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Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories- graphical and algebraic techniques.

## Graphical Techniques

Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tail-to-head technique and the parallelogram method.

In describing the mathematical properties of vectors we used displacements and the tail-to-head graphical method of vector addition as an illustration. In the tail-to-head method of vector addition the following strategy is followed:

• Choose a scale and include a reference direction.
• Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length- remember to put an arrowhead on the end to denote its direction.
• Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct direction and of the correct length.
• Continue until you have drawn each vector- each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tail-to-head.
• The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.

Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port B, before sailing 5.5km south-west to port C. Determine the ship's resultant displacement using the tail-to-head technique of vector addition.

Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question- you should choose a scale such that your vector diagram fits the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram.

Step 1 :

Let us draw a rough sketch of the situation

In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction.

Step 2 :

Next we choose a scale for our vector diagram. It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem )- the diagram will then take up a good fraction of an A4 page. We now start the accurate construction.

Step 3 :

Construction Step 1: Starting at the harbour H we draw the first vector 6cm long in the direction north (remember in the diagram 1cm represents 1km):

Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east:

Construction Step 3: Since the ship is now at port B we draw the third vector 5.5cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45o.

Construction Step 4: As a final step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction

Step 4 :

We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8.38   km. The direction can be specified in terms of the angle measured either as 75.4o east of north or on a bearing of 75.4o.

Step 5 :

Now we can quote the final answer: The resultant displacement of the ship is 8.38   km on a bearing of 75.4o!

Question: A man walks 40   m East, then 30   m North.

a) What was the total distance he walked?

b) What is his resultant displacement?

Step 1 :

What distance did the man travel? In the first part of his journey he traveled 40   m and in the second part he traveled 30   m. This gives us a total distance traveled of $40+30=70\ m$

Step 2 :

What is his resultant displacement? The man's resultant displacement is the vector from where he started to where he ended. It is the sum of his two separate displacements. We will use the tail-to-head method of accurate construction to find this vector. Firstly, we draw a rough sketch:

Step 3 :

Next we choose a scale suitable for the problem. A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction.

Step 4 :

We draw the first displacement as an arrow 8cm long (according to the scale $8cm=8\times 5m=40m$ ) in the direction east:

Step 5 :

Starting from the head of the first vector we draw the second displacement as an arrow 6cm long (according to the scale $6cm=6\times 5m=30m$ ) in the direction north:[/itex]

Step 6 :

Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant)

Step 7 :

Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36.9o north of east.

### The Parallelogram Method

When needing to find the resultant of two vectors another graphical technique can be applied- the parallelogram method. The following strategy is employed:

• Choose a scale and a reference direction.
• Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction.
• Draw the second vector as an arrow of the correct length in the correct direction from the tail of the first vector.
• Complete the parallelogram formed by these two vectors.
• The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.

#### Worked Example 6

Parallelogram Method of Graphical Addition I

Question: A force of F1 = 5 N is applied to a block in a horizontal direction. A second force F2 = 4 N is applied to the object at an angle of 30° above the horizontal.

Determine the resultant force acting on the block using the parallelogram method of accurate construction.

Step 1 :

Firstly we make a rough sketch of the vector diagram:

Step 2 :

Now we choose a suitable scale. In this problem a scale of 1 cm = 0.5 N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram.

Step 3 :

Let us draw F1 first. According to the scale it has length 10 cm:

Step 4 :

Next we draw F2. According to the scale it has length 8 cm. We make use of a protractor to draw this vector at 30° to the horizontal:

Step 5 :

Next we complete the parallelogram and draw the diagonal:

RIAAN NOTE: Image missing img155.png PDF page 51 File:Fhsst vectors43.png

Step 6 :

Finally we use the scale to convert the measured length into the actual magnitude. Since 1 cm = 0.5 N, 17.4 cm represents 8.7 N. Therefore the resultant force is 8.7 N at 13.3° above the horizontal.

The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point.

## Algebraic Addition and Subtraction of Vectors

### Vectors in a Straight Line

Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:

• Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
• Next simply add (or subtract) the vectors with the appropriate signs.
• As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).

Let us consider a couple of examples.

#### Worked Example 7

Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2.5m along the ground to the left, calculate algebraically the ball's resultant displacement.

(NOTE TO SELF: PGCE suggest a `more real looking' diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow))

Step 1 :

Let us draw a picture of the situation:

Step 2 :

We know that the resultant displacement of the ball ( ${\overrightarrow {s}}_{resultant}$ ) is equal to the sum of the ball's separate displacements (${\overrightarrow {s}}_{1}$  and ${\overrightarrow {s}}_{2}$  ):

${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&{\overrightarrow {s}}_{1}+{\overrightarrow {s}}_{2}\end{matrix}}$

Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained.

