Download Everyday Forces - Friction

1. Use coefficients of friction to
calculate frictional force.
2. Describe air resistance as a
form of friction.
3. Understand and explain terminal
velocity and drag.
The Force of Friction
 The force that opposes motion between two
surfaces that are touching
 Two types
 Static Friction
 Kinetic Friction
Friction Force
 Friction depends on the
surfaces in contact
 Every surface has microwelds
 Microwelds are microscopic
grooves in the surface of every
object.
 The more points or grooves in
contact, the more friction is
present
Static Friction (Fs)
 The force that keeps an object from moving
even if there is an applied force
 Keeps objects still
 Must be overcome to make objects move
Static Friction (Fs)
 Static friction reaches its maximum when the
applied force is as great as it can be without
moving the object
Kinetic Friction (Fk)
 The force that opposes the motion of two
surfaces sliding past each other
 The frictional force of an object in motion
 To keep objects moving, a force must be
continually applied to overcome kinetic friction
Kinetic Friction (Fk)
 Right before an object starts moving, Fk is less
than Fs(max)
Coefficient of friction (μ)
 μ – the lowercase Greek letter mu
 The ratio of the force of friction to the normal
force acting between two objects
 The normal force is the force that opposes
gravity
Coefficient of Kinetic Friction
 Ratio of the force of kinetic friction to
the normal friction
Fk = μkFN
Coefficient of Static Friction
 Ratio of the force of static friction to
the normal friction
Fs = μsFN
Coefficients of Friction
(don’t write this – you are getting a copy. Make sure
you keep it!)
Coefficients of Friction
 If you are not looking for the coefficient of
friction in your problem, look up the value for it
in the table
 Make sure you use the correct coefficient
 There are differences between the coefficients
for kinetic friction and static friction
A 24 kg crate initially at rest on a horizontal floor
requires a 75 N horizontal force to set it in
motion. Find the coefficient of static friction
between the crate and the floor.
Once the crate in the previous problem is in
motion, a horizontal force of 53 N keeps the
crate moving with a constant velocity. Find μk
between the crate and the floor.
Drag and Terminal Velocity
A fluid is anything that can flow,
generally a gas or liquid.
When a body moves through a fluid or a
fluid moves around a body there is a drag
force ( D ) that opposes motion and points
in the direction in which the fluid flows
relative to the body.
Air Resistance
 Although we often ignore it, air
resistance, R, is usually significant in
real life.
 R depends on:
R
 speed (approximately proportional
m
mg
to v 2 )
 cross-sectional area
 air density
 other factors like shape
 R is not a constant; it changes as
the speed changes
Volume & Cross-sectional Area
2z
z
Area
y
x
Volume = xyz
Area = xy
Area
2x
2y
Volume = 8 xyz
Area = 4 xy
If all dimensions of an object are doubled the
cross-sectional area gets 4 times bigger, but
the volume goes up by a factor of 8.
Falling in Air
R
m
A
mg
With all sides doubled, the area exposed
to air is quadrupled, so the resistance
force is 4 times greater. However, since
the volume goes up by a factor of 8, the
weight is 8 times greater (as long as
we’re dealing with the same materials).
Conclusion: when the only difference is
size, bigger objects fall faster in air.
4R
8m
4A
8 mg
D = ½ C ρ Av2
ρ is the density of air (mass per volume)
A is the effective cross sectional area of
the body (cross section taken
perpendicular to the velocity)
C is the drag coefficient (typically valued
b/w 0.4 - 1.0) but varies based off velocity
When a blunt object falls from rest
through the air, drag is directed upward,
and its mag gradually increases from zero
as the object speeds up.
D opposes Fg therefore
D – Fg = ma
If the body falls long enough, D
eventually equals Fg; therefore a = 0 and
the body falls at constant speed called
terminal velocity
 Terminal velocity (vt)
 To find vt we set a = 0 in the drag eqtn’
0 = ½ C ρ Av2 – Fg
Which gives
vt =
The terminal speed of a sky diver is
160 km/h in a spread-eagle
position and 310 km/h in the
nosedive position. Assuming that
the diver’s drag coefficient C does
not change from one position to
the other, find the ratio of the
effective cross-sectional area A in
the slower position to that in the
faster position
Terminal Velocity
Suppose a daredevil frog jumps out of a
skyscraper window. At first v = 0, so R = 0 too,
and a = -g. As the frog speeds up, R increases,
and his acceleration diminishes. If he falls long
enough his speed will be big enough to make R
as big as mg. When this happens the net force
is zero, so the acceleration must be zero too.
R
This means this frog’s velocity can’t
change any more. He has reached
his terminal velocity. Small objects,
like raindrops and insects, reach
terminal velocity more quickly than
mg
large objects.