Resolution of vectors

Breaking vectors down into their components—or resolving them—makes it easier to add or subtract them, especially when dealing with vectors that don't act along the same line. You will encounter this in many areas of STEM, like when analysing forces involved in robotics, studying the projectile motion of objects launched into the air, and in fluid mechanics. Use this resource to learn how to find the resolution of vectors.

To help simplify calculations, vectors can be thought of as the sum of several other vectors. For example, imagine that a single force is made up of several forces. If we break the force into imaginary vertical and horizontal components that are at right angles to each other, it is much less scary to work with.

Once we break the forces down into their components, we can just group the components that are parallel to each other and complete the required operations. Let's see how this works with some examples.

Components of a vector

If you pull on an object with a force \(\vec{F}\), the effect is the same as when two people pull with forces \(\vec{a}\) and \(\vec{b}\) at right angles to each other – one horizontal and the other vertical.

\(\vec{a}\) and \(\vec{b}\) are the components of the force vector. The horizontal component is \(\vec{a}\) and the vertical component is \(\vec{b}\). They are often called rectangular components because they are at right angles to each other.

From \(\vec{a}\) and \(\vec{b}\), we get a vector triangle, where \(\theta\) is the angle, \(\vec{F}\) is the hypotenuse, \(\vec{a}\) is the adjacent and \(\vec{b}\) is the opposite.

From this, we can find the magnitude of the component vectors using trigonometry. You might like to go back and review right-angled triangle trigonometry first.

We can find the length of the horizontal component using:
\[\begin{align*} \cos\left(\theta\right) & = \frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}\\
& = \frac{\left|\vec{a}\right|}{\left|\vec{F}\right|}
\end{align*}\]

Rearranging this in terms of \(\left|\vec{a}\right|\) gives:
\[\vec{\left|a\right|} = \vec{\left|F\right|}\cos\theta\]

We can also find the length of the vertical component using:
\[\begin{align*} \sin\left(\theta\right) & = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}\\
& = \frac{\left|\vec{b}\right|}{\left|\vec{F}\right|}
\end{align*}\]

Rearranging this in terms of \(\left|\vec{b}\right|\) gives:
\[\vec{\left|b\right|} = \vec{\left|F\right|}\sin\theta\]

Example 1 – resolving components of force vectors

Calculate the rectangular components of a force of \(10\textrm{ N}\) that acts in a direction of North \(30^{\circ}\) East.
First, we should always draw the vector triangle. North \(30^{\circ}\) East is the same as East \(60^{\circ}\) North, or \(60^{\circ}\) from the horizontal.

We can label the horizontal component of the vector triangle as \(\vec{a}\) and the vertical component as \(\vec{b}\). \(\vec{a}\) is in the East direction and \(\vec{b}\) is in the North direction.

The magnitudes of these forces can be calculated using trigonometry.
\[\begin{align*} \vec{\left|a\right|} & = \vec{\left|F\right|}\cos\left(\theta\right)\\
& = 10\cos\left(60^{\circ}\right)\\
& = 5\textrm{ N}
\end{align*}\] \[\begin{align*} \vec{\left|b\right|} & = \vec{\left|F\right|}\sin\left(\theta\right)\\
& = 10\sin\left(60^{\circ}\right)\\
& = 8.7\textrm{ N}
\end{align*}\]

Thus, the \(10\) newtons force can be resolved into two rectangular components: \(5\textrm{ N}\) to the East and \(8.7\textrm{ N}\) to the North.

Your turn – resolving components of force vectors

1. Determine the horizontal component of a force of \(11\textrm{ N}\) acting at \(\theta=60^{\circ}\) to the horizontal.
2. Determine the northerly component of wind blowing at \(15\textrm{ m s}^{-1}\) from the South East, to the nearest whole number.
3. Define the vector whose components are \(3\textrm{ N}\) horizontally and \(4\textrm{ N}\) vertically.
4. Determine the vertical and horizontal components of a force of \(24\textrm{ N}\) acting at \(\theta=30^{\circ}\) to the horizontal, to the nearest whole number.
5. A plane taking off leaves the runway at \(\theta=32^{\circ}\) to the horizontal, traveling at \(180\textrm{ km h}^{-1}\). Determine how long it will take for the plane to reach an altitude of \(1000\textrm{ m}\).

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