Step 3 :

First we choose a positive direction. Let's make to the right the positive direction. This means that to the left becomes the negative direction.

Step 4 :

With right positive:

${\begin{matrix}{\overrightarrow {s}}_{1}&=&+10.0m\\&and&\\{\overrightarrow {s}}_{2}&=&-2.5m\end{matrix}}$

Step 5 :

Next we simply add the two displacements to give the resultant:

${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&(+10m)+(-2.5m)\\&=&(+7.5)m\end{matrix}}$

Step 6 :

Finally, in this case right means positive so:

${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&7.5m{\rm {\ to\ the\ right}}\end{matrix}}$

Let us consider an example of vector subtraction.

#### Worked Example 8

Subtracting vectors algebraically I

Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m.s−1 to the right. After striking the wall, the ball returns to the thrower at 2m.s−1. Determine the change in velocity of the ball.

Step 1 :

Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:

${\begin{matrix}\Delta {\overrightarrow {v}}&=&{\overrightarrow {v}}_{final}-{\overrightarrow {v}}_{initial}\end{matrix}}$

Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.

Step 2 :

Let's make to the right the positive direction. This means that to the left becomes the negative direction.

Step 3 :

With right positive:

${\begin{matrix}{\overrightarrow {v}}_{initial}&=&+3m.s^{-1}\\&and&\\{\overrightarrow {v}}_{final}&=&-2m.s^{-1}\end{matrix}}$

Step 4 :

Thus, the change in velocity of the ball is:

${\begin{matrix}\Delta {\overrightarrow {v}}&=&(-2m.s^{-1})-(+3m.s^{-1})\\&=&(-5)m.s^{-1}\end{matrix}}$

Remember that in this case right means positive so:

${\begin{matrix}\Delta {\overrightarrow {v}}&=&5m.s^{-1}{\rm {\textbf {\ to\ the\ {\emph {left}}}}}\end{matrix}}$

Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line.

### A More General Algebraic technique

In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled first east and then north. However, the man's resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example.

#### Worked Example 9

An Algebraic solution to Worked Example 3

Question: A man walks 40 m East, then 30  m North.

1. Calculate the man's resultant displacement.

Step 1 :

As before, the rough sketch looks as follows:

Step 2 :

Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use Pythogoras' theorem to determine the length of the resultant. If the length of the resultant vector is called s then:

${\begin{matrix}s^{2}&=&(40m)^{2}+(30m)^{2}\\s^{2}&=&2500m^{2}\\s&=&50m\\\end{matrix}}$

Step 3 : Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle $\alpha$  between the resultant displacement vector and East.

We can do this using simple trigonometry:

${\begin{matrix}\tan \alpha &=&{\frac {opposite}{adjacent}}\\\tan \alpha &=&{\frac {30}{40}}\\\alpha &=&\arctan(0.75)\\\alpha &=&36.9^{o}\\\end{matrix}}$

Step 4 :

• Resultant Displacement: 50 m at 36.9o North of East

This is exactly the same answer we arrived at after drawing a scale diagram!

In the previous example we were able to use simple trigonometry to calculate a man's resultant displacement. This was possible since the man's directions of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this.

#### Worked Example 10

Further example of vector addition by calculation

Question: A man walks from point A to point B which is 12km away on a bearing of 45o. From point B the man walks a further 8km east to point C. Calculate the man's resultant displacement.

Step 1 : Let us begin by drawing a rough sketch of the situation

RIAAN NOTE: Image on page 56 is missing File:Fhsst vectors46.png

$B{\hat {A}}F=45^{o}$  since the man walks initially on a bearing of 45o.Then, $A{\hat {B}}G=B{\hat {A}}F=45^{o}$  (alternate angles parallel lines). Both of these angles are included in the rough sketch.

Step 2 :

Now let us calculate the length of the resultant (AC). Since we know both the lengths of $AB$  and $BC$  and the included angle $A{\hat {B}}C$ , we can use the cosine rule:

${\begin{matrix}AC^{2}&=&AB^{2}+BC^{2}-2\cdot AB\cdot BC\cos(A{\hat {B}}C)\\&=&(12)^{2}+(8)^{2}-2\cdot (12)(8)\cos(135^{o})\\&=&343.8\\AC&=&18.5\ km\end{matrix}}$

Step 3 :

Next we use the sine rule to determine the angle $\theta$ :

${\begin{matrix}{\frac {\sin \theta }{8}}&=&{\frac {\sin 135^{0}}{18.5}}\\\sin \theta &=&{\frac {8\times \sin 135^{o}}{18.5}}\\\theta &=&\arcsin(0.3058)\\\theta &=&17.8^{o}\end{matrix}}$

Thus, $F{\hat {A}}C=62.8^{o}$

Step 4 :