DO PHYSICS ONLINE
FORCES & NEWTON’S LAWS OF
MOTION
The study of what causes changes in the motion of an object is known as
dynamics. The central concept of dynamics is the physical quantity force.
The term force can’t be defined or explained in a simple sentence. You will
learn what a force is and the ideas associated with forces – however, you
just will not be able to state what a force is in a clear brief statement.
Sometimes force is said to be a pull or push, but a force is more than just a
push or pull. Forces acts in pairs – it is not possible to have a single force
acting. The interaction between objects is often described in terms of forces.
For example, (1) two objects will attract each other because of the
gravitational force between them, and (2) two electrons will repel each
other because electrons are negatively charged.
The principles of dynamics are described by the three Newton’s laws of
motion. These three laws are very simple to state, but extremely difficult to
interpret and apply. This is because of your experience in walking, throwing,
pushing, etc has given you “mental” models of how things work based upon
common sense. But, the way the world works is not necessarily based upon
logic, intuition or common sense. After careful observation and
measurement, physicists have developed scientific principles which often
are at odds to our “mental models” based upon our experience of the
physical world. Newton’s laws are not the product of mathematical
derivations, but rather they have been synthesized from a multitude of
experimental data about how objects move. These laws are truly
fundamental because they can’t be deduced from other principles.
PROPERTIES OF FORCES

A force is a push or pull.

A force is an interaction between two objects.

Forces can be attractive or repulsive.

Forces act in pairs.

An interaction between two objects gives rise to two forces of equal
magnitude but which act in opposite directions (Newton’s 3rd law).

A force is a vector quantity with magnitude and direction.

Multiple forces can be replaced by a single force using vector addition
to give the resultant or net force acting on an object.

A force can be resolved into two components at right angles to each
other.

If the resultant force on an object is zero, the object will remain
stationary or move with a constant velocity (Newton’s 1st law:
F  0

 a  0 ).
If the resultant force on an object is non-zero, the object will
experience an acceleration in the direction of the force (Newton’s 2nd
law:

a
1
 F ).
m
The S.I. Unit for force is the newton [N]
1 N = 1 kg.m.s-2
NEWTON’S LAWS OF MOTION
In studying motion the first step was defining a frame of reference. The
fame of reference chosen is one in which its acceleration is zero. This is
called an inertial frame of reference. You are an observer in a car and
the car represents your frame of reference. If you have your eyes closed,
then you can’t feel how fast you’re travelling. This is an inertial frame of
reference. You feel the effects of acceleration, if the car suddenly speeds up
you feel pushed backwards, if the car brakes abruptly you will be thrown for
forward and if the car turns sharply then you appear to be thrown sideways.
When the car accelerates, it is a non-inertial frame of reference. The
Earth although it is orbiting the Sun and rotating about its axis, these
effects are small and so we can take the Earth’s surface to be an inertial
frame of reference. For our study of the motion we always take an inertial
frame of reference.
Newton’s 1st law of motion
In an inertial frame of reference, if the resultant (net) force
acting upon an object is zero, then the object will be at rest
or moving with a constant velocity.
This law implies that the natural state of motion of an object is that it has
zero acceleration (a = 0), therefore, the object is at rest ( v = 0) or moving
in a straight line with a constant speed ( v = constant). The tendency for an
object to remain at rest or keep moving is called inertia. When the
resultant (net) force acting on a particle is zero, the object is in equilibrium
(1)
F  0 Fx  0 Fy  0
 a  0 ax  0 ay  0
equilibrium
Figure (1) helps us understand Newton’s 1st law. You are standing in a tram
travelling at a constant velocity when the tram suddenly brakes. You feel
that someone has pushed you in the back and you topple forward. But, you
do not experience a force. By Newton’s 1st law you upper part of the body
will continue to move forward while your legs attached to the floor slow
down with the tram. Again, you are standing in a stationary tram. The tram
starts moving forward. Your feet move forward with the tram but your
upper part of the body tends to remain at rest according to Newton’s 1st law.
tram moving
with constant
velocity v
tram suddenly stops
a
a 0
upper body keeps
moving forward
with velocity v
feet stop with tram
tram starts moving forward
tram stopped v = 0
a 0
a
upper body tends to
stay at rest v= 0
feet move forward with tram
Fig. 1.
Riding an accelerating tram.
Many injuries suffered in car accidents can be explain by Newton’s 1st law.
When a car is moving forward and then suddenly slows down as in a
collision, the people in the car will tend to keep moving forward with a
constant velocity until they are brought to rest by smashing into the
steering wheel or the windscreen. The faster the car is travelling before a
collision, the greater the speed before impact which brings the occupants to
rest and the more substantial are the injuries incurred. In many such
collisions occupants often suffer massive internal injuries. You internal
organs must obey Newton’s 1st law – they continue moving forward
damaging the diaphragm and the organs themselves which results in
excessive internal bleeding. Many young drivers have died from internal
injuries in car accidents – you should always remember Newton’s 1st law –
the faster you are going before an accident the more likely you are to die –
slowing down does save lives. Seat belts, collapsible steering wheels and air
bags are used to limit the injuries in accidents caused by inertia.
Similarly, a car moving off suddenly from rest can cause a person’s upper
parts of their body namely the head to remain stationary whilst the lower
parts of the body seated and in contact with the car floor move forward.
Serve neck injuries are very common in rear end collisions when the neck
muscles are severely stretched. The stretched muscles act a like a spring
and the head is whipped back to catch-up with the rest of the body. This is
often referred to whip-lash injuries of the neck and back. Head restraints on
car seats are used to carry the head forward to prevent this type of injury
to the neck.
Fig.2. Seat belts, air bags and head restraints can reduce
injuries in a collision.
Sir Isaac Newton (1642-1727) was by many standards the most important
person in the development of modern science. Newton, who was born the
same year that Galileo Galilei (1564-1642) died, built on Galileo's ideas to
demonstrate that the laws of motion in the heavens and the laws of motion
on the Earth were one and the same. The work of Galileo and Newton would
banish the notions of Aristotle that had stood for 2000 years. Galileo’s ideas
challenged the views of the Church on the Aristotelian conception of the
Earth centred Universe. He got him into trouble with the Inquisition and was
forced to recant publicly his Copernican views. In the later years of his life
he was under house arrest. His story is one of many that highlighted the
conflict between the scientific approach and "science" based on
unquestioned authority. Even today, there still are many forces in modern
society (religious, cultural, political, etc) that are against the scientific
approach of open enquiry.
Perhaps Galileo's greatest contribution to physics was his formulation of the
concept of inertia: an object in a state of motion possesses an “inertia” that
causes it to remain in that state of motion unless a resultant force acts on it.
Most objects in a state of motion do not remain in that state of motion. For
example, a block of wood pushed across table quickly comes to rest when
we stop pushing. Thus, Aristotle held the view that objects in motion did not
remain in motion unless a force acted constantly on them. Galileo, by virtue
of a series of experiments (many with objects sliding down inclined planes),
realized that the analysis of Aristotle was incorrect because it failed to
account properly for a hidden force: the frictional force between the surface
and the object. As we push the block of wood across the table, there are
two opposing forces that act: the force associated with the push, and a
force that is associated with the friction and that acts in the opposite
direction. Galileo deduced that as the frictional forces are decreased (for
example, by placing oil on the table) the object would move further and
further before stopping. From this he abstracted a basic form of the law of
inertia: if the frictional forces could be reduced to exactly zero a moving
object would continue moving at a constant speed in a straight line unless a
new force acts on it at a later time. Figure (3) illustrates the difference
between the Aristotle view motion and the views of Galileo and Newton.
Aristotle view: an
object keeps
moving only if it
continually pushed
Galileo’s
deduction: a
moving object will
slow down and
stop because of
friction
v
v
Ff
Newton’s 1st law: The
natural state of
motion: object at rest
or moving with
constant velocity
when F  0

