Frictional Forces
Chapter 4, Section 4
Pg. 141-149
Friction
 There
are two types of frictional forces
in the physical world that effect us.
- Static/Kinetic Friction
- Air Resistance
Air Resistance
 Is
a resistance force that acts in the
opposite direction of gravitational
forces.
Help!!
Force of
Gravity
Force of
Gravity
v1 > v2
Air
Resistance
SPLAT!!
Air
Resistance
Static and Kinetic Friction
 Forces
that oppose motion between two
surfaces that are touching each other.
Fx
Fr
Fr (non-moving) = Static Friction (Fs)
Fr (moving) = Kinetic Friction (Fk)
Frictional force depends on the size and
mass of the object moving across a
surface.
Fk
Fk
Fx
Fx
The amount of friction occurring
between an object and the surface
depends on the surface type.
Fk
Fx
Tile Floor
Fk
Fx
Carpet
The value that expresses the dependence
of frictional forces on the surface type
the object is in contact with is called the
coefficient of friction (µ).
It is the ratio between the force of
friction and the normal force.
Coefficient of Kinetic Friction
μk = Fk/Fn
coefficient of kinetic friction
μs = Fs,max/Fn
coefficient of static friction
Ff = μFn
frictional force
Sample problem
While redecorating her apartment, Suzy slowly pushes an
82 kg cabinet across a wooden dining room floor, which
resists the motion with a force of friction of 320 N. What is
the coefficient of kinetic friction between the cabinet and
the floor?
Fn
m = 82 kg
g = 9.81 m/s²
Fk = 320 N
w = mg
Fapplied
mg
w = (82 kg) (9.81 m/s²)
w = 804 N w = Fn = 804 N
μk = Fk/Fn
μk = 320 N/804 N
μk = 0.40
Sample Problem 2 (Pg. 146)
A student moves a box of books by
attaching a rope to the box by pulling
with a force of 90 N at an angle 30° to the
horizontal. The box of books has a mass
of 20.0 kg and the coefficient of kinetic
friction between the bottom of the box
and the sidewalk is 0.50. What is the
acceleration of the box?
Step 1: Solve for Weight and X & Y components of
applied force
Fn
Fk
Fapplied = 90 N
m = 20.0 kg g = 9.81 m/s²
μk = 0.50
30°
w = mg = 196 N
mg
90 N
30°
Fapp,y
Fapp, x
Fapp, y = (90N) (sin 30°) = 45.0 N
Fapp, x = (90N) (cos 30°) = 77.9 N
Step 2: Find the Kinetic Friction
Fapp, y = 45.0 N
μk = 0.50
Fapp, x = 77.9 N
w = 196 N
90 N
30°
Fapp, x
Fapp,y
∑Fy = Fn + w + Fapp, y = 0
0 = Fn – 196 N + 45.0 N
Fn = 196 N – 45 N = 151 N
μk = Fk/Fn
Fk = (μk)(Fn) = (0.50)(151 N) = 75.5 N
Step 3: Solve for acceleration
m = 20.0 kg
90 N
30°
Fapp, x
Fapp,y
Fk = 75.5 N
∑F = ma
a = ∑F/m
Fapp, x = 77.9 N
a = (Fapp, x – Fk) / m
a = (77.9 N – 75.5 N) / 20.0 kg
a = 0.12 m/s² to the right