MCV4UW

Vectors

Vectors

A vector is a quantity that involves both magnitude and direction.55 km/h [N35E]A downward force of 3 Newtons

A scalar is a quantity that does not involve direction.55 km/h18 cm long

Column Vectors

a

Vector a

4 RIGHT

2 up

4

2

x

y

Column Vectors

b

Vector b

3

2

3 LEFT

2 up

Describe these vectors

c

d

2

3

1

3

4

1

4

3

b

a

Alternative labelling

CD��������������

EF��������������

AB��������������

A

B

C

DF

E

G

H

GH��������������

General Vectors

k

k

k

k

A vector has both Length and direction

All 4 vectors have same length and same direction

General Vectors

kA

B

C

D

-k

2k

F

E

Line CD is Parallel to AB

CD is TWICE length of AB

Line EF is Parallel to AB

EF is equal in length to AB

EF is opposite direction to AB

Write these Vectors in terms of k

k

A

B

C

D

E

F G

H

2k1½k ½k

-2k

Vector Notation

Vectors are often identified with arrows in graphics and labeled as follows:

We label a vector with a variable.This variable is identified as a vector either by an arrow above itself : A

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OrBy the variable being BOLD: A

Displacement

Displacement is an object’s change in position. Distance is the total length of space traversed by an object.

5m

3m

1m

Distance:

Displacement:

6.7m

StartFinish

= 500 m

m

m

005 Distance

0Displacement

=

=

2 26 3 6.7m m m

5 3 1 9m m m m

Vector Addition

AB

C

DE AB

C

D

E

A

B

C

DE

R

A + B + C + D + E = DistanceR = Resultant = Displacement

R

R

Rectangular Components

A

B

R2 2R A B

sin

cos

tan

A opp

R hyp

B adj

R hyp

A opp

B adj

Quadrant IQuadrant II

Quadrant III Quadrant IV

y

-y

x-x cosR

sinR

Resultant of Two Forces• force: action of one body on another;

characterized by its point of application, magnitude, line of action, and sense.

• Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force.

• The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.

• Force is a vector quantity.

Vectors• Vector: parameters possessing magnitude and direction

which add according to the parallelogram law. Examples: displacements, velocities, accelerations.

• Vector classifications:- Fixed or bound vectors have well defined points of

application that cannot be changed without affecting an analysis.

- Free vectors may be freely moved in space without changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their line of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.• Negative vector of a given vector has the same magnitude

and the opposite direction.

• Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature

P

Q

P+Q

P

-P

Addition of Vectors• Trapezoid rule for vector addition

2 2 2 2 cosR P Q PQ B

R P Q

• Law of cosines,

• Law of sines,sin sin sinA B C

Q R A

• Vector addition is commutative,P Q Q P

• Vector subtraction

P Q P Q ��������������������������������������������������������

P

Q

P+Q

P P+Q

Q

• Triangle rule for vector addition

PP+Q

Q

PQ-QP-Q

Addition of Vectors• Addition of three or more vectors through

repeated application of the triangle rule

• The polygon rule for the addition of three or more vectors.

• Vector addition is associative,

P Q S P Q S P Q S

• Multiplication of a vector by a scalar increases its length by that factor (if scalar is negative, the direction will also change.)

P

QSQ+S

P+Q+S

P

QS

P+Q+S

P2P

-1.5P

Resultant of Several Concurrent Forces• Concurrent forces: set of forces

which all pass through the same point.

A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

• Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

Sample Problem

Two forces act on a bolt at A. Determine their resultant.

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.

• Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

Sample Problem Solution

98 N 35R

Q

P

R

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.

Sample Problem Solution• Trigonometric solution

From the Law of Cosines,

2 2 2

2 2

2 cos

40N 60N 2 40N 60N cos155

R P Q PQ B

sin sin

sin sin

60Nsin155

97.73N15.04

20

35.04

A B

Q R

QA B

R

A

A

97.73NR

From the Law of Sines,

Sample Problem

a) the tension in each of the ropes for α = 45o,

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine

• Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N.

• Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.

Sample Problem• Graphical solution -

Parallelogram Rule with known resultant direction and magnitude, known directions for sides.

• Trigonometric solution - Triangle Rule with Law of Sines

T1

T2

30

45

45

30

5000N

1 23700 2600T N T N

5000N

T2 T1

45 30

105 1 23700 2600T N T N

1 2 5000

sin 45 sin30 sin105

T T N

Rectangular Components of a Force: Unit Vectors

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

Fx and Fy are referred to as the scalar components of

ˆ ˆx yiF jF F

F

• May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components

and x yF F

• Define perpendicular unit vectors which are parallel to the x and y axes.

an ˆ ˆd i j

and x yF F

Addition of Forces by Summing Components

R P Q S

• Wish to find the resultant of 3 or more concurrent forces,

x y x y x y x y

x x x y y y

R i R j P i P j Q i Q j S i S j

P Q S i P Q S j

• Resolve each force into rectangular components

x x x x

x

R P Q S

F

• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

y y y y

y

R P Q S

F

2 2 1tan yx y

x

RR R R

R

• To find the resultant magnitude and direction,

P

Q

S yP j

xP i

yQ j

yS j

xS i xQ i

yR j

xR i

R

Sample Problem

Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

Plan:

• Resolve each force into rectangular components.

• Calculate the magnitude and direction of the resultant.

• Determine the components of the resultant by adding the corresponding force components.

