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Forces
Things to recall
• A force is usually simply defined as a push or
pull.
• Alt: An influence which causes motion.
• A force acting on a body can cause the object
to accelerate.
• When multiple forces act on a body, the
resultant (or net) force is a superposition of
the forces.
• We start this discussion by recalling Newton’s
2nd law.
• Definition: The net force on a body is equal to
the product of its mass and acceleration.


F  ma
(As an equation)
• We start this discussion by recalling Newton’s
2nd law.
• Definition: The net force on a body is equal to
the product of its mass and acceleration.


F  ma
(As an equation)
• Note that we can write this law in component
form for each axis.
 Fx  max , Fy  may , Fz  maz
Application of Newton’s laws
Example
Example
• A jumbotron weighing 48000N is suspended
above the 3Ws field by cables shown in the
diagram. Find the tension in the cables if there
is equilibrium.
20
20
• We start by drawing a free body diagram.
• We start by drawing a free body diagram.
• A free body diagram is one where all of the
forces acting on a body are shown while all
other forces and bodies are removed.
• Using a free body diagram helps you to
visualise the problem!
• Identify the system of interest .
– In this case the jumbotron is the object of interest.
• Identify and draw of the external forces acting
on the body.
• Identify the system of interest .
– In this case the jumbotron is the object of interest.
• Identify and draw of the external forces acting
on the body.
– The forces on the jumbotron are its weight and
the tension due to the cables.
• Write the equations for each axis.
Example
• Free body diagram:
T1
T2
20
20
W=48000N
• Sol: writing Newton’s 2nd law:
T1
Fnet , x  max  0
(1)
Fnet , y  ma y  0
(2)
Fnet = 0 in each case as there is no acceleration in equilibrium
T2
20
20
W
• Sol: writing Newton’s 2nd law:
T1
Fnet , x  max  0
(1)
Fnet , y  ma y  0
(2)
T2
20
Fnet = 0 in each case as there is no acceleration in equilibrium
20
W
• Resolving components,
T2 cos 20  T1 cos 20  max  0
(3)
T1 sin 20  T2 sin 20  W  may  0
(4)
+ve dir
T2 cos 20  T1 cos 20  max  0
T1 sin 20  T2 sin 20  W  ma y  0
• From equation 3,
T1  T2
T1 sin 20  T1 sin 20  48000
48000
 T1 
 70171N
2 sin 20
T2  70171N
Special types of Forces
Some special forces
•
•
•
•
Normal force
Friction
Tension
Gravitational force
Friction
• A simple explanation of friction (or formally
the frictional force) is the resistance to
motion.
• The friction is attributed to a single force.
• The friction acts in the direction opposing the
intended motion.
T
fr
a
• The friction force is usually:
– Proportional to the force pressing to surfaces
together (the normal force)
– Depends on the “roughness” of the surfaces.
Normal force
• The normal force is the force which occurs
when an object is in contact with a stable
object.
• It is an example of an applied force.
• It can be considered the “supporting” force.
• The normal force acts perpendicular to the
surface to which the applied the force acts.
• From Newton’s third law, a book placed on a
table pushes down on the table and the table
pushes up on the book.
N
Halliday/ Resnick/ Walker
Fundamentals of Physics
W
The forces acting on the
book: The weight of the
book and the normal
force N. The normal force
acts perpendicular to the
surface of table.
Tension
• The tension is a force which is transmitted
through a string, rope, cable or wire when it is
pulled tight by forces acting from opposite
ends.
• The tension force is directed along the length
of the wire and pulls equally on the objects on
the opposite ends of the wire.
Tension (T)
Tension (T)
Rope attached to a
wall
Pull
Newton’s
rd
3
law
Applications of Action-Reaction
• When an object interacts with another object
they exert a force on each other.
• These forces are equal in magnitude and are
called action-reaction forces.
• Definition: when two bodies interact, the
forces on the bodies from each other are
always equal in magnitude and opposite in
direction.
• Consider the given situation:
The table-book interaction
FBT (The force that the table exerts
on the book)
Halliday/ Resnick/ Walker
Fundamentals of Physics
FTB (The force that the book exerts
on the table)
Example
• Two blocks attached by a “massless” cord which
slides over a frictionless pulley as shown below. The
hanging block falls causing the large block to slide.
Find the (a) acceleration of the blocks and (b) tension
in the cord.
a
M  4kg m  2kg
M
Frictionless
surface
m
Solution
• Show the forces acting on the blocks.
N
a
M
T
a
Mg
N
m
mg
T
Mg
T
mg
(The free body diagrams
for each block)
Sliding block
• Consider the sliding block:
N
a
• Newton’s 2nd law for the x and y direction:
T  Ma (x-direction)
N  Mg  Ma y  0 (y-direction)
T
Mg
Sliding block
• Consider the sliding block:
N
a
• Newton’s 2nd law for the x and y direction:
T  Ma (x-direction)
N  Mg  Ma y  0 (y-direction)
 N  Mg
T  Ma N  Mg
T
Mg
Falling block
