The first step in adding vectors of different angles is to break them down
into their horizontal and vertical components. This sounds difficult, but it is
really very easy, and it will be even easier if you have already taken some form
of a Trigonometry course.
A problem with different angles will look something like this:
Add 6m at 30° and 10m at 45°.
Breaking angled vectors down in to their separate components is just a matter
of Let's break down the first vector of the problem: 6m at 30°
x component : 6m cos 30° = 5.192m
y component : 6m sin 30° = 3m
Now, let's break down the other one: 10m at 45°
x component : 10m cos 45° = 7.071m
y component : 10m sin 45° = 7.071m
Now that we have these numbers, what do we do with them? Very simple, we add
them together like we learned in the last lesson, being careful to keep horizontal
and vertical separate, and then transform them back into an angled vector.
First, let's add:
x 5.192m + 7.071m = 12.263m
y 3m + 7.071m = 10.071m
To get the magnitude of the resultant vector, we must plug it into a simple
formula:
x2 + y2 = r2, where r is the resultant vector
sqrt(r2) = r
Now, to get the direction, we find the inverse tangent of the two sides of the
triangle:
tan^-1 (y÷x) = q
Let's try that on our problem:
12.263m2 + 10.071m2 = 251.806m2
square root (251.806m2) = 15.868m
tan^-1 (10.071m÷12.263m) = 39.395°
So, the resultant vector in this problem is <15.868m at 39.395°>
If you are not familiar with some of the basic trigonometric properties of
triangles, then you might want to brush up, because it will help you to know why
we are using the formulas that we are using, not just how to "plug & chug" the
answers out. Here are some links you might want to try:
Another method of representing vectors, i, j, k notation, will be covered
in the next lesson.
