Physics » Two-Dimensional Kinematics » Vector Addition and Subtraction: Analytical Methods

Resolving a Vector Into Perpendicular Components

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 \(\mathbf{A}\) in the figure below, we may wish to find which two perpendicular vectors, \({\mathbf{A}}_{x}\) and \({\mathbf{A}}_{y}\), 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.

The vector \(\mathbf{A}\), with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, \({\mathbf{A}}_{x}\) and \({\mathbf{A}}_{y}\). These vectors form a right triangle. The analytical relationships among these vectors are summarized below.

\({\mathbf{A}}_{x}\) 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:

\({\mathbf{A}}_{x}{\mathbf{ + A}}_{y}\mathbf{ = A}\mathbf{.}\)

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 \({\mathbf{\text{A}}}_{x}=3\text{ m}\) east, \({\mathbf{\text{A}}}_{y}=4\text{ m}\) north, and \(\mathbf{\text{A}}=5\text{ m}\) north-east, then it is true that the vectors \({\mathbf{A}}_{x}{\mathbf{ + A}}_{y}\mathbf{ = A}\). However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

\(3\text{ m} + 4\text{ m} \ne 5\)

Thus,

\({A}_{x}+{A}_{y}\ne A\)

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.

\({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \)

and

\({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{.}\)

]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.

The magnitudes of the vector components \({\mathbf{A}}_{x}\) and \({\mathbf{A}}_{y}\) can be related to the resultant vector \(\mathbf{A}\) and the angle \(\theta \) with trigonometric identities. Here we see that \({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \).

Suppose, for example, that \(\mathbf{A}\) is the vector representing the total displacement of the person walking in a city considered in Two-Dimensional Motion: Walking in a City and Vector Addition: Head-to-Tail Method.

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.

We can use the relationships \({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) to determine the magnitude of the horizontal and vertical component vectors in this example.

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

\({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{cos}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{9.0 blocks}\)

\({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{sin}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{5.0 blocks}\text{.}\)

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