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Chapter 4
The Laws of Motion
The Laws of Motion

Kinematics
 Math of HOW things move
○ Position, velocity, acceleration

Dynamics
 WHY do things move?
 What causes a body to accelerate?
○ Forces => Acceleration

The properties of force and the
relationships between force and
acceleration are given by Newton’s Three
Laws of Motion
The Laws of Motion

The First Law describes the natural
state of motion of a body on which no
forces are acting. The other two laws
describe the behavior of a body under
the influence of forces.

Early 1600’s, theories of object’s
tendency to be at rest
Laws of Motion

Galileo Galilei
 Developed first correct ideas of motion
○ Gravity and constant acceleration
○ Forces acting on bodies

Sir Isaac Newton
 1687 Principia Mathematica
○ Laws of Motion
○ Law of Universal Gravitation
 Invented calculus to further describe speed,
acceleration
Forces

Force
 A central concept in all of physics
 A vector quantity
○ Magnitude
○ Direction
 Force is used to describe push or a pull
 Forces on objects
○ springs
○ rubber bands
○ ropes
○ Cables
Forces (cont.)

Force
 Bouyant Forces
○ liquids
 Friction
○ Surfaces

All examples above are known as
“contact forces”
Newton’s First Law

Newton’s First Law
“An object moves with a velocity that is
constant in magnitude and direction, unless
acted on by a nonzero net force.”
The net force on an object is defined as the
vector sum of all external forces exerted on
the object.
 Often called the Law of inertia

 Tendency of object in motion to stay in motion
ConcepTest 4.2 Cart on Track I
Consider a cart on a
horizontal frictionless
table. Once the cart has
1) slowly come to a stop
2) continue with constant acceleration
been given a push and
3) continue with decreasing acceleration
released, what will
4) continue with constant velocity
happen to the cart?
5) immediately come to a stop
ConcepTest 4.2 Cart on Track I
Consider a cart on a
horizontal frictionless
table. Once the cart has
1) slowly come to a stop
2) continue with constant acceleration
been given a push and
3) continue with decreasing acceleration
released, what will
4) continue with constant velocity
happen to the cart?
5) immediately come to a stop
After the cart is released, there is no longer a force in
the x-direction. This does not mean that the cart stops
moving!! It simply means that the cart will continue
moving with the same velocity it had at the moment of
release. The initial push got the cart moving, but that
force is not needed to keep the cart in motion.
Newton’s Second Law

Newton’s Second Law
“The acceleration a of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.”
 What it means:
a = ΣF / m
or conversely,
ΣF = ma
Newton’s Second Law
ΣF = ma
 A LAW
of nature!
 Precise definition of FORCE
 a and F are in the same
direction
 after F is completely summed
Newton’s Second Law (cont.)
ΣF = ma

ΣF
 ΣFx = max
 ΣFy = may
 ΣFz = maz

No net force, means acceleration is zero
 Velocity is constant
Newton’s Second Law (cont.)

Units
 Mass
○ Kilograms kg
 Acceleration
○ m/s2
 Force
○ Newtons 1N = 1 kg x m/s2 (m x a)
○ Pounds
1N = 0.225lb
○ Pound is defined as F = ma = slug x ft/s2
Newton’s Second Law
Definition of Mass (physics)
A measure of resistance a body
offers to changes in its velocity
(acceleration)
Standard is kilogram
 Masses can be compared by balances
 Mass vs. Weight

Newton’s 2nd Law proves that different masses
accelerate to the earth at the same rate, but with
different forces.
We know that objects
with different masses
accelerate to the
ground at the same
rate.
 However, because of
the 2nd Law we know
that they don’t hit the
ground with the same
force.

