5.2 Vector Addition and Subtraction: Analytical Methods - Physics | OpenStax (2024)

Section Key Terms

Components of Vectors

For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17.

Figure 5.17 For a right triangle, the sine, cosine, and tangent of θ are defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, x is the adjacent side, y is the opposite side, and h is the hypotenuse.

Since, by definition, $\text{cos}\theta =x/h$, we can find the length x if we know h and $\theta$ by using $x=h\text{cos}\theta$ . Similarly, we can find the length of y by using $y=h\text{sin}\theta$ . These trigonometric relationships are useful for adding vectors.

When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components.

For example, given a vector like $A$ in Figure 5.18, we may want to find what two perpendicular vectors, $A_x$ and $A_y$, add to produce it. In this example, $A_x$ and $A_y$ form a right triangle, meaning that the angle between them is 90 degrees. This is a common situation in physics and happens to be the least complicated situation trigonometrically.

Figure 5.18 The vector $A$, with its tail at the origin of an x- y-coordinate system, is shown together with its x- and y-components, $A_x$ and $A_y.$ These vectors form a right triangle.

$A_x$ and $A_y$ are defined to be the components of $A$ along the x- and y-axes. The three vectors, $A$, $A_x$, and $A_y$, form a right triangle.

$$A_x+A_y=A$$

If the vector $A$ is known, then its magnitude $A$ (its length) and its angle $\theta$ (its direction) are known. To find $A_x$ and $A_y$, its x- and y-components, we use the following relationships for a right triangle:

$$A_x=A\text{cos}\theta$$

and

$$A_y=A\text{sin}\mathrm{\theta ,}$$

where $A_x$ is the magnitude of A in the x-direction, $A_y$ is the magnitude of A in the y-direction, and $\theta$ is the angle of the resultant with respect to the x-axis, as shown in Figure 5.19.

Figure 5.19 The magnitudes of the vector components $A_x$ and $A_y$ can be related to the resultant vector $A$ and the angle $\theta$ with trigonometric identities. Here we see that $A_x=A\text{cos}\theta$ and $A_y=A\text{sin}\theta .$

Suppose, for example, that $A$ is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20.

Figure 5.20 We can use the relationships $A_x=A\text{cos}\theta$ and $A_y=A\text{sin}\theta$ to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A = 10.3 blocks and $\theta ={29.1}^{\circ }$, so that

This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly,

indicating that the walker has traveled 5 blocks to the north—a 5-block northward displacement.

Analytical Method of Vector Addition and Subtraction

Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components $A_x$ and $A_y$ of a vector $A$ are known, then we can find $A$ analytically. How do we do this? Since, by definition,

$$\text{tan}\theta =y/x\text{(orinthiscasetan}\theta =A_y/A_x\text{),}$$

we solve for $\theta$ to find the direction of the resultant.

$$\theta ={\tan }^{-1}(A_y/A_x)$$

Note that ${\text{tan}}^{-1}(\theta )$ gives an angle in the first quadrant if $A_y/A_x>0$ and in the fourth quadrant if $A_y/A_x<0$. If, in fact, both $A_x$ and $A_y$ are negative, or if $A_x$ is negative and $A_y$ positive, then $\theta$, measured from the positive $x$ direction, is ${\text{tan}}^{–1}(\theta )+180\text{°}$.

Since this is a right triangle, the Pythagorean theorem (x² + y² = h²) for finding the hypotenuse applies. In this case, it becomes

$$A^2=A_x^2+A_y^2\text{.}$$

Solving for A gives

$$A=\sqrt{A_x{}^2+A_y{}^2}\text{.}$$

In summary, to find the magnitude $A$ and direction $\theta$ of a vector from its perpendicular components $A_x$ and $A_y$, as illustrated in Figure 5.21, we use the following relationships:

$$\begin{array}{ccc}\hfill A& =& \sqrt{A_x{}^2+A_y{}^2}\hfill \\ & & \theta ={\tan }^{-1}(A_y/A_x)\hfill \end{array}$$

Figure 5.21 The magnitude and direction of the resultant vector $A$ can be determined once the horizontal components $A_x$ and $A_y$ have been determined.

Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors $A$ and $B$ are added to produce the resultant $R\text{,}$ as illustrated in Figure 5.22.

Figure 5.22 Vectors $A$ and $B$ are two legs of a walk, and $R$ is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of $R$ .

If $A$ and $B$ represent two legs of a walk (two displacements), then $R$ is the total displacement. The person taking the walk ends up at the tip of $R$ . There are many ways to arrive at the same point. The person could have walked straight ahead first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, $R_x$ and $R_y.$ If we know $R_x$ and $R_y$, we can find $R$ and $\theta$ using the equations $R=\sqrt{R_{\text{x}}{}^2+R_{\text{y}}{}^2}$ and $\theta =ta{n}^{–1}(R_y/R_x)$ .

