Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like $\mathbf{A}$ in Figure 19.1, we may wish to find which two perpendicular vectors, ${\mathbf{A}}_x$ and ${\mathbf{A}}_y$, add to produce it.

${\mathbf{A}}_x$ and ${\mathbf{A}}_y$ are defined to be the components of $\mathbf{A}$ along the x– and y-axes. The three vectors $\mathbf{A}$, ${\mathbf{A}}_x$, and ${\mathbf{A}}_y$ form a right triangle:

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if ${\text{A}}_x=3m$ east, ${\text{A}}_y=4m$ north, and $\text{A}=5m$ north-east, then it is true that the vectors ${\mathbf{A}}_x{\text{ + A}}_y\text{ = A}$. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

Thus,

If the vector $\mathbf{A}$ 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.

and

Suppose, for example, that $\mathbf{A}$ is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Then $A=10.3$ blocks and $\theta =29.1º$, so that

If the perpendicular components ${\mathbf{A}}_x$ and ${\mathbf{A}}_y$ of a vector $\mathbf{A}$ are known, then $\mathbf{A}$ can also be found analytically. To find the magnitude $A$ and direction $\theta$ of a vector from its perpendicular components ${\mathbf{A}}_x$ and ${\mathbf{A}}_y$, we use the following relationships:

Note that the equation $A=\sqrt{A_x^2+A_y^2}$ is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if $A_x$ and $A_y$ are 9 and 5 blocks, respectively, then $A=\sqrt{9^2{\text{+5}}^2}\text{=10}\text{.}3$ blocks, again consistent with the example of the person walking in a city. Finally, the direction is $\theta ={\text{tan}}^{–1}\left(\text{5/9}\right)=29.1º$, as before.

To see how to add vectors using perpendicular components, consider Figure 19.5, in which the vectors $\mathbf{A}$ and $\mathbf{B}$ are added to produce the resultant $\mathbf{R}$.

If $\mathbf{A}$ and $\mathbf{B}$ represent two legs of a walk (two displacements), then $\mathbf{R}$ is the total displacement. The person taking the walk ends up at the tip of $\mathbf{R}.$ There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x– and y-components of the resultant, ${\mathbf{R}}_x$ and ${\mathbf{R}}_y$. If we know ${\text{R}}_x$ and ${\mathbf{R}}_y$, we can find $R$ and $\theta$ using the equations $A=\sqrt{{A_x}^2+{A_y}^2}$ and $\theta ={\text{tan}}^{–1}\left(A_y/A_x\right)$. When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations $A_x=A\phantom{}\text{cos}\phantom{}\theta$ and $A_y=A\phantom{}\text{sin}\phantom{}\theta$ to find the components. In Figure 19.6, these components are $A_x$, $A_y$, $B_x$, and $B_y$. The angles that vectors $\mathbf{A}$ and $\mathbf{B}$ make with the x-axis are ${\theta }_{\text{A}}$ and ${\theta }_{\text{B}}$, respectively.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 19.7,

and

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of $\mathbf{R}$ are known, its magnitude and direction can be found.

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

Step 4. To get the direction of the resultant:

The following example illustrates this technique for adding vectors using perpendicular components.

Example 19.1 Adding Vectors Using Analytical Methods

Add the vector $\mathbf{A}$ to the vector $\mathbf{B}$ shown in Figure 19.8, using perpendicular components along the x– and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively. Vector $\mathbf{A}$ represents the first leg of a walk in which a person walks $\text{53}\text{.}\text{0 m}$ in a direction $\text{20}\text{.}0\text{º}$ north of east. Vector $\mathbf{B}$ represents the second leg, a displacement of $\text{34}\text{.}\text{0 m}$ in a direction $\text{63}\text{.}0\text{º}$ north of east.

Strategy

The components of $\mathbf{A}$ and $\mathbf{B}$ along the x– and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of $\mathbf{A}$ and $\mathbf{B}$ along the x– and y-axes. Note that $A=53.0m$, ${\theta }_{\text{A}}=20.0º$, $B=34.0m$, and ${\theta }_{\text{B}}=63.0º$. We find the x-components by using $A_x=A\phantom{}\text{cos}\phantom{}\theta$, which gives

$\begin{array}{lll}{A}_x& =& A\phantom{}\text{cos}\phantom{}{\theta }_A=\left(\text{53.}0m\right)\left(\text{cos 20.0º}\right)\\ & =& \left(\text{53.}0m\right)\left(0\text{.940}\right)=\text{49.}8m\end{array}$

and

$\begin{array}{lll}{B}_x& =& B\phantom{}\text{cos}\phantom{}{\theta }_B=\left(\text{34}\text{.}0m\right)\left(\text{cos 63.0º}\right)\\ & =& \left(\text{34}\text{.}0m\right)\left(0\text{.}\text{454}\right)=\text{15}\text{.}4m\text{.}\end{array}$

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

$\begin{array}{lll}{A}_y& =& A\phantom{}\text{sin}\phantom{}{\theta }_A=\left(\text{53}\text{.}0m\right)\left(\text{sin 20.0º}\right)\\ & =& \left(\text{53}\text{.}0m\right)\left(0\text{.}\text{342}\right)=\text{18}\text{.}1m\end{array}$

and

$\begin{array}{lll}{B}_y& =& B\phantom{}\text{sin}\phantom{}{\theta }_B=\left(\text{34}\text{.}0m\right)\left(\text{sin 63}\text{.}0º\right)\\ & =& \left(\text{34}\text{.}0m\right)\left(0\text{.}\text{891}\right)=\text{30}\text{.}3m\text{.}\end{array}$

The x– and y-components of the resultant are thus

$R_x=A_x+B_x=\text{49}\text{.}8m+\text{15}\text{.}4m=\text{65}\text{.}2m$

and

$R_y=A_y+B_y=\text{18}\text{.}1m+\text{30}\text{.}3m=\text{48}\text{.}4m\text{.}$

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

$R=\sqrt{R_x^2+R_y^2}=\sqrt{{\left(\text{65}\text{.}2\right)}^2+{\left(\text{48}\text{.}4\right)}^2\phantom{}\text{m}},$

so that

$R=81.2m.$

Finally, we find the direction of the resultant:

$\theta ={\text{tan}}^{-1}\left(R_y/R_x\right)\text{=+}{\text{tan}}^{-1}\left(\text{48}\text{.}4/\text{65}\text{.}2\right)\text{.}$

Thus,

$\theta ={\text{tan}}^{-1}\left(0\text{.}\text{742}\right)=\text{36}\text{.}6º\text{.}$

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, $\mathbf{A}-\mathbf{B}\equiv \mathbf{A}+\left(\mathbf{–}\mathbf{B}\right)$. Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of $\text{–B}$ are the negatives of the components of $\mathbf{B}$. The x– and y-components of the resultant $\mathbf{A}-\text{B = R}$ are thus

$R_x=A_x+\left(–B_x\right)$

and

$R_y=A_y+\left(–B_y\right)$

and the rest of the method outlined above is identical to that for addition. (See Figure 19.10.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.