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Title: Chapter 3 Two-Dimensional Motion and Vectors


1
Chapter 3 Two-Dimensional Motion and Vectors
2
3-1 Introduction to Vectors
  • Vectors indicate direction scalars do not.

3
Scalar
  • A physical quantity that has only magnitude, but
    not direction.
  • Ex speed 30 km/hr

4
Vector
  • A physical quantity that has both magnitude and
    direction.
  • Ex velocity 12 km/hr, North

5
Resultant
  • A vector representing the sum of two or more
    vectors.
  • It starts from where the 1st one begins and the
    last one ends.

6
The book uses symbols to distinguish between them.
7
  • Vectors by boldface type
  • v 3.5 m/s, north
  • Scalars by italicized type
  • v 3.5 m/s

8
Vectors are represented as arrows pointing in the
direction of the vector.
  • The length is a representation of the vectors
    magnitude.

9
When drawing vectors you need to establish a
scale.
  • Ex 4500 m, west.
  • You might chose
  • 1 cm 500 m ? 9 cm
  • or 1 in 1500 m ? 3 in

10
We are going to add vectors by placing them head
to tail.
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Adding vectors along the same line
  • If two or more forces are acting in the same
    direction, we simply add them.

13
Example A force of 30 N is acting toward the
right and a force of 40 N is also applied to the
right, what is the Net Force?
14
30 N 40 N 70 N
  • To draw a representation of the of the forces or
    a vector diagram, we need to establish a scale.
  • Scale 1 inch 10 N

15
So the the first vector would be 3 inches.
  • And the second vector would be 4 inches.

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So the the first vector would be 3 inches.
  • And the second vector would be 4 inches.

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So the the first vector would be 3 inches.
  • And the second vector would be 4 inches.

70 N
18
When vectors are added in opposite directions we
subtract them.
  • Example A 40 N force acting east and a 60 N
    force acting west.

19
Establish a scale and place head to tail.
  • Pick a starting point and draw vector to scale.
  • Scale 1 inch 10 N

20
Establish a scale and place head to tail.
  • Pick a starting point and draw vector to scale.
  • Scale 1 inch 10 N

21
Establish a scale and place head to tail.
  • Pick a starting point and draw vector to scale.
  • Scale 1 inch 10 N

20 N west
22
Vectors acting at right angles
  • 1. Pick a scale.
  • 2. Draw first vector to scale.
  • 3. Draw second vector, starting at the head of
    the first one.

23
  • The resultant is the arrow that is created by
    drawing a line from the tail of the first vector
    to the head of the second vector.

24
Example A planes engine applies a force of 200
N, north and a wind blows the plane with a force
of 40 N east. What is the net force?
25
Scale 2 inches 40 N
26
Scale 2 inches 40 N
27
Scale 2 inches 40 N
  • We can measure the length of the gray arrow in
    inches and convert it to Newtons, using the
    conversion scale.

28
Another way to calculate the net force or
resultant is to use Pythagorean theorem.
  • c2 a2 b2
  • c
  • b
  • a

29
c2 a2 b2
  • c2 2002 402
  • c2 40000 N 1600 N
  • c2 41600 N
  • c square root of 41600 N
  • c 203.96 N
  • Hypotenuse of a triangle is always the longest
    side.

30
Ex 1 A, B, C Add the following vectors.
  • 240 m/s, West 600 m/s, West
  • B) 300 m, East 120 m, West
  • C) 360 m/s, South 270 m/s, West

31
Right Triangles
B
c
a
A
b
32
  1. Side c is called the hypotenuse.
  2. Sides a and b called legs.
  3. The leg across from an angle is called the
    opposite side.
  4. The leg next to an angle is called the adjacent
    side.

