Module 1.1 – Displacement and
Velocity Vectors
Displacement and velocity vectors were studied in
physics 11 for one dimensional motion. This module
extends the study of kinematics to two dimensions,
allowing us to study a much wider range of situations.
RRHS Physics Unit 1
Slide #1
2D Vectors
• Vector Magnitude and Direction
• Components
RRHS Physics Unit 1
Slide #2
d
Trigonometry Review
adjacent
cos  
hypotenuse
opposite
sin  
hypotenuse
opposite
tan  
adjacent
Pythagorean Theorem
c  a b
2
RRHS Physics Unit 1
Slide #3
2
2
Example
Assume that the angle in the diagram below is
30.0° and the magnitude of the position vector is
7.2 cm. Calculate the components.
d
RRHS Physics Unit 1
Slide #4
Solution
adj
hyp
d
cos   x
d
d
cos 30.0  x
7.2
sin  
cos  
sin  
sin 30.0 
opp
hyp
dy
d
dy
7.2
d y  3.6cm
d x  6.2cm
RRHS Physics Unit 1
Slide #5
Vector Parts
RRHS Physics Unit 1
Slide #6
Check Your Learning
• Calculate the components for each of the
following vectors:
RRHS Physics Unit 1
Slide #7
Check Your Learning
a.
dx
cos 55 
12.2
d x  7.0m
sin 55 
dy
12.2
d y  10. m
Taking the signs into consideration (to the right
and down), we get
d x  7.0m
d y  10. m
RRHS Physics Unit 1
Slide #8
Check Your Learning
b.
dx
sin 43 
8.5
d x  5.8cm
cos 43 
dy
8.5
d y  6.2cm
Taking the signs into consideration (to the left and up), we get
d x  5.8cm
d y  6.2cm
RRHS Physics Unit 1
Slide #9
Expressing Direction
Assume that θ=30.0°
• The angle is measured from east
d
• Must go north to get to the vector
30.0° N of E
(Could also be expressed as 60.0° E of N)
RRHS Physics Unit 1
Slide #10
Check Your Learning
State the direction for each of the following
vectors:
a.
b.
c.
d.
e.
f.
20o S of E
80o W of S
40o W of N
50o N of W
S
75o E of N
RRHS Physics Unit 1
Slide #11
Vector Addition
a
We want to find
b
c  a b
RRHS Physics Unit 1
Slide #12
Vector Addition Diagram
• Must draw vectors being
added head to tail
c
b
a
• Resultant (c) goes from tail of
first vector to head of second vector
a b  c
RRHS Physics Unit 1
Slide #13
Check Your Learning
Given the following vectors, draw a vector diagram to
represent each of the following vector equations (where
is an unknown vector in each case):
c
a
b
RRHS Physics Unit 1
Slide #14
a.
b.
c.
d.
e.
a b  x
b c  x
a b c  0
ax b
x c  a
Check Your Learning
x
c
c
b
a
b
b
a
x
x
x
b
c
a
a
RRHS Physics Unit 1
Slide #15
Adding With Components
a b  c
c
b
b
a
a
a  (ax , a y )
c  (ax  bx , a y  by )
b  (bx , by )
RRHS Physics Unit 1
Slide #16
Example
A person walks 5.0 km east and then 8.0
km in a direction 75° N of E. What is his
displacement?
d1  5.0km E
d 2  8.0km, 75 N of E
dt  ?
dt  d1  d2
RRHS Physics Unit 1
Slide #17
Solution
dt
d 2 y
d 2
d 2 x
d1
d 2 x
8.0
d 2 x  2.1km
cos 75 
dt  dtx2  dty2
dt
d 2
 (7.1) 2  (7.7) 2
 10. km
sin 75 
d 2 y
tan  
8.0
 7.7 km
dty
dtx
7.7
7.1
 47

dt  10. km, 47 N of E
RRHS Physics Unit 1
Slide #18
d 2 y
Check Your Learning
A person walks 4.0 km south and then 7.2
km in a direction 21o W of N. What was
the person’s displacement?
d1  4.0km, S  4.0km
dt
d 2  7.2km, 21 W of N
dt  ?
