Web Sites
http://hyperphysics.phyastr.gsu.edu/hbase/vect.html#vec1
Physics I Unit 4
VECTORS
&
Motion in TWO
Dimensions
• Objectives:
Calculate and / or measure
Velocities Vectors using
graphical means
Calculate and / or measure
Velocities Vectors using
algebraic means
Do Now
 Page
117 #1
Unit 4 Lesson 1
Homework
 Pg 121 – 125
Practice Problems:
1-9 ODD:
 Pg 141 – 142
Problems:
80 – 89 all

80 – 88 EVEN

IN CLASS:
Page 121 Example 1
Vectors and Scalars
A vector has magnitude as
well as direction.
Some vector quantities:
displacement, velocity,
force, momentum
A scalar has only a
magnitude.
Some scalar quantities:
mass, time, temperature
Addition of Vectors – Graphical Methods
8 km + 6 km = 14 km East
For vectors in one
dimension, simple
addition and subtraction
are all that is needed.
8 km - 6 km = 2 km East
You do need to be
careful about the signs,
as the figure indicates.
Addition of Vectors – Graphical Methods
Even if the vectors are not at right angles, they can be added
graphically by using the “Head
to Tail” method:
1. Draw V1 & V2 to scale.
2. Place tail of V2 at tip of V1
3. Draw arrow from tail of V1 to tip of V2
This arrow is the resultant V
(measure length and the angle it makes with the x-axis)
Same for any number of vectors involved.
Addition of Vectors – Graphical Methods
The parallelogram method may also be used;
1. Draw V1 & V2 to scale
from common origin.
2. Construct parallelogram
using V1 & V2 as 2 of the 4
sides.
Resultant V = diagonal of
parallelogram from
common origin (measure
length and the angle it
makes with the x-axis)
Addition of Vectors – Graphical Methods
If the motion is in two dimensions, the situation is somewhat
more complicated. Here, the actual travel paths are at right
angles to one another; we can find the displacement by using
the Pythagorean Theorem.
Objectives:
Word Problems with VECTORS
Do Now
Unit 4 Lesson 2
Homework
• Pg 141 – 142 Problems:
80 – 89 all
• Page 117 #2
• Page 121 #4
IN CLASS:
Pg 125 Section Review Problems: 11 -15 ALL
http://hyperphysics.phyastr.gsu.edu/hbase/vect.html#vec1
Adding Vectors by Components
Unit 4 Lesson 2
Any vector can be expressed as the
sum of two other vectors, which are
called its components. Usually the
other vectors are chosen so that they
are perpendicular to each other.
Adding Vectors by Components
If the components are
NOT perpendicular,
they can be found using
trigonometric functions.
Vx = V Cosθ
1) __________
2) ___________
3) ___________
+
Total Vx
Vy = V Sin θ
_________
_________
_________
Total Vy
Objectives:
Unit 4 Lesson 3 Nov 2
Homework
-Determine Equilibrium Forces
 Page 133-135 Practice
-Determine the motion of an
Problems: 33- 41 odd
object on an inclined plane
with and without friction
IN CLASS:
Pg 133 Example 5
Pg 134 Example 6
Do Now
 Page
128 #17
 Next SLIDE(s)
DO NOW 3 Vector Problem Graphically Analytically
Find the resultant vector
(Magnitude and Direction of the
THREE FORCE vectors acting
on the Object at the origin
Unit 4 Lesson 3
Unit 4 Lesson 3
3 Vector Problem
•
•
•
•
•
X Components Vx = V Cosθ
Ax = 44.0 * COS (28.0) = 38.85
Bx = 26.5 *COS (124) = - 14.82
Cx = 31.0 * COS (270.0) = 0.00
X total = 24.031
V = √{[24.031]2 + [11.63]2}
V = √{[24.031]2 + [11.63]2}
V = √{[712.748]} = 26.6973 m/s
= 26.7 m/s
Solving the components
•
•
•
•
•
Y Components Vy = V SINθ
Ay = 44.0 * SIN(28.0) = 20.66
By = 26.5 *SIN(124) = 21.97
Cy = 31.0 * SIN(270) = - 31.0
Y total = 11.63
Θ = Tan-1 {11.63/24.031}
Θ = Tan-1 {0.483958}
Θ = 25.8225 degrees
Θ = 25.8 degrees
Objectives:
Unit 4 Lesson 5 NOV 5
Two Dimensions in Motion
Do Now





