# FHSST Physics/Vectors/Components

Vectors The Free High School Science Texts: A Textbook for High School Students Studying Physics. Main Page - << Previous Chapter (Waves and wavelike motion) - Next Chapter (Forces) >> PGCE Comments - TO DO LIST - Introduction - Examples - Mathematical Properties - Addition - Components - Importance - Important Quantities, Equations, and Concepts

# Components of Vectors

In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components.

While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into infinitely many sets of components. In the diagrams below the same black vector is resolved into different pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components).

In practice it is most useful to resolve a vector into components which are at right angles to one another.

## Worked Example 11

Resolving a vector into components

Question: A motorist undergoes a displacement of 250km in a direction 30o north of east. Resolve this displacement into components in the directions north ($\overrightarrow{s}_N$ and east ($\overrightarrow{s}_E$).

Step 1 :

Firstly let us draw a rough sketch of the original vector

Step 2 :

Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a right-angled triangle with the original displacement as its hypotenuse:

Notice how the two components acting together give the original vector as their resultant.

Step 3 :

Now we can use trigonometry to calculate the magnitudes of the components of the original displacement:

$\begin{matrix}s_N &=& 250\sin30^o\\&=& 125\ km\end{matrix}$

and

$\begin{matrix}s_E &=& 250\cos30^o\\&=& 216.5\ km\end{matrix}$

Remember sN and sE are the magnitudes of the components- they are in the directions north and east respectively.

## 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 $\theta$ to the horizontal. The block will obviously slide down the incline, but what causes this motion?

The forces acting on the block are its weight mg 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 as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. 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:

$\begin{matrix}W_{\|} &=& mg\sin\theta\\W_{\perp} &=& mg\cos\theta\end{matrix}$

The component of the weight perpendicular to the slope W$\perp$ exactly balances the normal force N exerted by the surface. The parallel component, however, $W_{\|}$ is unbalanced and causes the block to slide down the slope.

In Figure 3.3 two vectors are added in a slightly different way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong.

In Figure 3.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Figure 3.2 using the less complicated looking method.

Figure 3.2:An example of two vectors being added to give a resultant

Each vector can be broken down into a component in the x-direction and one in the y-direction. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in figure 3.3. If you add up the two red dotted ones in the x-direction and y-direction you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour.

But if we look carefully, 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 important property of vectors.

### Worked Example 12

Question: Lets work through the example shown in Figure 3.3 to determine the resultant.

Step 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.

Step 2 :

Let us start with the bottom vector. If you are told that this vector has a length of 5.385 units and an angle of 21.8o to the horizontal then we can find its components. We do this by using known trigonometric ratios. First we find the vertical or y component:

$\begin{matrix}\sin \theta & = & \frac{y}{\mbox{hypotenuse}} \\\sin (21.8) & = &\frac{y}{5.385}\\y & = & 5.385 \sin (21.8)\\y & = & 2\end{matrix}$

Secondly we find the horizontal or x component:

$\begin{matrix}\cos \theta & = & \frac{x}{\mbox{hypotenuse}} \\\cos (21.8) & = &\frac{x}{5.385}\\x & = & 5.385 \cos (21.8)\\x & = & 5\end{matrix}$

We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to find the final answer we can just add all the dotted vectors because they would add up to the two vectors we want to add.

Step 3 :

Now we move on to considering the second vector. The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can find its components.

$\begin{matrix}\sin \theta & = & \frac{y}{\mbox{hypotenuse}} \\\sin (53.13) & = &\frac{y}{5}\\y & = & 5 \sin (53.13)\\y & = & 4\end{matrix}$

$\begin{matrix}\cos \theta & = & \frac{x}{\mbox{hypotenuse}} \\\cos (53.13) & = &\frac{x}{5}\\x & = & 5 \cos (53.13)\\x & = & 3\end{matrix}$

Step 4 :

Now we have all the components. If we add all the x-components then we will have the x-component of the resultant vector. Similarly if we add all the y-components then we will have the y-component of the resultant vector.

The x-components of the two vectors are 5 units right and then 3 units right. This gives us a final x-component of 8 units right.

The y-components of the two vectors are 2 units up and then 4 units up. This gives us a final y-component of 6 units up.

Step 5 :

Now that we have the components of the resultant, we can use Pythagoras' theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value

$\begin{matrix}l^2&=&(6)^2 + (8)^2\\l^2&=&100\\l&=&10.\\\end{matrix}$

The resultant has length of 10 units 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 visualize 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 $\alpha$

$\begin{matrix}\tan \alpha & = & \frac{6}{8} \\\alpha & = & \arctan \frac{6}{8}\\\alpha & = & 36.8^o.\end{matrix}$

Step 6 :

Our final answer is a resultant of 10 units at 36.8o to the positive x-axis.