5.14: Vector Addition
Vector Addition
Adding Vectors in Two Dimensions
In the following image, vectors \begin{align*}A\end{align*} and \begin{align*}B\end{align*} represent the two displacements of a person who walked 90. m east and then 50. m north. We want to add these two vectors to get the vector sum of the two movements.
The graphical process for adding vectors in two dimensions is to place the tail of the second vector on the arrow head of the first vector as shown at left.
The sum of the two vectors is the vector that begins at the origin of the first vector and goes to the ending of the second vector, as shown below.
If we are using totally graphic means of adding these vectors, the magnitude of the sum would be determined by measuring the length of the sum vector and comparing it to the original standard. We would also use a compass to measure the angle of the summation vector.
If we are using calculation means, we can determine the inverse tan of 50 units divided by 90 units and get the angle of 29° north of east. The length of the sum vector can also be determined mathematically by the Pythagorean theorem, \begin{align*}a^2 + b^2 = c^2\end{align*}. In this case, the length of the hypotenuse would be the square root of (8100 + 2500) or 103 units.
If three or four vectors are to be added by graphical means, we would continue to place each new vector head to toe with the vectors to be added until all the vectors were in the coordinate system and then the sum vector would be the vector goes from the origin of the first vector to the arrowhead of the last vector. The magnitude and direction of the sum vector would be measured.
Mathematical Methods of Vector Addition
We can add vectors mathematically using trig functions, the law of cosines, or the Pythagorean theorem.
If the vectors to be added are at right angles to each other, we would assign them to the sides of a right triangle and calculate the sum as the hypotenuse of the right triangle. We would also calculate the direction of the sum vector by using an inverse sin or some other trig function.
Suppose, however, that we wish to add two vectors that are not at right angles to each other. Let’s consider the vectors in the following images.
The two vectors we are to add is a force of 65 N at 30° north of east and a force of 35 N at 60° north of west.
We know that vectors in the same dimension can be added by regular arithmetic. Therefore, we can resolve each of these vectors into components that lay on the axes – pictured below.
We can resolve each of the vectors into two components. The components are on the axes lines. The resolution of vectors creates each vector into a component on the northsouth axis and a component on the eastwest axis.
We can now mathematically determine the magnitude of the components and add then arithmetically because they are in the same dimension. Once we have added the components, we will once again have only two vectors that are perpendicular to each other and can be the legs of a right triangle.
The eastwest component of the first vector is (65 N)(cos 30°) = (65 N)(0.866) = 56.3 N north
The northsouth component of the first vector is (65 N)(sin 30°) = (65 N)(0.500) = 32.5 N north
The eastwest component of the second vector is (35 N)(cos 60°) = (35 N)(0.500) = 17.5 N west
The northsouth component of the second vector is (35 N)(sin 60°) = (35 N)(0.866) = 30.3 N north
The sum of the two eastwest components is 56.3 N  17.5 N = 38.8 N east
The sum of the two northsouth components is 32.5 N + 30.3 N = 62.8 N north
We can now consider those two vectors to be the sides of a right triangle and use the Pythagorean Theorem to find the length of the hypotenuse and use a trig function to find its direction.
\begin{align*}c=\sqrt{38.8^2+62.8^2}=74 \ \text{N}\end{align*}
\begin{align*}\sin \ x=\frac{62.8}{74} \ \text{so} \ x=\sin^{1} 0.84 \ \text{so} \ x=58^\circ\end{align*}
The direction of the sum vector is 74 N at 58° north of east.
Perpendicular vectors have no components in the other direction. For example, if a boat is floating down a river due south, and you are paddling the boat due east, the eastward vector has no component in the northsouth direction and therefore, has no effect on the northsouth motion. If the boat is floating down the river at 5 miles/hour south and you paddle the boat eastward at 5 miles/hour, the boat continues to float southward at 5 miles/hour. The eastward motion has absolutely no effect on the southward motion. Perpendicular vectors have NO effect on each other.
Example Problem: A motorboat heads due east at 16 m/s across a river that flows due north at 9.0 m/s.
(a) What is the resultant velocity of the boat?
(b) If the river is 135 m wide, how long does it take the boat to reach the other side?
(c) When the boat reaches the other side, how far downstream will it be?
Solution:
Sketch:
(a) Since the two motions are perpendicular to each other, they can be assigned to the legs of a right triangle and the hypotenuse (resultant) calculated.
\begin{align*}c=\sqrt{a^2+b^2}=\sqrt{(16 \ \text{m/s})^2+(9.0 \ \text{m/s})^2}=18 \ \text{m/s}\end{align*}
\begin{align*}\sin \theta=\frac{9.0}{18}=0.500 \ \text{and therefore} \ \theta=30^\circ\end{align*}
The resultant is 18 m/s at 30° north of east.
(b) The boat is traveling across the river at 16 m/s due to the motor. The current is perpendicular and therefore has no effect on the speed across the river. The time required for the trip can be determined by dividing the distance by the velocity.
\begin{align*}t=\frac{d}{v}=\frac{135 \ \text{m}}{16 \ \text{m/s}}=8.4 \ \text{s}\end{align*}
(c) The boat is traveling across the river for 8.4 seconds and therefore, it is also traveling downstream for 8.4 seconds. We can determine the distance downstream the boat will travel by multiplying the speed downstream by the time of the trip.
\begin{align*}d_{\text{downstream}} = (v_{\text{downstream}})(t) = (9.0 \ \text{m/s})(8.4 \ \text{s}) = 76 \ \text{m}\end{align*}
Summary
 Vectors can be added mathematically using geometry and trigonometry.
 Vectors that are perpendicular to each other have no effect on each other.
 Vector addition can be accomplished by resolving into axial components those vectors that are to be added, adding up the axial components, and then combining the two axial components.
Practice
A video demonstrating the component method of vector addition.
http://www.youtube.com/watch?v=nFDzRWw08Ew
Review
 A hiker walks 11 km due north from camp and then turns and walks 11 km due east.
 What is the total distance walked by the hiker?
 What is the displacement (on a straight line) of the hiker from the camp?
 While flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind. What is the plane’s resultant velocity?
 Two students push a heavy crate across the floor. John pushes with a force of 185 N due east and Joan pushes with a force of 165 N at 30° north of east. What is the resultant force on the crate?
 An airplane flying due north at 90. km/h is being blown due west at 50. km/h. What is the resultant velocity of the plane?
Image Attributions
Description
Learning Objectives
 Describe the independence of perpendicular vectors.
 Resolve vectors into axial components.
 Define resultant.
 Add vectors using geometric and trigonometric methods.
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Date Created:
Sep 26, 2012Last Modified:
Feb 26, 2015Vocabulary

axial component: A component situated in or on an axis.
 resolution of vectors: Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components) that lie on the axes (one horizontal and one vertical). The process of identifying these two components is known as the resolution of the vector.