*Before you read this, I suggest you read* post 16.50.

In post 16.50, we saw that the effect of a force depends on its direction, as well as how big it is – force is a *vector*. So, if we add a force of 5 N to another force of 5 N, the result can be anything between 0 (when they act in opposite directions) to 10 N (when they act in the same direction). If you’ve forgotten, force is measured in newtons (abbreviated to N – see post 16.13).

This means that we can’t add forces (vectors) in the same way as we add ordinary numbers (scalars – see post 16.50). To avoid confusion, we need to make it clear that something is a vector – we do this by writing it in **bold**, for example ** F**. In handwriting, we usually

__underline__vectors, for example

*. You may have been taught to add vectors by drawing triangles or parallelograms. This is very tedious; fortunately,*

__F__*vector algebra*was invented to make adding vectors much easier.

To define a vector, we need to define an *X*-axis direction (like the horizontal axis of a graph – see appendix 2 of post 16.50). A *Y*-axis is then automatically defined at 90^{o} to the *X*-axis (like the vertical axis of a graph – see appendix 2 of post 16.50), where anti-clockwise angles are defined to be positive. In the picture above, the length of the vector representing the force ** F** is

*F*; this length is called the

*magnitude*or the

*modulus*of

**. The projection of**

*F***on the**

*F**X*-axis is

*F*and its projection on the

_{x}*Y*-axis is

*F*;

_{y}*F*and

_{x}*F*are called the

_{y}*components*of

**and are scalars. According to Pythagoras’ theorem (post 16.50, appendix 1)**

*F** F*^{2} = *F _{x}*

^{2}+

*F*

_{y}^{2}. (1)

Now we define two vectors ** i** and

**which have a length equal to the number 1;**

*j***points in the**

*i**X*-axis direction and

**points in the**

*j**Y*-axis direction. It is important to understand that

**and**

*i***are no more than the number 1 pointing in particular directions; they are not forces, so they are not measured in newtons.**

*j*Now let’s think of ** F** acting on an object at O, in the picture above. This is equivalent to a force of

*i**F*, acting in the

_{x}*X*-axis direction, and a force of

*j**F*acting in the

_{y}*Y*-axis direction. So

** F** =

*i**F*+

_{x}

*j**F*. (2)

_{y}** F** makes an angle

*θ*with the

*X*-axis. From the definition of the cosine and sine of an angle (appendix 3 of post 16.50)

* F _{x}* =

*F*cos

*θ*and

*F*=

_{y}*F*sin

*θ*. (3)

The angle *θ* defines the direction of ** F**, with respect to the direction of the

*X*-axis. We can calculate

*θ*from the values of the components of

**. According to equation (3), dividing the**

*F**Y*-component of

**by the**

*F**X*-component gives

(*F _{y}*/

*F*) = (

_{x}*F*sin

*θ*)/(

*F*cos

*θ*) =(sin

*θ*)/(cos

*θ*) (4)

(The final step is just like dividing by a number but the number is represented by the symbol *F*; it’s the same idea as that if 20 items cost 20 dollars, 1 item costs 1 dollar.) The brackets in this equation mean that we calculate whatever is in the brackets first. In appendix 5 of post 16.50, we defined the tangent of an angle by tan *θ *= sin *θ*/cos *θ.*

So we can write equation (4) in the form tan *θ* = (*F _{y}*/

*F*). This means that

_{x}*θ*is the angle whose tangent is given by (

*F*/

_{y}*F*). The previous statement can be written as the equation

_{x}* θ*= arctan(*F _{y}*/

*F*) (5)

_{x}that enables us to calculate the direction of a vector from its components. We shall see how this is used soon. (Sometimes arctan is written as tan^{-1}).

Now let’s try adding two vectors together. To do this we will use the vectors

*F*_{1} = 2.0(** i**cos30

^{o}) + 2.0(

**sin30**

*j*^{o}) N and

*F*_{2}= 5.0(

**cos60**

*i*^{o}) + 5.0(

**sin60**

*j*^{o}) N.

Notice that we now need to specify the units in which *F*_{1} and *F*_{2} are measured, because otherwise the equations don’t tell us that their magnitude is measured in newtons (and not, for example, kN). In equation (2), we don’t need to specify any units because ** F** takes on whatever units

*F*and

_{x}*F*are measured in.

_{y}We can add forces in the ** i**-direction and forces in the

**-direction independently because**

*j***and**

*i***are at right angles to each other – so what happens in one direction is independent of what happens in the other direction. Adding**

*j***components and**

*i***components is then just like adding apples and oranges – the answer is a number of apples and a number of oranges. When we do this, the**

*j**resultant*vector

**is given by**

*R*** R** =

*F*_{1}+

*F*_{2}=

**(2cos30**

*i*^{o}+ 5cos60

^{o}) +

**(2sin30**

*j*^{o}+ 5 sin60

^{o})

= ** i**(2 × 0.8660 + 5 × 0.5) +

**(2 × 0.5 + 5 × 0.8660) N.**

*j*The values of the cosines and sines of the angles can be found at:

- http://www.rapidtables.com/calc/math/Cos_Calculator.htm
- http://www.rapidtables.com/calc/math/Sin_Calculator.htm.

When we do the arithmetic, we get the answer that

** R** = 4.2

**+ 5.3**

*i***N.**

*j*According to equation (1), *R*^{2} = 4.2^{2} + 5.3^{2} = 45.73 N. This means that

* R* = √45.73 = 6.8 N.

Remember that 4.2^{2} means 4.2 × 4.2 and that √45.73 =6.76 because 6.76 × 6.76 = 45.73. But I have given the answer as 6.8 N because we only know the values of the components of *F*_{1} and *F*_{2} (2.0 N and 5.0 N) to one decimal place (see post 16.7).

The angle that ** R** makes with the

*X*-axis (the direction of

*R*) is given, according to equation (5), by arctan(5.3/4.2)= arctan(1.26) = 52

^{o}.

How do I know the value of arctan(1.26)? I looked it up at: http://www.rapidtables.com/calc/math/Arctan_Calculator.htm.

You could also use a scientific calculator or a spreadsheet, like Excel; if you use Excel it will give you an answer measured in radians (post 16.50) – divide by 3.14159 and multiply by 180 to get an answer in degrees.

If you try to repeat my calculations, you will need to use the results of the intermediate calculations to several decimal places to prevent errors from reducing the number of decimal places in the presentation of the results.

If vector algebra seems a bit strange – don’t worry. I once met someone who had studied maths at a good British university for a year and didn’t understand it. But, once you get the idea, it makes dealing with vectors much easier than the geometrical constructions that are often used. I will use it, in a later post, to explain the idea of centripetal force. When I was first taught this, without using vector algebra, I found it almost impossible to understand. But I hope you will find it easy!

*Related posts*

Teaching vectors to a girl studying highers and when she looks slightly bamboozled I tell her my old (not age David!) boss used vectors for All his maths problems-probably not completely true, but close. It always makes me smile as I remember you and the group giving me so much help and support. Xx

LikeLike