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FORCES
What is a force?
Intuitively, a force is like
a push or a pull
which produces or tends to produce motion
Forces experienced in Daily Life
• Weight
• Normal reaction
• Friction
• Viscous force
• Tension
• Upthrust
• Lift
• Electrical force
• Magnetic force
Weight
W
• weight is not the same as mass; it is a force
• it is the gravitational force exerted by the Earth
• it passes through the centre of gravity of the body
Normal reaction
N
• two bodies in contact with each other
• perpendicular to the surface of contact
Friction
• friction is exerted two surfaces slide across one another
• direction is along the surface of contact
Cause of friction
movement
F
• hollows and humps all over the surface
• actual contact area only a fraction 1/10000 of total area
• extreme high pressure at contact points causes welding
of surfaces
• forces are needed to overcome these adhesive forces
when trying to slide over the surface
Static and kinetic friction
• It is harder to move a stationary object than to move
the object while it is moving
• Static friction is the friction exerted by the ground in
order to prevent the object from moving
• Kinetic friction is the friction exerted by the ground to
oppose the motion of the object while it is moving
Limiting friction
• Static friction is not constant; it varies in magnitude
• Suppose a force P is applied trying to move the object
2 N1 N F
•
•
•
•
•
P
1 N2 N
If P is 1 N, F will also be 1 N to prevent object moving
If P increased to 2 N, F also increased to 2 N
But there is limit to how much F can increase to
Maximum possible static friction is called limiting friction
P must exceed limiting friction in order to move object
Example 1
In Fig 1.1, an object was moving to the right on a rough
surface. In Fig 1.2, an object rests in equilibrium on a
rough slope. In both cases, draw the friction force acting
on each object.
Fig 1.1
friction
Fig 1.2
Viscous force
• When body moves in fluid, it experiences resistance
• such resistance is known as viscous force
• examples: air resistance and water resistance
• viscous force depends on the speed of the body
• the greater the speed, the greater the viscous force
Terminal velocity
release
v
F
W
F
W
V
F
W
VT
W
gathering
speed
gathering
more speed
finally reaches
constant terminal
velocity
Tension
• Tension is exerted by a stretched rope, string or spring.
• When a body is attached to a string, the tension in the taut
string would tend to pull the body.
Hooke’s Law
F
F  (l - lo)
=> F  e
=> F = ke
where k is force constant
(elastic constant, spring
constant or stiffness F
constant)
Strain energy in a Deformed wire
Assume that Hooke’s Law is obeyed.
=>For a force-extension graph, it will be a straight line.
In general, work done by a
force F in extending a wire
from x1 to x2 is the area
under the force-extension
graph.
F
e
x
=>Work done in extension
or strain energy stored in
wire, W = ½ Fe = ½ ke2
Example 2
A vertical wire suspended from one end is
stretched by attaching weight of 20 N to the
lower end. If the extension is 1 x 10-3 m,
what is
(a) the force constant;
(b) the energy stored in the wire;
(c) the gravitational potential energy loss
by the weight in dropping a distance of
1 x 10-3 m?
F
20
 3  2 x10 4
e 10
Assuming Hooke’s law is obeyed,
(a) F = ke
k = F/e = 20/(1x10-3) Nm-1
= 2 x 104 Nm-1
(b) energy stored in wire,
W = ½ Fe
= ½ (20)(1x10 -3)
= 1 x10-2 J
(c)Gravitational potential energy lost by weight
= mgh = 2 x 10-2 J
By conservation of energy,
P.E. lost = Energy stored in wire
+
heat dissipated when weight at
end of wire comes to rest after
vibrating.
Upthrust
upthrust
• Upthrust is an upward push on a body when it is immersed
in a fluid (gas or liquid)
• Upthrust is exerted by the fluid
• Upthrust is due to pressure difference of fluid at the top
and bottom of immersed portion of the body
Example 3
Consider an object partially immersed in a fluid of density
. The area of the top surface of the object is A and the
immersed depth is h.
h
h
(a) What is the pressure difference across the
immersed portion of the object?
hg
(b) Hence write down the expression for the
upthrust acting on the object.
hgA
(c) What is the volume of fluid displaced by the h A
object?
(d) Hence write down the expression for the
weight of fluid displaced.
(e) Comment on your answers to (b) and (d).
They are the same.
hAg
Example 3 shows that
Upthrust = weight of fluid displaced
This is actually the Achimedes’ Principle
Archimedes’ Principle states that
the upthrust on a body in a fluid
is equal and opposite to the
weight of the fluid displaced by the body.
Lift
What helps birds and aeroplanes maintain its flight?
The answer is the upward lift force exerted on their
wings when in motion
Electric force
Electric force is exerted between two electric charges
+
+
like charges repel
+
-
unlike charges attract
Magnetic force
Magnetic force is exerted between two magnetic
materials or between electric currents
N
N
like poles repel
N
S
unlike poles attract
Forces experienced in Daily Life
• Weight
• Normal reaction
• Friction
• Viscous force
• Tension
• Upthrust
• Lift
• Electrical force
• Magnetic force
Different forces
normal reaction
upthrust
weight
weight
normal reaction
weight
Different forces
lift
tension
weight
weight
Different forces
How did this ‘forward force’ come about?
normal
reaction
normal
reaction
speed
air
resistance
forward
force
friction
weight
Different forces
How did this ‘thrust’ come about?
lift
air
resistance
thrust
weight
Who exerts on who
A force is always exerted by some body on some other body.
What makes a car move?
Friction exerted by
ground on tires
What makes a rocket fly?
Gases expelled by rocket
Test Yourself. Identify the forces
normal reaction
weight
weight
FGM
weight
Fundamental types of force
When scientists examined all the forces,
they found that many of them are similar in nature.
Scientists have identified 4 fundamental types of force:
• gravitational force
• electromagnetic force
• nuclear force
• weak force
All forces in our daily life can be classified into one of
the fundamental types. In the following table, identify
the nature of each force:
Force
Nature
Weight (W) of an object
gravitational
Gravitational
Attraction between two oppositely-charged bodies
electromagnetic
Electromagnetic
Attraction by a magnet on a piece of iron
electromagnetic
Electromagnetic
Tension (T) in a string pulling an object
electromagnetic
Electromagnetic
Pushing a person with your hands
electromagnetic
Electromagnetic
Normal reaction (N) by the table on an object resting on it
electromagnetic
Electromagnetic
Friction (F) experienced by an object moving on a rough surface
Electromagnetic
electromagnetic
2
Addition of Vectors
Parallelogram Rule
A
B
A
R
B
Triangle Rule
A
B
B
A
R
Finding resultant force
The magnitude of resultant force can be
found by
• drawing vector diagram to scale
• calculation (pythagoras theorem, cosine
rule, etc)
• resolution
Example 4
Two forces are given below:
70º
5N
4N
30º
Find the magnitude of the resultant force.
Method 1 Drawing vector diagram to scale
Scale used is 1 cm : 1 N
5N
(5 cm)
4N
(4 cm)
R
(5.8 cm)
From the vector diagram,
magnitude of resultant R is 5.8 N
What is missing in the answer?
Method 2: By calculation
5N
30

