# 10.3 Components  (Page 2/3)

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## Worked example

Determine the force needed to keep a 10 kg block from sliding down a frictionless slope. The slope makes an angle of 30 ${}^{\circ }$ with the horizontal.

1. The force that will keep the block from sliding is equal to the parallel component of the weight, but its direction is up the slope.

2. $\begin{array}{ccc}\hfill {F}_{g\parallel }& =& mgsin\theta \hfill \\ & =& \left(10\right)\left(9,8\right)\left(sin{30}^{\circ }\right)\hfill \\ & =& 49\mathrm{N}\hfill \end{array}$
3. The force is 49 N up the slope.

Components can also be used to find the resultant of vectors. This technique can be applied to both graphical and algebraic methods of finding the resultant. The method is simple: make a rough sketch of the problem, find the horizontal and vertical components of each vector, find the sum of all horizontal components and the sum of all the vertical components and then use them to find the resultant.

Consider the two vectors, $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ , in [link] , together with their resultant, $\stackrel{\to }{R}$ .

Each vector in [link] can be broken down into one component in the $x$ -direction (horizontal) and one in the $y$ -direction (vertical). These components are two vectors which when added give you the original vector as the resultant. This is shown in [link] where we can see that:

$\begin{array}{ccc}\hfill \stackrel{\to }{A}& =& {\stackrel{\to }{A}}_{x}+{\stackrel{\to }{A}}_{y}\hfill \\ \hfill \stackrel{\to }{B}& =& {\stackrel{\to }{B}}_{x}+{\stackrel{\to }{B}}_{y}\hfill \\ \hfill \stackrel{\to }{R}& =& {\stackrel{\to }{R}}_{x}+{\stackrel{\to }{R}}_{y}\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{But},\phantom{\rule{1.em}{0ex}}{\stackrel{\to }{\mathrm{R}}}_{\mathrm{x}}& =& {\stackrel{\to }{A}}_{x}+{\stackrel{\to }{B}}_{x}\hfill \\ \hfill \mathrm{and}\phantom{\rule{1.em}{0ex}}{\stackrel{\to }{\mathrm{R}}}_{\mathrm{y}}& =& {\stackrel{\to }{A}}_{y}+{\stackrel{\to }{B}}_{y}\hfill \end{array}$

In summary, addition of the $x$ components of the two original vectors gives the $x$ component of the resultant. The same applies to the $y$ components. So if we just added all the components together we would get the same answer! This is another importantproperty of vectors.

If in [link] , $\stackrel{\to }{A}=5,385m·{s}^{-1}$ at an angle of 21.8 ${}^{\circ }$ to the horizontal and $\stackrel{\to }{B}=5m·{s}^{-1}$ at an angle of 53,13 ${}^{\circ }$ to the horizontal, find $\stackrel{\to }{R}$ .

1. The first thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order.

2. We find the components of $\stackrel{\to }{A}$ by using known trigonometric ratios. First we find the magnitude of the vertical component, ${A}_{y}$ :

$\begin{array}{ccc}\hfill sin\theta & =& \frac{{A}_{y}}{A}\hfill \\ \hfill sin21,{8}^{\circ }& =& \frac{{A}_{y}}{5,385}\hfill \\ \hfill {A}_{y}& =& \left(5,385\right)\left(sin21,{8}^{\circ }\right)\hfill \\ & =& 2m·{s}^{-1}\hfill \end{array}$

Secondly we find the magnitude of the horizontal component, ${A}_{x}$ :

$\begin{array}{ccc}\hfill cos\theta & =& \frac{{A}_{x}}{A}\hfill \\ \hfill cos21.{8}^{\circ }& =& \frac{{A}_{x}}{5,385}\hfill \\ \hfill {A}_{x}& =& \left(5,385\right)\left(cos21,{8}^{\circ }\right)\hfill \\ & =& 5m·{s}^{-1}\hfill \end{array}$

The components give the sides of the right angle triangle, for which the original vector, $\stackrel{\to }{A}$ , is the hypotenuse.

3. We find the components of $\stackrel{\to }{B}$ by using known trigonometric ratios. First we find the magnitude of the vertical component, ${B}_{y}$ :

$\begin{array}{ccc}\hfill sin\theta & =& \frac{{B}_{y}}{B}\hfill \\ \hfill sin53,{13}^{\circ }& =& \frac{{B}_{y}}{5}\hfill \\ \hfill {B}_{y}& =& \left(5\right)\left(sin53,{13}^{\circ }\right)\hfill \\ & =& 4m·{s}^{-1}\hfill \end{array}$

Secondly we find the magnitude of the horizontal component, ${B}_{x}$ :

$\begin{array}{ccc}\hfill cos\theta & =& \frac{{B}_{x}}{B}\hfill \\ \hfill cos21,{8}^{\circ }& =& \frac{{B}_{x}}{5,385}\hfill \\ \hfill {B}_{x}& =& \left(5,385\right)\left(cos53,{13}^{\circ }\right)\hfill \\ & =& 5m·{s}^{-1}\hfill \end{array}$

4. Now we have all the components. If we add all the horizontal components then we will have the $x$ -component of the resultant vector, ${\stackrel{\to }{R}}_{x}$ . Similarly, we add all the vertical components then we will have the $y$ -component of the resultant vector, ${\stackrel{\to }{R}}_{y}$ .

$\begin{array}{ccc}\hfill {R}_{x}& =& {A}_{x}+{B}_{x}\hfill \\ & =& 5m·{s}^{-1}+3m·{s}^{-1}\hfill \\ & =& 8m·{s}^{-1}\hfill \end{array}$

Therefore, ${\stackrel{\to }{R}}_{x}$ is 8 m to the right.

$\begin{array}{ccc}\hfill {R}_{y}& =& {A}_{y}+{B}_{y}\hfill \\ & =& 2m·{s}^{-1}+4m·{s}^{-1}\hfill \\ & =& 6m·{s}^{-1}\hfill \end{array}$

Therefore, ${\stackrel{\to }{R}}_{y}$ is 6 m up.

5. Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, $R$ .

$\begin{array}{ccc}\hfill {R}^{2}& =& {\left({R}_{x}\right)}^{2}+{\left({R}_{y}\right)}^{2}\hfill \\ \hfill {R}^{2}& =& {\left(6\right)}^{2}+{\left(8\right)}^{2}\hfill \\ \hfill {R}^{2}& =& 100\hfill \\ \hfill \therefore R& =& 10m·{s}^{-1}\hfill \end{array}$

The magnitude of the resultant, $R$ is 10 m. So all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicitly below and the angle we will calculate is labeled $\alpha$ .

Using our known trigonometric ratios we can calculate the value of $\alpha$ ;

$\begin{array}{ccc}\hfill tan\alpha & =& \frac{6m·{s}^{-1}}{8m·{s}^{-1}}\hfill \\ \hfill \alpha & =& {tan}^{-1}\frac{6m·{s}^{-1}}{8m·{s}^{-1}}\hfill \\ \hfill \alpha & =& 36,{8}^{\circ }.\hfill \end{array}$
6. $\stackrel{\to }{R}$ is 10 m at an angle of $36,{8}^{\circ }$ to the positive $x$ -axis.

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