1. Newton's first law of motion states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. 2. Inertia is the resistance of an object to changes in its motion. According to the first law, objects have a natural tendency to resist changes to their state of motion. 3. Examples are given demonstrating inertia, including a tablecloth trick, passengers in a abruptly stopping car, and a ladder on a abruptly stopping truck. In each case, objects in motion continue in motion according to their inertia unless an external force stops them.

1. VECTORS AND SCALARS
A scalar quantity has only magnitude and is completely
specified by a number and a unit.
Examples are: mass (2 kg), volume (1.5 L), and
frequency (60 Hz).
Scalar quantities of the same kind are added by using
ordinary arithmetic.

2. A vector quantity has both magnitude and direction.
Examples are displacement (an airplane has flown 200
km to the southwest), velocity (a car is moving at 60
km/h to the north), and force (a person applies an
upward force of 25 N to a package). When vector
quantities are added, their directions must be taken into
account.

3. A vector is represented by an arrowed line whose length
is proportional to the vector quantity and whose
direction indicates the direction of the vector quantity.

6. The resultant, or sum, of a number of vectors of a
particular type (force vectors, for example) is that single
vector that would have the same effect as all the original
vectors taken together.
R

10. VECTOR COMPONENTS
A vector in two dimensions may be resolved into two
component vectors acting along any two mutually
perpendicular directions.

13. 2.1 Draw and calculate the components of the vector
F (250 N, 235o
)
Fx
= F cos θ
= 250 cos (235o
)
= - 143.4 N
Fy
= F sin θ
= 250 sin (235o
)
= - 204.7 N
Fx
Fy
F

14. VECTOR ADDITION: COMPONENT METHOD
To add two or more vectors A, B, C,… by the component
method, follow this procedure:
1. Resolve the initial vectors into components x and y.
2. Add the components in the x direction to give Σx
and
add the components in the y direction to give Σy
.
That is, the magnitudes of Σx
and Σy are given by,
respectively:
Σx
= Ax
+ Bx
+ Cx…
Σy
= Ay
+ By
+ Cy…

15. 3. Calculate the magnitude and direction of the
resultant R from its components by using the
Pythagorean theorem:
and
2 2
x yR = Σ + Σ
1
tan
y
x
θ −
Σ
=
Σ

16. 2.2 Three ropes are tied to a stake and the following forces are
exerted. Find the resultant force.
A (20 N, 0º)
B (30 N, 150º)
C (40 N, 232º)

17. A (20 N, 0°)
B (30 N, 150°)
C (40 N, 232°)
x-component
20 cos 0°
30 cos 150°
40 cos 232°
Σx = - 30.6 N
= 34.7 N2 2
( 30.6) ( 16.5)R = − + −
y-component
20 sin 0°
30 sin 150°
40 sin 232°
Σy = -16.5 N

19. = 28.3°
Since Σx = (-) and Σy = (-)
R is in the III Quadrant:
1
tan
y
x
θ −
∑
=
∑
1 16.5
tan
30.6
−
=
therefore:
180° + 28.3° = 208.3°
R (34.7 N, 208.3°)

20. 2.3 Four coplanar forces act on a body at point O as shown in the
figure. Find their resultant with the component method.
A (80 N, 0°)
B (100 N, 45°)
C (110 N, 150°)
D (160 N, 200°)

21. A (80 N, 0°) B (100 N, 45°)
C (110 N, 150°) D (160 N, 200°)
x-component y-component
80 0
100 cos 45° 100 sin 45°
110 cos 150° 110 sin 150°
160 cos 200° 160 sin 200°
Σx = - 95 N Σy = 71 N
= 118.6 N2 2
( 95) (71)R = − +

22. = 36.7°
Since Σx = (-) and Σy = (+)
R is in the II Quadrant,
therefore:
180° - 36.7°= 143.3°
R (118.6 N, 143.3°)
1 71
tan
95
θ −
=

23. 22
)2.3()7.5( −+=R = 6.5 N
= 29°
Since Σx = (+) and Σy = (-)
R is in the IV Quadrant,
therefore:
360° - 29°= 331°
R (6.5 N, 331°)
1 3.2
tan
5.7
θ −
=

24. AP PHYSICS LAB 2. VECTOR ADDITION
Objective:
The purpose of this experiment is to use the force table to
experimentally determine the equilibrant force of two and
three other forces. This result is checked by the component
method.
A system of forces
is in equilibrium
when a force
called the
equilibrant force is
equal and opposite
to their resultant
force.

