Friction
Friction: A contact force tangent to the two contacting surfaces.
It is opposite the direction of motion or intended motion.
Three Types of Friction:
Internal: Friction between atoms in a solid.
We will NOT study this.
Fluid: Friction between layers and atoms in a fluid.
We will NOT study this.
Dry (or Coulomb): Friction between two contacting solids.
We WILL study this.
Simple Example
Given: A horizontal force P is applied to a block
of mass m initially at rest on horizontal table.
P
m
Required: Analyze for various values of P.
y
^
j
FBD
P
^ x
i
KD
may
=
W
max
F
N
∑F
x
= m ax ⇒ P − F = + m ax
∑F
y
= m ay ⇒ N − W = + m ay = 0
⇒
N = W
1
y
^
j
FBD
^ x
i
Simple Example Cont.
KD
may
P
∑F
= m ax ⇒ P − F = + m ax
x
=
W
∑F
max
y
= m ay ⇒ N = W
F 1) If P is small the block will NOT move. ∴ a = 0 ⇒
x
N
2) If P is continuously increased the
block will reach a point where it is about F
to move. At this point F has reached its
maximum value.
⇒ P = F = Fmax
Fmax - about to slip
(impending motion)
F
3) If P is increased again the block will
move. At this point F will decrease and
essentially remain constant.
45o
⇒ F < Fmax & F < P
Fmax - about to slip
(impending motion)
F
F
∴ Fmax = μ s N
Once the body is in motion we can
calculate the ratio of the friction force to
the normal force:
∑F
x
=
F < Fmax & F < P
P
45o
P
No Motion
F
μk ≡
N
FBD
The coefficient of kinetic friction is
defined as the kinetic friction force
divided by the normal force.
during motion
= m ax ⇒ P − μk N = + m ax
Motion
Simple Example Cont.
The coefficient of static friction is defined
as the maximum friction force divided by
the normal force.
∴ F = μk N
P
No Motion
At the point of slip (impending motion) if
we measure Fmax and N (the normal force)
and calculate their ratio:
F
μs ≡ max
N
F < Fmax & F < P
P
=
P
Motion
y
^
j ^
KD
i x
may
=
W
max
F
N
2
Another Example Problem
Given: The 100 lb crate is
carefully set down on the
incline with a zero velocity.
Assume 2 SF.
μ s = 0 .30
μ k = 0 . 25
Required: What happens if,
a)
θ = 15 °
b)
θ = 20 °
θ
Step 1: What do we isolate?
Problem Cont.
Step 2: Draw FBD & KD
Step 3: C.S.
y
^
j
^
i
The crate, Why?
FBD
may
KD
F
θ
N
x
=
θ
max
100 lb
Step 4: a) External Forces & Moments ON FBD b) Mass * Acceleration ON KD
Step 5: Dimensions
∑F
∑F
Step 6: Enforce Newton’s 2nd Law
x
= m ax ⇒ + 100 lb sinθ − F = + m ax
y
= m a y ⇒ + N − 100 lb cosθ = + m a y = 0 ⇒
Step 7: a) Set up Kinematic equations
∑F
= ma
N = + 100 lb cosθ
b) Solve equations:
Not necessary here
3
y
FBD
^
j
F
θ
N = + 100 lb cosθ
θ
If it does NOT move ax = 0
∴ Fno slip = + 100lb sinθ
but Fmax = μ s N
∴
100 lb
θ = 20 °
θ = 15 °
N
96.6 lb
94.0 lb
Fno slip
25.9 lb
34.2 lb
Fmax
29.0 lb
28.2 lb
∴
Yes
NO
no slip
slips
∴ F = μ k N = 0.25 (100 lb cos 20°) = 23.5 lb
⇒
a x = + 3.45 m / s 2 ⇒
Fmax = 0.3 (100 lb cosθ )
y
^
j
F
θ
+ 100 lb sinθ − F = + m ax
a = 3.4 m / s 2 iˆ
Step 8: Check Everything
FBD
N
N = + 100 lb cos θ
If it does NOT move ax = 0
∴ Fno slip = + 100 lb sinθ
It will NOT move provided
Fno slip ≤ Fmax
but Fmax = μ s N
∴
max
Is Fno slip ≤ Fmax
It will NOT move provided
Fno slip ≤ Fmax
^
i x
=
N
+ 100 lb sinθ − F = + m ax
may
KD
may
KD
^
i x
max
=
θ
100 lb
What is the maximum θ without slipping?
⇒ Fno slip = Fmax
∴ + 100 lb sinθmax = μs N = μs (100lb cosθmax )
⇒ μs = tanθmax
For our problem,
θ max = tan−1 μs = tan−1 (0.30) = 16.7 °
How would you calculate μ ?
k
Fmax = 0.3 (100 lb cos θ )
4