3 Addition of vectors

Vectors are added in a particular way known as the triangle law . To see why this law is appropriate to add them this way consider the following example:

3.1 Example: The route taken by an automated vehicle

An unmanned vehicle moves on tracks around a factory floor carrying components from the store at $A$ to workers at $C$ as shown in Figure 9.

The vehicle may arrive at $C$ either directly or via an intermediate point $B$ . The movement from $A$ to $B$ can be represented by a displacement vector $\overrightarrow{AB}$ . Similarly, movement from $B$ to $C$ can be represented by the displacement vector $\overrightarrow{BC}$ , and movement from $A$ to $C$ can be represented by $\overrightarrow{AC}$ . Since travelling from $A$ to $B$ and then $B$ to $C$ is equivalent to travelling directly from $A$ to $C$ we write

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

This is an example of the triangle law for adding vectors. We add vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ by placing the tail of $\overrightarrow{BC}$ at the head of $\overrightarrow{AB}$ and completing the third side of the triangle so formed ( $\overrightarrow{AC}$ ).

Figure 10 :

Consider the more general situation in Figure 10. Suppose we wish to add $\underset{̲}{b}$ to $\underset{̲}{a}$ . To do this $\underset{̲}{b}$ is translated, keeping its direction and length unchanged, until its tail coincides with the head of $\underset{̲}{a}$ . Then the sum $\underset{̲}{a}+\underset{̲}{b}$ is defined by the vector represented by the third side of the completed triangle, that is $\underset{̲}{c}$ in Figure 11. Note, from Figure 11, that we can write $\underset{̲}{c}=\underset{̲}{a}+\underset{̲}{b}$ since going along $\underset{̲}{a}$ and then along $\underset{̲}{b}$ is equivalent to going along $\underset{̲}{c}$ .

Task!

Using vectors $\underset{̲}{a}$ and $\underset{̲}{b}$ shown below, draw a diagram to find $\underset{̲}{a}+\underset{̲}{b}$ . Find also $\underset{̲}{b}+\underset{̲}{a}$ . Is $\underset{̲}{a}+\underset{̲}{b}$ the same as $\underset{̲}{b}+\underset{̲}{a}$ ?

Answer

3.2 Example: Resultant of two forces acting upon a body

A force ${\underset{̲}{F}}_1$ of 2 N acts vertically downwards, and a force ${\underset{̲}{F}}_2$ of 3 N acts horizontally to the right, upon the body shown in Figure 12.

We can use vector addition to find the combined effect or resultant of the two concurrent forces. (Concurrent means that the forces act through the same point.) Translating ${\underset{̲}{F}}_2$ until its tail touches the head of ${\underset{̲}{F}}_1$ , we complete the triangle ABC as shown. The vector represented by the third side is the resultant, $\underset{̲}{R}$ . We write

$\underset{̲}{R}=\underset{̲}{F_1}+\underset{̲}{F_2}$

and say that $\underset{̲}{R}$ is the vector sum of $\underset{̲}{F_2}$ and $\underset{̲}{F_1}$ . The resultant force acts at an angle of $\theta$ below the horizontal where $tan\theta =2∕3$ , so that $\theta =33.7^{\circ }$ , and has magnitude (given by Pythagoras’ theorem) $\sqrt{13}$ N.

3.3 Example: Resolving a force into two perpendicular directions

In the previous Example we saw that two forces acting upon a body can be replaced by a single force which has the same effect. It is sometimes useful to reverse this process and consider a single force as equivalent to two forces acting at right-angles to each other.

Consider the force $\underset{̲}{F}$ inclined at an angle $\theta$ to the horizontal as shown in Figure 13.

$\underset{̲}{F}$ can be replaced by two forces, one of magnitude $Fcos\theta$ and one of magnitude $Fsin\theta$ as shown. We say that $\underset{̲}{F}$ has been resolved into two perpendicular components . This is sensible because if we re-combine the two perpendicular forces of magnitudes $Fcos\theta$ and $Fsin\theta$ using the triangle law we find $\underset{̲}{F}$ to be the resultant force.