• Consider the falling block:
T
• Newton’s 2nd law for the y direction:
mg  T  may  ma
T  mg  a
a
mg
List of equations
T  Ma
(1)
(2)
N  Mg
T  mg  a (3)
T  Ma
(1)
T  mg  a (3)
• Using equations 1 and 3, we eliminate T:
Ma  mg  a
M  ma  mg
mg
a
M  m 
T  Ma
(1)
T  mg  a (3)
• Using equations 1 and 3, we eliminate T:
Ma  mg  a
M  ma  mg
mg
a
M  m 
• Substituting into equation 1:
Mmg
T
M  m
2  9.8
 3.3ms  2
• Therefore: a 
4  2
• and, T  4  3.3  13.1N
More about friction
• Here we will be considering the friction
between dry solid surfaces.
• Here we will be considering the friction
between dry solid surfaces.
• We recall that there are two types of friction
for this case: kinetic and static friction.
• Here we will be considering the friction
between dry solid surfaces.
• We recall that there are two types of friction
for this case: kinetic and static friction.
• Consider the following example when each
case occurs.
• Consider a book resting on a table. The forces
acting on it are shown below.
FN
Halliday/ Resnick/ Walker
Fundamentals of Physics
W
• An external force F is applied to the book.
Again the forces acting on it are shown below.
FN
F
Halliday/ Resnick/ Walker
Fundamentals of Physics
W
fs
(Stationary)
The applied force is
not large enough to
overcome friction
• The friction at this point is static friction.
• As the magnitude of the external force F is
increased so does the friction.
• NB: The applied force must balance the
friction.
F
N
F
Halliday/ Resnick/ Walker
Fundamentals of Physics
W
fs
(Stationary)
The applied force is still
not large enough to
overcome friction
• The friction increases with the applied force
until it reach a maximum.
• At this point the book will move when a
greater external force is applied.
• When the external force is greater than the
friction the book is moving. At this point the
friction acting on the book is kinetic friction.
a
F
FN
Halliday/ Resnick/ Walker
Fundamentals of Physics
W
fk
(Moving)
The book begins to
accelerate when the
friction is overcome.
Definition
• Static Friction: the friction that occurs
between the two surface when the two
surfaces are at rest relative to each other.
• Kinetic Friction: when there is relative motion
between surfaces.
Properties of Friction
• For static friction, the static frictional force
balances the component of the net external
force parallel to the surface.
• When the static friction reaches a maximum:
f s ,max   s N
• Where  s is the coefficient of static friction.
– a measure of the relative amount of adhesion
between the surfaces
Properties of Friction
• When the body is moving the kinetic friction is
given by: f k  k N
• Where  k is the coefficient of kinetic friction.
Example
• A 75kg roller is pulled at angle of 42° along a
cricket pitch at constant velocity. If the
coefficient of friction between the roller and
pitch is 0.1, find the tension T in the handle.
• The free body diagram:
N
T
42
fr
mg
• The free body diagram:
T sin42
N
T
42
fr
T cos42
mg
• Newton’s
2nd
T sin42
for each axis:
N
T cos 42  f r  max  m  0
T sin 42  N  mg  may  m  0
T
42
fr
T cos42
mg
• Newton’s
2nd
T sin42
for each axis:
N
T cos 42  f r  max  m  0
T sin 42  N  mg  may  m  0
T cos 42  f r
 N  mg  T sin 42
T
42
fr
T cos42
mg
• Newton’s
2nd
T sin42
for each axis:
N
T cos 42  f r  max  m  0
T sin 42  N  mg  may  m  0
T cos 42  f r
 N  mg  T sin 42
• For a moving body, f r  k N
T
42
fr
T cos42
mg
• Newton’s
2nd
T sin42
for each axis:
N
T cos 42  f r  max  m  0
T sin 42  N  mg  may  m  0
T cos 42  f r
 N  mg  T sin 42
• For a moving body, f r  k N
T cos 42  k N  0
T sin 42  N  mg  0
T cos 42  k N  0
T
42
fr
T cos42
mg
T sin 42  N  mg  0
T cos 42  k N  0
T sin42
• Solving for T:
T cos 42  k mg  T sin 42  0
 k mg
T 
cos 42   k sin 42
0.1 75  9.8
T
cos 42  0.1 sin 42
 91N
N
T
42
fr
T cos42
mg
Application of friction
Uniform circular motion
• Michael Schumacher travels in his Ferrari of
mass 600kg travels around a bend of radius
100m. The coefficient of static friction is 0.75.
(Note that a negative lift helps to keep the car
on the track.) Find the negative lift if when
v=28.6m/s it is about to slide out of the turn.
• Diagram,
v
R
• Diagram and free body diagram for the car:
v
fs
N
fs
R
a
mg
FL
• Newton’s 2nd for each axis,
mv 2
f s  max 
r
N  mg  FL  may  0
N
fs
a
mg
FL
• Newton’s 2nd for each axis,
mv 2
f s  max 
r
N
fs
a
N  mg  FL  may  0
• For a body on the verge of sliding f s   s N
mg
FL
• Newton’s 2nd for each axis,
N
fs
mv 2
f s  max 
r
a
N  mg  FL  may  0
FL
• For a body on the verge of sliding f s   s N
mv2
 s N 
r
mv 2
s N 
r
mg
N  mg  FL
mv 2
s N 
r
N  mg  FL
• Substituting for N,
mv 2
 s mg  FL  
r
N
fs
a
mg
FL
mv 2
s N 
r
N  mg  FL
• Substituting for N,
mv 2
 s mg  FL  
r
• Solving for the negative lift,
 mv2

FL  
  s mg   s
 r

 v2

 FL  m
 g 
 s r

N
fs
a
mg
FL
mv 2
s N 
r
N  mg  FL
• Substituting for N,
mv 2
 s mg  FL  
r
N
fs
a
• Solving for the negative lift,
 mv2

FL  
  s mg   s
 r

 v2

 28.62

 9.8   664 N
 FL  m
 g   600
 0.75 100

 s r

mg
FL