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Lecture 2
Acceleration due to gravity
Forces
Newton’s Laws
Free Falling Objects
We have worked out mathematical
relationships arising from the definitions
of velocity and acceleration.
v  v0  at
1 2
s  v0t  at
2
v  v  2as
2
2
0
Probably the most familiar system where
we observe acceleration is that due to
gravity.
Free Fall
Definition:
Freely falling object is one moving freely
under the influence of gravity alone,
Objects considered to be freely falling
propelled upwards
propelled downwards
released from rest
The “free fall condition” considers gravity only:
•neglects other effects such as air resistance
Hammer and a paper tissue
 Dropped from same height,
 Hammer will hit the ground first
 Air resistance will slow the tissue down.
However if we neglect air resistance both
objects will hit the ground at the same time
Free Falling Objects
A time delay image of
two spheres of very
different mass
falling in a vacuum. .
It can be seen that, in
the absence of air
resistance both
accelerate at the
same rate,
independent of mass.
The acceleration due to gravity near the
earths surface is known as g. This has
been measured to be g = 9.80 ms-2.
Free Falling Objects
If we assume objects falling near the
earths surface are affected only by
gravity (air resistance is negligible)
two basic facts govern their motion:
1. Objects accelerate at the same rate,
independent of their
 mass,
 size
 composition.
2 . This gravitational acceleration is
constant and so does not change as
the object falls.
Acceleration due to gravity
Ignoring air resistance, an object in free fall
experiences an acceleration of magnitude
9.8 ms-2.
In other words the downward directed velocity
increases by 9.8 ms-1 each second.
So if released from rest an object has a velocity
downwards
after 1 second of 9.8m/s.
after 2 seconds of 19.6m/s
after 3 seconds of 29.4m/s
v  v0  at
Since by convention displacement upwards is
positive, but gravity acts downwards, then
g = -9.8 ms-2.
Acceleration due to gravity
Since acceleration due to gravity is constant,
motion under the action of gravity is
uniformly accelerated motion,
so we can use the equations relating position,
displacement, velocity and acceleration
already derived
Example:
A ball is dropped from a window 10m above
ground. What will be its velocity just before it
hits the ground?
2
2
0
v  v  2as
v   v02  2as
v   02  (2)  (9.8ms 2 )(10m)
v  14ms 1
v  14ms1
Force, Acceleration &
Newton’s Laws
Up to now we have discussed kinematics i.e.
methods for describing motion (without
reference to the causes).
We will now study motion and the causes of
motion – dynamics.
The basic physical quantities used in dynamics
are
force, mass and acceleration.
Force: push or a pull
Characteristics:
•Strength or magnitude
•Direction
Force is a vector quantity
Force
A force resulting from direct contact with
another object is called a contact force.
For example when you push or pull an
object you exert a force on it.
Orthodontics:
contact force applied
Force is a vector quantity:
magnitude and direction
Force pushes or pulls
teeth in a particular direction
There are also non-contact forces.
Gravitational, electrical and magnetic forces
act through empty space.
You don’t have to be standing on the
surface of the earth to experience the
effects of gravity.
Force
Tooth extraction
Application of force
 break the periodontal ligaments
 Extends width of the socket
this web site concerns the mechanics of
extraction including the Physics forceps
http://www.dentistrytoday.com/oralmedicine/oral-surgery/1536
Force
The force due to gravity exerted on an
object is known as its weight.
The SI unit of force is the Newton, N.
Force can be measured with a spring balance.
When a force pulls on the spring, the spring
extends. A pointer attached to the end of the
spring can indicate the force on a scale.
Force causes Acceleration
Questions:
1/ What happens to an object when there
is no net force exerted on it.?
2/ What links force and acceleration?
3/ What happens to an object that
exerts a force on another object?
Answers are contained in
NEWTON’S three LAWS.
Isaac Newton ( 1643-1727)
Credited with establishing a mathematical
basis for the laws of motion
Earlier Galileo Galilei (1564 –1642) established
theories concerning moving (falling) objects
Newton’s Laws
All dynamics is based on Newton’s Laws.
These are three empirical laws which cannot
be derived from anything more fundamental.
1. When the vector sum of forces on an
object is zero then the acceleration of that
object is zero.
Force must be applied to an object to change
its velocity.
2. When the vector sum of forces is NOT zero
force is related to acceleration.
Force = mass x acceleration.
3.The third law describes the pairs of forces
that interacting objects exert on each other.
If we push an object it pushes back with an
equal force but in the opposite direction.
NEWTON’S FIRST LAW
“Any object will remain at rest or in motion
in a straight line with constant velocity
unless acted upon by an outside force”
There is no distinction between an object at
rest and an object moving with constant
velocity.
v=0
v = const.
}
No net
Force
Constant velocity means both constant
magnitude (speed) and constant direction.
Newton’s First Law
“Any object will remain at rest or in motion in
a straight line with constant velocity unless
acted upon by an outside force”
This is not as self evident as it may seem.
It actually seems counter intuitive because it
means that:
once an object is set in motion with
uniform velocity, no force is needed to
keep it moving.
This seems contrary to everyday
experience.
For example,
If you push a book across a table, the book
does not keep moving indefinitely after it has
left your hand. It slows down.
BUT as we will see later, this is due to
frictional forces slowing it down.
Equilibrium and Newton’s First Law
Imagine a tug of war match with each team
equally matched.
Both teams pull on the rope with equal strength
they each exert the same magnitude of force
on it, but in opposite directions.
Fleft
Fright
In this case the knot in the middle of the
rope does not move. It does not
accelerate. The rope is in equilibrium.
We can write - Fleft = Fright
This means Fleft + Fright = 0
In equilibrium
ΣF=0
Greek letter “S” represents the sum
Newton’s Second Law
Acceleration produced by forces acting on
an object is

