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PHYSICS 149: Lecture 4
•  Chapter 2
–  2.3 Inertia and Equilibrium: Newton’s First Law of
Motion
–  2.4 Vector Addition Using Components
–  2.5 Newton’s Third Law
Lecture 4
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Net Force
The net force is the vector sum of all the forces
acting on an object.
= 10 N
East
= 10 N
West
If the book is at rest, it will continue
to stay at rest because Fnet=0
(translational equilibrium).
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= 10 N
East
= 12 N
West
If the book is at rest, it will start
moving toward the west since
|F2|>|F1| (Fnet ≠ 0).
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Equilibrium
•  When an object is in equilibrium (either at rest or
moving with constant velocity), the net force
acting on it zero. A vector can only have zero
magnitude if all of its components are zero.
•  Nonzero Net Force
–  If the net force on an object is not zero, it is not in
equilibrium (that is, a change in velocity will occur), and
thus Newton’s Second Law must be used to determine
its motion.
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Vectors
•  Two vectors A and B are
equal if they have the same
magnitude and direction,
whether or not they have the
same initial points.
•  A vector having the same
magnitude as A but in the
opposite direction to A is
denoted by –A.
A
B
A=B
A
-A
–  Note that A - A = A + (-A) = 0
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Vector Components
Note that
Ax and Ay are
scalars.
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Adding Vectors in 2-D
•  (a) Draw an arrow to
represent one of the vectors.
•  (b) Draw the second vector
with placing its “tail” at the
“tip” of the first vector.
•  (c) Draw an arrow starting
from the “tail” of the first and
ending at the “tip” of the
second. The new arrow is the
vector sum.
•  Note: it does not matter in
what order vectors are added
F1 + F2 = F2 + F1
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Vector Addition Using Components
D=A+B+C
if and only if
D x = Ax + Bx + C x
D y = Ay + By + C y
Adding x-components of vectors gives the xcomponent of the vector sum.
Adding y-components of vectors gives the ycomponent of the vector sum.
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Example
Two forces A and B of magnitudes |A| = 3 N and |B| =
5 N are applied to an object at the origin. We want to
find the total force. First, we have to calculate the
components of A:
B
60°
45°
A
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A) (3cos45°, 3sin60°) N
B) (5cos45°, 5sin45°) N
C) (3cos45°, 3sin45°) N
D) (3cos45°, –3sin45°) N
E) (–3cos45°, –3sin45°) N
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Example
Two forces A and B of magnitudes |A| = 3 N and |B| =
5 N are applied to an object at the origin. We want to
find the total force. The components of B are:
B
60°
45°
A
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A) (–5cos60°, 5sin60°) N
B) (–5cos60°, 3sin60°) N
C) (5cos60°, 5sin60°) N
D) (5cos60°, –5sin60°) N
E) (–3cos60°, 5sin60°) N
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Example
We now found A = (2.12 N, –2.12 N) and B = (–2.5
N, 4.33 N). The components of the total force A+B
are:
A) (–0.38, 6.45) N
B) (–4.62, 6.45) N
C) (–0.38, 2.21) N
D) (–4.62, 2.21) N
E) (–0.38, -2.21) N
B
60°
45°
A
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Example
•  We now found A + B = (–0.38 N, 2.21 N)
•  The magnitude is
The angle θ is given by
A+B
θ = tan-1|2.21/0.38|
θ
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Vector Subtraction
•  A – B = A + (–B)
•  Thus, A – B is equivalent to adding A and –B.
•  (Note that –B is a vector having the same magnitude as B but in the
opposite direction to B.)
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Example
Vector A in the figure is equal to:
(A)  B - D
(B)  B + C
(C)  C + D + E
(D)  (D - C)/2
(E)  D + C
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Example
Vector D in the figure is equal to:
(A)  A + B
(B)  C + A
(C)  C – A
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Example
Vector E in the figure is equal to:
(A)  –2(B+C)
(B)  –(C+D)
(C)  –2A
(D)  –(A+B+C)
(E)  All of these
choices
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C
C+D
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Newton’s Laws of Motion
1. 
2. 
An object at rest remains at rest
and an object in motion remains
in motion unless acted upon by
an unbalanced force (law of Inertia)
An unbalanced force is equal to
the rate of change of momentum
( F = ma )
3. 
For every action, there is an equal
and opposite reaction.
Isaac Newton
(1643-1727)
Humans could explain/understand the most complex
phenomena with only 3 assumptions.
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Newton’s First Law of Motion
•  An object’s velocity (a vector) does not change if
and only if the net force acting on the object is
zero.
•  In other words, if there is no net force on an
object, its speed and direction of motion do not
change (including if it is at rest).
•  Also called “the law of inertia.” Inertia means
resistance to changes in velocity.
–  Example: the start or stop of a car’s motion, dusts on
clothes, a quarter on top of an index card on a cup.
–  Note that circular motion is NOT a constant motion,
that is, net force is not zero. Why?
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Galileo’s Experiment
•  Galileo postulated that, if he could eliminate all resistive
forces, a ball rolling on a horizontal surface would never stop.
h 1 = h2
h 1 = h2
•  Newton was influenced by some of his predecessors including
Galileo.
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ILQ 1
The net force on a moving object suddenly
becomes zero. The object then:
A)
B)
C)
D)
E)
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Stops abruptly
Stops during a short time interval
Changes direction
Continues at constant velocity
Slows down gradually
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ILQ 2
Is it possible for an object to be in motion and
yet have zero net force acting on it?
A)  YES
B)  NO
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Second Law
• 
This is a vector relation.
• 
The direction of the net force is the same as
the direction of the acceleration.
• 
Example in one dimension:
Sum of forces:
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2N
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Third Law
•  All forces come in pairs
•  Third law forces involve TWO OBJECTS.
•  The two forces are:
–  the force object one exerts on object two
–  the force object two exerts on object one
•  Three ways to state the 3rd law:
–  Forces on each other are equal and opposite
–  For every action there is an equal and opposite reaction
–  You can’t push on something without it pushing back on
you
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Newton’s Third Law of Motion
•  In an interaction between two objects, each
object exerts a force on the other. These two
forces are equal in magnitude and opposite in
direction.
–  To every action, there is always opposed an equal
reaction.
–  Forces always come in equal but opposite actionreaction pair.
•  Note that these two forces act on different
objects; they do not cancel in any way.
•  Don’t forget that forces always exist in pairs.
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Free-Body Diagram (FBD)
•  A simplified sketch of a single object with force
vectors drawn to represent every force acting “on”
that object. (It must not include any forces that
act on other objects.)
•  FBD is useful to find the net force acting on an
object.
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ILQ
A book is sitting on a desk top. Identify the 3rd law
partner of the weight of the book.
A)
B)
C)
D)
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The force of the desk on the book.
The force of the book on the desk.
The force of the earth on the book.
The force of the book on the earth.
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Why Doesn’t the Book Move?
Free-Body Diagram
Forces
acting
“on” book
Pair
-N
-W
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Pair
FBD includes all forces acting
“on” the book “only.”
The book does not move
because N + W = 0.
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