 # Chapter 1. Vectors and Coordinate Systems

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Chapter 1. Vectors and Coordinate Systems
Our universe has three dimensions, so some quantities also need a direction for a full description. For example, wind has both a speed and a direction; hence the motion of the wind is described by a vector. Chapter Goal: To learn how vectors are represented and used.

Student Learning Objectives – Ch. 1
• To understand the basic properties of vectors. • To add and subtract vectors both graphically and using components. • To be able to resolve a vector into its components and to reassemble vector components into a magnitude and a direction. • To work with tilted coordinate systems.

A scalar quantity is one that can be completely described by its magnitude (size).
Examples include temperature, energy, and speed Some scalars can be described with negative numbers (1st two examples, above, but not the third). In that case the negative sign means the quantity is less than a positive quantity. +30°C -30° C

A vector quantity has both magnitude and direction.
Examples include velocity, force, and displacement Vectors are never negative. A vector has a magnitude, which is an absolute value and a direction, which is usually given as an angle. Vector components may be described with negative signs. In that case the negative sign never means less than. It means “left” or “down” on a graph, and west or south on a map. The y component of velocity is – 40 m/s. The y component of velocity is + 40 m/s. Both cars are going at the exact same speed!

Arrows are used to represent vectors
Arrows are used to represent vectors. The direction of the arrow relative to some reference point (north or + x axis) gives the direction of the vector. The length of a vector arrow is proportional to the magnitude of the vector. 8 m 4 m

This is an example of a displacement vector with magnitude of 2 km and direction 30° north of east

Graphical Addition: Put the tail of one vector after the head of the other. The resultant vector is an arrow that starts where the first vector starts and ends where the second vector ends.

This is easy if the two vectors are co-linear, especially if they point in the same direction. These vectors add just like scalars. 3 m 5 m 8 m

If the two vectors are not in the same direction, then the graphical process is the same, but the magnitudes of the vectors can no longer be added like scalars. In this case, vector A has a magnitude of 27.5 cm due east and vector B has a magnitude of 12.5 cm in a direction 55° north of west. In order to get a value for the resultant vector, one needs a ruler and a protractor.

Multiplication by a scalar
Multiplication by a scalar serves to stretch or shrink the vector by a factor equal to that of the scalar.

1.6 Multiplication by -1 allows subtraction
When a vector is multiplied by -1, the magnitude of the vector remains the same, but the direction of the vector is reversed (i.e. the angle is increased by 180°).

Note that A + B is not the reverse of A – B.

Multiplication by a negative scalar
Vector A has a magnitude of +2 and a direction of 30° from the positive x axis. Vector -3A has: A. Magnitude -6; direction - 30° B. Magnitude -6, direction 210° C. Magnitude 6, direction 210° D. Magnitude 6, direction 30°

1.4 Trigonometry Review “SOH-CAH-TOA”

(Pythagorean theorem)
1.4 Trigonometry Review (Pythagorean theorem)

1.7 The Components of a Vector

1.7 The Components of a Vector

1.7 The Components of a Vector

Help! My calculator won’t tell me when a scalar component of a vector is negative if I use these acute angles as given. And inverse tangent doesn’t work right in quadrants other than the 1st . What do I do? 1. You can convert all angles from acute to their proper values. However, this doesn’t solve the problem for inverse tangent. 2. Better, use absolute values for all components. Draw the picture and determine which components should be negative and then manually insert the negative sign when needed. 3 Remember: “All Students Take Calculus” – whether they want to or not! Use this in conjunction with inverse tangent to determine the appropriate quadrant for θ. S A C T

Finding the components of an acceleration vector

Finding the components of an acceleration vector

If the angle is given with respect to the y-axis, “opposite” and “adjacent” are switched:
C1x= - C1sin φ (x-comp. is opposite); negative by inspection C1y = - C1 cos φ (y-comp. is adjacent); negative by inspection φ = tan-1 (|C1x |/|C1y|). Answer D

In what quadrant is vector , for purposes of finding components and θ?
1st 2nd 3rd 4th Both 2nd and 4th Answer D

What are the x- and y-components Cx and Cy of vector ?
Cx = 1 cm, Cy = –1 cm Cx = –3 cm, Cy = 1 cm Cx = –2 cm, Cy = 1 cm Cx = –4 cm, Cy = 2 cm Cx = –3 cm, Cy = –1 cm Answer D

Angle φ that specifies the direction of is given by
tan–1(F1y /F1x) tan–1(|F1y |/|F1x|) tan–1(F1x /F1y) tan–1(|F1x |/|F1y|) φ Answer D

Tilted axes Often it is convenient to use a tilted axis system (to represent an object on an incline for example). The axes stay perpendicular to each other. The components correspond to axes, not to “horizontal and vertical” so they are also tilted. The gravitational force always points in the “true vertical” direction.

Tilted axes Often it is convenient to use a tilted axis system (to represent an object on an incline for example). The axes are always perpendicular to each other. The components correspond to axes, not to “horizontal and vertical” so they are also tilted. The gravitational force always points in the “true vertical” direction, regardless of axis orientation.

To find components of a vector parallel and perpendicular to an inclined surface, as shown in (a):
Draw a set of coordinate axes so that one axis (usually x) is parallel and the other is perpendicular to the surface. Place the tail of the vector at the origin of the axes. Draw components parallel to each axis as shown in (b). Ax = A cos Ay = A sin Note that is defined relative to the tilted x-axis and not to “horizontal”

Finding the force perpendicular to a surface
A horizontal force, F, of 10 Newtons is applied to an object on a surface inclined at a 20° angle (object not pictured). Draw the component of the force vector which is perpendicular to the surface and find its magnitude. Inclined surface

Finding the force perpendicular to a surface
In this case the coordinate axes are drawn right on the inclined surface. The y- axis is perpendicular to the inclined surface.

Example: EOC #47 A foot ball player runs the pattern shown by the 3 displacement vectors A, B and C. The magnitudes of these vectors are A = 5.00 m, B = 15.0 m and C = 18.0 m. Find the magnitude and direction of the resultant vector.

Example: EOC #52 Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the location of the first team as 38 km away, 19° north of west, and the second team as 29 km away, 35° east of north. When the first team uses its GPS to check the position of the second team, what does the GPS give for the distance between the teams and direction,θ, measured from due east? q Second team Base camp E (x) N (y) W 19° 35° A B = A + C C First team

Graph of a line The vertical axis value comes first when naming graphs: i.e. this is a position vs time. The standard format is: y = mx + b, where m is the slope and b is the y-intercept. Note that the normal variable for position is x and the quantity on the horizontal axis is time (t) in which case the equation format is: x = mt + b

Graph of a line m = “rise over run”, in this case Δx/Δt, where:
Δx = x2-x1 and Δt = t2-t1 for any two points on the line. Note that the slope has units, in this case: Δx/Δt = 8m/2s = +4 m/s. In this case the slope has units of velocity and represents the velocity of the object being graphed.

Graph of a line Find the equation of a line if you know 2 points:
use Δx/Δt to find m substitute the values of one of the points into the equation to find b: x1 = mt1 + b b = x1 – mt1 or b = x2 – mt2

Equation of a line What is the equation of this velocity vs time graph? What are the units of the slope? v = 6 (m/s2) t + 5m/s or v = 6t + 5