Find the missing force from a northward resultant

Key concept: When a resultant is given, the easiest move is to treat it like a vector equation. Write the unknown as resultant minus the known force, then resolve into east and north components. The direction usually becomes clear after that.

Step by step

Let the resultant be $\vec R = 8\,\text{N north}$ and let $\vec f_1 = 5\,\text{N northeast}$

We use $\vec f_1 + \vec f_2 = \vec R$ so $\vec f_2 = \vec R - \vec f_1$

1) Resolve $\vec f_1$ into components

Northeast means $45^\circ$ north of east. So the east and north components are: $f_{1x}=5\cos45^\circ=\frac{5}{\sqrt2}$ $f_{1y}=5\sin45^\circ=\frac{5}{\sqrt2}$

2) Write the resultant in components

Since $\vec R$ is 8 N due north: $\vec R=(0,8)$

3) Subtract components

$\vec f_2=(0,8)-\left(\frac{5}{\sqrt2},\frac{5}{\sqrt2}\right)$ $\vec f_2=\left(-\frac{5}{\sqrt2},\,8-\frac{5}{\sqrt2}\right)$

Numerically, $\vec f_2\approx(-3.54,4.46)$

4) Find magnitude

$|\vec f_2|=\sqrt{(-3.54)^2+(4.46)^2}$ $|\vec f_2|\approx\sqrt{32.10}\approx 5.66\,\text{N}$

5) Find direction

The vector has a negative east component and positive north component, so it points to the northwest.

Angle west of north: $\tan\phi=\frac{3.54}{4.46}$ $\phi\approx 38.5^\circ$

Answer

$\boxed{5.66\,\text{N},\ 38.5^\circ\text{ west of north}}$

Pitfall alert

A very common slip is to read northeast as one of the axes instead of a 45° direction. Another one is forgetting that the unknown force must cancel the eastward part of $\vec f_1$, so $\vec f_2$ has a westward component.

Try different conditions

If the 5 N force were due east instead of northeast, then only the north component would remain to be balanced by $\vec f_2$. The magnitude and direction would change a lot, so the angle information in the wording matters.

Further reading

resultant vector, component form, vector subtraction