FORCES AND NEWTON’S LAWS OF MOTION IN 1d
1. Types of Force
Forces arise when objects interact. There are many different kinds of forces, but they are all
vector quantities because they all have both a size and a direction of action. The units for
measuring a force are Newtons (N).
Activity 1
Each pack contains the names of six types of force, with a description of what causes them and
their line of action. Match them up. (Types of Forces Pack)
Consider a man attached to a parachute falling from a plane. The person is subjected to a number
of forces that interact to determine their motion. Tension forces in the parachute strings (caused
by the parachute), gravity from the earth, wind forces (buffeting them), air resistance (on their
body).
At the same time the parachutist is exerting forces on the parachute. The forces on the parachute
are: Gravity from the earth, the tension in the parachute strings caused by the man, wind forces
(buffeting it) and air resistance (on their body)
Air resistance
Air resistance
Wind forces
Tensions
Wind forces
Tensions
Weight
Weight
In general the forces that we encounter in M1 and M2 can be broken down into the following
types:
Name of Force
Weight
Cause
Gravity pulling
Normal Contact
Something touching
Friction
Where there are rough
surfaces in contact
Driving
Motors or Engines
Resistance
Moving through a fluid
(liquid or gas)
Direction
Towards the ground from the centre of gravity
Perpendicular to the surface
In the opposite direction to the movement
In the same direction as the movement
In the opposite direction to the movement
Tension
Pulling with a string/bar
Outwards, along the line of the string/bar
Compression
Squeezing and object/bar
Inwards, along the line of the string/bar
The actual dynamics of these forces is very complicated; the reaction force, for example, is really
a collection of forces between the molecules that make up the two surfaces in contact, and the
weight of an object depends on the density of its component parts and their distribution.
However, in most cases it is perfectly reasonable to combine all the components contributing to a
particular force so that they act together in one place. This is made easier still if we model the
objects involved as (0 dimensional) particles because then all the forces act at the same point.
Surprisingly, although these assumptions seem fairly liberal, they still make for a highly
workable model for most applications.
2. Calculating the weight of an object
Many people confuse the terms mass and weight.
The mass of an object is a scalar quantity and remains the same for every object wherever it is in
the universe. The mass of a kilogram of apples is the same on Earth as it is in space or on the
surface of the moon. However, if we were to drop the apples on the Earth they would accelerate
downwards faster than on the moon. Whilst in space they would not fall at all! The forces that
cause this action are the varying gravitational fields that the apples are exposed to and it is found
by multiplying the mass of the apples by the accelerating effect of gravity. The quantity you end
up with is called the weight of the object and it will differ depending on the gravitational field.
In space the apples do not accelerate downwards
so the force acting on the apples is: 1 × 0 = 0 N (The apples are weightless)
On the moon the apples accelerate downwards at about 1.6ms-2
so the force acting on the apples is: 1 × 1.6 = 1.6 N (The apples have a weight of 1.6 N)
On the Earth the apples accelerate downwards at about 9.8 ms-2
so the force acting on the apples is: 1 × 9.8 = 9.8 N (The apples have a weight of 9.8 N)
Every object exerts a ‘gravitational pull’ based on its mass but unless we think about very
‘massive’ objects (like planets) the pull is negligible. In addition the gravitational pull lessens the
further apart the items are.
When calculating the gravitational force or weight of objects we will always use the formula
Weight = mass × g (where g is the gravitational constant declared in the question). We will only
consider the weight exerted by the Earth and we will assume that it acts downwards.
Exercise 3C page 52 questions 1 – 6
3. Newton’s Third Law
Newton’s third law states that:
When one object exerts a force on another there is always a reaction of the same kind which is
equal and opposite in direction to the acting force
When solving a problem relating to forces it is important to draw a clear diagram of the
object being considered and to label all the forces that are acting on the object being modelled.
For every force modelled there will always be an equal and opposite force that could be added
that is acting on the object causing the force but it is generally best to avoid adding these forces
on the diagram unless they belong to an object that we also need to model.
Eg The weight of the parachutist is the force that the earth (as a mass) exerts on the parachutist
through gravity, but the parachutist exerts an equal and opposite force on the earth (as a mass)
through his own gravitational field!
Activity 2
Look at the diagrams and add all the forces that act on the object stated by drawing arrows
Slides 4 to 13
Exercise 3a page 43 questions 1 - 11
4. Equilibrium and Newton’s First Law
Newton’s first law states that:
Every particle continues in a state of rest or in uniform motion in a straight line unless acted on
An
is in equilibrium
if the forces acting on it balance meaning that there is no resultant
byobject
a resultant
force
What this means is that if the forces acting on an object balance it will not accelerate or change
its direction. The converse is also true. If an object is not accelerating or changing its direction
then the forces acting on the object in every direction must balance. An object for which the
forces balance is said to be in equilibrium.
Consider a water skier moving at a constant speed behind a powerboat:
Normal reaction force, R
Air Resistance, R
Tension force, T
Friction, F
Weight, W
The water-skier is in equilibrium so we can write down the following equations:
Forces acting in the upwards direction ↑ = Forces acting in the downwards direction ↓
So:
R=W
Forces acting in to the left ← = Forces acting to the right →
So:
A+F=T
You can see from this example how given the mass of the water-skier it would be possible to
calculate the Reaction force of the Water on the skis, or how we could calculate the overall
resistance effect if we could measure the Tension in the tow rope.
Example 1 : A coin of mass 10g rests on your finger. What is the size of the reaction force of
your finger on the coin? What is the size of the reaction force of the coin on your finger?
