how to add vector components
Adding Vectors Using Components
The following example shows how to add vectors using components. We must realise that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order.
Adding vectors using components.
Example: Adding Vectors Using Components
Contents
- Example: Adding Vectors Using Components
- Question
- Step 1: Decide how to tackle the problem
- Step 2: Resolve \(\vec{F}_1\) into components
- Step 3: Resolve \(\vec{F}_2\) into components
- Step 4: Determine the components of the resultant vector
- Step 5: Determine the magnitude and direction of the resultant vector
- Step 6: Quote the final answer
Question
If in the figure above, \(\vec{F}_1\)=\(\text{5.385}\) \(\text{N}\) at an angle of \(\text{21.8}\) \(\text{°}\) to the horizontal and \(\vec{F}_2\)=\(\text{5}\) \(\text{N}\) at an angle of \(\text{53.13}\) \(\text{°}\) to the horizontal, find the resultant force, \(\vec{R}\).
Step 1: 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. We also draw up the following table to help us work through the problem:
| Vector | \(x\)-component | \(y\)-component | Total |
| \(\vec{F}_1\) | |||
| \(\vec{F}_2\) | |||
| Resultant |
Step 2: Resolve \(\vec{F}_1\) into components
We find the components of \(\vec{F}_1\) by using known trigonometric ratios. First we find the magnitude of the vertical component, \({F}_{\mathrm{1y}}\): \begin{align*} \sin(\theta) &= \frac{F_{1y}}{F_1} \\ \sin(\text{21,8}\text{°}) &= \frac{F_{1y}}{\text{5,385}} \\ F_{1y} &= \left(\sin(\text{21,8}\text{°})\right)\left( \text{5,385}\right) \\ &= \text{2,00}\text{ N} \end{align*}
Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{1x}}\): \begin{align*} \cos(\theta) &= \frac{F_{1x}}{F_1} \\ \cos(\text{21,8}\text{°}) &= \frac{F_{1x}}{\text{5,385}} \\ F_{1x} &= \left(\cos(\text{21,8}\text{°})\right)\left( \text{5,385} \right) \\ &= \text{5,00}\text{ N} \end{align*}
The components give the sides of the right angle triangle, for which the original vector, \(\vec{F}_1\), is the hypotenuse.
| Vector | \(x\)-component | \(y\)-component | Resultant |
| \(\vec{F}_1\) | \(\text{5.00}\) \(\text{N}\) | \(\text{2.00}\) \(\text{N}\) | \(\text{5.385}\) \(\text{N}\) |
| \(\vec{F}_2\) | |||
| Resultant |
Step 3: Resolve \(\vec{F}_2\) into components
We find the components of \(\vec{F}_2\) by using known trigonometric ratios. First we find the magnitude of the vertical component, \({F}_{\mathrm{2y}}\):
\begin{align*} \sin(\theta) &= \frac{F_{2y}}{F_2} \\ \sin( \text{53,13}\text{°}) &= \frac{F_{2y}}{5} \\ F_{2y} &= \left(\sin( \text{53,13}\text{°})\right)\left( 5\right) \\ &= \text{4,00}\text{ N} \end{align*}
Secondly we find the magnitude of the horizontal component, \({F}_{\mathrm{2x}}\):
\begin{align*} \cos(\theta) &= \frac{F_{2x}}{F_2} \\ \cos (\text{53,13}\text{°}) &= \frac{F_{2x}}{5} \\ F_{2x} &= \left(\cos( \text{53,13}\text{°})\right)\left( 5\right) \\ &= \text{3,00}\text{ N} \end{align*}
| Vector | \(x\)-component | \(y\)-component | Total |
| \(\vec{F}_1\) | \(\text{5.00}\) \(\text{N}\) | \(\text{2.00}\) \(\text{N}\) | \(\text{5.385}\) \(\text{N}\) |
| \(\vec{F}_2\) | \(\text{3.00}\) \(\text{N}\) | \(\text{4.00}\) \(\text{N}\) | \(\text{5}\) \(\text{N}\) |
| Resultant |
Step 4: 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, \({\stackrel{\to }{R}}_{x}\). Similarly, we add all the vertical components then we will have the \(y\)-component of the resultant vector, \({\stackrel{\to }{R}}_{y}\).
\begin{align*} R_x &= F_{1x} + F_{2x} \\ & = \text{5,00}\text{ N} + \text{3,00}\text{ N} \\ &= \text{8,00}\text{ N} \end{align*}
Therefore, \({\stackrel{\to }{R}}_{x}\) is \(\text{8}\) \(\text{N}\) to the right.
\begin{align*} R_y &= F_{1y} + F_{2y} \\ & = \text{2,00}\text{ N} + \text{4,00}\text{ N} \\ &= \text{6,00}\text{ N} \end{align*}
Therefore, \({\stackrel{\to }{R}}_{y}\) is \(\text{6}\) \(\text{N}\) up.
| Vector | \(x\)-component | \(y\)-component | Total |
| \(\vec{F}_1\) | \(\text{5.00}\) \(\text{N}\) | \(\text{2.00}\) \(\text{N}\) | \(\text{5.385}\) \(\text{N}\) |
| \(\vec{F}_2\) | \(\text{3.00}\) \(\text{N}\) | \(\text{4.00}\) \(\text{N}\) | \(\text{5}\) \(\text{N}\) |
| Resultant | \(\text{8.00}\) \(\text{N}\) | \(\text{6.00}\) \(\text{N}\) |
Step 5: 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{align*} R^2 &= (R_y)^2 + (R_x)^2 \\ & = (\text{6,00})^2 + (\text{8,00})^2 \\ &= \text{100,00}\\ R &= \text{10,00}\text{ N} \end{align*}
The magnitude of the resultant, R is \(\text{10.00}\) \(\text{N}\). 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 labelled α.
Using our known trigonometric ratios we can calculate the value of \(\alpha\):
\begin{align*} \tan\alpha &= \frac{\text{6,00}}{\text{8,00}}\\ \alpha &= \tan^{-1}\frac{\text{6,00}}{\text{8,00}}\\ \alpha &= \text{36,9}\text{°} \end{align*}
Step 6: Quote the final answer
\(\stackrel{\to }{R}\) is \(\text{10}\) \(\text{N}\) at an angle of \(36.{9}^{°}\) to the positive \(x\)-axis.
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how to add vector components
Source: https://nigerianscholars.com/tutorials/vectors-and-scalars/adding-vectors-using-components/
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