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Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
8/1 Introduction
• In preceding chapters, it was assumed that surfaces in contact were either
frictionless (surfaces could move freely with respect to each other) or rough
(tangential forces prevent relative motion between surfaces).
• Actually, no perfectly frictionless surface exists. For two surfaces in contact,
tangential forces, called friction forces, will develop if one attempts to move
one relative to the other.
• However, the friction forces are limited in magnitude and will not prevent
motion if sufficiently large forces are applied.
• The distinction between frictionless and rough is, therefore, a matter of degree.
• There are two types of friction: dry or Coulomb friction and fluid friction.
Fluid friction applies to lubricated mechanisms. The present discussion is
limited to dry friction between non lubricated surfaces.
8/2 The Laws of Dry Friction. Coefficients of Friction
Block of weight W placed on horizontal surface.
Forces acting on block are its weight and reaction
of surface N.
Small horizontal force P applied to block.
For block to remain stationary, in equilibrium,
a horizontal component F of the surface reaction
is required. F is a static-friction force.
As P increases, the static-friction force F
increases as well until it reaches a maximum
value Fm.
Fm = µ s N
Further increase in P causes the block to begin to
move as F drops to a smaller kinetic-friction force Fk.
Fk = µ k N
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Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
Maximum static-friction force:
Fm = µ s N
Kinetic-friction force:
Fk = µ k N
µ k ≅ 0.75µ s
Maximum static-friction force and kinetic-friction force are:
- proportional to normal force
- dependent on type and condition of contact surfaces
- independent of contact area
Four situations can occur when a rigid body is in contact with a horizontal
surface:
• No friction,
(Px = 0)
• No motion,
(Px < Fm)
• Motion impending,
(Px = Fm)
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• Motion,
(Px > Fm)
Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
8/3 Angles of Friction:
It is sometimes convenient to replace normal force N and friction force F by
their resultant R:
- No friction
- No motion
µ N
F
tan φ s = m = s
N
N
tan φ s = µ s
- Motion impending
- Motion
µ N
F
tan φk = k = k
N
N
tan φk = µk
Consider block of weight W resting on board with variable inclination angle θ.
- No friction
- Motion impending
- No motion
F
µ N
tan φ s = m = s
N
N
tan φ s = µ s
- Motion
Fk µ k N
=
N
N
tan φ k = µk
tan φ k =
81
Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
8/4 Problems Involving Dry Friction:
1. All applied forces
1. All applied forces
known
known
2. Coefficient of static
friction is known
3. Determine whether
body will remain at rest
friction is known
2. Motion is impending
2. Motion is impending
3. Determine value of
3. Determine magnitude
coefficient of static
or direction of one of
friction.
the applied forces
or slide
EXAMPLE 1. A 100 lb force acts as shown on
a 300 lb block placed on an inclined plane.
The coefficients of friction between the block
and plane are ms = 0.25 and mk = 0.20.
Determine whether the block is in equilibrium
and find the value of the friction force.
SOLUTION:
• Determine values of friction force and
normal reaction force from plane required
to maintain equilibrium.
•
∑ Fx = 0 :
1. Coefficient of static
100 lb - 3 (300 lb ) − F = 0
5
F = −80 lb
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Chapter Eight
∑ Fy = 0 :
A.Lecturer Saddam K. Kwais
N - 4 (300 lb ) = 0
5
N = 240 lb
• Calculate maximum friction force and compare with friction force required
for equilibrium. If it is greater, block will not slide.
Fm = µ s N
Fm = 0.25(240 lb ) = 48 lb
The block will slide down the plane.
• Calculate maximum friction force and compare with friction force required
for equilibrium. If it is greater, block will not slide.
Factual = Fk = µ k N
= 0.20(240 lb)
Factual = 48 lb
EXAMPLE 2. The moveable bracket shown
may be placed at any height on the 3-in.
diameter pipe.
If the coefficient of friction
between the pipe and bracket is 0.25, determine
the minimum distance x at which the load can
be supported. Neglect the weight of the bracket.
SOLUTION:
• When W is placed at minimum x,
the bracket is about to slip and
friction forces in upper and lower
collars are at maximum value.