d 2
dt  d1  d2
RRHS Physics Unit 1
Slide #19
d1
Check Your Learning
d tx
dt
d 2 y
d 2
dt
d ty
d1
d 2 x
d 2 x
cos 69 
7.2
d 2 x  2.6km
sin 69 
d 2 y
dtx  d 2 x  d1x
d 2 y
7.2
 6.7 km
 2.6  0
 2.6km
RRHS Physics Unit 1
Slide #20
d ty  d 2 y  d1 y
 6.7  (4.0)
 2.7 km
Check Your Learning
d tx
dt
dt  (dtx )2  (dty ) 2
d ty
tan  
 (2.6)2  (2.7) 2
 3.7km
dt  3.7km, 44 W of N
RRHS Physics Unit 1
Slide #21
dtx
dty
2.6
2.7
 44

Vector Subtraction
a
b
a b  c
a  (b )  c
b
a
RRHS Physics Unit 1
Slide #22
Vector Subtraction
a  (b )  c
ax
ay
a
b y
b
c
bx
RRHS Physics Unit 1
Slide #23
Check Your Learning
Given the following vectors, draw a vector diagram to
represent each of the following vector equations (where
is an unknown vector in each case):
c
a
b
RRHS Physics Unit 1
Slide #24
a.
b.
c.
d.
a b  x
b c  x
ax b
x c  a
Check Your Learning
c
x
a
b
x
x
b
b
a
a
x
c
RRHS Physics Unit 1
Slide #25
Velocity Vectors
d
v
t
v  (v x , v y )
d x
vx 
t
RRHS Physics Unit 1
Slide #26
vy 
d y
t
Relative Velocity
V of boat with
respect to water
V of boat with
respect to shore
vws
vbw
V of water with
respect to shore
vbs  vbw  vws
vbs
Same inside subscripts
RRHS Physics Unit 1
Slide #27
Example
A boat that has a speed of 5.0 m/s in still water heads
north directly across a river that is 250 m wide. The
velocity of the river is 3.5 m/s east.
a. What is the velocity of the boat with respect to the
shore?
b. How long does it take the boat to cross the river?
c. How far downstream does the boat land?
d. What heading (direction) would the boat need in order
to land directly across from its starting point?
RRHS Physics Unit 1
Slide #28
Solution
a.
vbw  5.0m / s N
vbs  vbw  vws
vws  3.5m / s E
vbs  ?
vws
3.5m / s
2
2
vbs  vbw
 vws
 (5.0) 2  (3.5) 2
vbw
vbs
5.0m / s
vbs
 6.1m / s
3.5
tan  
5.0
 35
vbs  6.1m / s,35 E of N
RRHS Physics Unit 1
Slide #29
Solution
b.
d  250m N
vbw  5.0m / s N
t ?
d
t
250
5.0 
t
v
t  50. s
c.
vws  3.5m / s E
t  50. s
d  ?
d
t
d
3.5 
50.
v
d  180m
RRHS Physics Unit 1
Slide #30
Solution
d.
vbs  vbw  vws
vws
vbw
vbs
where vbs is north
vws
sin  
vbw
3.5

5.0
  44
RRHS Physics Unit 1
Slide #31
44 W of N
Check Your Learning
In the example just completed, how long will it take the
boat to cross the river in part (d)? In other words, how
long will it take to cross the river when corrective action
is taken so that the boat lands directly across from its
starting point? How does this answer compare with the
time obtained in part (a) of the example?
RRHS Physics Unit 1
Slide #32
Check Your Learning
vws
vbw
vbs
v
tan   ws
vbs
3.5
tan 44 
vbs
vbs  3.6m / s N
d
v
t
250
3.6 
t
t  69 s
It takes more time to cross the river when correcting
for the flow of the river than was calculated in part
(a) since some of the boat’s velocity is being used to
compensate for the river and stop the boat from
moving downstream.
RRHS Physics Unit 1
Slide #33
Module Summary
•
In this module you have learned:
•
To represent two dimensional vectors using two methods:
– components and
– magnitude and direction.
•
How to add and subtract displacement vectors using vector diagrams and
components:
– When adding vectors, they must be drawn head to tail. The resultant vector goes
from the tail of the first vector to the head of the last vector.
