# Newton`s Laws Powerpoin ```Newton’s First &amp; Second
Law
AP Physics C
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Unit is the NEWTON(N)
Is by definition a push or a pull
Can exist during physical
contact(Tension, Friction, Applied
Force)
Can exist with NO physical contact,
called FIELD FORCES ( gravitational,
electric, etc)
INERTIA – a quantity of matter, also called MASS.
Italian for “LAZY”. Unit for MASS = kilogram.
Weight or Force due to Gravity is how your MASS
is effected by gravity.
W  mg
NOTE: MASS and WEIGHT are NOT the same thing. MASS never changes
When an object moves to a different planet.
What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s/s)?
W  mg  W  ( 85 . 3 )( 9 . 8 )  835 . 94 N
W MARS  ( 85 . 3 )( 3 . 2 )  272 . 96 N
Example Problem

Sam weighs 270 N on Mercury, and Andy
weighs 530 N on Venus. Who has a larger
mass? The acceleration due to gravity on
Mercury is 3.59 m/s2 and the acceleration
due to gravity for Venus is 8.87 m/s2.
An object in motion remains in motion in a
straight line and at a constant speed OR an
object at rest remains at rest, UNLESS acted
upon by an EXTERNAL (unbalanced) Force.
There are TWO conditions here and one constraint.
Condition #1 – The object CAN move but must be at a CONSTANT SPEED
Condition #2 – The object is at REST
Constraint – As long as the forces are BALANCED!!!!! And if all the forces
are balanced the SUM of all the forces is ZERO.
The bottom line: There is NO ACCELERATION in this case AND the object
must be at EQILIBRIUM ( All the forces cancel out).
acc  0 
F
0
Explain Inertia in each of the following situations:
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Crash Dummy
http://www.cleanvideosearch.com/media/action/yt/watch?videoId=d7iYZPp2
zYY
Football /Ref
Cat
Cheetah
Working in Space and Inertia
http://www.sciencewithmrnoon.com/physics/unit03.htm
Straw in Plywood
http://www.physics.umd.edu/lecdem/services/demos/demosc3/c3-12.htm
Masses on String
http://www.physics.umd.edu/lecdem/services/demos/demosc3/c3-03.htm
A pictorial representation of forces complete
with labels.
FN
Ff
T
T
W1,Fg1
or m1g
m2g
•Weight(mg) – Always
drawn from the center,
straight down
•Force Normal(FN) – A
surface force always drawn
perpendicular to a surface.
•Tension(T or FT) – force in
ropes and always drawn
AWAY from object.
•Friction(Ff)- Always drawn
opposing the motion.
Ff
FN
mg
Normal force
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The force that keeps one object from
invading another object is called the normal
force
“Normal” means “perpendicular”
You can determine the normal force by
considering all forces that have components
perpendicular to a surface
A 10-kg box is being pulled across the table to the
right at a constant speed with a force of 50N.
a)
b)
Calculate the Force of Friction
Calculate the Force Normal
FN
Ff
mg
Fa
F a  F f  50 N
mg  Fn  (10 )( 9 . 8 )  98 N
Suppose the same box is now pulled at an angle of 30
degrees above the horizontal.
a)
Calculate the Force of Friction
b)
Calculate the Force Normal
Fax  Fa cos   50 cos 30  43 . 3 N
F f  Fax  43 . 3 N
FN
Ff
Fa
Fay
30
Fax
mg
F N  mg !
F N  F ay  mg
F N  mg  F ay  (10 )( 9 . 8 )  50 sin 30
F N  73 N
Tension
A pulling force.
 Generally exists in a rope, string, or
cable.
 Arises at the molecular level, when a
rope, string, or cable resists being
pulled apart.
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Tension (static 2D)
The sum of the horizontal and vertical
components of the tension are equal to
zero if the system is not accelerating.
30o
45o
1
2
3
15 kg
Problem: Determine the tension in all three
ropes.
30o
45o
1
2
3
15 kg
T1 = 131 N
T2 = 107 N
T3 = 147 N
Friction

Friction is a force that exists between
two surfaces that opposes a sliding
motion
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Fs: Static friction exists before sliding
occurs, it prevents movement
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Fk: Kinetic friction exists after sliding
occurs, it opposes motion once
objects are moving
• Ff is directly proportional to Fn (normal
force):
F f   Fn
 
