### Do now

```10/11 do now
• 2nd and 3rd period: 3-1 diagram skills
• 9th period: page 113 #1-13
Homework – posted online
• 10/11/13 Homework – due Tuesday 10/15
• Read text book page 88-97 and write an essay to
1. What is the best choice for orienting axes in a
problem?
2. How to determine the resultant magnitude and
direction?
3. How to resolve vectors into components?
4. How to add vectors that not perpendicular?
• Be sure to use examples in your essay to clarify ideas.
Go over homework
3.2 vector operations
Objectives
1. Identify appropriate coordinate systems for solving
problems with vectors.
2. Apply the Pythagorean theorem and tangent
function to calculate the magnitude and direction of
a resultant vector.
3. Resolve vectors into components using the sine and
cosine functions.
4. Add vectors that are not perpendicular.
• Use your knowledge of sign conventions from Chapter
2 to fill in the blanks below.
•
Movement to the right along the x-axis is considered
_____. Movement downward along the y-axis is
considered ______. Movement upward along the yaxis is considered ________. Movement to the left
along the x-axis is considered _______.(positive;
negative; positive; negative)
Coordinate systems in two dimensions
• Page 88 figure 3-6
• Question: which set of axes will give the
• In order to describe motion of an object in
two dimensions, we use two axes to describe
its movement.
Determine resultant magnitude and
direction
• Example: Joe leaves the base camp and hikes 11 km, north
and then hikes 11 km east. Determine Joe's resulting
displacement.
11 km
11 km
θ
R=?
Using Pythagorean theorem to
determine the magnitude of the
resultant
This procedure is restricted to the
addition of two vectors that make right
angles to each other.
Using tangent function to determine a
Vector's Direction
• Note: The measure of an angle as determined through
use of SOH CAH TOA is not always the direction of the
vector.
R2 = (5.0km)2 + (10km)2
R = 11 km
Or at 26 degrees
south of west
Class work
10/15 do now
• A 5.0-newton force and a 7.0-newton force act
concurrently on a point. As the angle between
the forces is increased from 0° to 180°, the
magnitude of the resultant of the two forces
changes from
a. 0.0 N to 12.0 N
b. 2.0 N to 12.0 N
c. 12.0 N to 2.0 N
d. 12.0 N to 0.0 N
homework
• 3.2 essay and 3.1 essay correction are due
• Chapter 3 Project is due
• Tonight’s homework – castle learning
objectives
• Class work – determine resultant of vectors
using Pythagorean Theorem and
trigonometry.
• Resolving vectors into components
• Presentation of project.
Class work
Page 90, Sample problem 3A
Page 91, practice 3A #1-4
1. a. 23 km; b. 17 km, E
2. 45.6 m at 9.5 degrees E of N
3. 16 m at 22 degrees to the side
of down field
4. 1.8 m at 49 degrees below the
horizontal
10/16 do now
• Which pair of concurrent forces could produce
a resultant force having a magnitude of 10.
Newtons?
1. 10. N, 10. N
2. 10. N, 30. N
3. 4.7 N, 4.7 N
4. 4.7 N, 50. N
10/17 Do now
• Forces F1 and F2 act concurrently on point P,
as shown in the diagram below. What is the
equilibrant of F1 and F2? Include magnitude
and direction.
objectives
• RESOLVING VECTORS INTO COMPONENTS
• Adding vectors that are not perpendicular
• Homework: castle learning
RESOLVING VECTORS INTO COMPONENTS
• In mathematics, the components of a vector are called
projections. You draw a coordinate system by the tail of the
vector. The x component is the projection of the vector along
the x-axis, and the y component is the projection of the
vector along the y-axis.
y
Ay
A
x
Ax
Appendix A, pp. 947-948
All Vectors can be broken into COMPONENTS
• X-Y system of components: θ is measured from 0o
• AX = A cos θ
• AY = A sin θ
• Example
– vi = 5.0 m/s at 30°
– vix = 5.0 m/s (cos 30°) = 4.33 m/s
– viy = 5.0 m/s (sin 30°) = 2.5 m/s
Any vector can be broken into unlimited sets of
components depends on the coordinates
Example
1. An arrow is shot from a bow at an angle of 25o
above the horizontal with an initial speed of 45 m/s.
Find the horizontal and vertical components of the
arrow’s initial velocity.
vx= vcosθ = (45 m/s) cos
25o
vy
= 41 m/s
25o
vx
2. The arrow strikes the target with the speed of 45
m/s at an angle of -25o with respect to the
horizontal. Calculate the horizontal and vertical
components of the arrow’s final velocity.
vx= vcosθ = (45 m/s) cos (-25o) = 41 m/s
vy= vsinθ = (45 m/s) sin (-25o) = -19 m/s
vx
vy
-25o
What’s wrong?
• A skyrocket travels 113 m at an angle of 82.4
degrees with respect to the ground and
toward the south. What is the rocket’s
horizontal displacement?
• vx = v cosθ
• vx = (113 m) cos 82.4o
• vx = 14.9 m
Pd. 4 - homework
• Read Page 93 – sample example 3B
• Page 94 - Practice 3B
• Problem 3A and 3B – choose 2 questions from
each set.
• Page 94 – practice 3B answers
1. 95 km/h
2. 44 km/h
3. 21 m/s, 5.7 m/s
4. 0 m, 5 m
5. 110 m, -53 m
6. 19.8 m, -11.7 m
7. 2.4 m/s/s, -0.77 m/s/s
perpendicular
Add the components of the original displacement vectors to find two components that
form a right triangle with the resultant vector.
1. Select a coordinate system, draw a sketch of the vectors to be
2. Find the x and y components of all vectors.
Ax=AcosθA;
Ay=AsinθA;
Bx=BcosθB;
By=BsinθB
3. Find the x and y component of the resultant vector:
Rx=Ax + Bx;
Ry=Ay + By
4. Use the Pythagorean theorem to find the magnitude of the
resultant vector.
R 
2
Rx  Ry
2
5. Use tan-1 function to find the angle the resultant vector
makes with the x-axis.
Ry
1
 R  tan
(
)
Rx
Sample Example 3C
• A hiker walks 25.5 km from her base camp at 35o south of
east. On the second day, she walks 41.0 km in a direction 65o
north of east at which point she discovers a forest ranger’s
tower. Determine the manignitude and direction of her
resultant displacement between the base camp and the
ranger’s tower.
R
d2
35o
d1
65o
Class work
• Page 97 – 3C
• Section 3-2 review – vector operations
10/11 do now
• A car travels down a road at a certain velocity,
vcar. The driver slows down so that the car is
traveling only half as fast as before. Which of
the following is the correct expression for the
resultant velocity?
a. 2vcar
c. -½vcar
b. ½ vcar
d. -2vcar
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