– Subtracting vectors is the same operation as adding a negative vector, where a
negative vector points in the direction opposite the direction of the original vector.
•
How to use vector diagrams and components to calculate relative
velocities.
RRHS Physics Unit 1
Slide #34
Module 1.2 – Force Vectors
The concept of two-dimensional vectors will be
applied to free body diagrams and Newton’s
Laws of Motion from Unit 3. Two-dimensional
situations that will be studied include forces
acting at an angle and inclined planes.
RRHS Physics Unit 1
Slide #35
Pulling at an Angle
Must break all force
down into horizontal
and vertical directions
RRHS Physics Unit 1
Slide #36
Example
A 52.0 kg sled is being pulled along a frictionless
horizontal ice surface by a person pulling a rope with a
force of 235 N. The rope makes an angle of 35.0o with
the horizontal. What is the acceleration of the sled?
RRHS Physics Unit 1
Slide #37
Solution
Fpy  Fp sin 
Fpx  Fp cos 
 (235) cos 35.0
 193 N
m  52.0kg
Fp  235 N
  35.0
ax  ?
 (235) sin 35.0
 135 N
max   F x
max  Fpx
(52.0)ax  193
ax  3.71m / s 2
RRHS Physics Unit 1
Slide #38
Horizontal Forces
max  Fx
max  Fpx  F f
max  Fpx  Ff
RRHS Physics Unit 1
Slide #39
Vertical Forces
ma y  Fy
ma y  FN  Fpy  Fg
ma y  FN  Fpy  Fg
0  FN  Fpy  Fg
RRHS Physics Unit 1
Slide #40
FN  Fg
Check Your Learning
A 52.0 kg sled is being pulled along a horizontal surface
by a person pulling a rope with a force of 235 N. The
rope makes an angle of 35.0o with the horizontal. If the
coefficient of friction between the sled and the surface is
0.25, What is the acceleration of the sled?
Fpx  Fp cos 
m  52.0kg
 (235) cos 35.0
Fp  235 N
 193 N
  35.0
  0.25
Fpy  Fp sin 
ax  ?
 (235) sin 35.0
 135 N
RRHS Physics Unit 1
Slide #41
Check Your Learning
ma y   Fy
F f   FN
 (0.25)(375)
ma y  FN  Fg  Fpy
 94 N
0  FN  Fg  Fpy
max   F x
FN  Fg  Fpy
 (52.0)(9.80)  135
max  Fpx  Ff
 375 N
max  Fpx  Ff
(52.0)ax  193  94
ax  1.9m / s 2
RRHS Physics Unit 1
Slide #42
Inclined Planes
No friction
Acceleration will
be parallel to the
plane
Choose new
coordinate
system
RRHS Physics Unit 1
Slide #43
Force of Gravity Components
Fgx  mg sin 
Fgy  mg cos 
RRHS Physics Unit 1
Slide #44
Perpendicular Forces
ma y  Fy
ma y  FN  Fgy
ma y  FN  Fgy
m(0)  FN  Fgy
FN  Fgy
RRHS Physics Unit 1
Slide #45
Parallel Forces
max  Fx
max  Fgx
RRHS Physics Unit 1
Slide #46
Example
A 1200 kg car is on an icy (frictionless) hill that is inclined
at an angle of 12o with the horizontal. What is the
acceleration of the car down the hill?
m  1200kg
  12
ay  0
max   F x
max  Fgx
max  mg sin 
ax  ?
ax  (9.80)(sin12)
 2.0m / s 2
RRHS Physics Unit 1
Slide #47
Check Your Learning
A 1200 kg car is on an icy hill that is inclined at an angle
of 12o with the horizontal. As the car starts sliding, the
driver locks the wheels. The coefficient of friction
between the locked wheels and the icy surface is 0.14.
What is the acceleration of the car down the hill?
m  1200kg
  12
  0.14
Fgx  mg sin 
 (1200)(9.80) sin12
 2450 N
Fgy  mg cos 
ay  0
 (1200)(9.80) cos12
ax  ?
