Warm-up
Rose is sliding down an ice covered hill inclined
at an angle of 15° with the horizontal. If Rose
and the sled have a combined mass of 54.0 kg,
what is the force pulling them down the hill?
(neglect friction)
m = 54 kg
g = 9.81 m/s²
θ= 15°
Sin θ = Fx/mg
Fx = mg sin θ
15°
Fn
mg
Fn
Fx
Fx
Fx = (54 kg)(9.81 m/s²) sin 15°
Fx = 137 N
θ = 15°
mg
Friction
 There
are two types of frictional forces
in the physical world that affects 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
SPLAT!!
Air
Resistance
Air Resistance
 Can
also be a horizontal force
Fr
(Air Resistance)
mg
Fx
Static and Kinetic Friction
 Are
forces that oppose motion between
two surfaces that are touching each
other.
Fa
Fr
Fr (non-moving) = Static Friction (Fs)
Fr (moving) = Kinetic Friction (Fk)
Fk = ∑F
The force of friction is proportional to
the size and mass of an object that you
are pushing across a surface.
Fk
Fk
Fapplied
Fapplied
The amount of friction that occurs
between an object and the surface can
depend on the type of surface that the
object is in contact with.
Fk
Fapplied
Tile Floor
Fk
Fapplied
Carpet
The quantity that expresses the
dependence of frictional forces on the
particular surface the object is in contact
with is called the coefficient of friction
(µ).
It is the ratio between the normal force
and the force of friction between two
surfaces.
Coefficient of Kinetic Friction
μk = Fk/Fn
coefficient of kinetic friction
μs = Fs,max/Fn
coefficient of static friction
Ff = μFn
frictional force
Practice 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
Practice Problem 2
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? (Pg. 146)
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