F = ma
F = ma
98 N = 10 kg x 9.8 m/s/s
9.8 N = 1 kg x 9.8
m/s/s
Newton’s Second Law (cont.)
 Example
A mass of 0.2kg slides along the
table with a velocity of v = 2.8m/s.
It stops in 1.0 m. What force is
acting on the mass (neglect
friction)?
Newton’s Third Law

Newton’s Third Law

“If object 1 and object 2 interact, the force
F12 exerted by object 1 on object 2 is equal
in magnitude but opposite to the force F21
exerted by object 2 on object 1.”
 What it means:
○ “for every action, there is an equal and
opposite reaction”
Newton’s Third Law

Action-Reaction Pair
F21 = -F12



Forces in nature always act in
pairs
No single isolated force
The mutual actions of two
bodies upon each other are
ALWAYS equal and directed
contrary to one another
Newton’s Third Law

Action-Reaction Pairs
 Newton’s Law uses the forces
acting on an object
 n and F are both acting on the
object
 n is referred as to the normal
force and is the force exerted by
the TV stand on the TV
Applications of Newton’s Laws

An object in equilibrium has no net
external force acting on it, and the
second law, in component form, implies
that
ΣFx = 0
and
ΣFy = 0
for such an object. These two equations
are useful for solving problems where
the object is at rest or moving at
constant velocity.
Applications of Newton’s Laws

An object under acceleration requires the
same two equations, but with the
acceleration terms included:
ΣFx = max
and
ΣFy = may
Applications of Newton’s Laws

Assumptions
 Objects behave as particles
○ can ignore rotational motion (for now)
 Masses of strings or ropes are negligible
○ No stretching – constant length
 Interested only in the forces acting on the
object
○ can neglect reaction forces
 Pulleys are massless and frictionless
○ Used to change direction
Applications of Newton’s Laws

Free Body Diagram
Diagram representing all the forces applied
to an object
 Represent the object as a dot
 Identify all the forces acting on the object,
not exerted
 Choose appropriate coordinate system
 Incorrect FBD means incorrect solution
Applications of Newton’s Laws
The force T is the tension
acting on the box
 n and F are the forces exerted
by the earth and the ground
 Only forces acting directly on
the object are included in the
free body diagram

 Reaction forces act on other
objects and so are not included
Newton’s Third Law

Free Body Diagram
n
ΣFy = N – Fg = n – mg = may
but the TV is not moving, so ay = 0
n – mg = 0
Therefore,
n = mg
Fg
Solving Newton’s Second Law Problems


Read the problem at least once
Draw a picture of the system
 Identify the object of primary interest
 Indicate forces with arrows

Label each force
 Use labels that bring to mind the physical quantity
involved

Draw a free body diagram
 If additional objects are involved, draw separate free body
diagrams for each object
 Choose a convenient coordinate system for each object

Apply Newton’s Second Law
 The x- and y-components should be taken from the vector
equation and written separately

Solve for the unknown(s)
ConcepTest 4.11 On an Incline
Consider two identical blocks,
1) case A
one resting on a flat surface,
2) case B
and the other resting on an
incline. For which case is the
normal force greater?
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
ConcepTest 4.11 On an Incline
Consider two identical blocks,
1) case A
one resting on a flat surface,
2) case B
and the other resting on an
incline. For which case is the
normal force greater?
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
In Case A, we know that N = W.
y
In Case B, due to the angle of
the incline, N < W. In fact, we
N
f
can see that N = W cos(q).
q
W
q
Wy
x
Newton’s Law review

1st – Law of Inertia
 Objects at rest or motion will stay that way unless acted
on by a force
○ Sailboat
○ Target

2nd – F = ma
 The acceleration is proportional to the force and inversely
proportional to the mass