v 0
v  constant
friction
frictionless
surface between
box & floor
no air drag, etc
Fig.3. Aristotle view of motion compared to the view of Galileo
and Newton.
Web search
1. View several movie clips on Newton’s 1st law of motion.
2. Find at least eight examples of physical situations that can be
explained by Newton’s 1st law, e.g. spin dryer, sauce bottle, …
John Ogborn (Difficulties in dynamics, Physics at Secondary School: Physics
and Applications Vol II, Proceedings for the Conference 1983. Institute of
Physics, Slovak Academy of Sciences.) has summarised peoples difficulties
with Newton's Laws of Motion:
 the first law is unbelievable
 the second law is incomprehensible
 the third is merely religious incantation
Learning or teaching Physics is not a simple task. A Koyre (Journal of
Historical Ideas, 4,400,p.405, 1943) has made an appropriate comment:
"What the founders of modern science, among them Galileo, had to do, was
not to criticise and combat faulty theories, but to replace them by better
ones. They had to do something quite different. They had to destroy one
world and replace it by another. They had to reshape and reform its
concepts, to evolve a new approach to being, a new concept of knowledge,
a new concept of science - and even to replace a pretty natural approach,
that of common sense, by another which is not natural at all."
Newton’s 2nd law of motion
In an inertial frame of reference, the resultant (net) force acting
upon an object produces an acceleration of the object in the
direction of the force. The acceleration
resultant force
a is proportional to the
 F and inversely proportional to the mass m of
the object.
(2)
 F  ma
F
a
m
Newton’s 2nd law
Equation (2) tells us:

The greater the resultant (net) force acting on an object the greater
its acceleration.

The greater the mass of the object the less its acceleration.

A definition of the concept of force. If the object of mass m has an
acceleration a then the force acting on it is defined as the product of
its mass and acceleration. The product m a is not a force but gives a
number equal to the magnitude of the force.
Newton’s 3rd law of motion
In an inertial frame of reference, if an object A exerts on object B
a force
FBA , then object B exerts on object A a force FAB which is
equal in magnitude but opposite in direction
FBA  FAB
(3)
Newton’s 3rd law
Newton’s 3rd law is illustrated by the example of the repulsion
between like charges and the attraction between opposite charges
as shown in figure (4). Whenever a force acts on an object there
must be simultaneously be an equal in magnitude and opposite
direction force acting on some other object. Forces always act in
pairs. Note: this pair of forces act on different objects.
Action and reactions are equal and opposite.
A
B
FAB
FAB  FBA
FBA
repulsion between two negatively charged objects
FAC  FCA
A
FAC
FCA
+
C
attraction between a negative charge and a positive charge
Fig. (4)
charges.
Newton’s 3rd law – repulsion and attraction between
Discuss how Newton’s 3rd law explains the following examples:

The rotation of a lawn sprinkler.

A boat being propeller through water.

The recoil of a firing gun.

The operation of a rocket and jet engine.

Walking.

Driving a car.

Forces between Sun, Moon and Earth give rise to the tides of the
oceans.

A plane flying through the air.
The following list are common statements of Newton's Third Law found in
Physics textbooks, all of which are misleading.

Every action has an equal opposite reaction.

To every force there is an equal and opposite force.

