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ME 201Engineering Mechanics: Statics
Unit 8
Friction
Friction
A retarding force that resists the relative movement
of two bodies in contact with each other
Always acts in oppose direction of motion – actual motion
or impending motion
Always acts parallel or tangent to surface(s) in contact
Magnitude is mainly dependent on surface roughness
(other contributing factors: temperature, molecular
adhesion, electrostatic attraction, lubrication, relative
velocities)
Friction Theory
Consider a block on a horizontal surface
No Frictional resistance
N is referred to as the Normal force, perpendicular to the
contact surface
W
N
Friction Theory
Next an external force is applied
If no friction, block would immediately move
If force is applied, F = P, for equilibrium
At some point, P > F, motion occurs
W
N
P
F
Friction Theory
Ratio of F to N is called the Coefficient of Static
Friction or μ
NF
s Where
μs = coefficient of static friction (unitless)
F – Max or limiting frictional force resistance (lb, k, N)
N – Normal force, perpendicular to contact surface (lb, k, N)
Friction Theory
Stated another way:
Fs = μs N (for impending motion)
Fk = μk N (for sustaining motion, kinetic)
In general:
F ≤ μ N
Coefficient of Static Friction on Dry Surfaces
Materials Coefficient of Static Friction or μ
Metal on metal 0.15 – 0.60
Metal on wood 0.20 – 0.60
Metal on stone 0.30 – 0.70
Metal on leather 0.30 – 0.60
Wood on wood 0.25 – 0.50
Wood on leather 0.25 – 0.50
Stone on stone 0.40 – 0.70
Earth on earth 0.20 – 1.00
Rubber on concrete 0.60 – 0.90
Example Problem - PullingGiven:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
W Pθ
Solution - PullingGiven:
W = 400 lb.μ = 0.4θ = 30º
Find:P for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
W Pθ
FN
0 yF 0 xF
NF 4.
030sin400 PN
NF 4005. NP
04.866. NP
04.30cos NP
030cos FP
NF 4.
Substituting
4005. NP
04.866. NP
Solve Simult. Eqn.
W Pθ
FN
lbP 150
lbN 325
lbF 130
Friction
A retarding force that resists the relative movement
of two bodies in contact with each other
Always acts in oppose direction of motion – actual motion
or impending motion
Always acts parallel or tangent to surface(s) in contact
Magnitude is mainly dependent on surface roughness
(other contributing factors: temperature, molecular
adhesion, electrostatic attraction, lubrication, relative
velocities)
Coefficient of Static Friction on Dry Surfaces
Materials Coefficient of Static Friction or μ
Metal on metal 0.15 – 0.60
Metal on wood 0.20 – 0.60
Metal on stone 0.30 – 0.70
Metal on leather 0.30 – 0.60
Wood on wood 0.25 – 0.50
Wood on leather 0.25 – 0.50
Stone on stone 0.40 – 0.70
Earth on earth 0.20 – 1.00
Rubber on concrete 0.60 – 0.90
Friction
Two types of friction
Dry or Coulomb friction
Static – surfaces are at rest with respect to each other
Kinetic/dynamic – surfaces are moving with respect to each other
Fluid Friction
Viscosity, friction developed between layers of a fluid moving at different velocities
We’ll limit our discussion to dry friction
Friction Theory
Consider a block on a horizontal surface
No Frictional resistance
N is referred to as the Normal force, perpendicular to the
contact surface
W
N
Friction Theory
Next an external force is applied
If no friction, block would immediately move
If force is applied, F = P, for equilibrium
At some point, P > F, motion occurs
W
N
P
F
Friction Theory
Ratio of F to N is called the Coefficient of Static
Friction or μ
NF
s Where
μs = coefficient of static friction (unitless)
F – Max or limiting frictional force resistance (lb, k, N)
N – Normal force, perpendicular to contact surface (lb, k, N)
Friction Theory
Stated another way:
Fs = μs N (for impending motion)
Fk = μk N (for sustaining motion, kinetic)
In general:
F ≤ μ N
Friction
Which of the following best explains the friction
force developed in a boulder at rest vs. a boulder
being acted on by a force:
A. The friction forces are equal
B. The friction force is greater in the left boulder
C. The friction force is greater in the right boulder
D. There is no friction force in either boulder
E. None of the above
WPθ
W
Example ProblemTo Push or to Pull, That is the Question
Given:
W = 400 lb.
μ = 0.4
θ = 30ºWP
θ
Find Force P required to cause motion to impend
W Pθ
Example Problem - Pushing
Given:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
WP
θ
PushingGiven:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
WPθ
Solution - PushingGiven:
W = 400 lb.μ = 0.4θ = 30º
Find:P for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
WPθ
0 yF
030cos FP
NF 4.
030sin400 PN
0 xF
NF 4005. NP
NF 4.04.30cos NP
04.866. NP
4005. NP
04.866. NP
Substituting
Solve Simult. Eqn.
WP
θ
FN
lbP 240
lbN 520
lbF 208
Example Problem - PullingGiven:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
W Pθ
PullingGiven:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
W Pθ
Solution - PullingGiven:
W = 400 lb.μ = 0.4θ = 30º
Find:P for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
W Pθ
FN
0 yF 0 xF
NF 4.
030sin400 PN
NF 4005. NP
04.866. NP
04.30cos NP
030cos FP
NF 4.
Substituting
4005. NP
04.866. NP
Solve Simult. Eqn.
W Pθ
FN
lbP 150
lbN 325
lbF 130
In Class Exercise – Up an InclineGiven:
W = 400 lb.
μ = 0.4
θ = 30º
Find:Force P required to cause motion to impend
W P
θ
lbP 6.338
Solution - Up an InclineGiven:
W = 400 lb.μ = 0.4θ = 30º
Find:P for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
400 lb P
θ
F
N
x’
y’
0' yF
030sin400 FP
θ
0' xF
030sin4006.138 P
NF 4.
030cos400 N
4.346*4.F
W P
θ
lbN 4.364
lbP 6.338
lbF 6.138
NF
In Class Exercise - Down an Incline
Given:
W = 400 lb.
μ = 0.4
θ = 30º
Find:
Force P required to cause motion to impend
W
Pθ
lbP 4.61
Solution - Down an InclineGiven:
W = 400 lb.μ = 0.4θ = 30º
Find:P for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
400 lb
Pθ
F
N
0' yF
x’
y’
θ
What does the “-” mean?
A “push” of at least 61.4 lb required
to keep block from sliding
0' xF
NF
030sin400 PF
NF 4.
030cos400 N
030sin4006.138 P
4.346*4.F
W
Pθ
lbN 4.346
lbF 6.138
lbP 4.61
In Class Exercise – Angle of Impending Motion
Given:
W = 400 lb.
μ = 0.4
Find:
θ such that it causes motion to impend
W
θ
8.21
Solution - Impending Motion
Given:μ = 0.4
Find:θ for impending motion
Solution:1-FBD2-Eqn of Equil
F=μN
W
θ
F
N
0' yFN
F
x’
y’
θ
Note that θ is independent of weight, dependent only on μ
sinWF NF
cosWN
0sin WF
4.tan
cos
sin
W
W
N
F
0cos WN
0' xF
tan
cos
sin
tan
Note the following
relationships:Dividing/Substituting:
8.21
W
θ
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