3rd – Action-Reaction
 if a force is acted on to an object, the object will exert an
equal and opposite force
○ Recoil
ConcepTest 4.1c Newton’s First Law
You put your book on
the bus seat next to
you. When the bus
stops suddenly, the
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
book slides forward off
4) it did not move, but only seemed to
the seat. Why?
5) gravity briefly stopped acting on it
ConcepTest 4.1c Newton’s First Law
You put your book on
the bus seat next to
you. When the bus
stops suddenly, the
1) a net force acted on it
2) no net force acted on it
3) it remained at rest
book slides forward off
4) it did not move, but only seemed to
the seat. Why?
5) gravity briefly stopped acting on it
The book was initially moving forward (since it was
on a moving bus). When the bus stopped, the book
continued moving forward, which was its initial state
of motion, and therefore it slid forward off the seat.
Follow-up: What is the force that usually keeps the book on the seat?
ConcepTest 4.9b Going Up II
A block of mass m rests on the
1) N > mg
floor of an elevator that is
2) N = mg
accelerating upward. What is
3) N < mg (but not zero)
the relationship between the
4) N = 0
force due to gravity and the
5) depends on the size of the
elevator
normal force on the block?
a
m
ConcepTest 4.9b Going Up II
A block of mass m rests on the
1) N > mg
floor of an elevator that is
2) N = mg
accelerating upward. What is
3) N < mg (but not zero)
the relationship between the
force due to gravity and the
normal force on the block?
4) N = 0
5) depends on the size of the
elevator
The block is accelerating upward, so
it must have a net upward force. The
N
m
a>0
forces on it are N (up) and mg (down),
so N must be greater than mg in order
to give the net upward force!
Follow-up: What is the normal force if
the elevator is in free fall downward?
mg
S F = N – mg = ma > 0
\ N > mg
Applications of Newton’s Laws

Equilibrium
 An object either at rest or moving with a
constant velocity is said to be in equilibrium
 The net force acting on the object is zero
(since the acceleration is zero)
ΣF = 0
 Should use components
ΣFx = 0
ΣFy = 0
Applications of Newton’s Laws

Equilibrium Example - FBD
Applications of Newton’s Laws

Equilibrium Example – FBD
 Choose the coordinate
system with x along the
incline and y perpendicular to
the incline
 Replace the force of gravity
with its components
Applications of Newton’s Laws

Equilibrium Example?
Applications of Newton’s Laws

Equilibrium – Multiple Objects
 When you have more than one object, the
problem-solving strategy is applied to each
object
 Draw free body diagrams for each object
 Apply Newton’s Laws to each object
 Solve the equations
Applications of Newton’s Laws

Example

A traffic light weighing
1.00x102 N hangs from a
vertical cable tied to two
other cables that are
fastened to a support, as in
the figure to the right. The
upper cables makes angles
of 37° and 53° with the
horizontal. Find the tension
in each of the three cables.
Applications of Newton’s Laws

Example:
An object with a mass m1 =
5.00kg rests on a frictionless
horizontal table and is
connected to a cable that
passes over a pulley and is
then fastened to a hanging
object with mass m2 = 10.0 kg,
as shown in Figure P4.30. Find
the acceleration of each object
and the tension in the table.
Forces of Friction
An object moving on a surface encounters
resistance as it interacts through its
surroundings. This resistance is called
friction.





Friction is a force
The force of friction is opposite of motion
Two types of friction – static and kinetic
Friction is proportional to the normal force
Friction Examples


Car on the road
Forces of Friction

Static Friction, ƒs
 Static friction acts to keep
the object from moving
 If F increases, so does ƒs
 If F decreases, so does
ƒs
 ƒs  µ n
Forces of Friction
• Kinetic Friction, ƒk
– The force of kinetic friction
acts when the object is in
motion
– ƒk = µ n
• Variations of the
coefficient with speed
will be ignored
Forces of Friction

Friction
Examples
Forces of Friction

Friction Example
 Axes are rotated as
usual on an incline
 The direction of
impending motion would
be down the plane

