force times moment arm.

You should find the torque biggest when the force is at right angles to the spanner, and zero when the force points straight along it, through the nut.

Watch out. A force whose line passes through the axis has no torque about it, however big it is.

A worked example

A mechanic pushes on a spanner $0.30\ \text{m}$ from the nut with $120\ \text{N}$, at $70^\circ$ to the spanner.

1. Moment arm. $0.30\ \text{m} \times \sin 70^\circ = 0.282\ \text{m}$.
2. Torque. $\tau = 120\ \text{N} \times 0.282\ \text{m} = 33.8\ \text{N m}$.
3. Best angle. At $90^\circ$ she would get $36\ \text{N m}$, only a little more.

Think first. Find the line of the force, then the shortest distance from the axis to that line. That distance times the force is the torque.

Practise with new numbers

Try it in Python

The program works out a vector product from the parts, and finds the torque on a spanner.

Remember these

• $|\vec{A} \times \vec{B}| = AB\sin\theta$, at right angles to both, by the right-hand rule.
• $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$; $\hat{i} \times \hat{j} = \hat{k}$.
• Torque $\vec{\tau} = \vec{r} \times \vec{F}$; size $rF\sin\theta$ = force times moment arm.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $60\ \text{N}$ force acts at right angles to a door, $0.80\ \text{m}$ from the hinges. What is its torque?

2. What is $\hat{k} \times \hat{j}$?

3. Why is it hard to open a door by pushing near its hinges?

4. Find $\vec{A} \times \vec{B}$ for $\vec{A} = 3\hat{i}$ and $\vec{B} = 2\hat{i} + 4\hat{j}$.