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Introduction to Newton’s Laws
Newton’s First Law.
Isaac Newton
Arguably the greatest scientific genius ever.
Came up with 3 Laws of Motion to explain the
observations and analyses of Galileo and Johannes
Kepler.
Discovered that white light was composed of many
colors all mixed together.
Invented new mathematical techniques such as
calculus and binomial expansion theorem in his
study of physics.
Published his Laws in 1687 in the book
Mathematical Principles of Natural Philosophy.
What is Force?
A force is a push or pull on an object.
Forces cause an object to
accelerate…



To speed up
To slow down
To change direction
Newton’s First Law
A body in motion stays in motion at
constant velocity and a body at rest
stays at rest unless acted upon by a
net external force.
This law is commonly referred to as
the Law of Inertia.
The First Law is
Counterintuitive
Aristotle firmly believed this.
But Physics B students know better!
Implications of Newton’s 1st
Law
If there is zero net force on a body,
it cannot accelerate, and therefore
must move at constant velocity, which
means



it cannot turn,
it cannot speed up,
it cannot slow down.
What is Zero Net Force?
The table pushes up
on the book.
A book rests on a table.
FT
Physics
Gravity pulls down on
the book.
FG
Even though there are forces on the book, they are balanced.
Therefore, there is no net force on the book.
SF = 0
Diagrams
Draw a force diagram and a free body
diagram for a book sitting on a table.
Force Diagram
N
Physics
W
Free Body Diagram
N
W
Sample Problem
a) A monkey hangs by its tail from a tree
branch. Draw a force diagram
representing all forces on the monkey
FT
FG
Sample Problem
b) Now the monkey hangs
by both hands from two
vines. Each of the
monkey’s arms are at a
45o from the vertical.
Draw a force diagram
representing all forces
on the monkey.
Fa1
Fa2
FG
Mass and Inertia
Chemists like to define mass as the
amount of “stuff” or “matter” a
substance has.
Physicists define mass as inertia,
which is the ability of a body to
resist acceleration by a net force.
Sample Problem
A heavy block hangs from a string
attached to a rod. An identical string
hangs down from the bottom of the
block. Which string breaks
a)
when the lower string is pulled with a
slowly increasing force?
Top string breaks due to its greater force.
a)
when the lower string is pulled with a
quick jerk?
Bottom string breaks because block has lots of inertia and
resists acceleration. Pulling force doesn’t reach top string.
Newton’s Second Law
Newton’s Second Law
A body accelerates when acted upon
by a net external force.
The acceleration is proportional to
the net force and is in the direction
which the net force acts.
Newton’s Second Law
∑F = ma
where ∑F is the net force
measured in Newtons (N)
 m is mass (kg)
 a is acceleration (m/s2)

Units of force
Newton (SI system)