80
4N
70
x
• Using Cosine rule: x2 = 52 + 42-2 (4) (5) cos800
=>x =5.84 N
0

sin 
• Using Sine rule: sin 80

   42.4 0
5.836
4
Method 3 By resolving vectors
4 cos 70°
5 sin 30°
70º
5N
30º
4 sin 70°
5 cos 30°
4N
Rx = 5 cos 30° + 4 cos 70° = 5.70 N
Ry = 5 sin 30° - 4 sin 70° = -1.26 N
Rx 5.70 N
Ry
1.26 N
Magnitude of resultant R is given by
R2 = (5.70)2 + (1.26)2  R = 5.8 N
R
Example 5
Two horizontal forces act at a point to produce a
resultant force of magnitude 40 N in the eastward
direction. Given that one of the forces is in the
northward direction and has a magnitude of 30 N,
find the magnitude and direction of the second force.
N

F
30 N
E
40 N
Magnitude of second force F =  302 + 402 = 50 N
Angle  = tan-1 (40/30) = 53°,
direction of F is 53° east of south (or bearing 127°)
Centre of gravity and
Free body diagram
Centre of Gravity
• The centre of gravity
of a body is the
single point at which
the entire weight of
the body can be
considered to act.
Free body diagram (Important)
• is a diagram showing all the forces
acting on a particular object
• is an important tool for solving problems
Example 6
An object A of weight w rests on top of another object
B of weight W placed on the ground, as shown.
Draw separate free body diagrams showing forces
acting on
(a)
A only
(b)
B only, and
(c)
A and B together.
Answer
N1
N2
w
N3
W
N3
W+w
N1 = normal reaction exerted by B on A
N2 = normal reaction exerted by A on B
N3 = normal reaction exerted by ground on B
N1 is numerically equal to N2 (action / reaction pair)
Common forces in free body diagrams
Force
Appli cable when
weight
W
Object has a mass
tension
T
Object is attached
to a string
normal
reaction
N
friction
F
Direction of force
vertically
downwards
through centre of
gravity
along the string
pulling towards
the centre of the
string
Object is in con tact
with a surface
normal to and
away from the
contact surface
Object tries to
move across a
rough surface
along the rough
surface
Exampl e
Force exerted by surface (only)
Total force R exerted by surface on moving
object consists of two components
- normal reaction N
- frictional force F
R is also known as the contact force
N
R
Motion
F
A Non lecture Note Example
An object of weight W, resting on a rough surface, is
connected to a suspended object of weight w by a
string over a smooth pulley. Draw and label the
forces acting on each object.
Normal
reaction
tension
tension
friction
w
W
Turning effect of a force
Consider a water wheel which is free to rotate about its
centre.
Water flowing to the right exerts force on lower blades.
This force causes the wheel to rotate about its centre.
We say that the force has a turning effect.
Turning effect of a force is also known as its moment.
Amount of moment depends on force and distance away.
Moment of a force
The moment of a force about an axis is defined as the
product of the force and the perpendicular distance
between the axis and the line of action of the forces.
The moment of a force is also known as the torque.
A
Moments about A
= F l (clockwise)
l
F
Moment of a force
Example: Not in lecture notes
A
l
300
Mtd 2
300
Mtd 1
F
A
A
l F cos 300
F
F sin 300
l
300
F
Moments about A
= F l sin 30 0 (clockwise)
= 1/2 F l
Example 7
Find the moments of the following forces about point A.
5m
4m
A
40
20 N
3m
30 N
40 N
60