83
Friction
Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
F A = µ s N A = 0.25 N A
FB = µ s N B = 0.25 N B
• Apply conditions for static equilibrium to find minimum x.
NB − NA = 0
NB = N A
∑ Fx = 0 :
∑ Fy = 0 :
F A + FB − W = 0
0.25 N A + 0.25 N B − W = 0
0.5 N A = W
N A = N B = 2W
N A (6 in.) − F A (3 in.) − W ( x − 1.5 in.) = 0
∑MB = 0:
6 N A − 3(0.25 N A ) − W ( x − 1.5) = 0
6(2W ) − 0.75(2W ) − W ( x − 1.5) = 0
x = 12 in
8/5 Wedges:
1.Wedges - simple machines
used to raise heavy loads.
2.Force required to lift block is
significantly less than block
weight.
3.Friction prevents wedge from
sliding out.
• Block as free-body
∑ Fx = 0 :
− N1 + µ s N 2 = 0
∑ Fy = 0 :
− W − µ s N1 + N 2 = 0
Or
r
r
v
R1 + R2 + W = 0
4.Want to find minimum force
P to raise block.
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• Wedge as free-body
∑ Fx = 0 :
− µ s N 2 − N 3 (µ s cos 6° − sin 6°)
+ P =0
∑ Fy = 0 :
− N 2 + N 3 (cos 6° − µ s sin 6°) = 0
r r
r
P − R2 + R3 = 0
Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
EXAMPLE 3. The position of the machine
400 Ib
block B is adjusted by moving the wedge A.
Knowing that the coefficient of static friction is
0.35 between all surfaces of contact, determine
B
the force P required (a) to raise block B, (b) to
8ᵒ
P
A
lower block B.
SOLUTION:
For each part, the free-body diagrams of block B and wedge A are drawn,
together with the corresponding force triangles, and the law of sines is used
to find the desired forces. We note that since µ s = 0.35, the angle of friction is
tan 0.35 19.3°
400 Ib
a)A force P to raise Block
27.3ᵒ
400 Ib
R2
Free body: block B
B
R1
ᵒ
180 - 27.3 - 109.3 ϕs = 19.3
Apply the law of sines
400
°
sin 109.3
sin 43.4°
ᵒ
ᵒ
ᵒ
ᵒ
= 43.3
8
R2
90ᵒ + 19.3ᵒ =109.3ᵒ
R1
8 + 19.3 =27.3ᵒ
549
ᵒ
P
Free body: Wedge A
ᵒ
Free Body : Block B
Force triangle
90ᵒ - 19.3ᵒ =70.7ᵒ
19.3ᵒ
Apply the law of sines
549
sin 46.6° sin 70.7°
ϕs = 19.3ᵒ
ᵒ
27.3ᵒ
R1 = 549 Ib
R3
27.3ᵒ
549 Ib
27.3ᵒ + 19.3ᵒ =46.6ᵒ
A
R3
5423 ← .
19.3ᵒ
Force triangle
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Free Body : Wedge A
P
Chapter Eight
A.Lecturer Saddam K. Kwais
Friction
b)A force P to Lower Block
400 Ib
Free body: block B
90ᵒ - 19.3ᵒ =70.7ᵒ
Apply the law of sines
R1
180ᵒ - 70.7ᵒ - 11.3ᵒ
=98.0ᵒ
19.3ᵒ - 11.3ᵒ
=8ᵒ
ᵒ
R1
11.3
400 Ib
400
sin 70.7° sin 98.0°
R2
381
R2
8ᵒ
Free Body : Block B
90ᵒ - 19.3ᵒ
= 70.7ᵒ
R1 = 381 Ib
11.3ᵒ
P
Apply the law of sines
381
sin 30.6° sin 70.7°
B
ϕs = 19.3ᵒ
Force triangle
Free body: Wedge A
ϕs = 19.3ᵒ
11.3ᵒ
A
19.3ᵒ
P
381 Ib
R3
19.3ᵒ + 11.3ᵒ
= 30.6ᵒ
206 → .
Force triangle
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19.3ᵒ
R3
Free Body : Wedge A