directly proportional to and in the same
direction as the net external force

inversely proportional to the mass of the
object
a
Fextnet
m
Fextnet  ma
Newtons (N)
m = mass of the object
F
a=F/m
(kg)
(ms-2)
Newton’s Second Law
Equation F=ma means that mass, in addition
to being a measure of the amount of matter
in an object, is a measure of how difficult it is
to move an object or its inertia.
Inertia is the tendency
of an object at rest to remain at rest
of an object in motion to remain in motion
with its original velocity.
The greater the mass of a body, the less effect
a given force has.
The unit of force, the Newton, is defined as
follows:
A force of 1N acting on a mass of 1kg
produces an acceleration of 1ms-2.
F = ma so (1N) = (1kg)(1ms-2) = 1kgms-2
Newton’s Second Law
If a number of forces act on an object at
the same time,
Newton’s second law applies to the sum
of the forces and is written.
SF  ma
Thus when working out problems involving
a number of forces, it is best to calculate
the resultant force and then set that
equal to ma.
Mass and Weight
Mass is a measure both of
how much matter an object contains
how difficult it is to move.
Weight however is the force exerted by
gravity on a body.
Thus a heavy truck is
 difficult to push because of its mass
 difficult to lift due to its weight.
When an object falls under the influence of
gravity it accelerates downwards at the rate:
a = g = 9.80 m/s2
Force produces an acceleration given by
F = ma
Mass and Weight
But if the force on an object due to gravity
is weight, w, and it accelerates at
g = 9.8 ms-2 then, we can write,
w  mg
This equation gives the gravitational force
on an object whether it is in freefall or not.
Any object with mass “near” the surface of
the earth feels a gravitational force (weight),
w = mg.
Newtonian Gravity
m1m2
G = 6.67  10-11 N m2 kg-2
F G 2
r
MEm
ME
F  G 2  ma  mg
g G 2
r
r
ME
g G 2
RE
g  9.81ms
2
Example
Mass of a person is 65kg. What is his weight?
His weight is given by w =mg
w = 65kg * 9.8m/s2
= 637kg m/s2 = 637N
Example
On the surface of the moon the force of
gravity is approximately 1/6 of that on earth.
What is the weight of the same person on
the moon.
Mass of a person on Earth is 65kg
Weight of this person on the moon is
w = mgm
weight (w) = 65kg * {(1/6) 9.8m/s2}
w =106.2kg.m/s2 = 106.2N
Example.
A tennis ball and a golf ball are simultaneously
dropped from a tall building of height 120m.
Neglecting air resistance, determine
(a) the speed with which each ball hits the
ground.
(b) The time taken for each to reach the
ground (g = 9.8 ms-2)
all objects regardless of their mass or size
fall freely with an acceleration g = 9.8ms-2
v  v  2as
2
2
0
v0 = 0
Acceleration and displacement
are in the direction such that
a = -g and s = -120m
(a) the speed with which each ball hits
the ground.
2
2
v  v0  2as
v2 = 2*(-g)(-120)
v = ±√2*(-9.8m/s2)(-120m)
= ±48.5m/s
But since the direction is downwards
v = - 48.5m/s
And since speed is the magnitude of velocity
Speed = 48.5m/s
(b) The time taken for each to reach the ground
v = v0+ at
-48.5m/s = 0 + (-9.8m/s2)t
t = (-48.5m/s )/ (-9.8m/s2) = 5s
Example
A car has a maximum acceleration of 4 ms-2.
What will its maximum acceleration be while
towing a second car of the same mass.
F=ma
a = 4ms-2
F=MaN
where M is the combined mass of the cars
F
F
aN 

M 2m
ma a
aN 

2m 2
4ms
aN 
2
2
 2ms 2
EXAMPLE
A ball is thrown upward at 20m/s from a
window 60m above the ground.
(a) How high does it go?
(b) When does it reach its highest point?
(c) When does it hit the ground?
Here we will take the upward direction as the
positive direction. This means any vector
quantities pointing upward (initial velocity) are
positive while vector quantities pointing down
(acceleration due to gravity) are negative.
(a) To find the highest point. We note at this
point the velocity is zero. We use
v 2  v02  2as
v=0
or
(v 2  v02 )
s
2a
so
s = [0-(20m/s)2]/[2(-9.80m/s2)] = 20.4 m
Highest point is 20.4m above the window
(b) When does it reach its highest point
v  v0  at
or t = (v-v0)/a
t = (0-20m/s)/(-9.80m/s2) = 2.04 s
It reaches its highest point after 2.04s
(c) The ball hits the ground when s = -60m.
t?
1 2
s  v0t  at
2
(-60m) = (20m/s) t+(1/2)(-9.8m/s2) (t)2
Rearranging gives
(4.9m/s2) (t)2 - (20m/s) (t) – (60m) = 0
This is a quadratic equation whose solution is.
(20m / s)  (20m / s) 2  4(4.9m / s 2 )(60m)
t
2(4.9m / s 2 )
Reducing to
(20m / s)  (40m / s)
t
(9.8m / s 2 )
The roots of this equation are then
t = -2s or 6.1s
As the ball could not hit the ground before it is
thrown, The correct answer must be the positive
one t = 6.1s