25. Equipment
Force Table Set

26. FORCE MASS (kg) FORCE
(N)
mg = m (9.8 m/s2
)
DIRECTION
F1
F2
Equilibrant
FE
Resultant
FR
DATA Table

27. An object that experiences a push or a pull has a force
exerted on it. Notice that it is the object that is
considered. The object is called the system. The world
around the object that exerts forces on it is called the
environment.
system

28. FORCE
Forces can act either through the physical contact of
two objects (contact forces: push or pull) or at a
distance (field forces: magnetic force, gravitational
force).

29. Type of Force and
its Symbol
Description of Force Direction of Force
Applied
Force
An applied force is a force that is applied to an
object by another object or by a person. If a
person is pushing a desk across the room, then
there is an applied force acting upon the desk. The
applied force is the force exerted on the desk by
the person.
FA
In the direction of
the pull or push.

30. Type of Force and
its Symbol
Description of Force
Direction of Force
Normal
Force
The normal force is the support force exerted upon an
object that is in contact with another stable object.
For example, if a book is resting upon a surface, then
the surface is exerting an upward force upon the book
in order to support the weight of the book. The normal
force is always perpendicular to the surface
Perpendicular to
the surface
FN

31. Type of Force and
its Symbol
Description of Force
Direction of Force
Friction
Force
The friction force is the force exerted by a surface as
an object moves across it or makes an effort to move
across it. The friction force opposes the motion of the
object. For example, if a book moves across the
surface of a desk, the desk exerts a friction force in
the direction opposite to the motion of the book.
Opposite to the
motion of the
object
FF

32. Type of Force and
its Symbol
Description of Force
Direction of Force
Air
Resistance
Force
Air resistance is a special type of frictional force that acts
upon objects as they travel through the air. Like all
frictional forces, the force of air resistance always
opposes the motion of the object. This force will frequently
be ignored due to its negligible magnitude. It is most
noticeable for objects that travel at high speeds (e.g., a
skydiver or a downhill skier) or for objects with large
surface areas.
Opposite to the
motion of the
object
FD

33. Type of Force and
its Symbol
Description of Force
Direction of Force
Tensional
Force
Tension is the force that is transmitted through a
string, rope, or wire when it is pulled tight by
forces acting at each end. The tensional force is
directed along the wire and pulls equally on the
objects on either end of the wire.
In the direction of
the pull
FT

34. Type of Force and
its Symbol
Description of Force
Direction of Force
Gravitational
Force (also
known as
Weight)
The force of gravity is the force with which the
earth, moon, or other massive body attracts an
object towards itself. By definition, this is the
weight of the object. All objects upon earth
experience a force of gravity that is directed
"downward" towards the center of the earth. The
force of gravity on an object on earth is always
equal to the weight of the object.
Straight downward
Fg

35. FORCES HAVE AGENTS
Each force has a specific
identifiable, immediate cause
called agent. You should be
able to name the agent of
each force, for example the
force of the desk or your
hand on your book. The
agent can be animate such
as a person, or inanimate
such as a desk, floor or a
magnet. The agent for the
force of gravity is Earth's
mass. If you can't name an
agent, the force doesn't
exist.
agent

36. Directions:
- Choose a coordinate system defining the positive
direction of motion.
- Replace the object by a dot and locate it in the center
of the coordinate system.
- Draw arrows to represent the forces acting on the
system.
A free-body-diagram (FBD) is a vector
diagram that shows all the forces that
act on an object whose motion is being
studied.

37. FN
FG

38. FN
FG

39. FN
FG
FF

40. FG
FN
FGX
FGY

41. FG
FN
FF

42. FT
FG

43. FT2
FG
FT1

44. FT2
FG
FT1

45. FG
FN
FT

46. FG
FN1
FN2

47. FN
FG
FF
FA

48. FN
FG
FF
FA
θ

49. FT
FG

50. FN
FG
FF
FA
θ

51. FG

52. FD
FG

53. FG
17.The ball has been punted by a football player.

54. ARISTOTLE studied motion and divided it
into two types: natural motion and violent
motion.
Natural motion: up or down.
Objects would seek their natural resting
places.
Natural for heavy things to fall and for
very light things to rise.
Violent motion: imposed motion.
A result of forces.