R
W
↑ R=W
R = 0.01 × 9.8
R = 0.098N
Example 2 : A parachutist of mass 80kg falls vertically. Air resistance on his body is 20N and he
is supported by a harness that consists of 4 ropes each attached to the parachute. He is travelling
at a constant speed.
Find the size of the tension force in each of the ropes.
Rp
4T
W
↑ 4T + Rp = Wp
4T + 20 =80 × 9.8
4T + 20 =784
4T
= 764
T
= 191N
If the mass of the parachute is 10kg, what is the resistance force on the parachute?
Rc
4T
W
↑ Rc = 4T + Wc
Rc = 764 + 98
Rc = 862N
Note, you could solve the second part of this question by considering the system as a whole. That
is by modelling the parachutist and the parachute together as a single object and ignoring the
tension forces that are now internal to the system.
Rp+Rc
↑ Rp + Rc = W p + W c
Rc + 20 = 784 + 98
Rc = 862N
Wp + W c
This is a useful method for dealing with connected particles. You can either consider the forces
on each particle individually by considering all the forces acting on each particle separately or
you can look at the forces acting on the system as a whole and ignore the internal forces.
Example 3 : Consider a pile of 3 books resting on a table. The book on the bottom has weight
1N, the middle book has weight 2N and the top book has weight 3N.
a) What is the reaction force on the table?
b) What is the normal contact force between the book on the top and the book on in the middle?
a) Consider the system as a whole:
R
↑ R=W
R=1+2+3
R = 6N
W
b) Consider the books separately:
Rtable = 6N
Botto
m
Book
Rb/m
1N
Rb/m
Middle
Book
2N
↑ 6N = 1N + Rb/m
Rb/m = 5N
↑ Rb/m = 2N + Rm/t
5N = Rm/t + 2N
Rm/t = 3N
Rm/t
5. Common Models
In mechanics we frequently model real objects as geometrical figures to simplify the mathematics
involved. In the process we make assumptions that may render our results unreliable. We may
then need to refine the model.
• Particle.
A particle is a body whose dimensions are so small compared with the other lengths involved
that its position can be represented by a single point. For example, in considering the motion
of the Earth relative to the Sun you may represent the Earth and the Sun by particles.
The implications are that there is no spin, no air resistance, and that the distance between
objects can be treated as the distance between their centres (clearly this is not good for
modelling car collisions, since the dimensions of the cars are important).
A person standing on a plank can be modelled as a particle so that the weight can be taken as
acting at a single point. This is much simpler than considering how the weight is distributed
over the area of contact between feet and plank.
• Bead
A bead is a particle which is assumed to have a hole drilled through it so that it may be
threaded onto a string or wire. For example, the plastic animal threaded on a wire as part of a
chid’s rattle may be modelled as a bead.
• Lamina
A lamina is a flat object whose thickness is small compared with its width and length. For
example, a piece of card, a sheet of paper or a thin metal sheet may be represented by a lamina.
• Uniform Lamina
A uniform lamina is one in which equal areas of the lamina have equal masses. This is clearly
the case when the whole of the lamina is made of the same material.
• Rigid Body
A rigid body is an object made up of particles, all of which remain in the same position
relative to one another. A rigid beam does not bend, and a snooker ball is rigid in the sense that
it does not deform when striking a cushion.
• Wire
A wire is a rigid body in the form of a thin thread of metal.
• Rod
A rod is a rigid body whose mass is concentrated along a line. It is assumed to have length
only, and its width and breadth are neglected.
If a rod is supported horizontally, it will not bend, and all normal reactions will be vertical.
This greatly simplifies the calculations.
R
R
R
R
• Uniform Rod
A uniform rod is one in which equal lengths have equal masses.
• Light
A light object is one whose mass is negligible compared to the masses of other objects under
consideration. We often talk about light rods or light strings connecting particles together. We
can ignore their weights in calculations.
• Inextensible or Inelastic String
This is a string whose length is the same no matter how much mass is hung from it. If two
particles are connected by an inextensible string in tension, the particles will have the same
speed and the same acceleration. This is very useful in solving problems involving connected
particles.
• Smooth Surface
A smooth surface is one which offers so little friction, that it may be neglected.
• Rough Surface
A rough surface is one where friction is significant, and cannot be neglected.
• Smooth Pulley
A smooth pulley offers no friction to a string passed over it. In particular, this means that the
tensions in the string either side of the pulley must be equal. This is very useful in solving
problems involving connected particles.
• Peg
A peg is a support on which an object (such as a loop of wire or a string) may be hung, or on
which an object (such as a plank with on end on the ground) may rest. The peg may be rough
or smooth, but is considered to have no actual dimensions so that any normal reaction acts at a
single point.
• Plane Surface
A plane surface is a completely flat surface.
• Earth’s Surface
The Earth's surface is usually modelled as a plane surface, as its curvature is negligible on a
small scale. Clearly this assumption cannot be made when considering orbiting satellites.
6. Assumptions and Refinements to mathematical models
If our mathematical model yields unacceptably inaccurate answers, we will need to refine it by
taking extra factors into consideration.
Example 1 : A cricket ball is thrown through the air. Make a mathematical model of the situation,
and then suggest refinements.
The model is...
•
•
•
•
•
The cricket ball is modelled as a particle.
The acceleration due to gravity is constant.
The motion of the ball takes place in a vertical plane.
The only force acting on the ball is its weight.
The ground is a plane surface.
The model could be refined by...
•
•
•
•
•
Considering the dimensions of the ball.
Considering the decrease of the Earth’s gravitational force as the ball climbs higher.
Considering sideways movement due to spin and crosswinds.
Considering air resistance.
Considering the curvature of the Earth.
The diagrams below illustrate the effect of the refinements to the model.
refined model
initial model
height
height
distance
Topic Review : Mathematical Models
distance