Friction acts up the
plane
 Opposes the motion

Apply Newton’s Laws
and solve equations
Static Friction...
 We
want to know how it acts in fixed or “static” systems:
the force provided by friction depends on the forces
applied on the system (magnitude: fs ≤ ms N)
 Opposes
motion that would occur if ms were zero
N
Fapplied
fS
y
x
mg
Static Friction...
 If
a = 0.
x:
y:
Fapplied  fS = 0
N = mg
the block is static: fS  Fapplied
(unlike kinetic friction)
While
N
Fapplied
fS
y
x
mg
Static Friction...
 The
maximum possible force that the friction
between two objects can provide is fMAX = mSN,
where ms is the “coefficient of static friction”.
 So fS  mS N.
 As one increases F, fS gets bigger until fS = mSN
and the object “breaks loose” and starts to move.
N
F
y
x
fS
mg
Static Friction...
mS
is discovered by increasing F until
the block starts to slide:
FMAX  mSN = 0
N = mg
x:
y:
mS  FMAX / mg
N
FMAX
mSmg
y
x
mg
Additional comments on
Friction:
The
force of friction does not depend
on the area of the surfaces in contact
(a relatively good approximation if
there is little surface deformation)

Generally mS > mK for any system
Kinetic Friction
Dynamics:
x-axis max = F  mKN
y-axis may = 0 = N – mg or N = mg
so
F  mKmg = m ax
fk
v
y
N
F
max
fk
mK mg
mg
x
Frictionless inclined plane
 A block
of mass m slides down a
frictionless ramp that makes angle q
with respect to horizontal. What is its
acceleration a ?
m
a
q
Angles of the inclined
plane
max = mg sin q
q
N
q + f  90
q
f
mg
q
Case 1 - Frictionless inclined plane...
Use
a FBD and consider x and y
components separately:
 Fx
 Fy
max = mg sin q
may = 0 = N – mgcos q
ax = g sin q
N = mg cos q
N
max
y
mg cos q
q
mg sin q
mg
q
x
Case 2 - Inclined plane...static friction
Use
a FBD and consider x and y
components separately:\
 Fx
max = 0 = mgsin q - f
 Fy may
Friction
Force
= 0 = N – mgcos q
N = mg cos q
N
y
Special case:
max= 0
mg cos q
0 = g sin q - f
At the breaking point
q
x
mg sin q
mg
q
f = ms N = ms mg cos q
g sin q = f = ms mg cos q
Example

A roller coaster reaches the top of the
steepest hill with a speed of 6.0 km/hr. It then
descends the hill, which is at an average of
45º and is 45.0m. What will its speed be at the
bottom? Assume µk = 0.18.
Air Resistance and Drag

So far we’ve “neglected air resistance” in physics
 Can be difficult to deal with

Affects projectile motion
 Friction force opposes velocity through medium
 Imposes horizontal force, additional vertical forces
 Terminal velocity for falling objects

Dominant energy drain on cars, bicyclists, planes
Drag Force Quantified

With a cross sectional area, A (in m2), coefficient of drag of 1.0
(most objects),  sea-level density of air, and velocity, v (m/s), the
drag force is:
D = ½ C  A v2  c A v2
c = ¼ kg/m3
○
in Newtons
Increases as v increases
In falling, when D = mg, then at terminal velocity
 Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of
0.5 m2 exerts ~30 Newtons
Requires (F v) of power  300 Watts to maintain speed
 Minimizing drag is often important

“Free” Fall
 Terminal
velocity reached when
Fdrag = Fgrav (= mg)
= ¼Av2 ≈ mg
 v ≈ √4mg/A
D
 For
75 kg person with a frontal area of
0.5 m2,
vterm  50 m/s, or 110 mph
which is reached in about 5 seconds, over
125 m of fall
Trajectories with Air Resistance
 Baseball
launched at 45° with v = 50
m/s:
 Without air resistance, reaches about 63 m
high, 254 m range
 With air resistance, about 31 m high, 122 m
range
Vacuum trajectory vs. air trajectory for 45° launch angle.
‘Free’ Fall

Indoor sky-diving