Thanks to PowerPoint fromPaul E. Tippens, Professor of Physics

Southern Polytechnic State University

Vectors

Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry.

Define and give examples of scalar and vector quantities.

Determine the components of a given vector.

Find the resultant of two or more vectors.

Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry.

Define and give examples of scalar and vector quantities.

Determine the components of a given vector.

Find the resultant of two or more vectors.

http://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/introduction-to-vectors-and-scalars

Surveyors use accurate measures of magnitudes and

directions to create scaled maps of large regions.

You must be able convert units of measure for physical quantities.

Convert 40 m/s into kilometers per hour.

40--- x ---------- x -------- = 144 km/h

m

s

1 km

1000 m

3600 s

1 h

College algebra and simple formula manipulation are assumed.

Example:

0

2fv v

x t

Solve for vo

Let’s do Table talks—I will call on one person at one table. They will tell me their groups answer. I will then call on each table and ask if they agree or disagree and why.

You must be able to work in scientific notation.

Evaluate the following:

(6.67 x 10-11)(4 x 10-3)(2)

(8.77 x 10-3)2 F = -------- = ------------

Gmm’

r2

Turn to your neighbor and explain what the exponent would be in this equation.

Let’s see who agrees and disagrees before you figure out the answer

-8 Now with your table figure out the answer-

see if you can do it without a calculator. F = 6.94 x 10-9 N = 6.94 nN

You must be familiar with SI prefixes

The meter (m) 1 m = 1 x 100 m

1 Gm = 1 x 109 m 1 nm = 1 x 10-9 m

1 Mm = 1 x 106 m 1 m = 1 x 10-

6 m

1 km = 1 x 103 m 1 mm = 1 x 10-

3 m

You are familiar with right-triangle trigonometry.

y

x

R

y = R sin y = R sin

x = R cos x = R cos

siny

R

tany

x R2 = x2 +

y2

R2 = x2 + y2

We begin with the measurement of length: its magnitude and its direction.

We begin with the measurement of length: its magnitude and its direction.

LengtLengthh

WeighWeightt

TimeTime

A scalar quantity:

Contains magnitude only and consists of a number and a unit.

(20 m, 40 mi/h, 10 gal)

A

B

DistanceDistance is the length of the actual is the length of the actual path taken by an object.path taken by an object.

DistanceDistance is the length of the actual is the length of the actual path taken by an object.path taken by an object.

s = 20 m

A vector quantity:

Contains magnitude AND direction, a number, unit & angle.

(12 m, 300; 8 km/h, N)

A

BD = 12 m, 20o

• DisplacementDisplacement is the straight-line is the straight-line separation of two points in a separation of two points in a specified direction. specified direction.

• DisplacementDisplacement is the straight-line is the straight-line separation of two points in a separation of two points in a specified direction. specified direction.

Net Net displacement:displacement:4 m,E4 m,E

6 6 m,Wm,W

D

What is the What is the distance traveled?distance traveled?

10 m !!

DD = 2 m, W= 2 m, W

• DisplacementDisplacement is the is the x x or or yy coordinate of position. Consider a coordinate of position. Consider a car that travels 4 m, E then 6 m, car that travels 4 m, E then 6 m, W.W.

• DisplacementDisplacement is the is the x x or or yy coordinate of position. Consider a coordinate of position. Consider a car that travels 4 m, E then 6 m, car that travels 4 m, E then 6 m, W.W.

xx = +4= +4xx = -2= -2

A common way of identifying direction is by reference to East, North, West, and South. (Locate points below.)

A common way of identifying direction is by reference to East, North, West, and South. (Locate points below.)

40 m, 5040 m, 50oo N of E N of E

EW

S

N

40 m, 60o N of W40 m, 60o W of S40 m, 60o S of E

Length = 40 m

5050oo60o

60o60o

Write the angles shown below by using references to east, south, west, north.Write the angles shown below by using references to east, south, west, north.

EW

S

N45o

EW

N

50o

S

Click to see the Answers . . .Click to see the Answers . . .500 S of E500 S of E

450 W of N450 W of N

Polar coordinates (Polar coordinates (R,R,) are an ) are an excellent way to express vectors. excellent way to express vectors. Consider the vector Consider the vector 40 m, 5040 m, 500 0 N of EN of E,, for example.for example.

Polar coordinates (Polar coordinates (R,R,) are an ) are an excellent way to express vectors. excellent way to express vectors. Consider the vector Consider the vector 40 m, 5040 m, 500 0 N of EN of E,, for example.for example.

0o

180o

270o

90o

0o

180o

270o

90o

RR

RR is the is the magnitudemagnitude and and is the is the directiondirection..