Moment of 30 N about A = 30 × 4 = 120 N m (anticlockwise)
Moment of 40 N about A = 40 × 3 sin 60
= 104 N m (clockwise)
Moment of 20 N about A = 20 × 5 cos 40
= 77 N m (clockwise)
Torque of a couple
The torque of a couple is equal to the product of one
of its forces
the perpendicular
between
Couple
= pairand
of equal
and oppositedistance
forces whose
lines
the lines of
forces.
of action
action of
dothe
nottwo
coincide
F
A
x
d
F
Taking moment about any arbitrary point, say A,
total anticlockwise moment = F × (d+x) - F × x
= Fd
Example 8
Calculate the torque acting on the rod 2.0 m long in
Figs 9.1 and 9.2.
10 N
10 N
Fig. 9.1
Fig. 9.2
2.0 m
30º
2.0 cos 30º
10 N
10 N
Fig. 9.2:
9.1:
Torque = F d distance
Perpendicular
= 10 × 2.0
between
= 20 N10
m N forces = 2.0 cos 30º
Torque = 10 × 2.0 cos 30º = 17 N m
5
System in equilibrium
A system is in equilibrium when there is
no resultant force and no resultant torque.
Second
First
condition:
condition:
Resultant
Resultant
force
torque
is zero
is zero
forces
would form
a closed
triangle
or polygon
• total
clockwise
moment
= total
anticlockwise
moment
sum
of components
resolved in any direction is zero
• if
there
are only 3 forces,
would
intersect
at translational
a common point
• they
system
is said
to be in
equilibrium
• system
in rotational
equilibrium
is eitherisatsaid
restto
orbe
moving
with constant
velocity
• is
atconstant
rest or rotating
with constant angular velocity
has
linear momentum
• has constant angular momentum
Example 9
A horizontal force F is exerted on the pendulum of weight
W, causing the pendulum to be suspended at an angle 
to the vertical, as shown. Find F in terms of W and .


T
T
W
F
F
W
From the vector diagram,
tan  = F / W  F = W tan 
Example 10
A body of weight 200 N is suspended by two cords, A and
B, as shown in the diagram. Find the tension in each cord.
60º
cord A
cord B
TB
TA
W
TB
W
60º
TA
From the vector diagram,
tan 60º = W / TA  TA = W / tan 60º = 200 / tan 60º = 115 N
sin 60º = W / TB  TB = W / sin 60º = 200 / sin 60º = 231 N
Example 11
A uniform rod is supported with the fulcrum exactly at the
centre of the rod. Two masses were placed on the rod and
the system is in equilibrium. Find m.
0.45 m
2.0 kg
2.0 × g
0.30 m
N
m
W
mg
Taking moments about the fulcrum,
clockwise moments = anticlockwise moments
m g × 0.30
=
2.0 × g × 0.45

m
=
3.0 kg
Example 12
A uniform rod XY of weight 20 N is freely hinged to a wall
at X. It is held horizontal by a string attached at Y at an
angle of 20º to the rod, as shown.
string
X
20º
Y
Find
(a)
the tension in the string,
(b)
the magnitude of the force exerted by the hinge.
Example 12
(continued)
string
R
R
X
T
20º
20 N
Y
70º
T
29 N
20 N
(a) Let R
 be
of the rod
(b)
bethe
thelength
force exerted
by and
the hinge
T be
tension
in theusing
stringcosine rule,
From
thethe
vector
diagram,
2 - 2(20)(29)
Taking
about
X,
R2moments
= 202 + 29
cos 70º
anticlockwise

R = 29 N moments = clockwise moments
T sin 20º × 
=
20 × ( / 2)

T
=
29 N
Example 13
A heavy uniform beam of length  is supported by two
vertical cords as shown.
cord A
TA
(3/10) 
cord B
TB
(7/10) 
weight
Taking momentstension
about the
centre
in cord
A of gravity,
Find
the ratio moments = anticlockwise moments
clockwise
tension in cord B
TA × (2/10) 
=
TB × (5/10) 
 ratio TA / TB
=
5/2
The End