2.2 Coordinate systems and components of a vector (Page 3/15)

University physics volume 1 Page 3 / 15

When we know the scalar components $A_{x}$ and $A_{y}$ of a vector $\vec{A}$ , we can find its magnitude A and its direction angle $θ_{A}$ . The direction angle —or direction, for short—is the angle the vector forms with the positive direction on the x -axis. The angle $θ_{A}$ is measured in the counterclockwise direction from the + x -axis to the vector ( [link] ). Because the lengths A , $A_{x}$ , and $A_{y}$ form a right triangle, they are related by the Pythagorean theorem:

A^{2} = A_{x}^{2} + A_{y}^{2} \Leftrightarrow A = \sqrt{A_{x}^{2} + A_{y}^{2}} .

This equation works even if the scalar components of a vector are negative. The direction angle $θ_{A}$ of a vector is defined via the tangent function of angle $θ_{A}$ in the triangle shown in [link] :

tan θ_{A} = \frac{A_{y}}{A_{x}} \Rightarrow θ_{A} = {tan}^{−1} (\frac{A_{y}}{A_{x}}) .

Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A. — For vector $\vec{A}$ , its magnitude A and its direction angle $θ_{A}$ are related to the magnitudes of its scalar components because A, $A_{x}$ , and $A_{y}$ form a right triangle.

When the vector lies either in the first quadrant or in the fourth quadrant, where component $A_{x}$ is positive ( [link] ), the angle $θ$ in [link] ) is identical to the direction angle $θ_{A}$ . For vectors in the fourth quadrant, angle $θ$ is negative, which means that for these vectors, direction angle $θ_{A}$ is measured clockwise from the positive x -axis. Similarly, for vectors in the second quadrant, angle $θ$ is negative. When the vector lies in either the second or third quadrant, where component $A_{x}$ is negative, the direction angle is $θ_{A} = θ + 180 °$ ( [link] ).

Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees. — Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I)have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is $θ_{A} = θ + 180 °$ .

Magnitude and direction of the displacement vector

You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What is the magnitude and direction of the displacement vector of the pointer?

Strategy

In [link] , we found the displacement vector

\vec{D}

of the mouse pointer (see [link] ). We identify its scalar components

D_{x} = −4.0 cm

and

D_{y} = + 2.9 cm

and substitute into [link] and [link] to find the magnitude D and direction

θ_{D}

, respectively.

Solution

The magnitude of vector

\vec{D}

D = \sqrt{D_{x}^{2} + D_{y}^{2}} = \sqrt{{(−4.0 cm)}^{2} + {(2.9 cm)}^{2}} = \sqrt{{(4.0)}^{2} + {(2.9)}^{2}} cm = 4.9 cm .

The direction angle is

tan θ = \frac{D_{y}}{D_{x}} = \frac{+ 2.9 cm}{−4.0 cm} = −0.725 \Rightarrow θ = {tan}^{−1} (−0.725) = −35.9 ° .

Vector $\vec{D}$ lies in the second quadrant, so its direction angle is

θ_{D} = θ + 180 ° = −35.9 ° + 180 ° = 144.1 ° .

Got questions? Get instant answers now!

Check Your Understanding If the displacement vector of a blue fly walking on a sheet of graph paper is $\vec{D} = (−5.00 \hat{i} - 3.00 \hat{j}) cm$ , find its magnitude and direction.

5.83 cm, $211 °$

Got questions? Get instant answers now!

In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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