Homework
 page 135
 #’s 38 -45 ALL
If F┴ = Fapp Cosθ
If F ⁄⁄ = Fapp SINθ
And
If Fg = 2.55 Kg * 9.8 m/s2 = 25 N
Θ = 15.0 °
 What
is the Velocity after 10
seconds of the object rolling
down a 15.0 degree ramp?
IN CLASS:
A 62 kg skier is going
down a 37 degree
slope. The coefficient
of friction is 0.15. How
fast is the skier going
after 5.0 sec?
Unit 4 Lesson 5A NOV 6
Grade quiz Now
solution
 A 62 kg skier is going  F = F = F Cosθ
N
┴
g

down a 37 degree
 = mg(Cosθ)
slope. The
coefficient of friction  = 62*9.8*(Cos37) = 485.2509N
is 0.15.
 If F ⁄⁄ = Fg SINθ
How fast is the skier
going after 5.0 sec?  = mg(Sinθ)


= 62*9.8*(Sin37) = 365.6628N
Vf = ______________  F = F
net
app - Ffric
 Fnet =
F ⁄⁄ - μmg(Cosθ)

Fnet = 365.6628N – μ 485.2509N
Fnet = 365.6628N – (0.15) 485.2509N
Fnet = 298.8752 N = ma
a = Fnet /m = 298.8752 /62 = 4.7238 m/s2
Vf = Vi + at = (4.7238 m/s2 )* (5.0)

Vf = 23.6190 = 2.4 X 101 m/s




Unit 4 Lesson 6 Nov 6
Objectives:
Homework
 Page 159 # 23, 25
Multiple Vectors
Do Now







Vector 1 = 42.5N @ 56.0 Degrees
Vector 2 = 75.1 N @ 245.0 deg
Vector 3 = 10.5 N @ 315.0 deg
Mass = 15.5 Kg
Find F net= ___________
Find Vel (5 sec)= _________
Find Distance (5 sec) = _________
IN CLASS: Solve**
Vx = Vi COS 
Vx Total =
Vy = Vi SIN 
Vy Total =
V resultant = √{[Vx Total]2 + [Vy Total]2}
Unit 4 Lesson 7 Nov 8
Objectives:
Adding Vectors ALGEBRAICALLY
Forces in Equilibrium
Homework


Do Now 1



A vector has a Magnitude
of 10.0 Newton’s, 30.0
from the horizontal.
Find the “X” component
of the vector :
Find the “Y” component
of the vector :

Page 142
#’s 88, 90, 92, 94, 95, 96
97 HONORS
DO NOW 2:
What is the actual
magnitude and the
direction of a boat if it is
heading due NORTH (0
degrees) at 3.0 m/s on a
river that is flowing due
EAST (90 degrees) at 2.0
m/s and a wind that is
blowing @ 5.0 m/s to the
North East (30 degrees)?
Unit 4 Lesson 7 Nov 9
IN CLASS 1:
Romac hangs his 50.0 Kg Repair sign
from two wires that make a 90.0 degree
angle between themselves and a 45.00
degree angle from the horizontal.
What is the tension on each wire?
Equilibrium
90.0
45.00
50.0*9.801 = 490.0 N
490 / 2 = 245.0
Sin (45) = 174.94785 / Hyp
Hyp = 174.94785 / Sin (45) =
DO NOW: What is
the Frictional Force
required to keep a
10.0 Kg block from
sliding down a 25
degree incline?
Objectives:
Exam Review
Do Now
Unit 4 Lesson #
Homework
 Putting it all
together
IN CLASS: