VECTORSHonors Section 1.5 – 1.9

Regular Physics Chapter 2

Objectives and Essential Questions

Objectives Distinguish between

basic trigonometric functions (SOH CAH TOA)

Distinguish between vector and scalar quantities

Add vectors using graphical and analytical methods

Essential Questions What is a vector

quantity? What is a scalar quantity? Give examples of each.

SCALARA SCALAR quantityis any quantity in physics that has MAGNITUDE ONLY

Number valuewith units

ScalarExample Magnitude

Speed 35 m/s

Distance 25 meters

Age 16 years

VECTOR

A VECTOR quantityis any quantity in physics that hasBOTH MAGNITUDE and DIRECTION

Vector Example

Magnitude and

Direction

Velocity 35 m/s, North

Acceleration 10 m/s2, South

Force 20 N, East

An arrow above the symbolillustrates a vector quantity.It indicates MAGNITUDE andDIRECTION

r x ,

r v ,

r a ,

r F

VECTOR APPLICATIONADDITION: When two (2) vectors point in the SAME direction, simply add them together.

EXAMPLE: A man walks 46.5 m east, then another 20 m east. Calculate his displacement relative to where he started.

66.5 m, E

MAGNITUDE relates to thesize of the arrow and DIRECTION relates to the way the arrow is drawn

46.5 m, E + 20 m, E

VECTOR APPLICATIONSUBTRACTION: When two (2) vectors point in the OPPOSITE direction, simply subtract them.

EXAMPLE: A man walks 46.5 m east, then another 20 m west. Calculate his displacement relative to where he started.

26.5 m, E

46.5 m, E

-

20 m, W

NON-COLLINEAR VECTORSWhen two (2) vectors are PERPENDICULAR to each other, you mustuse the PYTHAGOREAN THEOREM

Example: A man travels 120 km eastthen 160 km north. Calculate his resultant displacement.

120 km, E

160 km, N

the hypotenuse iscalled the RESULTANT

HORIZONTAL COMPONENT

VERTICALCOMPONENT

FINISH

FINISH

c 2 a2 b2 c a2 b2

c resul tan t 120 2 160 2 c 200km

WHAT ABOUT DIRECTION?In the example, DISPLACEMENT asked for and since it is a VECTOR quantity,we need to report its direction.

N

S

EW

N of E

E of N

S of W

W of S

N of W

W of N

S of E

E of S

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.

N of E

NEED A VALUE – ANGLE!Just putting N of E is not good enough (how far north of east ?). We need to find a numeric value for the direction.

N of E

160 km, N

120 km, E

To find the value of the angle we use a Trig function called TANGENT.

Tan opposite side

adjacent side160

1201.333

Tan 1(1.333)53.1o

200 km

So the COMPLETE final answer is : 200 km, 53.1 degrees North of East

What are your missing components?

Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components?

65 m25

H.C. = ?

V.C = ?

The goal: ALWAYS MAKE A RIGHT TRIANGLE!

To solve for components, we often use the trig functions since and cosine.

EmCHopp

NmCVadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,47.2725sin65..

,91.5825cos65..

sincos

sinecosine

For tonight, find the following vertical and horizontal components

ExampleA bear, searching for food wanders 35 meters east then 20 meters

north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

35 m, E

20 m, N

12 m, W

6 m, S

- =23 m, E

- =14 m, N

23 m, E

14 m, N

3.31)6087.0(

6087.23

14

93.262314

1

22

Tan

Tan

mR

The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST

R

ExampleA boat moves with a velocity of 15 m/s, N in a river which

flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

15 m/s, N

8.0 m/s, W

Rv

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

The Final Answer : 17 m/s, @ 28.1 degrees West of North

ExampleA plane moves with a velocity of 63.5 m/s at 32 degrees South of

East. Calculate the plane's horizontal and vertical velocity components.

63.5 m/s

32

H.C. =?

V.C. = ?

SsmCVopp

EsmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,/64.3332sin5.63..

,/85.5332cos5.63..

sincos

sinecosine

ExampleA storm system moves 5000 km due east, then shifts

course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.

NkmCVopp

EkmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,2.96440sin1500..

,1.114940cos1500..

sincos

sinecosine

5000 km, E

40

1500 km

H.C.

V.C.

5000 km + 1149.1 km = 6149.1 km

6149.1 km

964.2 kmR

R 6149.12 964.22 6224.2km

Tan 964.2

6149.10.157

Tan 1(0.157)8.92o

The Final Answer: 6224.2 km @ 8.92 degrees, North of East

Homework #1Remember YOU CANNOT SOLVE A VECTOR

PROBLEM WITHOUT DRAWING A PICTURE!!!!! Use lots of space.

Honors Physics: P23 Q21 – 26

Regular Physics: P91 Q1,2,3 and P94 Q1,2,3,4,5,6,7

Adding vectors algebraically that are not at right angles:

So far the vectors have been “tidy” and at right angles to each other. In real life that is rarely the case.

Consider this problem:

A 200N force is placed on a box 15° W of N. Another 150N force is placed on the box 25° E of N. Find the net horizontal force on the box.

Strategy for dealing with adding vectors:

Draw an accurate picture of all vectors.

Add and /or subtract collinear vectors to reduce the problem.

Find x and y components of each remaining vector. Be consistent with + and – signs.

Add all x components for all vectors.

Add all y components for all vectors

Reconstruct the resultant vector from x and y component sums, STATING MAGNITUDE AND ANGLE IN ANSWER. .

A 200N force is placed on a box 15° W of N. Another 150N force is placed on the box 25° E of N. Find the net horizontal force on the box.

Projectiles – anything shot or thrown into the air that have only gravity as a net external force on

them

ProjectilesProjectilesAny object that is shot or thrown into the air.Any object that is shot or thrown into the air.Examples: Baseball, bullet, basketball, tennis Examples: Baseball, bullet, basketball, tennis

ball, etc. can you think of others.ball, etc. can you think of others.The following are not projectiles: Rockets, The following are not projectiles: Rockets,

aircraft, birds, helicopters etc. can you think aircraft, birds, helicopters etc. can you think why? why?

The path of a projectile throught the air is called The path of a projectile throught the air is called the TRAJECTORY.the TRAJECTORY.

Projectiles may be horizontal or launched at an Projectiles may be horizontal or launched at an angle.angle.

Horizontal ProjectilesHorizontal ProjectilesExamples: a ball rolled of the edge of a table, a Examples: a ball rolled of the edge of a table, a

car driving off a cliff etc. Can you think of any car driving off a cliff etc. Can you think of any more?more?

The time it takes a horizontal projectile to reach The time it takes a horizontal projectile to reach the ground is the same time taken for an object the ground is the same time taken for an object to fall straight down from the same height.to fall straight down from the same height.

The range (how far it lands from the base of cliff The range (how far it lands from the base of cliff etc) depends on its speed at time of launch and etc) depends on its speed at time of launch and the height of the drop.the height of the drop.

Let’s do the following problem... Let’s do the following problem...

Projectiles launched at an angleProjectiles launched at an angleExamples would be throwing a ball far, throwing Examples would be throwing a ball far, throwing

a javelin. Can you think of more?a javelin. Can you think of more?These projectiles have a horizontal and vertical These projectiles have a horizontal and vertical

speed at the launch.speed at the launch.The range depends on speed of launch and The range depends on speed of launch and

angle alone.angle alone.The angle for maximum range is 45 degreesThe angle for maximum range is 45 degreesAngles > 45 make it stay in the air longer.Angles > 45 make it stay in the air longer.Angles < 45 make it land sooner. Angles < 45 make it land sooner.