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Concept 1. Friction
C ONCEPT
1
Friction
In this lesson students will learn about friction, the difference between static and kinetic friction and how to solve
problems involving friction.
In this lesson students will learn about friction, the difference between static and kinetic friction and how to solve
problems involving friction.
Key Equations
Static and Kinetic Friction
(
fs  µs |FN | Opposes potential motion of surfaces in contact
fk = µk |FN | Opposes motion of surfaces in contact
Guidance
• The force of friction can actually be described in terms of the coefficient of friction, µ. It is determined
experimentally and varies depending upon the two surfaces in contact.
• Static friction (µs ) acts between two surfaces that are in contact but not in motion with respect to each other.
This force prevents objects from sliding. It always opposes potential motion, and it rises in magnitude to a
maximum value given by the formula below.2
• Kinetic friction (µk ) acts between two surfaces that are in contact and in motion with respect to each other.
This force reduces the acceleration and it always opposes the direction of motion.2
Example 1
Calculate the force necessary to slide a 4.7-kg chair across a room at a constant speed if the coefficient of kinetic
friction between the chair and the floor is 0.68.
Question:F =? [N]
Given:m = 4.7 kg
µk = 0.68
g = 10.0 m/s2
Equations:Â F = ma
 Fy = N
mg = 0, so N = mg
 Fx = max
Fpull
fk = 0 (because the chair is moving at constant speed, so a = 0)
Fpull = µk N
Plug n’ Chug:Fg ) of the chair.
N = mg = (4.7 kg)(10 m/s2 ) = 47 N
Fpull = µk N = (0.68)(47 N) = 32 N
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Answer:
32 N
Example 2
MEDIA
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Watch this Explanation
MEDIA
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Simulation
• http://simulations.ck12.org/Friction/
RampForces and Motion (PhETSimulation)
Time for Practice
1. Nathan pulls his little brother in a wagon, using a rope inclined at 30 above the horizontal. The wagon and
brother have a total mass of 80 kg, the average coefficient of friction between the wagon wheels and the floor
is 0.08, and Nathan pulls on the rope with a force of 100 N.
a. Draw a force diagram for the wagon, labeling all forces.
b. Calculate the horizontal and vertical components of Nathan’s pull. Label them on your diagram (use
dotted lines for components so as not to confuse them with other forces).
c. Calculate the normal force acting on the wagon. (HINT: It is NOT equal to the weight! Use your FBD
above).
d. Calculate the force of friction on the wagon.
e. Calculate the horizontal acceleration of the wagon.
2. When the 20 kg box to the right is pulled with a force of 100 N, it just starts to move (i.e. the maximum value
of static friction is overcome with a force of 100 N). What is the value of the coefficient of static friction, µS ?
3. A different box, this time 5 kg in mass, is being pulled with a force of 20 N and is sliding with an acceleration
of 2 m/s2 . Find the coefficient of kinetic friction, µK .
4. The large box on the table is 30 kg and is connected via a rope and pulley to a smaller 10 kg box, which is
hanging. The 10 kg mass is the highest mass you can hang without moving the box on the table. Find the
coefficient of static friction µS .
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Concept 1. Friction
5. A block has a little block hanging out to its side, as shown: As you know, if the situation is left like this, the
little block will just fall. But if we accelerate the leftmost block to the right, this will create a normal force
between the little block and the big block, and if there is a coefficient of friction between them, then the little
block won’t slide down! Clever, eh?
a. The mass of the little block is 0.15 kg. What frictional force is required to keep it from falling? (State a
magnitude and direction.)
b. If both blocks are accelerating to the right with an acceleration a = 14.0 m/s2 , what is the normal force
on the little block provided by the big block?
c. What is the minimum coefficient of static friction required?
Answers
1.
2.
3.
4.
5.
b. Fx = 87 N, Fy = 50 N c. N = 750 N d. f = 60 N e. 0.34 m/s2
0.5
0.2
0.33
a. 1.5 N; 2.1 N; 0.71
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