Action and reaction are equal and opposite.
Why are these three statements misleading?
The donkey and cart problem
If the above are statements of
Newton's Third Law are true, how
the donkey pull the cart?
can
Hit by a truck problem
You have just been hit by a speeding bus.
What force does the bus exert on you?
What force do you exert on the bus?
Head-On problem
Consider the consequences of
driving a car into a head-on
collision with an identical car
travelling toward you at the
same speed, as opposed to
colliding at the same speed
into a massive concrete wall.
Which of these two situations
would result in the greatest
impact force? (Lewis Epstein & Paul Hewitt, Thinking Physics)
Newton's laws of motion: weight
If you review many introductory Physics textbooks you will find two
conflicting definitions of weight.
What are these two definitions?
Use these two definitions to answer the following questions.
 How can you explain the "weightlessness" of astronauts orbiting the
earth?
 What is the weight of a high jumper when standing on the ground
and in flight?
 You are in a lift and the cable snaps. What is your weight?
 A mass is suspended from a spring balance? What is the force on the
spring balance? Will the balance read different values at different
latitudes?
Discuss these questions.
How can we walk across the room?
How can an automobile move forward
or stop?
TYPES OF FORCES
Electrostatic force (long range: action at a distance) – the attractive or
repulsion between two stationary charged objects.
A
B
FAB
FAB  FBA
FBA
repulsion between two negatively charged objects
FAC  FCA
A
FCA
FAC
+
C
attraction between a negative charge and a positive charge
Gravitational force (long range: action at a distance) – attraction between
two objects because of their mass.
Weight (long range: action at a distance) – the gravitational force acting
on an object due to its attraction to the Earth. Near the Earth’s surface the
weight of an object is
(4)
FG  mg
weight of an object near Earth’s surface
+y
where m is the mass of the object and
g is the local acceleration due to gravity
box
at the Earth’s surface.
+x
FG = mg
Beware: there are other definitions of weight.
Contact forces occur when two objects are in direct contact with each
other. Three examples of contact forces are: normal force, friction and
tension.
+y
Normal force – the force acting
FN
on an object in contact with a
box
surface which acts in a direction
+x
at right angles to the surface.
+y
FN
+x
Friction force – the force acting on the object which acts in a direction
parallel to the surface. A simple model for friction Ff is that it is proportional
to the normal force FN and the constant of proportionality is called the
coefficient of friction . The maximum frictional force is
(5)
Ff max   FN
simple model for friction
+y
v
FN
normal
box
max frictional force
Ff max   FN
+x
constant of proportionality
 is the coefficient of friction
between the two surfaces
Ff
friction
front wheel drive car
FN
normal
v
Ff
front
driving wheels
Ff driving
friction
Equation (5) is only valid for a simplified model. There are two coefficients
of friction known as the static coefficient of friction
 s (no movement
between surfaces in contact) and the kinetic coefficient of friction
(relative movement between the surfaces) where
difference between the values for
 s  k
 s and k is only small.
k
. Usually the
In many texts, it is stated that friction opposes the motion of all objects and
eventually slows them down. But this type of statement is very misleading.
How can we walk across the room, how can cars move? It is because of
friction. In walking, you push against the floor and the frictional force of the
floor on you is responsible for the forward movement. The drive wheels of a
car turn because of the engine. The driven wheels exert a force on the road,