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Pythagorean Theorem
35
Q Greek letter Theta
  • This is the symbol used in Trig and Geometry to
    represent an angle.
  • Q 23O

36
B
c
a
A
b
37
B
c
a
A
b
38
B
c
a
A
b
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When given a vector at any angle, use the trig
functions to find the x and y components.
44
Components of Vectors
45
Ex 4 Find the x and y components for 72 m/s _at_
30O (pick scale)
72 m/s
30o
Scale 1 inch 12 m/s
46
Ex 2 Find the x and y components for 72 m/s _at_
30O (pick scale)
72 m/s
y
30o
x
Scale 1 inch 12 m/s
47
x-component
  • Use the cosine function, because you have the
    angle and the hypotenuse and its the side
    adjacent.

48
x-component
  • Use the cosine function, because you have the
    angle and the hypotenuse and its the side
    adjacent.

49
x-component
  • Use the cosine function, because you have the
    angle and the hypotenuse and its the side
    adjacent.

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y-component
  • Use the sine function, because you have the angle
    and the hypotenuse and its the side opposite.

54
y-component
  • Use the sine function, because you have the angle
    and the hypotenuse and its the side opposite.

55
y-component
  • Use the sine function, because you have the angle
    and the hypotenuse and its the side opposite.

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Or since we knew the hypotenuse and x-component,
we could have used the Pythagorean Theorem.
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Ex 5 140 m _at_ 145o. Find the x and y components.
145o
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145o
66
Still use the given angle, even though the
triangle formed doesnt have this angle. This
will ensure the proper sign for the x and y
components.
67
x-component
68
x-component
69
x-component
70
y-component
71
y-component
72
y-component
73
When adding 2 or more vectors at different angles.
  1. Pick a scale for both vectors, draw the1st
    vector.
  2. Draw the 2nd vector to scale, starting where the
    1st one ended.

74
  • 3. Find the x and y components for each vector.
  • 4. Draw the final vector, from where the 1st
    vector started to where the 2nd one ends.

75
  • 5. Add the x components together, then the y
    components.
  • 6. Use the added x and y components to find
    the resultant (Pythagorean Theorem) and its angle
    (Inverse tangent).

76
v1
77
v1
v2
78
v1
v2
q2
v1,y
v1,x
q1
79
v1
v2
v2,y
q2
v2,x
v1,y
v1,x
q1
80
v1
v2
v2,y
vR
q2
v2,x
v1,y
v1,x
q1
81
v2,y
vR
v1,y
v1,x
v2,x
qR
82
vRx v1,x v2,x
vRy v1,y v2,y
vR
qR
83
Ex 4 Find the resultant vector and its angle for
4500 N at 70o 7500 N at 40o.
  • Scale 1 cm 500 N

84
  • v1,x hyp cosQ
  • v1,x 4500cos70
  • v1,x 1539.1 N

85
  • v1,y hyp sinQ
  • v1,y 4500sin70
  • v1,y 4228.6 N

86
  • v2,x hyp cosQ
  • v2,x 7500cos40
  • v2,x 5745.3N

87
  • v2,y hyp sinQ
  • v2,y 7500sin40
  • v2,y 4820.9 N

88
  • vR,x v1,x v2,x
  • vR,x 1539.1 N 5745.3 N
  • vR,x 7284.4 N

89
  • vR,y v1,y v2,y
  • vR,y 4228.6 N 4820.9 N
  • vR,y 9049.5 N

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Ex 5 Find the resultant and its angle for 4 m/s
_at_ 47o east of south 8 m/s _at_ 15o south of west.
98
v1
Q1
99
v1
Q2
v2
100
QR
v1
vR
v2
101
  • v1,x hyp sinQ
  • v1,x 4sin47
  • v1,x 2.93 m/s

102
  • v1,y hyp cosQ
  • v1,y 4cos47
  • v1,y -2.73 m/s

103
  • v2,x hyp cosQ
  • v2,x 8cos15
  • v2,x -7.73 m/s

104
  • v2,y hyp sinQ
  • v2,y 8sin15
  • v2,y -2.07 m/s

105
  • vR,x v1,x v2,x
  • vR,x 2.93 m/s -7.73 m/s
  • vR,x -4.80 m/s

106
  • vR,y v1,y v2,y
  • vR,y -2.73 m/s -2.07 m/s
  • vR,y -4.80 m/s

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