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Lecture 3
Newton’s laws
Forces
Friction
NEWTON’S THIRD LAW
If one object exerts a force on a
second object, the second object exerts a
force back on the first that is equal in
magnitude and opposite in direction.
“For every action there is an equal and
opposite reaction”
Freaction of the table
object
table
Fweight object
Faction
Magnitude of force
Fweight object = Freaction table
Freaction
Application: Newton’s 2nd Law
Fa
Define system and external force
System 1 (trolley, patient, and attendant):
External force provided by the floor: reaction to
the force the attendant exerts on the floor.
System 2 (trolley with patient): External force
provided by the attendant.
define the system and
determine the external forces
Application: system & external force
Mass of patient = 50 kg
Mass of trolley = 20 kg
Mass of attendant=85 kg
Fa
Mass System 1 = 155 kg
Mass System 2 = 70 kg
1/ If the attendant exerts a force of 100 N on the
floor, what is the magnitude of acceleration of
the trolley and of the attendant?
System 1: F=ma
100 N = (Mass System 1)a
 a = 100N/155kg = 0.645 ms-2
Application: system & external force
Mass System 2
= 70 kg
Fa
2/ What force does the attendant then exert on
the trolley?
System 2:
Fattn = (Mass System 2)*a
= (70kg)(0.645ms-2) = 45.2 N
3/ Why is the force exerted by the attendant on
the floor bigger than the force he exerts on the
trolley?
Because he also accelerates.
Mass of attendant = 85 kg
F=ma = 85kg.0.645ms-2=54.8N
Adding Forces
F is the Resultant force
F1
F=F1+F2
“co-linear”
F2
Length of arrow represents magnitude of force
“perpendicular”
F2
Magnitude
Direction
In general
F
q
F1
F  F12  F22
 F2 
tan q   
 F1 
*
 F2 
q  tan  
 F1 
1
we use vectors or graphical method
to determine the resultant force
Projection of a force
Often need to split a force into two
perpendicular components.
Projection
parallel to
the plane
a mgCosa
a
W=mg
F2
Projection
perpendicular
to the plane
Cosa = F1/mg
a
F1
w=mg
Sina = F2/mg
F1 = mgCosa
 F = mgSina
2
Example
A gardener pushes a lawnmower with a force
of 500N directed along its handle which
makes an angle of 60º with the ground in order
for it to move at a constant velocity. Determine
(a). The component of this force which is directed
horizontally H, in order to overcome the frictional
forces and maintain a constant velocity, and,
(b). the component of the force which is
directed vertically downwards, V .
Use method of components
500N
V
600
H
H = 500 cos600
= 250N
V =500Sin600
= 433N
Classes of Forces
WEIGHT (w=mg) due to gravity
w=mg
N=w=mg
NORMAL FORCE (N) normal
to a plane
w = mg
FRICTION (f ) parallel to a plane
f
TENSION (T) is any force
carried by a flexible string,
rope etc. acts all along the
string
F
T= w
w
Normal Force
The normal force is the reaction by the plane to
the perpendicular (or normal) component of
force applied.
N=mg
W=mg
a
mgCosa
Origin:
Plane deforms under
a W=mg
load. The elastic
force attempting to
restore the plane to its original shape is the
“normal force”.
EXAMPLE
A constant horizontal force F = 4N to the right,
is applied to a box of mass 2kg resting on a
level, frictionless surface.
What is the acceleration of the box?
Let the x axis (right) and y axis (up) be the
positive directions.
We assume the positive x axis is the
direction of the force.
Mark the forces acting on the box in the
diagram. There are three forces acting
on this box
1. The force F pulling the box
2. The weight (w) of the box
3. The normal reaction (N) (The third law says
that if the box pushes down on the table, the
table pushes up on the box)
Exercise:
Two children pull in opposite directions on a
toy wagon of mass 8.0kg. One exerts a
force of 30N, the other a force of 45N. Both
pull horizontally and friction is negligible.
What is the acceleration of the wagon?
N=mg
System: the toy
wagon;
x positive to the
right
Fext, net
F2=45N
F1=30N
w=mg
=F2 - F1 = 45N - 30N = 15N (right)
a = Fext, net/m = (15/8)ms-2=1.875ms-2
to the right
Biting Force
Force between bottom and top teeth
Bite-force estimation for Tyrannosaurus rex
Nature 382, 706 - 708 (22 August 1996);
6,410 N to 13,400 N
Muscles in the human jaw
can provide a typical biting
force of ≈ 600N
muscle
pivot
force
Biting Force
Average maximum human biting force ≈ 750 N
Biting force varies depending on region of mouth
Molar region
400-850 N
Premolar region 220-440 N
Cuspid region
130-330 N
Incisor region
90-110 N
force
Energy of a bite is absorbed by
Food
Teeth
Periodontal ligament
bone
Tooth design is such that it can absorb large
static and impact energies
Applications of Newton’s Laws
Friction is a force that always acts to oppose
the motion of one object sliding on another.