 11500 N
RRHS Physics Unit 1
Slide #48
Check Your Learning
Perpendicular Forces
ma y   F y
Parallel Forces
F f   FN
ma y  FN  Fgy
 (0.14)(11500)
ma y  FN  Fgy
 1610 N
0  FN  11500
FN  11500
max   F x
max  Fgx  Ff
max  Fgx  Ff
(1200)ax  2450  1610
ax  0.70m / s 2
RRHS Physics Unit 1
Slide #49
Module Summary
In this module you learned that
• Force vectors can be broken into components so that dynamics
situations can be analyzed using free body diagrams and Newton’s
Second Law.
• Situations involving inclined planes (ramps) can be analyzed by
rotating the coordinate system so that the x-axis is parallel to the
ramp and the y-axis is perpendicular to the ramp. The components
for the force of gravity can then be given by
Fgx  mg sin 
Fgy  mg cos 
RRHS Physics Unit 1
Slide #50
Module 1.3 – Equilibrium
This module is a continuation of the previous module,
which introduced two dimensional force vectors. In this
module, force vectors will be incorporated into studies of
systems in equilibrium. We will look at both translational
equilibrium (not accelerating linearly) and rotational
equilibrium (not rotating). This module will also introduce
the concept of a torque. Real world situations such as
using support cables, cranes, scaffolding, and many
other structures require an understanding of force
vectors and torques.
RRHS Physics Unit 1
Slide #51
Translational Equilibrium
RRHS Physics Unit 1
Slide #52
Translational Equilibrium
F
FT 1  FT 2  Fg  0
x
0
F1x  F2 x  0
F1x  F2 x  0
F1x  F2 x
F
y
0
F1 y  F2 y  Fg  0
F1 y  F2 y  Fg  0
F1 y  F2 y  Fg
RRHS Physics Unit 1
Slide #53
Equilibrant Force
• If the vector sum of all of the forces acting on an object is
not zero, there will be a net force in some direction.
• There is a single additional force that can be applied to
balance this net force. This additional force is called the
equilibrant force.
• The equilibrant force is equal in magnitude to the sum of
all of the forces acting on the object, but opposite in
direction.
RRHS Physics Unit 1
Slide #54
Example
A 20.0 kg sack of potatoes is suspended by a rope. A
man pushes sideways with a force of 50.0 N and
maintains this force so that the sack is in equilibrium.
What is the tension in the rope and what angle does the
rope make with the vertical?
m  20.0kg
Fp  50.0 N
FT  ?
RRHS Physics Unit 1
Slide #55
Fg  mg
 (20.0)(9.80)
 196 N
Solution
FT  Fg  Fp  0
FT  Fg2  Fp2
 (196) 2  (50.0) 2
 202 N
tan  

Fp
Fg
50.0
196
  14.3
RRHS Physics Unit 1
Slide #56
Check Your Learning
Joe wishes to hang a sign weighing 750.0 N so that
cable A attached to the store makes a 30.0° angle as
shown in the picture below. Cable B is attached to an
adjoining building and is horizontal. Calculate the
necessary tension in cable B.
RRHS Physics Unit 1
Slide #57
Check Your Learning
Fg  750.0 N
  30.0
FTA  Fg  FTB  0
FTB  ?
tan  
FTB
Fg
FTB
tan 30.0 
750.0
FTB  433N
RRHS Physics Unit 1
Slide #58
Torque
• Even though forces
balance, object will spin
The size of a torque depends on two things:
1. The size of the force being applied (a larger force will
have a greater effect)
2. The distance away from the pivot point (the further
away from this pivot, the greater the effect).
RRHS Physics Unit 1
Slide #59
Torque
  F r
where torque is in Nm if force is in N and r is in m
RRHS Physics Unit 1
Slide #60
Example 1
Suppose that you are trying to open a door that is 70.0
cm wide. You are applying a force of 68 N 10.0 cm from
the outer edge of the door, but you are pushing at an
angle of 75o from the surface of the door. What torque
are you applying on the door?
RRHS Physics Unit 1
Slide #61
Solution
r  0.600m
F  68 N
  75
 ?
