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Posts Tagged ‘torque’

In this previous blog I discussed some of the basics of vectors, including how to add them. As I discussed, because vectors have both size and direction, we have to take their directions into account when adding them. What about multiplying vectors together, how do we do this?

It turns out that vectors can be multiplied together in two different ways. The first way is called the scalar or dot product, and produces a result which is a scalar (i.e. size but no direction). The second way is called the vector or cross product, and the result is a new vector with a different direction to either of the two vectors being multiplied together.

The scalar or dot product

This is the easier of the two vector multiplications to understand. To give an example, in physics the definition of work is the force multiplied by the displacement (W=Fd). But force and displacement are both vectors, so strictly speaking we are multiplying two vectors together, and for work we multiply them using the scalar or do product.

W = \vec{F} \cdot \vec{s} \; \; \text{ measured in Joules (J)}

where I have used the usual \vec{s} to denote displacement, although we will also use \vec{d}. The way the scalar product works is that the resultant scalar is the product of the size (magnitudes) of the two vectors multiplied by the cosine of the angle between them, i.e.

W = \lvert F \rvert \; \lvert d \rvert \; \cos( \theta) where \lvert F \rvert represents the size (or absolute value) of the force vector \vec{F} etc, and \theta is the angle between the direction of the force vector and the displacement vector.


Work is the scalar (or dot) product of the force vector and the displacement vector. It is measured in Joules (J).

Work is the scalar (or dot) product of the force vector and the displacement vector. It is measured in Joules (J).


As an example, suppose we have a force of 10N being applied at an angle of 60^{\circ} below the horizontal to push a lawnmower a distance of 5m. The work done is \vec{F} \cdot \vec{d} = (10)(5)\cos(60^{\circ}) = (50)(0.5) = 25 \; \text{Joules}.

The vector or cross product

The other way to multiply two vectors together produces a result which is also a vector. Again, to use a real example, displacement and force can be multiplied together using the vector product to produce something called the torque. Mathematically we write this as

\vec{\tau} = \vec{d} \times \vec{F}.

Normally, it does not matter in which order one multiplies things. With numbers, 3 \times 2 is the same as 2 \times 3. This is not true with the vector product. The rule for finding the direction is the so-called “right hand rule shown below. If we wish to multiply the vectors \vec{a} \; \text{and} \; \vec{b} together using the vector product, then one points one’s first finger in the direction of the first vector, and one’s second finger in the direction of the second vector, and one’s thumb shows the direction of the new vector.


To find the direction of the vector (or cross) product of two vectors "a" and "b", we point the first finger in the direction of "a", the second finger in the direction of "b", and the thumb shows the direction of the new vector.

To find the direction of the vector (or cross) product of two vectors “a” and “b”, we point the first finger in the direction of “a”, the second finger in the direction of “b”, and the thumb shows the direction of the new vector.


For the vector product, they are in opposite directions.

For the vector product, a \times b \neq b \times a, they are in opposite directions.


Angular momentum and torque

Two examples of the vector product are angular momentum and torque. Angular momentum is the rotational equivalent of linear momentum. It is defined as

\vec{L} = \vec{r}  \times \vec{p}

where \vec{r} is the radius vector and \vec{p} is the linear momentum (defined as \vec{p} = m \vec{v}). By convention, the radius vector is from the centre of the circle outwards, and the linear momentum vector is just in the direction of the motion. So, using the right-hand rule above, aas \vec{L} = \vec{r} \times \vec{p} we can see that, for clockwise motion, the angular momentum vector is pointing vertically upwards.


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