Friction is a contact force
resulting from surface roughness
Surfaces slide over each other,
rough bits catch each other
and impede the motion.
Surfaces not perfectly smooth.
Schematic showing the nature
of forces on a very small (atomic)
scale. Surface roughness results
in frictional forces
Applications of Newton’s Laws
The normal reaction force, N, is described
by Newton’s third law.
The third law says that for every force there
is an equal but opposite force.
Friction and Normal reaction force are two
types of contact forces
If one object is sliding on another,
the frictional force, F, always acts in the
direction opposed to the motion,
the normal force always acts in a direction
perpendicular to the surface.
Applications of Newton’s Laws
Imagine an object lying on a frictional surface
and a very small horizontal force, A, acts on it.
The force diagram is drawn as follows.
N
A
F
W
The forces acting on the object are its
weight w, the normal force N, the frictional
force F and the applied force A.
Applications of Newton’s Laws
To a good approximation the frictional force (F)
of an object on a surface is proportional to the
normal force of the surface on the object
F
N
F = µN
where the constant of proportionality is known
as the Coefficient of friction µ.
The most familiar thing about friction is that the
heavier an object the greater the friction. It
is more difficult to slide a full box than an empty
one.
Large force required to
move heavy object
N=mfg
Wf=mfg
Small force required to
move light object
N=meg
We=meg
Kinetic Friction
Kinetic friction is the force that opposes
motion due to physical contact between
substances, and is parallel to the contact
surfaces.
It is proportional to the normal force: F= kN
k is the coefficient of kinetic friction
(between moving substances)
N=mg
F= kN
A
w=mg
Const. speed: (A=F)
Acceleration: (A>F)
Deceleration: (A<F)
Static Friction
Static friction is the force that prevents motion
due to physical contact between substances, and
which is parallel to the contact surface.
F=A
A
It behaves like the normal force:
F=A up to a maximum of sN
s is the coefficient of static friction
(between immobile substances)
If A< sN: static friction prevents motion
If A> sN: static friction isn’t strong enough 
motion  kinetic friction.
Generally k < s
Applications of Newton’s Laws
Friction– Graphical representation
F
Fs,max = sN
F = kN
A
Static
Kinetic
Values of coefficients of friction s and k
depend on the nature of the surfaces
Typically 0<k<s<1
Typical values for coefficient of static friction s
of some substances
Rubber on concrete s = 1
Synovial joints s ≈0.016
EXAMPLE
You press a book flat against a vertical wall.
In what direction is the frictional force exerted
by the wall on the book?
F
(a)downward
(b) upward
(c) into the wall
H
N
(d) out of the wall
w
EXAMPLE
A crate is sitting in the centre of a flatbed truck.
The truck accelerates eastwards, and the crate
moves with it (not sliding on the bed of the
truck). In what direction is the frictional force
exerted by the bed of the truck on the crate?
(a)to the west
(b) to the east
(c) there is no frictional force
because the crate does
not slid
Friction
Dental considerations
Abrasion
Abrasion or wear due to surface roughness
Dental restorations and appliances should
have smooth surfaces
Abrasive resistance of material depends on
Hardness
Surface roughness
Example
Excessive wear of tooth enamel by opposing
ceramic crown caused by
High biting force
Rough ceramic surface
Remedy
Adjust occlusion•broader contact areas---reduces stresses
Polish ceramic to reduce abrading surface
Friction
Lubrication in the mouth is provided by saliva
Saliva greatly reduces the frictional force and
helps prevent excessive wear of teeth
Saliva reduces friction force
by up to a factor of 20.
Lubricating properties of saliva are maintained
over the range of jaw biting forces (0→600N)
muscle
pivot
force
Reduced friction→ reduces wear
ensures good contact between
opposite teeth is not impaired and surfaces
used for chewing are maintained.
Saliva, also protects the tongue and other soft
tissues inside the mouth
EXAMPLE
A crate slides at constant velocity down a plane
inclined at an angle a to the horizontal. If the
coefficient of kinetic friction between the surfaces
of the crate and the plane is 0.6,
determine the angle a.
a
mgCosa
a W=mg
F  k N  k  mgCosa 
F  k N   mgSina  Constant velocity
mgSina
k 
 Tana
k mgCosa  mgSina
mgCosa
a  tan 1  k   tan 1  0.6   310
EXAMPLE
You need to move an object across a
horizontal surface by pushing it. Its mass is
50kg and the coefficients of friction are µs=0.5
and µk=0.4.
How much force A do you have apply
horizontally to a) get it moving?
b) To keep it moving?
When the object is stationary the friction that
impedes motion is determined by µs. The
forces on the box are outlined below
N = mg
Fmax = sN
A
w=mg
The object will start to move when the applied
force A becomes very slightly greater than the
maximum frictional force Fmax=µsN. For our
purposes we will say this happens when A=Fmax