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/h
 18 cm long

Column Vectors
Vector a
a
2 up
4 RIGHT
 4  x 
  
 2  y 
Column Vectors
Vector b
2 up
b
3 LEFT
 3 
 2
 
Describe these vectors
 4
1
 
1
a  
 3
c
b
 2 
 3
 
d
 4 
 3 
 
Alternative labelling
B
F
EF
D
E
AB
CD
G
C
A
GH
H
General Vectors
A vector has both Length and direction
k
k
k
k
All 4 vectors have same length and same direction
General Vectors
Line CD is Parallel to AB
CD is TWICE length of AB
B
k
D
A
2k
Line EF is Parallel to AB
E
C
-k
F
EF is equal in length to AB
EF is opposite direction to AB
Write these Vectors in terms of k
B
k
D
2k
F
½k
1½k
E
A
C
-2k
H
G
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
Or
By 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.
1m
6.7m
3m
Start
= 500 m
Finish
5m
Displacement:
 6m   3m
2
2
 6.7m
Distance: 5m  3m  1m  9m
Displacement = 0 m
Distance = 500 m
Vector Addition
A
R
E
B
R
E
C
B A
D
D
E
D
R
C
A
C
B
A + B + C + D + E = Distance
R = Resultant = Displacement
Rectangular Components
Quadrant II
R A B
2
2
R sin  
y
A
R

-x
A opp
sin   
R hyp
B adj
cos   
R hyp
A opp
tan   
B adj
Quadrant I
B
R cos  
Quadrant III
Quadrant IV
-y
x
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.
• Scalar: parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
P
Q
• 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.
P
-P
• Equal vectors have the same magnitude and direction.
• Negative vector of a given vector has the same magnitude
and the opposite direction.
Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
P
• Law of cosines,
Q
R 2  P 2  Q 2  2 PQ cos B
Q
R  PQ
P
P
Q
• Law of sines,
sin A sin B sin C


Q
R
A
• Vector addition is commutative,
PQ  Q P
-Q
Q
P-Q
P
• Vector subtraction
 
P  Q  P  Q
Addition of Vectors
Q
S
• Addition of three or more vectors through
repeated application of the triangle rule
P
Q
S
P
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,



PQ S  PQ  S  P Q S
P
2P
-1.5P

• Multiplication of a vector by a scalar
increases its length by that factor (if scalar
is negative, the direction will also change.)
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
• 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.
Two forces act on a bolt
at A. Determine their
resultant.
• 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
R
Q

P
R  98 N   35
• 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,
R 2  P 2  Q 2  2PQ cos B
  40N    60N   2  40N  60N  cos155
2
2
R  97.73N
From the Law of Sines,
sin A sin B

Q
R
sin A  sin B
Q
R
 sin155
A  15.04
  20  A
 35.04
60N
97.73N
Sample Problem
• 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.
A barge is pulled by two
• Find a trigonometric solution by
tugboats. If the resultant of the
applying the Triangle Rule for vector
forces exerted by the tugboats
addition. With the magnitude and
is 5000 N directed along the
direction of the resultant known and the
axis of the barge, determine
directions of the other two sides parallel
to the ropes given, apply the Law of
a) the tension in each of the
Sines to find the rope tensions.
o
ropes for α = 45 ,
Sample Problem
• Graphical solution Parallelogram Rule with known
resultant direction and
magnitude, known directions for
sides.
T1  3700 N T2  2600 N
T1
30
5000N
45
• Trigonometric solution Triangle Rule with Law of Sines
30
45
T2
T1
T2
5000 N


sin 45 sin 30 sin105
5000N
45
T2
30
105
T1
T1  3700 N T2  2600 N
Rectangular Components of a
Force: Unit Vectors
• May resolve a force vector into perpendicular
components so that the resulting parallelogram is
a rectangle. Fx and Fy are referred to as
rectangular vector components
• Define perpendicular unit vectors iˆ and ˆj which
are parallel to the x and y axes.
• Vector components may be expressed as
products of the unit vectors with the scalar
magnitudes of the vector components.
F  Fxiˆ  Fy ˆj
Fx and Fy are referred to as the scalar
components of Fx and Fy

F
Addition of Forces by Summing Components
• Wish to find the resultant of 3 or more
concurrent forces,
P
Py j
S
Sy j
Qx i Px i
S xi
Qy j
Q
R  PQS
• Resolve each force into rectangular
components
Rxi  Ry j  Pxi  Py j  Qxi  Qy j  S xi  S y j
  Px  Qx  S x  i   Py  Qy  S y  j
Ry j
R