Ff
Fn
• Coefficient of friction ():
– Determined by the nature of the
surfaces
– s is for static friction.
– k is for kinetic friction.
– s &gt; k
two
μ
Typical Coefficients of Friction
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Values for  have no units and are approximate
 Now Try These Phun Phree Body Diagrams
A block pulled to the right on a rough horizontal
surface
Fn
Fk
Fn
A block pulled up a rough incline
Fa
Fg
Fa
Fk
Fg
Two blocks in contact with each other,
pushed to the right on a frictionless surface
P’
Fn
Fn
Fa
P
Fg
Two blocks connected by a cord,
one is on top of the table, the other
is hanging off the side. The table
surface is rough, the pulley is
frictionless
Fg
Fn
Fk
T
T
Fg
Fg
Classroom Practice Problem
What is the Frictional force in this situation?
EXPLAIN
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Mass of the block is 500
g
Force applied is 3 N.
 = 0.75
The Block is stationary
Sample problem: A 1.00 kg book is held against
a wall by pressing it against the wall with a force of
50.00 N. What must be the minimum coefficient of
friction between the book and the wall, such that
the book does not slide down the wall?
μs = .196
F
Since the Fnet = 0, a system moving at a
constant speed or at rest MUST be at
EQUILIBRIUM.
TIPS for solving problems
• Draw a FBD
• Resolve anything into COMPONENTS
• Write equations of equilibrium
• Solve for unknowns
If an object is NOT at rest or moving at a
constant speed, that means the FORCES are
UNBALANCED. One force(s) in a certain
direction over power the others.
THE OBJECT WILL THEN ACCELERATE.
The acceleration of an object is directly
proportional to the NET FORCE and
inversely proportional to the mass.
a  F NET
a
F NET
m
a
1
F NET 
F
m
 F NET  ma
Tips:
•Draw an FBD
•Resolve vectors into components
•Write equations of motion by adding and
subtracting vectors to find the NET FORCE.
•Solve for any unknowns
A 10-kg box is being pulled across the table to
the right by a rope with an applied force of
50N. Calculate the acceleration of the box if a
12 N frictional force acts upon it.
FN
Ff
mg
Fa
In which direction,
is this object
accelerating?
The X direction!
So N.S.L. is worked
out using the forces
in the “x” direction
only
F Net  ma
F a  F f  ma
50  12  10 a
a  3 .8 m / s
2
Problem: A hockey puck has an initial velocity of 20 m/s. If
the puck slide 115m before coming to a stop, what is the
coefficient of kinetic friction between the puck and the ice?
μk = .196
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected
to a cable that passes over a pulley and then is fastened to a hanging mass,
m2 = 11.0 kg as shown below. Find the acceleration of each mass and the
tension in the cable.
FN
F Net  ma
T
m2 g  T  m2a
T  m1a
T
m 2 g  m1a  m 2 a
m1g
m 2 g  m 2 a  m1a
m 2 g  a ( m 2  m1 )
m2g
a 
m2g
m1  m 2

(11 )( 9 . 8 )
14
 7 .7 m / s
2
F Net  ma
T  ( 3 )( 7 . 7 )  23 . 1 N
m2 g  T  m2a
T  m1a
F Net  ma 
Slope 
Rise
Run
F NET
a
m
Non-constant Forces
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Up until this time, we have mainly dealt with forces that
are constant. These produce a uniform, constant
acceleration.
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Kinematic equations can be used with these forces.
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However, not all forces are constant.
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Forces can vary with time, velocity, and with
position.
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The Derivative
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The derivative yields tangent lines and slopes
We use the derivative to go from
•
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The Integral
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The integral yields the area under a curve
Use the integral to go from
•
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position -&gt; velocity -&gt; acceleration
acceleration -&gt; velocity -&gt; position
In these two approaches, F = ma, so
methods for determining or using
acceleration are also used to find force
Where does the calculus fit in?
2


dv
d x
F  ma  m
m
dt
dt
There could be situations where you are
given a displacement function or velocity
function. The derivative will need to be
taken once or twice in order to get the
acceleration. Here is an example.
You are standing on a bathroom scale in an elevator in a tall
building. Your mass is 72-kg. The elevator starts from rest
and travels upward with a speed that varies with time
according to:
2
v ( t )  3t  0 . 20 t
When t = 4.0s , what is the reading on the bathroom scale
(a.k.a. Force Normal)?
F net  ma
a 
dv
dt
d ( 3 t  0 . 20 t )
2

 3  0 . 40 t
dt
a ( 4 )  3  0 . 40 ( 4 )  4.6 m/s/s
F N  mg  ma  F N  ma  mg
F N  ( 72 )( 9 . 8 )  ( 72 )( 4 . 6 ) 
1036.8 N


Consider a force that is a function of time:
◦ F(t) = (3.0 t – 0.5 t2)N
If this force acts upon a 0.2 kg particle at rest for 3.0 seconds,
what is the resulting velocity and position of the particle?
vf = 45 m/s
xf = 50.6 m
Sample problem:
 Consider a force that is a function of time:
 F(t) = (16 t2 – 8 t + 4)N
 If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is
the resulting change in velocity of the particle?
Δv = 1.33 m/s
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