For example, Figure 14 shows a force of 5 N acting at an angle of $30^{\circ }$ to the $x$ axis. It can be resolved into two components, one directed along the $x$ axis with magnitude $5cos30^{\circ }$ and one perpendicular to this of magnitude $5sin30^{\circ }$ . Together, these two components have the same effect as the original force.

Task!

Consider the force shown in the diagram below.

Resolve this force into two perpendicular components, one horizontally to the right, and one vertically upwards.

Answer

The need to resolve a vector along a given direction occurs in other areas. For example, as a police car or ambulance with siren operating passes by the pitch of the siren appears to increase as the vehicle approaches and decrease as it goes away. This change in pitch is known as the Doppler effect This effect occurs in any situation where waves are reflected from a moving object.

A radar gun produces a signal which is bounced off the target moving vehicle so that when it returns to the gun, which also acts as a receiver, it has changed pitch. The speed of the vehicle can be calculated from the change in pitch. The speed indicated on the radar gun is the speed directly towards or away from the gun. However it is not usual to place oneself directly in front of moving vehicle when using the radar gun (Figure 15(a).) Consequently the gun is used at an angle to the line of traffic (Figure 15(b).)

Figure 15

(a)

(b)

This means that it registers only the component of the velocity towards the gun. Suppose that the true speed along the road is $v$ . Then the component measured by the gun ( $vcos\theta$ ) is less than $v$ .

Example 2

A safety inspector wishes to check the speed of a train along a straight piece of track. She stands 10 m to the side of the track and uses a radar gun. If the reading on the gun is to be within 5% of the true speed of the train, how far away from the approaching train should the reading be taken?

Solution

Figure 16

For an error of 5%, the gun should read $0.95\left|\underset{̲}{v}\right|$ . So

$\left|\underset{̲}{v}\right|cos\theta =0.95\left|\underset{̲}{v}\right|$ or $cos\theta =0.95$

If the distance to the side of the track at which the gun is used is $s$ m and the distance between the radar gun and the train is $d$ m, then

$sin\theta =\frac{s}{d}$

Here $s=10$ so

$\sqrt{1-{cos}^2\theta }=\frac{10}{d}$ .

So, if $cos\theta =0.95$

$d=\frac{10}{\sqrt{1-{\left(0.95\right)}^2}}=32.03$

This means that the reading should be taken when the train is over 32 m from the radar gun to ensure an error of less than 5%.

3.4 The force vectors on an aeroplane in steady flight

The forces acting on an aeroplane are shown in Figure 17.

Figure 17

The magnitude (strength) of the forces are indicated by

$T$ : the thrust provided by the engines,

$W$ : the weight,

$D$ : the drag (acting against the direction of flight) and

$L$ : the lift (taken perpendicular to the path.)

In a more realistic situation force vectors in three dimensions would need to be considered. These are introduced later in this Workbook.

As the plane is in steady flight the sum of the forces in any direction is zero. (If this were not the case, then, by Newton’s second law, the non-zero resultant force would cause the aeroplane to accelerate.)

So, resolving forces in the direction of the path:

$Tcos\alpha -D-Wsin\beta =0$

Then, resolving forces perpendicular to the path:

$Tsin\alpha +L-Wcos\beta =0$

If the plane has mass 72 000 tonnes, the drag is 130 kN, the lift is 690 kN and $\beta =6^o$ find the magnitude of the thrust and the value of $\alpha$ to maintain steady flight. From these two equations we see:

$Tcos\alpha =D+Wsin\beta =130000+\left(72000\right)\left(9.81\right)sin{6}^o=203830.54$

and

$Tsin\alpha =Wcos\beta -L=\left(72000\right)\left(9.81\right)cos{6}^o-690000=12450.71$

hence

$tan\alpha =\frac{Tsin\alpha }{Tcos\alpha }=\frac{12450.71}{203830.54}=0.061084\to \alpha =3.50^0$

and consequently, for the thrust:

$T=204210$ N.