  F r
 Fy r
 Fr sin 
 (68)(0.600) sin 75
 39 N  m
RRHS Physics Unit 1
Slide #62
Rotational Equilibrium
Rotational Equilibrium
  0
 cw   ccw  0
 cw   ccw
When calculating torques, all distances must be
measured from the pivot point.
RRHS Physics Unit 1
Slide #63
Static Equilibrium
There are two conditions for static
equilibrium:
1. The sum of the forces is zero (providing
translational equilibrium), and
2. The sum of the torques is zero (providing
rotational equilibrium).
RRHS Physics Unit 1
Slide #64
Static Equilibrium
• Equilibrant force must provide both translational and
rotational equilibrium
• Center of gravity – the point at which we could apply a
single upward force to balance the object. For a mass
with a uniform distribution of mass (such as a ruler), the
center of gravity would be at the geometric center (the
middle of the ruler).
RRHS Physics Unit 1
Slide #65
Example 2
A 2.0 kg board serves as a see-saw for two children.
One child has a mass of 30.0 kg and sits 2.5 m from the
pivot point. At what distance from the pivot must a 25.0
kg child sit on the other side to balance the see-saw?
Assume that the board is uniform and centred over the
pivot.
mb  2.0kg
m1  30.0kg
m2  25.0kg
r1  2.5m
Fg1  m1 g
 (30.0)(9.80)
 294 N
r2  ?
RRHS Physics Unit 1
Slide #66
Fg 2  m2 g
 (25.0)(9.80)
 245 N
Solution
• Using the centre of
the board as the
pivot point
 cw   ccw
Fg 2 r2  Fg1r1
245r2  (294)(2.5)
r2  3.0m
RRHS Physics Unit 1
Slide #67
Example 3
A 4.0 m platform with a uniform distribution of mass has
a 3.2 kg box 0.80 m from the left end. The mass of the
platform is 2.0 kg. Calculate the size and location of the
required equilibrant force.
RRHS Physics Unit 1
Slide #68
Solution
mb  3.2kg
m p  2.0kg
rb  0.80m
rp  2.0m
req  ?
Fgb  mb g
 (3.2)(9.80)
 31.4 N
Translational Equilibrium
Rotational Equilibrium
  0
Fy  0
0  Feq  Fgb  Fgp
Feq  Fgb  Fgp
 31.4  19.6
 51N
 cw   ccw  0
 cw   ccw
Fgb rb  Fgp rp  Feq req
(31.4)(0.80)  (19.6)(2.0)  (51) req
req  1.3m
Feq  51N up, 1.3 m from the left end
Fgp  m p g
 (2.0)(9.80)
 19.6 N
RRHS Physics Unit 1
Slide #69
Check Your Learning
A uniform 1500 kg bridge, 20.0 m long, supports a 2200
kg truck whose centre of mass is 5.0 m from the right
support column as shown in the diagram below.
Calculate the force on each of the vertical support
columns.
RRHS Physics Unit 1
Slide #70
Check Your Learning
Using the left end as the pivot,
mb  1500kg
mt  2200kg
F2  ?
F1  ?
rb  10.0m
Fgb  mb g
 (1500)(9.80)
 14700 N
Fy  0
F1  F2  Fgb  Fgt  0
Fgt  mt g
rt  15.0m
 (2200)(9.80)
r2  20.0m
 21600 N
F1  F2  14700  21600
F1  F2  36300 N
r1  0
RRHS Physics Unit 1
Slide #71
Check Your Learning
  0
F1  F2  36300
 cw   ccw  0
 cw   ccw
Fgb rb  Fgt rt  F2 r2
(14700)(10.0)  (21600)(15.0)  F2 (20.0)
F2  23600 N
 24000 N
RRHS Physics Unit 1
Slide #72
F1  23600  36300
F1  12700 N
 13000 N
Module Summary
In this module you learned that
• An object is said to be in static equilibrium if it is in both
translational equilibrium and rotational equilibrium.
• Translational Equilibrium is achieved when the net force
is zero:
Fx  0
Fy  0
RRHS Physics Unit 1
Slide #73
Module Summary
• Torque can be calculated using the equation
  F r
• Rotational Equilibrium is achieved when the
net torque is zero:
  0
RRHS Physics Unit 1
Slide #74