Vectors (Page 9/10)

Physics - grade 10 [caps 2011]

Page 9 / 10

Block on an incline

As a further example of components let us consider a block of mass $m$ placed on a frictionless surface inclined at some angle $θ$ to the horizontal. The block will obviously slide down the incline, butwhat causes this motion?

The forces acting on the block are its weight $m g$ and the normal force $N$ exerted by the surface on the object. These two forces are shown in the diagram below.

Now the object's weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown asdashed arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the samepoint. Applying the parallelogram method, the two components of the block's weight sum to the weight vector.

To find the components in terms of the weight we can use trigonometry:

\begin{matrix} F_{g ∥} & = & m g sin θ \\ F_{g ⊥} & = & m g cos θ \end{matrix}

The component of the weight perpendicular to the slope $F_{g ⊥}$ exactly balances the normal force $N$ exerted by the surface. The parallel component, however, $F_{g ∥}$ is unbalanced and causes the block to slide down the slope.

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.

Draw a diagram of the situation

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.
Calculate $F_{g ∥}$
$\begin{matrix} F_{g ∥} & = & m g sin θ \\ = & (10) (9, 8) (sin 30^{\circ}) \\ = & 49 N \end{matrix}$
Write final answer
The force is 49 N up the slope.

Vector addition using components

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, $\vec{A}$ and $\vec{B}$ , in [link] , together with their resultant, $\vec{R}$ .

An example of two vectors being added to give a resultant

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{matrix} \vec{A} & = & {\vec{A}}_{x} + {\vec{A}}_{y} \\ \vec{B} & = & {\vec{B}}_{x} + {\vec{B}}_{y} \\ \vec{R} & = & {\vec{R}}_{x} + {\vec{R}}_{y} \end{matrix}

\begin{matrix} But, {\vec{R}}_{x} & = & {\vec{A}}_{x} + {\vec{B}}_{x} \\ and {\vec{R}}_{y} & = & {\vec{A}}_{y} + {\vec{B}}_{y} \end{matrix}

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] , $\vec{A} = 5, 385 m \cdot s^{- 1}$ at an angle of $21,8^{\circ}$ to the horizontal and $\vec{B} = 5 m \cdot s^{- 1}$ at an angle of $53,13^{\circ}$ to the horizontal, find $\vec{R}$ .

Decide how to tackle the problem
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.
Resolve $\vec{A}$ into components
We find the components of $\vec{A}$ by using known trigonometric ratios. First we find the magnitude of the vertical component, $A_{y}$ :

$\begin{matrix} sin θ & = & \frac{A_{y}}{A} \\ sin 21, 8^{\circ} & = & \frac{A_{y}}{5, 385} \\ A_{y} & = & (5, 385) (sin 21, 8^{\circ}) \\ = & 2 m \cdot s^{- 1} \end{matrix}$

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

$\begin{matrix} cos θ & = & \frac{A_{x}}{A} \\ cos 21 . 8^{\circ} & = & \frac{A_{x}}{5, 385} \\ A_{x} & = & (5, 385) (cos 21, 8^{\circ}) \\ = & 5 m \cdot s^{- 1} \end{matrix}$

The components give the sides of the right angle triangle, for which the original vector, $\vec{A}$ , is the hypotenuse.
Resolve $\vec{B}$ into components
We find the components of $\vec{B}$ by using known trigonometric ratios. First we find the magnitude of the vertical component, $B_{y}$ :

$\begin{matrix} sin θ & = & \frac{B_{y}}{B} \\ sin 53, 13^{\circ} & = & \frac{B_{y}}{5} \\ B_{y} & = & (5) (sin 53, 13^{\circ}) \\ = & 4 m \cdot s^{- 1} \end{matrix}$

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

$\begin{matrix} cos θ & = & \frac{B_{x}}{B} \\ cos 21, 8^{\circ} & = & \frac{B_{x}}{5, 385} \\ B_{x} & = & (5, 385) (cos 53, 13^{\circ}) \\ = & 5 m \cdot s^{- 1} \end{matrix}$
Determine the components of the resultant vector
Now we have all the components. If we add all the horizontal components then we will have the $x$ -component of the resultant vector, ${\vec{R}}_{x}$ . Similarly, we add all the vertical components then we will have the $y$ -component of the resultant vector, ${\vec{R}}_{y}$ .

$\begin{matrix} R_{x} & = & A_{x} + B_{x} \\ = & 5 m \cdot s^{- 1} + 3 m \cdot s^{- 1} \\ = & 8 m \cdot s^{- 1} \end{matrix}$

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

$\begin{matrix} R_{y} & = & A_{y} + B_{y} \\ = & 2 m \cdot s^{- 1} + 4 m \cdot s^{- 1} \\ = & 6 m \cdot s^{- 1} \end{matrix}$

Therefore, ${\vec{R}}_{y}$ is 6 m up.
Determine the magnitude and direction of the resultant vector
Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, $R$ .

$\begin{matrix} R^{2} & = & {(R_{x})}^{2} + {(R_{y})}^{2} \\ R^{2} & = & {(6)}^{2} + {(8)}^{2} \\ R^{2} & = & 100 \\ ∴ R & = & 10 m \cdot s^{- 1} \end{matrix}$

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 $α$ .

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

$\begin{matrix} tan α & = & \frac{6 m \cdot s^{- 1}}{8 m \cdot s^{- 1}} \\ α & = & {tan}^{- 1} \frac{6 m \cdot s^{- 1}}{8 m \cdot s^{- 1}} \\ α & = & 36, 8^{\circ} \end{matrix}$
Quote the final answer
$\vec{R}$ is 10 m at an angle of $36, 8^{\circ}$ to the positive $x$ -axis.

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Source: OpenStax, Physics - grade 10 [caps 2011]. OpenStax CNX. Jun 14, 2011 Download for free at http://cnx.org/content/col11298/1.3

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