1 N = 1 kg m /s2
1 N is the force required to
accelerate a 1 kg mass at a rate of 1
m/s2
Pound (British system)

1 lb = 1 slug ft /s2
Working 2nd Law Problems
1. Draw a force or free body diagram.
2. Set up 2nd Law equations in each
dimension.
SFx = max and/or
SFy = may
3. Identify numerical data.
x-problem and/or y-problem
4. Substitute numbers into equations.
“plug-n-chug”
5. Solve the equations.
Sample Problem
In a grocery store, you push a 14.5-kg cart with a
force of 12.0 N. If the cart starts at rest, how far
does it move in 3.00 seconds?
m = 14.5 kg, SF = 12 N, t = 3 s, vo = 0 m/s
x=?
x = vot + ½at2
Need to find ‘a’, use SF = ma
a = SF/m = 12/14.5 = 0.83 m/s2
x = (0)(3) + ½(0.83)(3)2
x = 3.72 m
Sample Problem
A catcher stops a 92 mph pitch in his glove,
bringing it to rest in 0.15 m. If the force
exerted by the catcher is 803 N, what is the
mass of the ball?
vo = 92 mph = 41.1 m/s, v = 0 m/s,
x = 0.15 m, SF = -803 N
m=?
need to find ‘a’, use v2 = vo2 +2ax
a = v2 - vo2/(2x) = - (41.1)2 / [(2)(0.15)] = -5630.7 m/s2
m = SF/a m = -803/(-5630.7) = 0.14 kg
Sample Problem
A 747 jetliner lands and begins to slow to a stop
as it moves along the runway. If its mass is 3.50 x
105 kg, its speed is 27.0 m/s, and the net braking
force is 4.30 x 105 N
a) what is its speed 7.50 s later?
m = 3.50 x 105 kg, vo = 27 m/s, SF= -4.30 x105 N
t = 7.5 s
v=?
v = vo + at
need to find ‘a’, use SF = ma
a =SF/m  a = -4.30 x105/ 3.50 x 105 = -1.23 m/s2
v = 27 + (-1.23)7.5
v = 17.78 m/s
b) How far has it traveled in this time?
x=?
x = vot + ½at2
x = 27.5(7.5) + ½(-1.23)(7.5)2
x = 201.64 m
Newton’s Third Law
Newton’s Third Law
For every action there exists an equal
and opposite reaction.
If A exerts a force F on B, then B
exerts a force of -F on A.
Examples of
Newton’s
3rd Law
Copyright James Walker, “Physics”, 1st edition
Sample Problem
You rest an empty glass on a table.
a) How many forces act upon the glass?
b) Identify these forces with a free body diagram.
c) Are these forces equal and opposite?
d) Are these forces an action-reaction pair?
a) 2
b)
c) No
d) No
Sample Problem
A force of magnitude 7.50 N pushes three boxes with masses m1
= 1.30 kg, m2 = 3.20 kg, and m3 = 4.90 kg as shown. Find the
contact force between (a) boxes 1 and 2 and (b) between boxes
2 and 3.
a) m1 = 1.3 kg, m2 = 3.2 kg, m3 = 4.9 kg, SF = 7.5 N
Find total ‘a’ for all boxes. mT = 9.4 kg, using SF = ma
a =SF/m
a =7.5 /9.4
a = 0.79 m/s2
Copyright James Walker, “Physics”, 1st edition
Looking at box 1 only, there is only
1 force acting on it, from box 2.
F2 on 1
SF = F2 on 1 = m1a
= 1.3(0.79)
=
b.) Looking at box 3 only, there are
2 forces acting on it. Box 2 on 3 and
the applied force.
Fapp
F2 on 3
SF = Fapp - F2 on 3 = m3a
F2 on 3 = Fapp - m3a
= 7.5 – 4.9(0.79)
=
Newton’s Laws in 2D
Newton’s 2nd Law in 2-D
The situation is more complicated
when forces act in more than one
dimension.
You must still identify all forces and
draw your force diagram.
You then resolve your problem into an
x-problem and a y-problem
(remember projectile motion????).
Forces in 2-D
Copyright James Walker, “Physics”, 1st edition
Forces in 2-D
Copyright James Walker, “Physics”, 1st edition
Forces in 2-D
Copyright James Walker, “Physics”, 1st edition
Forces in 2-D
Copyright James Walker, “Physics”, 1st edition
Forces in 2-D
Copyright James Walker, “Physics”, 1st edition
Sample Problem
A surfer “hangs ten”, and accelerates down the sloping
face of a wave. If the surfer’s acceleration is 3.50 m/s2 and
friction can be ignored, what is the angle at which the face
of the wave is inclined above the horizontal?
We’ll do this in class.
Mass and Weight
Sample Problem
How long will it take a 1.0 kg block initially at rest to slide down a
frictionless 20.0 m long ramp that is at a 15o angle with the
horizontal?
Draw a free body diagram
vo = 0, g = 10 m/s2, 20 m or x = 20 m and xo = 0 m q = 15o
t=?
Dx = vot + ½at2
∑F = ma  a = ∑F/m,
From FBD ∑F is the force in the x direction Fgx
Fgx = mgsinq = (1)(10) sin 15 = 2.59 N
a = ∑F/m = 2.59 / 1 = 2.59 m/s2
t = √(2Dx/a)
t = √[2(20)/2.59]
Sample Problem
An object acted on by three forces moves with constant
velocity. One force acting on the object is in the positive x
direction and has a magnitude of 6.5 N; a second force has
a magnitude of 4.4 N and points in the negative y direction.
Find the direction and magnitude of the third force acting
on the object.
F1 = 6.5 N, x F2 = 4.4 N, -y
F3 =?
∑F = ma
F1→ + F2↓ + F3 = 0
Mass and Weight
Many people think mass and weight
are the same thing. They are not.
Mass is inertia, or resistance to
acceleration.
Weight can be defined as the force
due to gravitation attraction.
W = mg