55. GALILEO demolish the notion that a force
is necessary to keep an object moving.
He argued that ONLY when friction is
present, is a force needed to keep an
object moving.
In the absence of air resistance (drag)
both objects will fall at the same time.

56. A ball rolling down the incline rolls up the opposite
incline and reaches its initial height.
As the angle of the upward incline is reduced, the ball
rolls a greater distance before reaching its initial height.
How far will the ball roll along the horizontal?

58. FIRST LAW OF MOTION
According to Newton's
First Law of Motion:
" If no net force acts on it, a
body at rest remains at rest and
a body in motion remains in
motion at constant speed in a
straight line."
Isaac Newton
(1642-1727)

59. NEWTON'S FIRST LAW OF MOTION
"An object at rest tends to stay at rest and an object
in motion tends to stay in motion with the same speed
and in the same direction unless acted upon by an
unbalanced force."
There are two parts to this statement:
- one which predicts the behavior ofstationary
objects and
- the other which predicts the behavior of moving
objects.

61. The behavior of all objects can be described by saying
that objects tend to "keep on doing what they're doing"
(unless acted upon by an unbalanced force).
If at rest, they will continue in this same state of rest.
If in motion with an eastward velocity of 5 m/s, they will
continue in this same state of motion (5 m/s, East).

62. It is the natural tendency of objects to resist
changes in their state of motion.
This tendency to resist changes in their state of
motion is described as inertia.
Inertia is the resistance an object has to a change in
its state of motion.
The elephant at rest tends
to remain at rest.

64. Tablecloth trick:
Too little force, too little time to
overcome "inertia" of tableware.

65. If the car were to abruptly stop and the seat belts
were not being worn, then the passengers in
motion would continue in motion.
Now perhaps you will be convinced of the need to wear
your seat belt. Remember it's the law - the law of inertia.

66. If the truck were to abruptly stop and the straps were
no longer functioning, then the ladder in motion
would continue in motion.

67. If the motorcycle were to abruptly stop, then the
rider in motion would continue in motion. The rider
would likely be propelled from the motorcycle and
be hurled into the air.

71. FIRST CONDITION FOR EQUILIBRIUM
A body is in translational equilibrium if and only if the
vector sum of the forces acting upon it is zero.
Σ Fx
= 0 Σ Fy
= 0

72. Example of Free Body Diagram
300 600
50 N
A AB
B
Fg
300
600
Bx
By
Ax
Ay
1. Draw and label a sketch.1. Draw and label a sketch.
2. Draw and label vector force diagram. (FBD)2. Draw and label vector force diagram. (FBD)
3. Label x and y components opposite and3. Label x and y components opposite and
adjacent to angles.adjacent to angles.

73. 2.4 A block of weight 50 N hangs from a cord that is knotted to two
other cords, A and B fastened to the ceiling. If cord B makes an
angle of 60˚ with the ceiling and cord A forms a 30° angle, draw the
free body diagram of the knot and find the tensions A and B.
A B
50 N
ΣFx
= B cos 60º - A cos 30º = 0
cos30
cos60
A
B =
o
o = 1.73 A
ΣFy
= B sin 60º + A sin 30º - 50 = 0
1.73 A sin 60º + A sin 30º = 50
1.5 A + 0.5 A = 50
A = 25 N
B = 1.73 (25) = 43.3 N
A = 25 N B = 43.3 N
60º30º
Ax
Ay
By
Bx
N1L

74. 2.5 A 200 N block rests on a frictionless inclined plane of slope
angle 30º. A cord attached to the block passes over a frictionless
pulley at the top of the plane and is attached to a second block.
What is the weight of the second block if the system is in
equilibrium?
FN
200 N
FG2
FT
FT
x
yθ
ΣFx
= FT
- 200 sin 30º = 0
FT
= 200 sin 30º
= 100 N
ΣFy
= FT
- FG
= 0
FT
= FG2
FG2
= 100 N
N1L

75. To swing open a door, you
exert a force.
The doorknob is near the
outer edge of the door. You
exert the force on the
doorknob at right angles to
the door, away from the
hinges.
To get the most effect from
the least force, you exert
the force as far from the
axis of rotation (imaginary
line through the hinges) as
possible.

