Download Chapter 3 Kinematics in Two Dimensions

Chapter 3
Kinematics in Two Dimensions
3.1 Trigonometry
3.1 Trigonometry
ho
sin ! =
h
ha
cos! =
h
ho
tan ! =
ha
3.1 Trigonometry
ho
tan ! =
ha
ho
tan 50 =
67.2m
!
ho = tan 50 ( 67.2m ) = 80.0m
!
3.1 Trigonometry
" ho %
! = sin $ '
# h&
!1
" ha %
! = cos $ '
# h&
!1
" ho %
! = tan $ '
# ha &
!1
3.1 Trigonometry
3.1 Trigonometry
Pythagorean theorem:
3.2 Scalars and Vectors
!
!
Directions of vectors F1 and F2 appear to be the same.
!
Vector F1 , (bold + arrow over it)
!# magnitude = F (italics)
1
has 2 parts: "
#$ direction = up & to the right
!
Vector F2 , (bold + arrow over it)
!# magnitude = F (italics)
2
has 2 parts: "
#$ direction = up & to the right
F1 = 4 lb
!
F1
F2 = 8 lb
!
F2
!
!
Directions of vectors F1 and F2 appear to be the same.
3.2 Scalars and Vectors
!
Displacement vector S
!# S = 2 km (magnitude) , and
has 2 parts: "
#$ 30° North of East (direction)

S
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Often it is necessary to add one vector to another.
The “Resultant Vector”
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Two displacement vectors
with magnitudes A and B
A = 5m
B = 3m
Adding together
Resultant
A = 5m
B = 3m
R = 8m
Vector addition is "commutative" - order of the addition doesn't matter
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Two displacement vectors
!
A = 6.00m, East

B = 2.00m,
North
Resultant
Adding vectors together
Graphical “head to tail” method
But only a visual aid
Not a substitute for math
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Graphical Addition
Two displacement vectors
!
A = 6.00m, East

B = 2.00m,
North
Graphical addition is only qualitative
YOU need quantitative addition!
R
B = 2.00m
!
A = 6.00m
To do this addition of vectors requires trigonometry
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Apply Pythagorean Theroem
R = ( 2.00m ) + ( 6.00m )
2
2
R=
2
( 2.00m ) + (6.00m )
2
2
= 6.32m
R = 6.32m
2.00m
!
6.00m
3.2 Scalars and Vectors
(Vector Addition and Subtraction)
Use trigonometry to determine the angle
tangent (angle) =
R = 6.32m
opposite side
adjacent side
2.00m
!
6.00m
Also
θ = sin −1 ( 2.00 6.32 ) = 18.4
Also
θ = cos −1 ( 6.00 6.32 ) = 18.4
Clicker Question 3.1
 

Three vectors are given: A, B, and C.
!
!
!
C
B
A
!
A
  
Which of the choices below is equal to A + B + C ?
A)
B)
C)
! ! !
C+B+A
! ! !
B+A+C
! ! !
C+A+B
D)
None of the above
E)
All of the above
  
A+B+C

B
!
C
Clicker Question 3.1
 

Three vectors are given: A, B, and C.
!
!
!
C
B
A
  
A+B+C
!
A
  
Which of the choices below is equal to A + B + C ?
!
!
B
! ! !
C
A)
C+B+A
B)
C)
D)
E)
! ! !
B+A+C
! ! !
C+A+B
None of the above
All of the above
!
B
!
A
!
A
!
A
Red vector
is the same as
  
A+B+C
!
C
!
C
!
C

B
!
B
3.2 Scalars and Vectors
  
C= A+B
!
B
!
A

C
  
A = C−B

–B
(Vector Addition and Subtraction)
When a vector is multiplied
by -1, the magnitude of the
vector remains the same, but
the direction of the vector is
reversed.
3.2 Scalars and Vectors
(Vector Addition and Subtraction)


Add vectors A and B
!
B
! !
A+B
!
A
 
Now you are asked to find A − B
Instead of trying to do Vector Subtraction


add to vector A the negative of the vector B


Subtracting vector B from vector A
!
A

−B
 
A−B
!
B

−B


Adding vector A to vector − B
3.2 Vector Addition and Subtraction (The Components of a Vector)
!


x and y are called the x − component vector

and the y − component vector of r.
3.2 Vector Addition and Subtraction
+y
(The Components of a Vector)
+y

A

Ay
!

Ax

A
+x
!

