CHAPTER THREE

TWO DIMENSIONAL MOTION

AND

VECTORS

Objectives: After completing this module, you should be able

to:• 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.

Expectations (Continued)

• You must have mastered right-triangle trigonometry.

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

Physics is the Science of Measurement

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

Distance: A Scalar Quantity

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

Displacement—A Vector Quantity

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.

Distance and Displacement

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

Identifying DirectionA 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

Identifying DirectionWrite 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

500 S of E500 S of E

450 W of N450 W of N

Example 1: Find the height of a building if it casts a shadow 90 m long

and the indicated angle is 30o.

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

Finding Components of Vectors

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

Example 2: A person walks 400 m in a direction of 30o N of E. How far is the displacement east and how far north?

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

Example 2: A person walks 400 m in a direction of 30o N of E. How far is the displacement east and how far north?

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

Example 2 (Cont.): A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north?

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

Example 2 (Cont.): A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north?

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

Example 2 (Cont.): A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north?

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.

Resultant of Perpendicular VectorsFinding 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

Example 3: A 30-lb southward force and a 40-lb eastward force act on a donkey at

the same time. What is the NET or resultant force on the donkey?

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!

Finding Resultant: (Cont.)

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

= 323.1o

= 323.1o

Example 7. Find the components of the 240-N force exerted by the boy on the girl if his arm makes an angle of 280 with the

ground.

282800

FF = 240 N = 240 N

FFFFyy

FFxx

FFyy

FFxx = -|(240 N) cos 28= -|(240 N) cos 2800|| = = -212 -212 NN

FFyy = +|(240 N) sin 28= +|(240 N) sin 2800|| = = +113 N+113 N

Or in Or in i,ji,j notation:notation:

F F = -(212 N)= -(212 N)ii + (113 + (113 N)N)jj

Example 8. Find the components of a 300-N force acting along the handle of a lawn-mower. The angle with the ground is 320.

323200

FF = 300 N = 300 N

FF FFyy

FFxx

FFyy

FFxx = -|(300 N) cos 32= -|(300 N) cos 3200|| = = -254 -254 NN

FFyy = -|(300 N) sin 32= -|(300 N) sin 3200|| = = -159 -159 NN

3232oo

3232oo

Or in Or in i,ji,j notation:notation:

F F = -(254 N)= -(254 N)ii - (159 - (159 N)N)jj

Example 11: A bike travels 20 m, E then 40 m at 60o N of W, and finally 30 m at 210o.

What is the resultant displacement graphically?

60o

30o

R

Graphically, we use ruler and protractor to draw components, then measure the Resultant R,

A = 20 m, E

B = 40 m

C = 30 m

R = (32.6 m, 143.0o)

R = (32.6 m, 143.0o)

Let 1 cm = 10 m