Sample Problem Solution• Resolve each force into rectangular

components.

1

2

3

4

150 129.9 75.0

80 27.4 75.2

110 0 110.0

100 96.6 25.9

force mag x comp y comp

F

F

F

F

2 2199.1 14.3R 199.6NR

• Calculate the magnitude and direction.

14.3Ntan

199.1N

4.1

• Determine the components of the resultant by adding the corresponding force components.

199.1xR 14.3yR

Equilibrium of a Particle• When the resultant of all forces acting on a particle is zero, the

particle is in equilibrium.

• Particle acted upon by two forces:- equal magnitude- same line of action- opposite sense

• Particle acted upon by three or more forces:- graphical solution yields a closed

polygon- algebraic solution

0

0 0x y

R F

F F

• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.

Free-Body Diagrams

Space Diagram: A sketch showing the physical conditions of the problem.

Free-Body Diagram: A sketch showing only the forces on the selected particle.

TAB

TAC

736N

50 30A

Sample Problem

In a ship-unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?

Plan of Attack:

• Construct a free-body diagram for the particle at the junction of the rope and cable.

• Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle.

• Apply trigonometric relations to determine the unknown force magnitudes.

Sample Problem

SOLUTION:

• Construct a free-body diagram for the particle at A.

• Apply the conditions for equilibrium in the horizontal and vertical directions.

2

30A

TAB

3500lb

TAC

horizontal

0

cos 88 cos 30 0x xAB AC

AB AC

T T

T T

Vertical 0

cos 2 3500 sin 30 0

y yAB car AC

AB AC

T T T

T lb T

2

30A

TAB

3500lb

Sample Problem cos 88 cos 30 0AB ACT T

cos 2 3500 sin 30 0AB ACT lb T

cos 88

cos 30AB

AC

TT

sin 30 3500

cos 2AC

AB

T lbT

cos 88 sin 30 3500

cos 30 cos 2

0.02016015 141.12105

0.979839 141.12105

144

ACAC

AC

AC

AC

T lbT

T lb

T lb

T lb

• Solve for the unknown force magnitudes using Sine Law.

3500lb

sin120 sin 2 sin58AB ACT T

3570lbABT

144lbACT

3500lb

2

120TAB

TAC

58

Sometimes the Sine Law / Cosine Law is faster than component vectors. Intuition should tell you which is best.

Sample Problem

Sample Problem

It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE.

Determine the drag force exerted on the hull and the tension in cable AC.

PLAN OF ATTACK:

• Choosing the hull as the free body, draw a free-body diagram.

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

Sample Problem

SOLUTION:

• Choosing the hull as the free body, draw a free-body diagram.

7 fttan 1.75

4 ft60.25

1.5 fttan 0.375

4 ft20.56

• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.

0AB AC AE DT T TR F

60.2669.44

TAB=40

TAC

FD

TAE=60

A

Sample Problem• Resolve the vector equilibrium

equation into two component equations. Solve for the two unknown cable tensions.

cos 69.44 40sin 60.26

0.351188 34.73142

0.35 42.1188 34.7889 3142

19.66

AC

D ACF T

lb

T

TAB

TAC

FD

TAE=60

A

sin 69.44ACT

cos 69.44ACT

40sin 60.26

40cos 60.26

sin 69.44 40cos 60.26

60 0.936305 19.842597

42.889

AE AC

AC

AC

T T

T

T

Sample Problem

This equation is satisfied only if all vectors when combined, complete a closed loop.

Rectangular Components in Space

• The vector is contained in the plane OBAC.

F

sinh yF F

• Resolve into horizontal and vertical components.

F

cosy yF F

• Resolve into rectangular components

hF

cos

sin cos

sin

sin sin

x h

y

y h

y

F F

F

F F

F

Rectangular Components in Space

• With the angles between and the axes,

F

cos cos cos

cos cos cos

cos cos cos

x x y y z z

x y z

x y z

x y z

F F F F F F

F F i F j F k

F i j k

F

i j k

• is a unit vector along the line of action ofand are the direction cosines for

F

F

cos , cos , and cosx y z F

Rectangular Components in Space

Direction of the force is defined by the location of two points, 1 1 1 2 2 2, , and , ,M x y z N x y z

2 1 2 1 2 1

vector joining and

1

x y z

x y z

x y z

yx zx y z

d M N

d i d j d k

d x x d y y d z z

F F

d i d j d kd

FdFd FdF F F

d d d

d is the length of the vector F

Sample Problem

The tension in the guy wire is 2500 N. Determine:

a) components Fx, Fy, Fz of the force acting on the bolt at A,

b) the angles qx, qy, qz defining the direction of the force

PLAN of ATTACK:

• Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B.

• Apply the unit vector to determine the components of the force acting on A.

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

Sample ProblemSOLUTION:• Determine the unit vector pointing from A towards B.

2 2 2

40m 80m 30m

40m 80m 30m

94.3 m

AB i j k

AB

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• Determine the components of the force.

2500 N 0.424 0.848 0.318

1060 N 2120 N 795 N

F F

i j k

i j k

40 80 30

94.3 94.3 94.3

0.424 0.848 0.318

i j k

i j k

Sample Problem• Noting that the components of the unit vector are the direction

cosines for the vector, calculate the corresponding angles.

cos cos cos

0.424 0.848 0.318

x y zi j k

i j k

115.1

32.0

71.5

x

y

z