Download NEWTON`s FIRST LAW

JP
1
JP
2
JP
3
(1686)
Philosophiae Naturalis
Principia Mathematica
In this work, he proposed
three “laws” of motion:
JP
4
NEWTON’S FIRST LAW :
“Every body continues in a
state of rest or uniform
motion in a straight line
unless impressed forces
act upon it.”
JP
5
parked up
NEWTON’S FIRST LAW :
“Every body continues in a state of rest or
uniform motion in a straight line unless
impressed forces act upon it.”
JP
6
AAAAAAA
H
STATE OF REST
JP
UNLESS ……. !
7
NEWTON’s FIRST LAW
… constant velocity, unless ……..
YOU UNDERSTAND
WHAT I AM SAYING
JP
8
NEWTON’s FIRST LAW
INTRODUCES THE IDEA OF
INERTIA
THE RELUCTANCE OF A BODY AT REST TO MOVE
(OR OF A MOVING BODY TO CHANGE ITS STATE OF MOTION)
THE PRACTICAL MEASURE OF INERTIA IS
JP
MASS
9
The greater the mass of a body, the greater its INERTIA
Make me
JP
10
locking rod
pivots
seat belt
Car stops but
inertia carries
pendulum
forward
ratchet wheel
Direction of motion
JP
11
WHEN FORCES
ARE BALANCED
OBJECTS IN
MOTION, STAY
IN MOTION
OBJECTS AT
REST, STAY AT
REST
[SAME SPEED &
DIRECTION]
JP
12
A body is in a state of equilibrium if the
forces acting on it are in balance
8N
8N
8N
2 forces
balanced
8N
1200
1200
1200
3 forces
balanced
8N
JP
13
THERE ARE TWO CONDITIONS FOR
A BODY TO BE IN EQUILIBRIUM:
•
THE SUM OF THE FORCES IN
ANY DIRECTION IS ZERO
• THE SUM OF THE MOMENTS
ABOUT ANY POINT IS ZERO
JP
14
CAN YOU EXPLAIN WHAT HAPPENS
BELOW?
JP
15
WHEN FORCES ARE NOT BALANCED
A RESULTANT FORCE CHANGES A
BODY’S VELOCITY
JP
16
NEWTON’S SECOND LAW :
“THE RATE OF CHANGE OF
MOMENTUM OF A BODY IS
DIRECTLY PROPORTIONAL TO
THE RESULTANT EXTERNAL
FORCES ACTING UPON IT, AND
TAKES PLACE IN THE
DIRECTION OF THAT FORCE”
JP
17
A RESULTANT FORCE PRODUCES A CHANGE
IN A BODY’S MOMENTUM
A RESULTANT FORCE
AN
JP
ACCELERATION
18
JP
19
NEWTON’S SECOND LAW
units
• F
in
Newtons
• m
in
kilograms
• a
in
metres per second2
JP
20
ACCELERATION IS DIRECTLY PROPORTIONAL TO THE APPLIED FORCE
acceleration / ms-2
force / N
N.B. - STRAIGHT LINE THROUGH THE ORIGIN
JP
21
GRAPH OF ACCELERATION VERSUS MASS
acceleration / ms-2
mass / kg
1
a
m
F=ma
JP
22
ACCELERATION IS INVERSELY PROPORTIONAL TO MASS OF THE BODY
acceleration / ms-2
1
/ kg 1
m
N.B. - STRAIGHT LINE THROUGH THE ORIGIN
JP
23
NEWTON’S SECOND LAW IS USED IN TWO FORMS
 (mv)
F
t
F  ma
Where F is the
RESULTANT FORCE
JP
24
JP
25
NEWTON’S THIRD LAW :
“ACTION AND REACTION ARE
ALWAYS EQUAL AND OPPOSITE”
“IF A BODY A EXERTS A FORCE ON
BODY B, THEN B EXERTS AN EQUAL
AND OPPOSITELY DIRECTED FORCE
ON A”
JP
26
I’LL
PULL
HIM
devishly
clever
WELL ACTION AND REACTION ARE
ALWAYS EQUAL AND OPPOSITE!!
JP
27
ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE!!
WELL ACTION AND REACTION ARE
SO WHY DOES THE GIRL MOVE FASTER?
ALWAYS EQUAL AND OPPOSITE!!
JP
28
NEWTON’S THIRD LAW PAIRS
• THEY ARE EQUAL IN MAGNITUDE
• THEY ARE OPPOSITE IN DIRECTION
• THEY ACT ON DIFFERENT BODIES
JP
29
NEWTON’S THIRD LAW PAIRS
SIMILARITIES
DIFFERENCES
The 2 forces act for the
same length of time
The 2 forces act on
different bodies
The 2 forces are the
same size
The 2 forces are in
opposite directions
The 2 forces act along
the same line
Both forces are of the
same type
JP
30
THE CLUB EXERTS A FORCE F ON THE BALL
THE BALL EXERTS A N EQUAL AND OPPOSITE
FORCE F ON THE CLUB
F
JP
31
Drawing Free-Body Diagrams
A free-body diagram singles out a body from its
neighbours and shows the forces, including
reactive forces acting on it.
Free-body diagrams are used to show the
relative magnitude and direction of all forces
acting upon an object in a given situation.
The size of an arrow in a free-body diagram is
reflective of the magnitude of the force. The
direction of the arrow reveals the direction in which
the force is acting. Each force arrow in the diagram
is labeled to indicate the exact type of force.
JP
32
Tug assisting a ship
Free body diagram for the ship
Upthrust
[buoyancy]
Thrust
from
engines
SHIP
Friction
Pull from
tug
weight
JP
33
EXAMPLE 1 - A LIFT ACCELERATING UPWARDS
If g = 10 ms-2, what “g force” does
the passenger experience?
The forces experienced by the
passenger are her weight, mg and
the normal reaction force R.
The resultant upward force which
gives her the same acceleration as
the lift is R – mg.
Apply F = ma
R
a = 20 ms-2
R – mg = ma
Hence the forces she “feels”, R = ma + mg
mg
The “g force” is the ratio of this force to her weight.
ma  mg m(a  g ) 20  10


3
mg
mg
10
JP ©
34