and the road exerts a forward force on the car because of friction between
the types and road. In icy conditions the driven wheels spin so that that
there is very little friction between the tyres and the road and the car can’t
be driven safely. When the foot is taken off the accelerator peddle, a car will
slow down because of friction.
Tension – the force acting along a stretched rope which is connected to an
object.
rope pulling box along floor
+y
tension FT
box
+x
tension – a pulling force exerted by
a rope, chain, chord, etc
box
rope under tension
pulley changes the
direction of the
force (tension)
box being pulled by the
rope  tension in the
rope
+y
tension FT
+x
tension FT
Springs – a common instrument for measuring force magnitudes is the
spring balance. When an object is connected to a spring it will experience a
force when the spring is compressed or extended. A simple model for a
spring is that the force Fs exerted on the object is proportional to the
compression or extension
x of the string from its natural length. This force
is known as the elastic restoring force. The constant of proportionality is
the spring constant k. This relationship is known as Hooke’s law
(6)
Fs  k x
Hooke’s law – elastic springs
+x
x=0
k
Fs = 0
x
x
F
Fs = -kx Fs < 0 x > 0
F
Fs = -kx Fs > 0 x < 0
Example 1
We will consider the forces acting on a stationary book on a floor and the
reaction forces. The book is at rest, therefore, from Newton’s 1st law the
sum of all the forces acting on the book must be zero. The book pushes
down on the floor due to its weight FFB and the floor deflects slightly pushing
back on the book FBF. The book is attracted towards the centre of the Earth
with a force FBE and the Earth is pulled by the book with a force FEB.
book at rest on floor
F 0
book
+y
floor pushes up on
book FN = FBF
FN
Earth pulls on
book – weight
FG = FBE
+x
FG
FN  FG
Forces acting ON book
normal force FN
weight FG
F N = FG
FN and FG are not a
action / reaction pair
since they act upon
the same object
Fig. 5.
book pushed down on
floor FG = FFB
Action / reaction pair
FFB  FBF
Book pulls on
Earth FEB
Action / reaction pair
FEB  FBE
Forces involved in a book resting on a floor.
Example 2
You need to study all the steps in this example of the box on an inclined
surface as shown in figure (6). Note the procedure. Choose the x-axis
parallel to the include (up) and the y-axis at right angle to the incline (up).
The object is represented as a dot. Show all the forces which act only on the
box. Resolve all these forces into components parallel and at right angles to
the incline. Apply Newton’s 2nd law to the forces acting along each
coordinate axis to find the acceleration up the incline ax. The box only
moves along the surface therefore, ay = 0.
The diagram showing the forces are often referred to as free-body
diagram. To gain the most from this very important example, work
through it many times after you have completed this Module on Forces and
Newton’s law.
We are interested only in the forces
that are acting on the box being
pulled up the inclined surface due to
the object attached to the pulley
system. The forces acting on the box
are its weight FG = mg, the tension FT
due to the rope and the contact force
FC between the box and surface.
The tension is due to the object at
the end of pulley rope, FT = Mg. The
box is replaced by a particle
represented as a dot. Choose a
coordinate system with the x-axis
acting up the inclined surface and
the y-axis acting upward and at right
angles to the surface.
The contact surface can be resolved
into two components: the normal
force FN at right angles to the incline
and the friction Ff parallel to the
surface. The frictional force is related
to normal force Ff =  FN where  is
the coefficient of friction.
The weight can also be resolved into
its x and y components, FGx and FGy
FGx = FG sin = mg sin
FGy = FG cos = mg cos
m
M