76. TORQUE
Torque is a measure of a force's ability to rotate an
object.

77. Torque is Determined by Three Factors:
• The magnitude of the applied force.
• The direction of the applied force.
• The location of the applied force.
• The magnitude of the applied force.
• The direction of the applied force.
• The location of the applied force.
20 N
Magnitude of force
40 N
The 40-N force
produces twice the
torque as does the
20-N force.
Each of the 20-N
forces has a different
torque due to the
direction of force. 20 N
Direction of Force
20 N
θ
θ
20 N20 N
Location of force
The forces nearer the end
of the wrench have
greater torques.
20 N
20 N

78. The perpendicular
distance from the axis
of rotation to the line of
force is called the lever
arm of that force. It is
the lever arm that
determines the
effectiveness of a given
force in causing
rotational motion. If the
line of action of a force
passes through the
axis of rotation (A) the
lever arm is zero.

79. Units for Torque
Torque
Depends on the magnitude of the applied
force and on the length of the lever arm,
according to the following equation.
r is measured perpendicular to the line of
action of the force F
Torque
Depends on the magnitude of the applied
force and on the length of the lever arm,
according to the following equation.
r is measured perpendicular to the line of
action of the force F
τ = Frτ = Fr
Units: N⋅m
6 cm
40 N
τ = (40 N)(0.60 m)
= 24.0 N⋅m

80. Applying a Torque

81. Sign Convention:
Torque will be positive if F tends
to produce counterclockwise
rotation.
Torque will be negative if F tends
to produce clockwise rotation.

82. ROTATIONAL EQUILIBRIUM
An object is in rotational equilibrium when the sum of
the forces and torques acting on it is zero.
First Condition of Equilibrium:
Σ Fx
= 0 and Σ Fy
= 0
(translational equilibrium)
Second Condition of Equilibrium:
Στ = 0
(rotational equilibrium)
By choosing the axis of rotation at the point of
application of an unknown force, problems may
be simplified.

83. CENTER OF MASS
The terms "center of mass" and "center of gravity" are
used synonymously in a uniform gravity field to
represent the unique point in an object or system that
can be used to describe the system's response to
external forces and torques.
The concept of the center of mass is that of an average
of the masses factored by their distances from a
reference point. In one plane, that is like the balancing of
a seesaw about a pivot point with respect to the torques
produced.

84. Center of Gravity
The center of gravity of an object is the point at
which all the weight of an object might be
considered as acting for purposes of treating
forces and torques that affect the object.
The single support force has line of action that passes
through the c. g. in any orientation.

85. Examples of Center of
Gravity
Note: C. of G. is not always inside material.

86. 2.6 A 300 N girl and a 400 N boy stand on a 16 m platform
supported by posts A and B as shown. The platform itself weighs
200 N. What are the forces exerted by the supports on the
platform?

87. 2200
12
A = = 183. 3 N
B = 900 - 183.3
= 716.7 N
ΣF = 0
A + B - 300 - 200 - 400 = 0
A + B = 900 N
Selecting B as the hinge
ΣτB
= 0
-A(12) +300(10) +200(4) - 400(4) = 0
- A12 + 3000 + 800 - 1600 = 0

88. = 716.7 N
A = 900 - 716.7
= 183. 3 N
Selecting A as the hinge
ΣτA
= 0
- 300(2) - 200(8) + B(12) - 400(16) = 0
- 600 - 1600 + B12 - 6400 = 0
8600
12
B =

89. 2.7 A uniform beam of negligible weight is held up by two
supports A and B. Given the distances and forces listed, find
the forces exerted by the supports.
ΣF = 0
A - 60 - 40 + B = 0
A + B = 100
ΣτA
= 0
= - 60 (3) - 40(9) + B(11) = 0A B
60 N 40 N
3 m 6 m 2 m
540
11
B = = 49.1 N
A = 100 - 49.1 = 50.9 N