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Chapter 3. Vectors and Coordinate SystemsPowerPoint Presentation

Chapter 3. Vectors and Coordinate Systems

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Chapter 3. 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.

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Chapter 3. 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.

- 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. 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 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.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 head of the other. The 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 head of the other. The 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.

Vector Addition Problem head of the other. The

- Which figure shows A1 + A2 + A3?

Which figure shows head of the other. The ?

Multiplication by a scalar head of the other. The

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

1.6 head of the other. The 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°).

1.6 head of the other. The Vector Addition and Subtraction

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

Multiplication by a negative scalar head of the other. The

- 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°

Multiplication by a negative scalar head of the other. The

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°

The magnitude is an absolute value

To reverse a vector, add 180°.

1.7 head of the other. The The Components of a Vector

1.7 head of the other. The The Components of a Vector

1.7 head of the other. The 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

In what quadrant is vector 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 1 , for purposes of finding components and θ?

1st

2nd

3rd

4th

Both 2nd and 4th

In what quadrant is vector 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 1 , for purposes of finding components and θ?

1st

2nd

3rd

4th

Both 2nd and 4th

What are the 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 1x- 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

What are the 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 1x- 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

If 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 1the 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 = -C1cos φ (y-comp. is adjacent); negative by inspection

φ = tan-1 (|C1x |/|C1y|).

Angle 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 1φthat specifies the direction of is given by

tan–1(F1y /F1x)

tan–1(|F1y|/|F1x|)

tan–1(F1x /F1y)

tan–1(|F1x |/|F1y|)

φ

Angle 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 1φthat specifies the direction of is given by

tan–1(F1y /F1x)

tan–1(|F1y|/|F1x|)

tan–1(F1x /F1y)

tan–1(|F1x |/|F1y|)

φ

Tilted axes 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 1

- 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 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 1

- 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”

EOC #38 an inclined surface, as shown in (a):

A pine cone falls straight down from a pine tree growing on a 20° slope the pine cone hits the gorund with a speed of 10 m/s. What is the component of the pine cone’s velocity

- Parallel to the ground
- Perpendicular to the ground?

EOC #38 an inclined surface, as shown in (a):

A pine cone falls straight down from a pine tree growing on a 20° slope the pine cone hits the gorund with a speed of 10 m/s. What is the component of the pine cone’s velocity

- Parallel to the ground
-3.4 m/s

- Perpendicular to the ground?
-9.4 m/s

More football an inclined surface, as shown in (a):

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.

More football an inclined surface, as shown in (a):

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.

Ans: 30.2 m at 10.2° below the +x axis

Second team an inclined surface, as shown in (a):

N (y)

First team

C

q

B = A + C

35°

A

19°

W

E (x)

Base camp

Field workTwo 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?

Second team an inclined surface, as shown in (a):

N (y)

First team

C

q

B = A + C

35°

A

19°

W

E (x)

Base camp

Field workTwo 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?

53.8 km, 12.2 ° from east

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