Sample Problem
A man weighs 150 pounds on earth at sea level. Calculate
his
a) mass in kg. (2.2 lb. = 1 kg) Convert.
b) weight in Newtons.
w = mg
Apparent weight
If an object subject to gravity is not
in free fall, then there must be a
reaction force to act in opposition to
gravity.
We sometimes refer to this reaction
force as apparent weight.
Elevator rides
When you are in an elevator, your actual
weight (mg) never changes.
You feel lighter or heavier during the
ride because your apparent weight
increases when you are accelerating up,
decreases when you are accelerating
down, and is equal to your weight when
you are not accelerating at all.
Going
Up?
v=0
a=0
v>0
a>0
v>0
a=0
v>0
a<0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Ground
floor
Just
starting up
Between
floors
Arriving at
top floor
Going
Down?
v<0
a<0
v=0
a=0
v<0
a>0
v<0
a=0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Beginning
descent
Between Arriving at
floors Ground floor
Top
floor
Normal Force
Sample Problem
An 85-kg person is standing on a bathroom scale
in an elevator. What is the person’s apparent
weight
a) when the elevator accelerates upward at 2.0
m/s2?
ay = 2 m/s2, m = 85 kg, g = 10 m/s2
Wapp = ?
SF = Wapp – W = may
Find W: W = mg
Wapp = W + may
W = (85)(10)
Wapp = (850) + 85(2)
W = 850 N
Wapp = 1020 N
b) when the elevator is moving at constant
velocity between floors?
Constant velocity  ay = 0 m/s2
SF = Wapp – W = may
Wapp = W = (85)(0) = 0 N
c) when the elevator begins to slow at the top
floor at 2.0 m/s2?
ay = -2 m/s2
Wapp = ?
SF = Wapp – W = may
Find W: W = mg
Wapp = W + may
W = (85)(10)
Wapp = (850) + 85(-2)
Wapp = 680 N
Sample Problem
A 5-kg salmon is hanging from a fish scale in an elevator. What is
the salmon’s apparent weight when the elevator is
a) at rest?
m = 5 kg, g = 10 m/s2, a = 0 m/s2
Wapp – W = may
Wapp = mg + ma = m(g + a)
Wapp = 5(10 + 0)
Wapp = 50 N
b) moving upward and slowing at 3.2 m/s2? a = -3.2 m/s2
Wapp = m(g + a)
Wapp = 5(10 + -3.2)
Wapp = 34 N
c) moving downward and speeding up at 3.2
m/s2? a = 3.2 m/s2
Wapp = m(g + a)
Wapp = 5(10 + 3.2)
Wapp = 66 N
Normal force
The normal force is a force that
keeps one object from penetrating
into another object.
The normal force is always
perpendicular a surface.
The normal exactly cancels out the
components of all applied forces that
are perpendicular to a surface.
Normal force on flat surface
The normal force is equal to the weight of an
object for objects resting on horizontal
surfaces.
N = W = mg
N
mg
Normal force on ramp
N = mgcosq
N
mgsinq
mg
q
The normal force is
perpendicular to
angled ramps as well.
It’s always equal to
the component of
weight perpendicular
to the surface.
mgcosq
q
Normal force not associated
with weight.
A normal force can exist that is
totally unrelated to the weight of an
object.
applied force
normal
friction
weight
N = applied force
The normal
force most
often equals
the weight of
an object…
…but this is by
no means
always the
case!
More on the Normal Force
Draw a free
body diagram
for the skier.
Sample problem
A 5.0-kg bag of potatoes sits on the bottom of a stationary
shopping cart. Sketch a free-body diagram for the bag of
potatoes. Now suppose the cart moves with a constant velocity.
How does this affect the free-body diagram?
FN
Fg
Sample problem
a) Find the normal force exerted on a 2.5-kg
book resting on a surface inclined at 28o
above the horizontal.
m = 2.5 kg, g = 10 m/s2, q = 28o
N=?
N = mgcosq
N = 2.5(10)cos(28)
N = 22.07 N
b) If the angle of the incline is reduced, do you
expect the normal force to increase, decrease,
or stay the same? INCREASED
Sample problem
A gardener mows a lawn with an old-fashioned push mower. The
handle of the mower makes an angle of 320 with the surface of the
lawn. If a 249 N force is applied along the handle of the 21-kg
mower, what is the normal force exerted by the lawn on the
mower?
m = 21 kg, Fapp = 249 N, q = 32o g = 10 m/s2
FN = ?
∑Fy = may
Fapp y + FN - Fg = 0 
FN = Fg - Fapp y = mg – Fappsin q
FN = (21)(10) – 249 sin (32)
FN = 78.05 N
Sample problem
Larry pushes a 200 kg block on a frictionless floor at a 45o angle
below the horizontal with a force of 150 N while Moe pulls the
same block horizontally with a force of 120 N.
a) Draw a free body diagram.
b) What is the acceleration of the block?
c) What is the normal force exerted on the block?
m = 200 kg, Fapp L = 150 N, q = 45o Fapp M = 120 N, g = 10 m/s2
ax = ?
∑Fx = max
Fapp Lx + Fapp M = max
ax = Fapp Lx + Fapp M = Fapp L cos q + Fapp M
ax = [150 cos (45) + 120]/ 200
ax = 1.13 m/s2
FN = ?
∑Fy = may
Fapp y + FN - Fg = 0 
FN = Fg - Fapp y = mg – Fappsin q
FN = (200)(10) – 150 sin (45)
FN = 1893.93 N