1. Draw in the x and y components of each vector (including the resultant) with a dashed line. Use the equations $A_x=A\text{cos}\theta$ and $A_y=A\text{sin}\theta$ to find the components. In Figure 5.23, these components are $A_x$, $A_y$, $B_x$, and $B_y.$ Vector $A$ makes an angle of ${\theta }_A$ with the x-axis, and vector $B$ makes and angle of ${\theta }_B$ with its own x-axis (which is slightly above the x-axis used by vector A).

   

   Figure 5.23 To add vectors $A$ and $B,$ first determine the horizontal and vertical components of each vector. These are the dotted vectors $A_x,$ $A_y$ $B_y$ shown in the image.

2. Find the x component of the resultant by adding the x component of the vectors

   $$R_x=A_x+B_x$$

   and find the y component of the resultant (as illustrated in Figure 5.24) by adding the y component of the vectors.

   $$R_y=A_y+B_y\text{.}$$

   

   Figure 5.24 The vectors $A_x$ and $B_x$ add to give the magnitude of the resultant vector in the horizontal direction, $R_{\text{x}}.$ Similarly, the vectors $A_y$ and $B_y$ add to give the magnitude of the resultant vector in the vertical direction, $R_{\text{y}}.$

   Now that we know the components of $R,$ we can find its magnitude and direction.

3. To get the magnitude of the resultant R, use the Pythagorean theorem.

   $$R=\sqrt{R_x^2+R_y^2}$$

4. To get the direction of the resultant

   $$\theta ={\tan }^{-1}(R_y/R_x)\text{.}$$

Using the Analytical Method of Vector Addition and Subtraction to Solve Problems

Figure 5.25 uses the analytical method to add vectors.

Worked Example

An Accelerating Subway Train

Add the vector $A$ to the vector $B$ shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and the y-axis is along the north–south directions. A person first walks $\text{53}\text{.0m}$ in a direction $\text{20}\text{.0°}$ north of east, represented by vector $A.$ The person then walks $\text{34}\text{.0m}$ in a direction $\text{63}\text{.0}°$ north of east, represented by vector $B.$

Figure 5.25 You can use analytical models to add vectors.

Strategy

The components of $A$ and $B$ along the x- and y-axes represent walking due east and due north to get to the same ending point. We will solve for these components and then add them in the x-direction and y-direction to find the resultant.

Solution

First, we find the components of $A$ and $B$ along the x- and y-axes. From the problem, we know that $A=53.0\text{m,}$ ${\theta }_{\text{A}}={20.0}^{\circ ,}$ $B$ = $\text{34}\text{.0m}$, and ${\theta }_{\text{B}}={63.0}^{\circ }$ . We find the x-components by using $A_x=A\cos \theta$, which gives

$$\begin{array}{ccc}\hfill A_x& =& A\cos {\theta }_A=(53.0\text{m})(\cos {20.0}^{\circ })\hfill \\ & =& (53.0\text{m})(0.940)=49.8\text{m}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill B_x& =& B\cos {\theta }_B=(34.0\text{m})(\cos {63.0}^{\circ })\hfill \\ & =& (34.0\text{m})(0.454)=15.4\text{m.}\hfill \end{array}$$

Similarly, the y-components are found using $A_y=A\sin {\theta }_A$

$$\begin{array}{ccc}\hfill A_y& =& A\sin {\theta }_A=(53.0\text{m})(\sin {20.0}^{\circ })\hfill \\ & =& (53.0\text{m})(0.342)=18.1\text{m}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill B_y& =& B\sin {\theta }_B=(34.0\text{m})(\sin {63.0}^{\circ })\hfill \\ & =& (34.0\text{m})(0.891)=30.3\text{m}\text{.}\hfill \end{array}$$

The x- and y-components of the resultant are

$$R_x=A_x+B_x=49.8\text{m}+15.4\text{m}=65.2\text{m}$$

and

$$R_y=A_y+B_y=18.1\text{m}+30.3\text{m}=48.4\text{m}.$$

Now we can find the magnitude of the resultant by using the Pythagorean theorem

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{{(65.2)}^2+{(48.4)}^2}\text{m}$$

5.8

so that

$$R=\sqrt{6601\text{m}}=81.2\text{m}\text{.}$$

Finally, we find the direction of the resultant

$$\theta ={\tan }^{-1}(R_y/R_x)=+{\tan }^{-1}(48.4/65.2).$$

This is

$$\theta ={\tan }^{-1}(0.742)={36.6}^{\circ }.$$

Discussion

This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, $\text{A}-\text{B}\equiv \text{A}+(-\text{B})$ . The components of – $\text{B}$ are the negatives of the components of $\text{B}$ . Therefore, the x- and y-components of the resultant $\text{A}-\text{B}=\text{R}$ are

$$R_x=A_x+-B_x$$

and

$$R_y=A_y+-B_y$$

and the rest of the method outlined above is identical to that for addition.

See Also

2.2 Coordinate Systems and Components of a Vector

Practice Problems

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