40 40 mm5050oo

(R,(R,) = 40 m, 50) = 40 m, 50oo

(R,(R,) = 40 m, ) = 40 m, 120120oo (R,(R,) = 40 m, 210) = 40 m, 210oo

(R,(R,) = 40 m, ) = 40 m, 300300oo

5050oo60o

60o60o

0o180o

270o

90o

120o

Polar coordinates (Polar coordinates (R,R,) are given for ) are given for each of four possible quadrants:each of four possible quadrants:Polar coordinates (Polar coordinates (R,R,) are given for ) are given for each of four possible quadrants:each of four possible quadrants:

210o

3000

Right, up = (+,+)

Left, down = (-,-)

(x,y) = (?, ?)

x

y

(+3, (+3, +2)+2)

(-2, +3)(-2, +3)

(+4, -3)(+4, -3)(-1, -3)(-1, -3)

Reference is made Reference is made to to xx and and yy axes, axes, with with ++ and and -- numbers to numbers to indicate position in indicate position in space.space.

++++

----

Application of Trigonometry to Vectors

y

x

R

y = R sin y = R sin

x = R cos x = R cos

siny

R

cosx

R

tany

x R2 = x2 +

y2

R2 = x2 + y2

TrigonometryTrigonometry

90 m

300

The height h is opposite 300

and the known adjacent side is 90 m.

h

h = (90 m) tan 30o

h = 57.7 mh = 57.7 m

0tan 3090 m

opp h

adj

A component is the effect of a vector along other directions. The x and y components of the vector (R, are illustrated below.

x

yR

x = R cos

y = R sin

Finding components:

Polar to Rectangular Conversions

x

yR

x = ?

y = ?400 m

E

N

The y-component (N) is OPP:

The x-component (E) is ADJ:

x = R cos y = R sin

E

N

x = R cos

x = (400 m) cos 30o

= +346 m, E

x = ?

y = ?400 m

E

N Note:Note: xx is the side is the side adjacentadjacent to angle to angle

303000

ADJADJ = HYP x = HYP x CosCos 303000

The x-component The x-component is:is:RRxx = = +346 m+346 m

y = R sin

y = (400 m) sin 30o

= + 200 m, N

x = ?

y = ?400 m

E

N

OPPOPP = HYP x = HYP x SinSin 303000

The y-component The y-component is:is:RRyy = = +200 m+200 m

Note:Note: yy is the side is the side oppositeopposite to angle to angle

303000

Rx = +346 m

Ry = +200 m

400 m

E

NThe x- and y- The x- and y- components components are are eacheach + in + in

the first the first quadrantquadrant

Solution: The person is displaced 346 m east and 200 m north of the original

position.

First Quadrant:

R is positive (+)

0o > < 90o

x = +; y = +x = R cos y = R sin

+

+

0o

90o

R

Second Quadrant:

R is positive (+)

90o > < 180o

x = - ; y = +x = R cos y = R sin

+R

180o

90o

Third Quadrant:

R is positive (+)

180o > < 270o

x = - y = - x = R cos y = R sin

-R

180o

270o

Fourth Quadrant:

R is positive (+)

270o > < 360o

x = + y = -

x = R cos y = R sin

360o+

R

270o

Finding resultant of two perpendicular vectors is like changing from rectangular to polar coord.

R is always positive; is from + x axis

2 2R x y 2 2R x y

tany

x tan

y

x x

yR

30 lb

40 lb

Draw a rough Draw a rough sketch.sketch.

Choose rough Choose rough scale:scale:Ex: 1 cm = 10 lb

4 cm = 40 lb

3 cm = 30 lb

40 lb

30 lb

Note: Force has direction just like length does. We can treat force vectors just as we have length vectors to find the resultant force. The procedure is the same!

Note: Force has direction just like length does. We can treat force vectors just as we have length vectors to find the resultant force. The procedure is the same!

40 lb

30 lb

40 lb

30 lb

Finding (Finding (R,R,) from given () from given (x,yx,y) = (+40, -) = (+40, -30)30)

R

Ry

Rx

R = x2 + y2 R = (40)2 + (-30)2 = 50 lb

tan = -30

40 = -36.9o

is S of E

= 323.1o

= 323.1o

40 lb

30 lbR

Ry

Rx40 lb

30 lb R

Ry

Rx

40 lb

30 lbR

Ry

Rx

40 lb

30 lb

R

Ry

Rx

= 36.9o; = 36.9o; 143.1o; 216.9o; 323.1o

= 36.9o; = 36.9o; 143.1o; 216.9o; 323.1o

R = 50 lb

R = 50 lb