Rx i
• The scalar components of the resultant are
equal to the sum of the corresponding
scalar components of the given forces.
Ry  Py  Qy  S y
Rx  Px  Qx  S x
  Fx
  Fy
• To find the resultant magnitude and
direction,
R R R
2
x
2
y
  tan
1
Ry
Rx
Sample Problem
Plan:
Four forces act on bolt A as
shown. Determine the
resultant of the force on the
bolt.
• Resolve each force into
rectangular components.
• Determine the components of
the resultant by adding the
corresponding force
components.
• Calculate the magnitude and
direction of the resultant.
Sample Problem Solution
• Resolve each force into rectangular
components.
force mag
F1 150
F2
F3
F4
80
110
100
x  comp
129.9
y  comp
75.0
27.4
0
96.6
75.2
110.0
25.9
Rx  199.1
Ry  14.3
• Determine the components of the
resultant by adding the
corresponding force components.
• Calculate the magnitude and direction.
R  199.1  14.3
2
2
tan  
14.3 N
199.1N
R  199.6N
  4.1
Equilibrium of a Particle
• When the resultant of all forces acting on a particle is zero, the
particle is in equilibrium.
• 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.
• 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
R  F  0
- algebraic solution
 Fx  0  Fy  0
Free-Body Diagrams
TAB
50
Space Diagram: A sketch
showing the physical
conditions of the problem.
A
TAC
30
736N
Free-Body Diagram: A sketch
showing only the forces on the
selected particle.
Sample Problem
Plan of Attack:
• Construct a free-body diagram
for the particle at the junction
of the rope and cable.
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?
• 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.
TAB
horizontal
2
A
 cos  88  TAB  cos  30  TAC  0
TABy  Tcar  TAC y  0
30
TAC
3500lb
TABx  TACx  0
Vertical cos  2  T  3500lb  sin  30  T  0
AB
AC
Sample Problem
 cos 88TAB  cos 30TAC  0
cos  2 TAB  3500lb  sin  30TAC  0
TAB
TAC
cos  88  TAB

cos  30 
TAB
sin  30  TAC  3500lb

cos  2 
2
TAC
A
30
cos  88   sin  30  TAC  3500lb 



cos  30  
cos  2 

 0.02016015TAC  141.12105lb
0.979839TAC  141.12105lb
TAC  144lb
3500lb
Sample Problem
• Solve for the unknown force
magnitudes using Sine Law.
TAB
TAC
3500lb


sin120 sin 2 sin 58
TAB  3570lb
TAC  144lb
TAC
58
120
TAB
3500lb
2
Sometimes the Sine Law / Cosine
Law is faster than component
vectors. Intuition should tell you
which is best.
Sample Problem
PLAN OF ATTACK:
• Choosing the hull as the free body,
draw a free-body diagram.
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.
• 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 ft
tan   
 1.75
4 ft
  60.25
TAC
TAB=40
60.26
69.44
A
FD
1.5 ft
 0.375
4 ft
  20.56
tan    
• Express the condition for
equilibrium for the hull by
writing that the sum of all forces
must be zero.
TAE=60
R  TAB  TAC  TAE  FD  0
Sample Problem
TAC cos  69.44
TAB
40sin  60.26
TAC
TAC sin  69.44
40cos  60.26
A
FD
• Resolve the vector equilibrium
equation into two component
equations. Solve for the two
unknown cable tensions.
TAE  TAC sin  69.44   40cos  60.26 
60  0.936305TAC  19.842597
TAE=60
TAC  42.889
FD  TAC cos  69.44   40sin  60.26 
 0.351188TAC  34.73142
   0.351188  42.889   34.73142
 19.66 lb
Sample Problem
This equation is satisfied only if all
vectors when combined, complete a
closed loop.
Rectangular Components in Space
• The vector F
is contained in
the plane
OBAC.
• Resolve Fh into
• Resolve F into
horizontal and vertical rectangular
components
components.
Fh  F sin  y
Fy  F cos  y
Fx  Fh cos 
 F sin  y cos 
Fy  Fh sin 
 F sin  y sin 
Rectangular Components in Space
• With the angles between F and
the axes, Fx  F cos x Fy  F cos y Fz  F cos z
F  Fxi  Fy j  Fz k

 F cos x i  cos y j  cos z k

 F
  cos xi  cos y j  cos z k

of F
•  is a unit vector along the line
action of F
and cos x , cos  y , and cos  z are the
direction cosines for F
Rectangular Components in
Space
Direction of the force is defined by the location
of two points, M  x1, y1, z1  and N  x2 , y2 , z2 
d  vector joining M and N
 d xi  d y j  d z k
d x  x2  x1 d y  y2  y1 d z  z2  z1
F  F
1
  d xi  d y j  d z k
d
Fd y
Fd x
Fd
Fx 
Fy 
Fz  z
d
d
d

d is the length of the vector F

Sample Problem
PLAN of ATTACK:
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 x, y, z defining
the direction of the force
• 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 Problem
SOLUTION:
• Determine the unit vector pointing from A towards B.
AB   40 m  i   80 m  j   30 m  k
AB 
 40 m   80 m    30 m 
2
2
 94.3 m
 40 
 80 
 30 
 
i  
 j 
k
 94.3   94.3 
 94.3 
 0.424 i  0.848 j  0.318k
• Determine the components of the force.
F  F

  2500 N  0.424 i  0.848 j  0.318k
  1060 N  i   2120 N  j   795 N  k

2
Sample Problem
• Noting that the components of the unit vector are the direction
cosines for the vector, calculate the corresponding angles.
  cos x i  cos y j  cos z k
 0.424 i  0.848 j  0.318k
 x  115.1
 y  32.0
 z  71.5