Components of Vectors

OpenStax; Heath Hatch

29 Components of Vectors

OpenStax and Heath Hatch

Summary

This material is presented both as videos by Heath Hatch and with the relevant OpenStax textbook. Either method is fine. Regardless, by the end of this section, you should be able to:

Apply analytical methods to determine vertical and horizontal component vectors.

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components

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 $\vec{A}$ in Figure 1, we may wish to find which two perpendicular vectors, ${\vec{A}_x}$ and ${\vec{A}_y}$ , add to produce it.

In the given figure a dotted vector A sub x is drawn from the origin along the x axis. From the head of the vector A sub x another vector A sub y is drawn in the upward direction. Their resultant vector A is drawn from the tail of the vector A sub x to the head of the vector A sub y at an angle theta from the x axis. On the graph a vector A, inclined at an angle theta with x axis is shown. Therefore vector A is the sum of the vectors A sub x and A sub y.

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

${\vec{A}_x +\vec{A}_y =\vec{A}.}$

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 ${\vec{A}_x=3\, \mathrm{m}}$ east, ${\vec{A}_y=4\, \mathrm{m}}$ north, and ${\vec{A}=5\, \mathrm{m}}$ north-east, then it is true that the vectors ${\vec{A}_x+\vec{A}_y=\vec{A}}.$ However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

${3\, \mathrm{m}+4\, \mathrm{m}\neq 5\, \mathrm{m}}$

Thus,

${\vec{A}_x+\vec{A}_y\neq\vec{A}}$

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

${\vec{A}_x=A \cos\:\theta}$

and

${\vec{A}_y=A \sin\:\theta}.$

]A dotted vector A sub x whose magnitude is equal to A cosine theta is drawn from the origin along the x axis. From the head of the vector A sub x another vector A sub y whose magnitude is equal to A sine theta is drawn in the upward direction. Their resultant vector A is drawn from the tail of the vector A sub x to the head of the vector A-y at an angle theta from the x axis. Therefore vector A is the sum of the vectors A sub x and A sub y.

Suppose, for example, that $\vec{A}$ is the vector representing the total displacement of the person walking in a city considered in Chapter 3.1 Kinematics in Two Dimensions: An Introduction and Chapter 3.2 Vector Addition and Subtraction: Graphical Methods.

In the given figure a vector A of magnitude ten point three blocks is inclined at an angle twenty nine point one degrees to the positive x axis. The horizontal component A sub x of vector A is equal to A cosine theta which is equal to ten point three blocks multiplied to cosine twenty nine point one degrees which is equal to nine blocks east. Also the vertical component A sub y of vector A is equal to A sin theta is equal to ten point three blocks multiplied to sine twenty nine point one degrees, which is equal to five point zero blocks north.

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

${\vec{A}_x=A \cos\:\theta=(10.3\mathrm{ blocks})(\cos 29.1^o)=9.0\mathrm{ blocks}}$

${\vec{A}_y=A \sin\:\theta=(10.3\mathrm{ blocks})(\sin 29.1^o)=5.0\mathrm{ blocks}.}$

Calculating a Resultant Vector

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

${\vec{A}=\sqrt{A_x\:^2\:+\:A_y\:^2}}$

${\theta=\tan ^{-1}(\vec{A}_y\:/\:\vec{A}_x).}$

Vector A is shown with its horizontal and vertical components A sub x and A sub y respectively. The magnitude of vector A is equal to the square root of A sub x squared plus A sub y squared. The angle theta of the vector A with the x axis is equal to inverse tangent of A sub y over A sub x

Note that the equation ${\vec{A}=\sqrt{A_x\:^2\:+\:A_y\:^2}}$ is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if ${\vec{A}_x}$ and ${\vec{A}_y}$ are 9 and 5 blocks, respectively, then ${\vec{A}=\sqrt{9^2\:+\:5^2}=10.3}$ blocks, again consistent with the example of the person walking in a city. Finally, the direction is ${\theta=\tan ^{-1}(5/9)=29.1^o},$ as before.

DETERMINING VECTORS AND VECTOR COMPONENTS WITH ANALYTICAL METHODS

Equations ${\vec{A}_x=A \cos\:\theta}$ and ${\vec{A}_y=A \sin\:\theta}$ are used to find the perpendicular components of a vector—that is, to go from $\vec{A}$ and ${\theta}$ to ${\vec{A}_x}$ and ${\vec{A}_y}.$ Equations ${\vec{A}=\sqrt{{\vec{A}_x}^2\:+\:{\vec{A}_y}^2}}$ and ${\theta=\tan ^{-1}(\vec{A}_y\:/\:\vec{A}_x)}$ are used to find a vector from its perpendicular components—that is, to go from ${\vec{A}_x}$ and ${\vec{A}_y}$ to $\vec{A}$ and ${\theta}.$ Both processes are crucial to analytical methods of vector addition and subtraction.

Conceptual Questions

1: Give an example of a nonzero vector that has a component of zero.

2: Explain why a vector cannot have a component greater than its own magnitude.

Problems & Exercises

1: Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 13.

A map of northern California with a circle with a radius of one hundred twenty three kilometers centered on San Francisco. Sacramento lies on the circumference of this circle in a direction forty-five degrees north of east from San Francisco. — **Figure 13.**

2: You drive ${7.50\vec{ km}}$ in a straight line in a direction ${15^o}$ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

3: You fly ${32.0\vec{ km}}$ in a straight line in still air in the direction ${35.0^o}$ south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction ${45.0^o}$ south of west and then in a direction ${45.0^o}$ west of north. These are the components of the displacement along a different set of axes—one rotated ${45^o}.$

Glossary

analytical method: the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities

Solutions

Problems & Exercises

1:

North-component 87.0 km, east-component 87.0 km

2:

30.8 m, 35.8 west of north

3:

(a) 18.4 km south, then 26.2 km west

(b) 31.5 km at ${45.0^o}$ south of west, then 5.56 km at ${45.0^o}$ west of north

Homework Problems

8. Resolve a vector into components.

9. Resolving a force into components.

License

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Components of Vectors by OpenStax and Heath Hatch is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.