Ax

Ay
+x
!
The vector components of A are two perpendicular
!
!
vectors A x and A y that are parallel to the x and y axes,
! !
!
and add together vectorially so that A = A x + A y .
3.2 Vector Addition and Subtraction
(The Components of a Vector)
It is often easier to work with the scalar components
of the vectors, rather than the component vectors themselves.
+y
+y
" j-hat"
Ax î
î
"i-hat"
" j-hat"
î
ĵ
î and ĵ are "unit vectors", magnitude is 1.

vector A expressed in terms of
!
its scalar components: A = Ax î + Ay ĵ
Ay ĵ
ĵ
"i-hat"
+x
+x
Ax and Ay are the scalar

components of the vector, A.
3.2 Vector Addition and Subtraction
(The Components of a Vector)
Example
A displacement vector has a magnitude of 175 m and points at
an angle of 50.0 degrees relative to the x axis. Find the x and y
components of this vector.
sin ! = y r
Given magnitude
r = 175 m
α

r
(
)
y = r sin ! = (175 m ) sin50.0! = 134 m

y
90°
50.0°

x
!
y-component of the vector r
!
vector x has magnitude x

vector y has magnitude y
!
x-component of the vector r
(
)
x = r cos! = (175 m ) cos50.0! = 112 m
!
r = xî + yĵ
= (112 m ) î + (134 m ) ĵ
Check: r = (112)2 + (134)2 m = 175m
Clicker Question 3.2

A = Ax î + Ay ĵ, where Ax = 3m, and Ay = 4m.

What is the magnitude of the vector A ?
A)
7m
B)
6m
C)
5m
D)
7m
E)
25m
Clicker Question 3.2

A = Ax î + Ay ĵ, where Ax = 3m, and Ay = 4m.

What is the magnitude of the vector A ?
A)
B)
C)
7m

A
(4m) ĵ
6m
5m
(3m) î

A = Ax î + Ay ĵ
D)
7m

magnitude of A, A =
E)
25m
=
(
A +A =
2
x
2
y
)
(3m ) + ( 4m )
2
9 + 16 m = 25 m = 5m
2
3.2 Vector Addition and Subtraction (using Components)
+y

B

A
θ1
θ2

Vector A has magnitude A and angle θ 1

Vector B has magnitude B and angle θ 2
What is the magnitude and direction
  
of vector C = A + B ?
+x
THIS IS A BIG PROBLEM

What is the magnitude of C,
and what angle θ does it make
relative to x-axis ?
Graphically no PROBLEM
+y
  
C = A+B

A
θ1

B
θ2
+x
3.2 Vector Addition and Subtraction (using Components)

What is the magnitude of C,
and what angle θ does it make
THIS IS A BIG PROBLEM
relative to x-axis ?
The only way to solve this problem
is to use vector components!
+y
Vector A
+y
Vector B

B

A
θ1
Ax î
Ay ĵ

B

A
+x
θ2
Bx î
By ĵ
+x
3.2 Vector Addition and Subtraction (using Components)


Get the components of the vectors A and B.


B : magnitude B and angle θ 2
A : magnitude A and angle θ 1
Ax = Acosθ 1
Ay = Asin θ 1
+y
B x = Bcosθ 2
By = Bsin θ 2
Vector A
+y
Vector B

B

A
θ1
Ax î
Ay ĵ

B

A
+x
θ2
Bx î
By ĵ
+x
3.2 Vector Addition and Subtraction (using Components)



Get components of vector C from components of A and B .
Ax = Acosθ 1
Ay = Asin θ 1
B x = Bcosθ 2
By = Bsin θ 2
  
C= A+B
By ĵ
θ
Ax î
Ay ĵ
Bx î
Cx = Ax + B x
C y = Ay + B y

C
C y ĵ
θ
Cx î
3.2 Vector Addition and Subtraction (using Components)

What is the magnitude of C,
and what angle θ does it make
relative to x-axis ?
Cx = Ax + B x
C y = Ay + B y
PROBLEM IS SOLVED

magnitude of C : C = Cx2 + C y2

C
C y ĵ
θ
Cx î
Cy
Angle θ : tan θ =
;
Cx
θ = tan −1 (C y Cx )
3.2 Addition of Vectors by Means of Components
Summary of adding two vectors together
+y

B

A
θ1
θ2

Vector A has magnitude A and angle θ 1

Vector B has magnitude B and angle θ 2
  
What is the vector C = A + B ?
+x


1) Determine components of vectors A and B : Ax , Ay and Bx , By
2) Add x-components to find Cx = Ax + Bx
3) Add y-components to find C y = Ay + By

4) Determine the magnitude and angle of vector C
magnitude C = Cx2 + C y2 ; θ = tan −1 (C y Cx )