Download SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) Page 1 of 10

SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
1.9.1 Introduction
The study of friction is called tribology.
The force of friction is defined as: the force that acts between objects in contact to
oppose their relative motion.
The force of friction acting between two surfaces has three properties:
i) It opposes their relative motion
ii) It depends on the strength of the force pushing the surfaces together (the normal).
iii) (don’t mention this one until after the lab is handed in) does not depend on the
surface area in contact.
Consider the case of a 100 N. mass on a horizontal surface as shown below:
If we apply a horizontal force on the mass as shown, a static frictional force from the
interlocking of the irregularities of two surfaces will resist the motion. If the applied force
is smaller than the maximum friction force that could be developed, the static friction
force will prevent any relative motion. When the applied force becomes sufficiently
large, there will be a net force and the mass will begin to accelerate.
A graph of this is shown below:
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
It is that threshold of motion (i.e. the starting friction force) which can be used to
calculate the coefficient of static friction. Once the block begins to move, the friction
force is reduced as a result of a smaller coefficient of friction. As a result, the applied
force that is required to maintain constant velocity becomes smaller.
In general, the force due to friction is calculated as:
for static friction and
for kinetic friction
The coefficients of friction depend on the nature of both surfaces in contact.
The static friction coefficient is larger than the corresponding kinetic friction coefficient.
A chart of friction coefficients is shown below:
Surfaces in Contact
Coefficient of
Static Friction
Coefficient of
Kinetic Friction
Oak on oak, dry
0.40
0.30
Waxed hickory on snow, dry
0.06
0.04
Waxed hickory on snow, wet
0.20
0.14
Steel on steel, dry
0.60
0.41
Steel on steel, greasy
0.15
0.06
Steel on ice
0.013
0.011
Rubber on concrete, dry
1.0
0.8
Rubber on concrete, wet
0.7
0.5
Rubber on asphalt, dry
1.2
0.8
Rubber on asphalt, wet
0.6
0.5
0.006
0.005
Ice on ice
0.1
0.03
Teflon on teflon
0.04
0.04
Synovial joints in humans
0.01
0.003
Glass on glass
0.94
0.40
Rubber on ice
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
The coefficient of friction allows us to calculate the magnitude of the friction force if we
know the magnitude of the normal force. We can do this using the formula:
i.e.
(1.9)
Note that this is not a vector equation. (Why not?)
Eg.#1 A 10. kg. block of steel is moving on steel
with motion to the right as shown. What is the force
due to friction?
First, we’ll find the magnitude of the Normal force:
Next, we determine the coefficient of kinetic friction from the table. For dry steel on
steel,
Now, we can determine the magnitude of the friction force using the formula above
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
Eg.#2 A 1.2 tonne car has its brakes locked and comes to a stop on an asphalt road.
a) Calculate the friction force if the
road surface is dry
b) Calculate the friction force if the road
surface is wet
c) Comment on the results obtained.
A car trying to stop on a wet road will require a greater distance in order to stop.
1.9.2 Kinetic Friction
There are three different types of kinetic friction:
1.9.2a Rolling friction
The resistance to motion which occurs between a rolling wheel and the surface on
which it rotates
Caused by small distortions of the two surfaces
1.9.2b Sliding Friction
The resistance to motion which occurs when one surface slides over another with which
it is in contact
1.9.2c Fluid Friction
Friction between an object and the liquid through which it moves, whether that fluid be a
liquid, gas or vapor.
To recap, static friction is present whenever the frictional force opposes placing a body
at rest into motion. Kinetic friction is present whenever the frictional force tends to slow
a body that is in motion. In general, the maximum static friction force is greater than the
kinetic friction force.
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
1.9.3 Inclined Planes
1.9.3a Static Equilibrium on an Inclined Plane
Consider a mass that remains stationary on an
inclined plane as shown. An interesting result is
obtained by applying Newton’s Second Law to this
situation.
In order that the mass doesn’t slide down the plane,
the surface must exert two forces:
•
•
a normal force perpendicular to the plane
a friction force parallel to the plane.
These two forces when combined will be equal in
magnitude and opposite in direction to the force of
gravity.
Before we apply Newton’s Second Law to this situation, we must resolve the weight into
two components, one parallel to the surface and one perpendicular to the surface.
The component of the weight acting at 90/ to the surface is
The component of the weight acting parallel to the surface is
A diagram that shows all of the forces we have discussed is shown below:
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
Newton’s Second Law, when applied to the forces perpendicular to the plane yields:
When applied to the forces acting along the plane, Newton’s Second Law yields:
As long as the block remains stationary,
This shows us that, if we know
for the surfaces in contact, we can elevate an
inclined plane until the tangent of the angle is equal to or less than the friction
coefficient and the object will not slide down the plane.
If we complete the same analysis with the block moving at a constant velocity (
the analysis will show that
Eg. #4. How does the angle of the inclined plane compare when the object moves at
constant velocity to the angle it has when the object is stationary?
The angle will be less because the coefficient is less.
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),
SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
Eg.#5 A 75.0 kg. crate sitting on a variable angle ramp remains stationary if the angle of
the ramp is 37° or less.
a) What type of friction is at work here? b) What is the coefficient of friction?
Static friction
c) Once the crate begins moving, it will continue moving only if the angle of the ramp is
33° or more. What is the type of friction at work here?
Kinetic friction
d) What is the new coefficient of friction?
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
Worksheet 1.9
1. A car accelerates southward because there is a force exerted on the tires by the
pavement.
a) What is the direction of this force?
b) Why does this force exist?
[S]
It is a reaction force to the force that is
applied by the tires onto the road
c) Is this friction force static or kinetic? Explain your answer.
It is a static friction force because the tires do not move relative to the road surface.
2. The coefficients of kinetic and static friction between a 23 kg. exercise mat and the
floor are:
and
.
a) Determine the magnitude of the
b) Once the mat is moving what
minimum horizontal force that would be
minimum force is required in order to
needed to get this mat moving.
keep it moving?
3. A musician applies a horizontal force of 17 N.[W] to an instrument case of mass
5.1kg. The case slides across a table with an acceleration of 0.39m./s.2[W]. What is the
coefficient of kinetic friction between the case and the table?
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
4. A small box is resting on a larger
box sitting on a horizontal surface.
When a horizontal force is applied
to the larger box, both boxes
accelerate together. The small box
does not slip on the larger box.
a) Draw a free body diagram of the small box b) What force causes the small box to
accelerate horizontally?
The force of static friction between the
boxes causes the small box to
accelerate horizontally.
c) If the two boxes accelerate at 2.5m./s.2,
determine the smallest coefficient of friction
between the boxes that will prevent slippage
d) Draw a free body diagram of the
larger box when it is accelerating
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SPH 4U Unit #1 Dynamics
Topic #9: Friction (Teacher)
5. An adult is pulling two smaller children in a sleigh up a snow covered hill. The sleigh
and children have a total mass of 47 kg. The hill makes an angle of 23 degrees with the
horizontal. The coefficient of kinetic friction between the sleigh and the snow is 0.11.
Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at
a constant velocity in the snow.
6. A race car accelerates on a level track. The coefficient of friction between the tires
and the track is 0.88 and the tires are not slipping.
a) Draw a free body diagram of the car
b) Determine the maximum possible
acceleration if the car is to travel without
slipping.
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