4. Friction
a) Kinetic friction:
f k  k N
fk – friction force
N – normal force
k – coefficient of kinetic friction
b) Static friction:
f s  f s max   s N
 s - coefficient of static friction
k   s
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Example: When you push a book against a wall, the static friction between
the wall and the book can prevent it from falling. If you press harder, the
friction force will be:
fS
A. Larger than before
B. The same
N
F
C. Smaller than before.
For the book not to fall down:
f s  mg
mg
This force is independent from the normal force.
f s  f s max   s N
Pushing harder (increasing F) increases N and therefore fS,MAX increases,
but not the actual value of fS that we had, which needs to continue to be
exactly mg.
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Example: Trying to move a trunk
friction
Fby you
ffSk by
by floor
floor
μsN
μkN
Fby you
Static friction
Kinetic friction
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Example: Trying to move a trunk
For “small” forces, the trunk
does not move. So there must Fby you
be a friction force fS = Fby you
fS by floor
You increase the applied force, but
the trunk still does not move:
static friction is increasing too.
Fby you
fS by floor
You increase the applied force even
more. Eventually, the trunk moves.
Static friction cannot be larger
than a certain value.
And then friction
becomes kinetic!
Fby you
fSk by
by floor
floor
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Question 1: Mass m1 sits at rest on a
horizontal surface. A string connects m1
over a pulley to a hanging mass m2.
The coefficient of static friction between
m1 and the horizontal surface is 0.25.
The maximum value that m2 can be
before m1 starts to slide is ___.
A) m1/4
B) m1/2
C) m1
D) 2 m1
E) 4 m1
m2 g  s m1 g  m2  s m1
Question 2: A box of weight 100 N is at rest on a floor where s = 0.4.
A rope is attached to the box and pulled horizontally with tension T = 30 N.
Which way does the box move?
m
T
The static friction force has a maximum of sN = 40 N.
The tension in the rope is only 30 N.
So the pulling force is not big enough to overcome friction.
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Example: Box on incline with friction
N
f
m g cos

mg
m gsin 
y
(y)
N  m g cos  0
(x)
m gsin   f  m a
1) Kinetic: f k   k N   k m g cos
m gsin   k m g cos  m a

a g (sin  k cos )
x
2) Static:
f s   s N   s m g cos
m g sin   f s  0
m g sin    s m g cos
t an   s
 s  t an critical
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5. Fluid resistance and terminal speed
• The force of fluid resistance is in the opposite direction of the object’s
velocity and will cause the object to slow down if no other force opposes it.
• The faster an object goes, the larger the force of fluid resistance.
• There is no simple equation for the force of air resistance, however,
f  kv ,
at high speed: f  Dv 2
• at low speed:
•
(v – speed, k and D – coefficients)
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v
• As a body falls through the air, the force of air resistance opposite to the
downward force of gravity, so the net force decreases.
• This continues as the body gains speed until the force of air resistance
acting upward equals the weight acting downward: f=mg.
• At this point, the net force is zero, so the speed stays constant from then on.
• This speed is called the terminal speed of the body.
t
Question: What is terminal speed of a 50 kg skydiver if
the expression for the air resistance (air drag) is f  Dv 2 ?
Use D=0.2 kg/m, and g = 10 m/s2.
Dv2  mg 
v  mg D 
50 kg 10 m / s 2  0.2kg / m   50 m / s
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