Download FRICTION

FRICTION - the force that present whenever two
surfaces are in contact and always acts opposite to the
direction of motion.
Depends on:
• Type of materials in
contact
• Surfaces of materials
Does NOT depend on:
•Surface area in contact
•Speed
Ff = μFn
Ff = Force of Friction in N
μ = Coefficient of Friction (depends on material
0<μ<1
Fn = Normal Force; Force acting perpendicular to
the two surfaces in contact. In level surfaces, this is
equal in magnitude but acting opposite to the
weight
TYPES OF FRICTION
Kinetic Friction - Force needed to keep it going at a
constant velocity.
Ff = μ kFn
Static Friction - Force needed to start motion.
Ff < μ sFn
Keeps the object at rest.
Only relevant when object is stationary.
Calculated value is a maximum
Fs ≥ Fk
Fs
Max
Fs  Fk
Static
Region
Fk
Kinetic Region
Force Causing the Object to Move
Frictional Forces Occur When
Materials are in Fn Contact
Fa
Fs
M1
Surfaces in
Contact
Spring Scale
Fw
Fa = Force Causing Motion (Pull on Scale)
Fs = Force of Static Friction (Resists Motion)
Fn = Normal Force (Perpendicular to Surfaces)
Fw = Weight of Object ( Mass x Gravity)
Friction is a Force That Always
Resists Motion
Fn
Fs
Fa
M
Surfaces in
Contact
Spring Scale
Fw
The Block M will only move if the Applied Force (Fa) is
greater the Force of Static Friction (Fs).
FRICTION converts kinetic energy
to heat energy.
Banana peel reduces friction.
INCREASING FRICTION
REDUCING FRICTION
The coefficient of friction (μ) is the ratio of
the Applied Force (Fa) over the Normal Force
(Fn).
F
n
Fs
Fa
M
Surfaces in
Contact
Spring Scale
μ = Fa
Fn
Fw
Determine the amount of force needed to
move 12.0 kg of rubber at a constant
velocity across dry concrete?
Solution:
F = ma
FFr = μkFN
FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 N
FFr = μkFN = (0.8)(117.6 N) = 94.08 N = 90 N
90 N
What is the force needed to slide a
stationary 150 kg rubber block across
wet concrete?
SOLUTION:
F = ma
FFr < μsFN
FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 N
FFr = μkFN = (.7)(1470) = 1029 N = 1000 N
1000 N
Given the following :
μs = .62, μk = .48
Determine the acceleration of a sliding 8.50
kg block if a force of 72 N is applied to it?
v
FFr
72 N
8.5 kg
FFr = μkFN
FN = mg
FFr = μkmg
FFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 N
F = ma
<72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 m/s2
3.8 m/s/s