+y
+x
FT
FC

+y
FG = mg
+x
FT
FN
Ff


+y
FG = mg
+x
FT
FN
The acceleration ax up the incline can
be found from Newton’s 2nd law
Fy = FN – FGy = FN – mg cos = may
ay = 0  FN = mg cos
 Ff =  FN =  mg cos
Ff
FGy
FGx

Fx = FT - Ff – FGx = max
max = Mg -  mg cos - mg sin
ax = (M/m -  cos - sin)g
Fig. 6. Forces acting on a box accelerating up the inclined
surface due to the weight of the object at the end of the rope of
the pulley system. Newton’s 2nd law can be used to find the
unknown acceleration of the box up the incline.
Example 3
A flower pot which has a mass of 5.55 kg is suspended by two ropes – one
attached horizontally to a wall and the other rope sloping upward at an
angle of 40o to the roof. Calculate the tension in both ropes.
Solution
How to approach the problem

Identify Setup Execute Evaluate
Visualize the situation – write down all the given and unknown
information.

Draw a free-body diagram – forces.

Draw a free-body diagram – x and y components for the forces.

Forces can be added using the head-to-tail method but it is best to
solve the problem using x and y components.

Object at rest: apply Newton’s 1st law to the x and y components.

Solve for the unknown quantities.
physical picture
+y
free-body diagrams
T1  ? N
40 o
T2  ? N
40 o
+x
m  5.55kg
FG
g  9.81m.s -2
free-body diagrams
 
Object is stationary:
Newton’s 1 st law
F T1  T2  FG  0
40 o
 
T1 cos 40
T1
FG
o
Fx  0
T1 sin 40o
T2
+x
FG  mg
Fy  0
T2
Applying Newton’s 1st law to the x and y components for the forces:
Fx  T2  T1 cos40o  0
 Fy  T1 sin40o  mg  0
T2  T1 cos40o
T1 
mg
(5.55)(9.81)

 84.7 N
o
sin40
sin40o
T2  T1 cos40o  (84.7)cos40o  64.9 N
Example 4
You are driving a car along a straight road when suddenly a car pulls out of
a driveway of 50.0 m in front of you. You immediately apply the brakes to
stop the car. What is the maximum velocity of your car so that the collision
could be avoided? Consider two cases: (1) the road is dry and (2) the road
is wet.
mass of car
m = 1500 kg
coefficient of friction
  0.800
rubber on dry concrete
  0.300
rubber on wet concrete
Solution
How to approach the problem

Identify Setup Execute Evaluate
Visualize the situation – write down all the given and unknown
information. Draw a diagram of the physical situation showing the
inertial frame of reference.

Type of problem – forces and uniform acceleration.

Draw a free-body diagram showing all the forces acting on the car.

Use Newton’s 2nd law to find normal force and frictional force and
acceleration.

Car slows down due to constant frictional force  acceleration of car
is uniform  use equations of uniform acceleration to find initial
velocity of car.

Solve for the unknown quantities.
u = ? m.s-1
v = 0 m.s-1
+x
x =0
x = 50.0 m
motion map – velocity vectors
a = ? m.s-2
constant acceleration
FN
m = 1500 kg
Ff = -  FN= -  m g
FN = FG = mg
g = 9.81 m.s-2
FG = mg
(dry) = 0.800 (wet) = 0.300
From Newton’s 2nd law the acceleration of the car is
 Fy  FN  mg  0
FN  mg
Ff   FN   mg
 Fx  Ff  max  ma
a
Ff
m
  g
Constant acceleration
v 2  u2  2 a s
u   v 2  2 a s   0  2     9.81 50    981 
Maximum initial velocity to avoid collision
dry conditions =0.800
u = +28.0 m.s-1
u = +(28.0)(3.6) km.h-1 = +101 km.h-1
wet conditions =0.300
u = +17.2 m.s-1
u = +(17.2)(3.6) km.h-1 = +61.9 km.h-1
When you are driving, whether an event occurs that could lead to a fatal
accident is very dependent upon the road conditions. In this example, to
stop and avoid the collision, there is a difference of 40 km.h-1 in the initial
velocities. Many occupants inside a car have been killed or severely injured
because the driver did not slow down in poor road holding conditions. Also,
in this example we ignored any reaction time of the driver. If the reaction
time of the driver was just 1 s, then a car travelling at 100 km.h-1 (28 m.s-1)
would travel 28 m before breaking.
Predict Observe Explain
Image that you are standing on a set of bathroom scales in an elevator (lift).
Predict how the scale reading changes when the elevator accelerates
upwards, downwards and travels with a constant velocity. Write down your
predictions and justify them. Observe: work through Example 5. Explain
any discrepancies between the answers given in Example 5 and your
prediction.
Example 5
A 60 kg person stands on a bathroom scale while riding an elevator. What is
the reading on the scale in the following cases:
(1)
The elevator is at rest.
(2)
The elevator is going up at 2.0 m.s-1.
(3)
The elevator is going up at 4.0 m.s-1.
(4)
The elevator is going down at 2.0 m.s-1.
(5)
The elevator starts from rest and goes up reaching a speed of 2.0 m.s-1
in 1.8 s.
(6)
The elevator is moving up and accelerates from 2.0 m.s-1 to 4.0 m.s-1 in
1.8 s.
(7)
The elevator starts from rest and goes down reaching a speed of 2.0
m.s-1 in 1.8 s.
(8)
The elevator is moving down and accelerates from 2.0 m.s-1 to 4.0 m.s-1
in 1.8 s.
(9)
The elevator is moving up and slows from 4.0 m.s-1 to 2.0 m.s-1 in 1.8 s.
(10)
The elevator is moving down and slows from 4.0 m.s-1 to 2.0 m.s-1 in 1.8 s.
Solution
How to approach the problem

