Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
Newton’s
rd
3
Law: More Practice
An Introduction to Friction:
Student Learning Goal
Students will be able to perform an investigation to
determine the factors affecting static and kinetic
friction, and to determine the corresponding coefficient
of friction between an everyday object and the surface
with which it is in contact.
(C2.3)
An Introduction to
Friction
SPH4C
Weight
Recall that the gravitational force acting on an
object is also called the weight.
Weight
Recall that the gravitational force acting on an
object is also called the weight.
Its magnitude is equal to: Fg  mg
where m is mass, measured in kg, and g = 9.8 m/s2.
Weight
Recall that the gravitational force acting on an
object is also called the weight.
Its magnitude is equal to: Fg  mg
where m is mass, measured in kg, and g = 9.8 m/s2.
Weight is therefore measured in Newtons.
The Normal Force
Recall that the normal force is the force acting to
keep objects apart.
The Normal Force
Recall that the normal force is the force acting to
keep objects apart. For an object on a
horizontal surface with no vertical applied force,
the normal force will be equal to __________.
FN
e.g. a box pushed
rightward across a
table
Ff
FA
Fg
The Normal Force
Recall that the normal force is the force acting to
keep objects apart. For an object on a
horizontal surface with no vertical applied force,
the normal force will be equal to the weight.
FN
e.g. a box pushed
rightward across a
table
Ff
FA
Fg
The Normal Force
Recall that the normal force is the force acting to
keep objects apart. For an object on a
horizontal surface with no vertical applied force,
the normal force will be equal to the weight
(since the vertical forces must balance).
FN
e.g. a box pushed
rightward across a
table
Ff
FA
Fg
FN not equal to Fg
If there is a vertical applied force, the normal force
will not equal the weight.
Example
A 5 kg book is resting on the table. If Ms.
Rosebery is pushing down on the book with a
force of 20 N, what is the normal force the table
is exerting on the book?
FN not equal to Fg
Example
A 5 kg book is resting on the table. If Ms.
Rosebery is pushing down on the book with a
force of 20 N, what is the normal force the table
is exerting on the book?
FN= ?
FA= 20 N
Fg=(5 kg)(9.8 m/s2) = 50 N
FN not equal to Fg
Example
A 5 kg book is resting on the table. If Ms.
Rosebery is pushing down on the book with a
force of 20 N, what is the normal force the table
is exerting on the book?
FN= ?
FA= 20 N
The normal force
must be 70 N [up].
Fg=(5 kg)(9.8 m/s2) = 50 N
Direction of FN
Note that the normal force is always perpendicular
to the surface. If the surface is vertical (e.g. a
magnetic board to which a magnet is attached),
the normal force must be horizontal.
Friction
Recall that friction is the force acting to oppose
any (attempted) motion and is therefore always
opposite in direction to the attempted motion.
The magnitude of the force of friction will depend
on the properties of the surfaces in contact
Friction
Recall that friction is the force acting to oppose
any (attempted) motion and is therefore always
opposite in direction to the attempted motion.
The magnitude of the force of friction will depend
on the properties of the surfaces in contact and
on the normal force acting on the object
attempting to move .
Mu
The magnitude of the frictional force is given by:
Ff  FN
Mu
The magnitude of the frictional force is given by:
Ff  FN
where  (mu) is the coefficient of friction,
determined by the properties of the surfaces in
contact.
Mu
The magnitude of the frictional force is given by:
Ff  FN
where  (mu) is the coefficient of friction,
determined by the properties of the surfaces in
contact.
 is dimensionless (no units) and is ≤ 1 (approx.).
Mu Examples
E.g.
 For steel on wood, k (kinetic) = 0.25
s (static) = 0.45
Note that s > k : it’s harder to get things moving than
keep things moving.
Mu Examples
E.g.
 For steel on wood, k (kinetic) = 0.25
s (static) = 0.45
 For steel on ice, k (kinetic) =
s (static) =
Mu Examples
E.g.
 For steel on wood, k (kinetic) = 0.25
s (static) = 0.45
 For steel on ice, k (kinetic) = 0.010
s (static) = 0.10
 For rubber on dry concrete, k (kinetic) =
s (static) =
Mu Examples
E.g.
 For steel on wood, k (kinetic) = 0.25
s (static) = 0.45
 For steel on ice, k (kinetic) = 0.010
s (static) = 0.10
 For rubber on dry concrete, k (kinetic) = 1.0
s (static) = 1.1
Example Problem
A 1200-kg car without ABS brakes is skidding on
dry concrete. What is the magnitude of the
force of friction acting on the car?
Example Problem
A 1200-kg car without ABS brakes is skidding on
dry concrete. What is the magnitude of the
force of friction acting on the car?
Givens: m  1200kg
g  9.8 sm2
 k  1 .0
Unknown: F f  ?
Example Problem
A 1200-kg car without ABS brakes is skidding on
dry concrete. What is the magnitude of the
force of friction acting on the car?
Givens: m  1200kg
g  9.8 sm2
 k  1 .0
Unknown: F f  ?
Select : F f  FN
F f  Fg  m g
Solve: F f  (1.0)(1200kg )(9.8 sm2 )
F f  11760 N
Example Problem
A 1200-kg car without ABS brakes is skidding on
dry concrete. What is the magnitude of the
force of friction acting on the car?
Givens: m  1200kg
g  9.8 sm2
 k  1 .0
Unknown: F f  ?
Select : F f  FN
F f  Fg  m g
Solve: F f  (1.0)(1200kg )(9.8 sm2 )
F f  11760 N
ABS Brakes
If the car does have ABS brakes, the car’s tires will
continue to roll and grip the road while stopping
and the relevant coefficient will be that of static
friction:
ABS Brakes
If the car does have ABS brakes, the car’s tires will
continue to roll and grip the road while stopping
and the relevant coefficient will be that of static
friction: F f  FN  Fg  m g
F f  (1.1)(1200kg )(9.8 sm2 )
F f  12936N
ABS Brakes
If the car does have ABS brakes, the car’s tires will
continue to roll and grip the road while stopping
and the relevant coefficient will be that of static
friction: F f  FN  Fg  m g
F f  (1.1)(1200kg )(9.8 sm2 )
F f  12936N
i.e. there is more force stopping the car
when the tires are rolling and not sliding
More Practice
Static and Kinetic Friction Lab