Algebraic Techniques of Vector Addition
Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique: Choose a positive direction. As an example, for situations involving ...
Algebraic Techniques: Vectors in a Straight Line
Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:
Method: Addition/Subtraction of Vectors in a Straight Line
- Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
- Next simply add (or subtract) the magnitude of the vectors using the appropriate signs.
- As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).
Let us consider a few examples.
Example 1:
Question
A tennis ball is rolled towards a wall which is \(\text{10}\) \(\text{m}\) away from the ball. If after striking the wall the ball rolls a further \(\text{2.5}\) \(\text{m}\) along the ground away from the wall, calculate algebraically the ball's resultant displacement.
Step 1: Draw a rough sketch of the situation
Step 2: Decide which method to use to calculate the resultant
We know that the resultant displacement of the ball (\(\vec{x_{R}}\)) is equal to the sum of the ball's separate displacements (\(\vec{x_{1}}\) and \(\vec{x_{2}}\)):
\(\vec{x_{R}} = \vec{x_{1}} + \vec{x_{2}}\)Since the motion of the ball is in a straight line (i.e. the ball moves towards and away from the wall), we can use the method of algebraic addition just explained.
Step 3: Choose a positive direction
Let's choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.
Step 4: Now define our vectors algebraically
With right positive:
\begin{align*} \vec{x_{1}} & = + \text{10,0}\text{ m} \\ \vec{x_{2}} & = -\text{2,5}\text{ m} \end{align*}Step 5: Add the vectors
Next we simply add the two displacements to give the resultant:
\begin{align*} \vec{x_{R}} & = \left(+\text{10,0}\text{ m}\right) + \left(-\text{2,5}\text{ m}\right) \\ & = \left(+\text{7,5}\text{ m}\right) \end{align*}Step 6: Quote the resultant
Finally, in this case towards the wall is the positive direction, so: \(\vec{x_{R}} = \text{7.5}\text{ m}\) towards the wall.
Example 2:
Question
Suppose that a tennis ball is thrown horizontally towards a wall at an initial velocity of \(\text{3}\) \(\text{m·s$^{-1}$}\) to the right. After striking the wall, the ball returns to the thrower at \(\text{2}\) \(\text{m·s$^{-1}$}\). Determine the change in velocity of the ball.
Step 1: Draw a sketch
A quick sketch will help us understand the problem.
Step 2: Decide which method to use to calculate the resultant
Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:
\(\Delta \vec{v} = \vec{v_{f}} - \vec{v_{i}}\)Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.
Step 3: Choose a positive direction
Choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.
Step 4: Now define our vectors algebraically
\begin{align*} \vec{v_{i}} & = +\text{3}\text{ m·s$^{-1}$} \\ \vec{v_{f}} & = -\text{2}\text{ m·s$^{-1}$} \end{align*}Step 5: Subtract the vectors
Thus, the change in velocity of the ball is:
\begin{align*} \Delta \vec{v} & = \left(-\text{2}\text{ m·s$^{-1}$}\right) - \left(+\text{3}\text{ m·s$^{-1}$}\right) \\ & = -\text{5}\text{ m·s$^{-1}$} \end{align*}Step 6: Quote the resultant
Remember that in this case towards the wall means a positive velocity,
so away from the wall means a negative velocity: \(\Delta \vec{v}=\text{5}\text{ m·s$^{-1}$}\) away from the wall.
Example 3:
Question
A man applies a force of \(\text{5}\) \(\text{N}\) on a crate. The crate pushes back on the man with a force of \(\text{2}\) \(\text{N}\). Calculate algebraically the resultant force that the man applies to the crate.
Step 1: Draw a sketch
A quick sketch will help us understand the problem.
Step 2: Decide which method to use to calculate the resultant
Remember that force is a vector. Since the crate moves along a straight line (i.e. left and right), we can use the algebraic technique of vector addition just discussed.
Step 3: Choose a positive direction
Choose the positive direction to be towards the crate (i.e. in the same direction that the man is pushing). This means that the negative direction is away from the crate (i.e. against the direction that the man is pushing).
Step 4: Now define our vectors algebraically
\begin{align*} \vec{F_{\text{man}}} & = +\text{5}\text{ N} \\ \vec{F_{\text{crate}}} & = -\text{2}\text{ N} \end{align*}Step 5: Subtract the vectors
Thus, the resultant force is:
\begin{align*} \vec{F_{\text{man}}} + \vec{F_{\text{crate}}} & = \left(\text{5}\text{ N}\right) + \left(\text{2}\text{ N}\right) \\ & = \text{7}\text{ N} \end{align*}Step 6: Quote the resultant
Remember that in this case towards the crate means a positive force: \(\text{7}\) \(\text{N}\) towards the crate.
Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at an angle to each other then simple geometric and trigonometric techniques can be used to find resultant vectors.
This lesson is part of:
Vectors and Scalars