Identify Setup Execute Evaluate
Visualize the situation – write down all the given and unknown
information. Draw a diagram of the physical situation showing the
inertial frame of reference.

Type of problem – forces and Newton’s laws.

Draw a free-body diagram showing all the forces acting on the person.

Use Newton’s 2nd law to give the relationship between the forces
acting on the person and the acceleration of the person.

Determine the acceleration of the person in each case.

Solve for the unknown quantities.
+ up is the positive direction
FN scale reading: force of scale on
person
m
m = 60 kg
g = 9.81 m.s-2
a
a > 0 acceleration up
a < 0 acceleration down
FG =mg weight of person
The person exerts a force on the bathroom scales and the bathroom scales
exerts a force on the person. This is an action / reaction pair. But, we are
only interested in the forces acting on the person which are the weight and
the normal force due to the scale on the person.
The scale reading FN is found from Newton’s 2nd law:
 Fy  FN  FG  FN  mg  ma
FN  m(g  a)
acceleration due to gravity g = 9.8 m.s-2 (scalar quantity in this example)
acceleration of person a > 0 if direction up and a < 0 if acceleration down
The weight of the person is FG = mg = (60)(9.81) N =588.6 N
We can assume when the velocity changes the acceleration a is constant
and equal to the average acceleration
a  aavg 
v
t
In cases (1), (2), (3) and (4) there is no change in the velocity, hence
v  0  a  0  FN  FG  mg
Therefore, the scale reading is FN = 588.6 N or 60 kg.
For cases (5), (6) and (10)
t  1.8 s
case (5)
v  2  0  m.s-1  2m.s-1
case (6)
v   4  2 m.s-1  2m.s-1
case (10)
v   2   4   m.s-1  2m.s-1
The acceleration is
a
v  2.0 
-2
-2

 m.s  1.11m.s
t  1.8 
The scale reading is
FN  m  g  a    60  9.81  1.11 N  655N
or
67 kg
This scale reading is often called the person’s apparent weight.
The person feels the floor pushing up harder than when the elevator is
stationary or moving with a constant velocity.
For cases (7), (8) and (9)
t  1.8 s
case (7)
v   2  0  m.s-1  2m.s-1
case (8)
v   4   2  m.s-1  2m.s-1
case (9)
v  2  4  m.s-1  2m.s-1
The acceleration is
a
v  2.0 
-2
-2

 m.s  1.11m.s
t  1.8 
The scale reading is
FN  m  g  a    60  9.81  1.11 N  522N
or
53 kg
The person feels their weight has decreased.
In the extreme case when the cable breaks and the elevator and the person
are in free-fall and the downward acceleration is a = -g. In this case the
normal force of the scales on the person is FN = m(g - g) = 0 N. The person
seems to be weightless. This is the same as an astronaut orbiting the Earth
in a spacecraft where they experience apparent weightlessness. The
astronaut and spacecraft are in free-fall and there are zero normal forces
acting on the person. The astronaut still has weight because of the
gravitational force acting on them.
The acceleration does not depend upon the direction of the velocity. What is
important is the change in the velocity. A good way to understand this
concept is to draw the appropriate motion maps:
(1)
(2)
(3)
(4)
(5)
+
a=0
a=0
a=0
a>0
a=0
(6)
(7)
a>0
a<0
(8)
a<0
red arrows: velocity vectors
(9)
a<0
(10)
a>0