One of the things you have to consider when doing calculations involving forces is the direction of the force. Forces not only act forwards and backward but can also act in a 2D or 3D space requiring us to be able to find the effect of a force acting at an angle. This is because forces are sectors so have direction and magnitude.

The easiest way to deal with this is to resolve forces into perpendicular components. For example if we are working in a 2D plane it might make sense to consider the horizontal and vertical parts of a force separately. We can then do two sets of calculations, one vertical and the other horizontal, to find out the affect of the force.

Consider the example below,

It shows how we can split a force into its two components and what they’re relative sizes would be. To work out the magnitude of the components we must know the magnitude of the force we are resolving and the angle between it and one of the two perpendicular component. We can then use trigonometry.

If we know the angle between the force and its component then the magnitude of the component is the magnitude of the force multiplied by the cos of the angle and the magnitude of the other, perpendicular component, is the magnitude of the force multiplied by the sin of the angle.

So in the above example where the magnitude of the force is M and that of the horizontal and vertical components is H and V:

V = Msin(a) and H = Mcos(a)
or
V = Mcos(b) and H = Msin(b)

Finding a resultant force from components

Once you have done calculations in two perpendicular directions you are likely to end up with two perpendicular forces which you want to combine to find the resultant force. To do this we must use pythagoras’s theorem and more trigonometry.

Again consider the example above. Suppose you know H and V and want to find M and the angles a and b. To find M we use Pythagoras so we get

and to find the angles a and b we must use the tan function to get
and

Example – finding the resultant force

Find the direction and magnitude of the resultant force below.

First find all the horizontal components (taking forces acting right as positive).
The horizontal component of the:
3N force is 3cos(30) = 2.598
2N force is 2cos(90) = 0 — note that this is expected because a vertical force has no horizontal component as they are perpendicular
4N force is -4cos(45) = -2.828

Then to find the horizontal component H of the resultant force we add these up to get
H = 2.598 + 0 – 2.828 = -0.23

And then we do the same in the vertical direction. The vertical components of the:
3N force is 3sin(30) = 1.5
2N force is 2sin(90) = -2
4N force is 4sin(45) = 2.828

And adding these up we find that the vertical component of the resultant force is
V = 1.5 -2 + 2.828 = 1.328

We now combine these to find the magnitude and direction of the resultant force.
The magnitude M is found using Pythagoras’s to get

and the angle, measured above the right horizontal, is
or 80.17 above the left horizontal