Ch. 4: Friction
4.0 Outline
 Introduction
 Types of Friction
 Dry Friction
203
203
204
205
206
4.0 Outline
Ch. 4: Friction
204
4.1 Introduction
In real situation, the forces of action and reaction
between contacting surfaces have their components
both in the tangential and normal directions to the
contacting surface. Tangential forces are known as
Friction forces. Whenever a tendency exists for one
contacting surface to slide along another surface,
the friction forces developed are always in a direction
to oppose this tendency.
In some systems, friction is undesirable since it normally
spoils the required behavior. But in many situations,
friction functions the systems.
In real case where sliding motion between parts occurs,
the friction forces result in a loss of energy.
4.1 Introduction
Ch. 4: Friction
205
4.2 Types of Friction
a) Dry (Coulomb) friction when unlubricated surfaces
are in contact under a condition of sliding
or tendency to slide.
Friction force tangent to the surfaces of contact is
developed both during the interval leading up to
impending slippage and while slippage takes place.
Its direction always opposes the motion or impending
motion which would occur if no friction were present.
b) Fluid friction
c) Internal friction
4.2 Types of Friction
Ch. 4: Friction
206
4.3 Dry Friction
Mechanism of friction
4.3 Dry Friction
Ch. 4: Friction
207
Three regions of static  motion transition
a) No motion is the region up to the point of slippage
or impending motion. Friction force is determined
by the equations of equilibrium because
the system is in equilibrium. When the motion is not
impending, F < Fmax
b) Impending motion is the moment where the body
is on the verge of slipping. Static friction force reaches
the max value. For a given pair of mating surfaces,
=
F F=
µs N .
max
c) Motion The body starts moving in the direction
of the applied force. Here, friction force drops to a
lower value called kinetic friction F = µk N .
It will drop further with higher velocity.
4.3 Dry Friction
Ch. 4: Friction
208
Friction cone Friction coefficient reflects the roughness
of a pair of mating surfaces. The smaller the
coefficient value, the smoother the surfaces.
Direction of resultant R is specified by tan α = F/N .
When the friction force reaches max value, tan φs = µs .
When slippage occurs, tan φk = µk .
The friction angle φs , φk defines the limiting position of
the total reaction force R. The friction cone of
vertex angle 2φs , 2φk represents the locus of possible
positions for the reaction force R.
Friction force is independent of the apparent
or projected area of contact.
4.3 Dry Friction
Ch. 4: Friction
209
Friction cone
4.3 Dry Friction
Ch. 4: Friction
210
Types of dry friction problems First step is to identify
which of these categories applies.
1) Condition of impending motion is known to exist
The body is in equilibrium and on the verge of slipping.
Friction force is the max static friction F = µs N
2) Relative motion is known to exist
Friction force is the kinetic friction F = µk N
3) Unknown status of the problem
Assume static equilibrium and solve for the required
friction force F. Then check and conclude the status.
4.3 Dry Friction
Ch. 4: Friction
211
Possible outcomes
a) F < µs N friction force for the assumed equilibrium
can be provided and so the body is in static equilibrium.
b) F = µs N max friction force is required for the
static equilibrium condition and so motion impends.
c) F > µs N surfaces cannot support more friction than
µs N. So the equilibrium assumption is invalid
and motion occurs instead. Friction force
is the kinetic friction F = µk N . Even with the correct
kinetic friction substituted, equilibrium equations
are still not hold  accelerated motion
4.3 Dry Friction
Ch. 4: Friction
212
P. 4/1 Determine the max angle θwhich the adjustable
incline may have with the horizontal before the
block of mass m begins to slip. The coefficient
of static friction between the block and the
inclined surface is μs.
4.3 Dry Friction
Ch. 4: Friction
213
P. 4/1
at the moment of slipping, friction is
 ∑ Fy =0 
 ∑ Fx = 0 
F = µs N
upward
N − mgcosθ =0
µs N − mgsinθ = 0
=
=
µs tan
θ or θ tan −1 µs
when the friction force reaches max value, tan φs
by equilibrium, R = W and φs
= µs
= θ ∴θ =
tan −1 µs
φs
R
4.3 Dry Friction
Ch. 4: Friction
214
P. 4/2 Determine the range of values which the mass
mo may have so that the 100 kg block shown
in the figure will neither start moving up the
plane nor slip down the plane. The coefficient
of static friction for the contact surface is 0.30.
4.3 Dry Friction
Ch. 4: Friction
215
P. 4/2
bounded mo values  block start moving 
 ∑ Fy =
0  N − 100gcos20 =
0, N =
922 N
F = µs N
Case I: max mo, start moving up, friction downward
 ∑ Fx= 0 
m o g − µs N − 100gsin20= 0, m o= 62.4 kg
Case II: min mo, start moving down, friction upward
 ∑ Fx= 0  m o g + µs N − 100gsin20= 0, m o= 6.0 kg
277 N up/downward
∴ 6.0 ≤ m o ≤ 62.4 kg and F ≤ Fmax =
4.3 Dry Friction
Ch. 4: Friction
216
P. 4/3 Determine the magnitude and direction of the
friction force acting on the 100 kg block shown
if, first, P = 500 N and, second, P = 100 N. The
coefficient of static friction is 0.20, and the
coefficient of kinetic friction is 0.17. The force
are applied with the block initially at rest.
4.3 Dry Friction
Ch. 4: Friction
217
P. 4/3
don’t know if the block is impending or is moving  assume static equilibrium
P = 500 N: assume the block tends to move up  friction downward
 ∑ Fy =
0  N − 500sin 20 − 100gcos20 =
0, N =
1092.85 N
= µ=
max supportable friction
218.6 N
sN
 ∑ F=
= 0, =
0  500cos20 − F − 100gsin20
F 134.3 N < µs N
x
∴ the assumption is valid
P = 100 N: assume the block tends to slide down  friction upward
 ∑ Fy =
0  N − 100sin 20 − 100gcos20 =
0, N=
956.04 N
= µ=
max supportable friction
191.21 N
sN
 ∑ Fx =
0  F + 100 cos 20 − 100gsin20 =
0, F =
241.55 N > µs N
∴ the assumption is invalid, block is moving downward
= µ=
162.5 N
kinetic friction upward
kN
4.3 Dry Friction
Ch. 4: Friction
P. 4/4
218
The homogeneous rectangular block of mass m, width b, and
height H is placed on the horizontal surface and subjected to a
horizontal force P which moves the block along the surface with
a constant velocity. The coefficient of kinetic friction between
the block and the surface is μk. Determine (a) the greatest
value that h may have so that the block will slide without tipping
over and (b) the location of a point C on the bottom face of the
block through which the resultant of the friction and normal
forces acts if h = H/2.
4.3 Dry Friction
Ch. 4: Friction
219
P. 4/4
a) On the verge of tipping over, reaction acts at the corner A
When slippage occurs, tan θ = µ k
Block moves w/ const. velocity  equilibrium
Three-force member: reaction at A must pass through B
tan θ = µk = b/2h, ∴ h = b/ ( 2µk )
b) When slippage occurs, tan θ = µ k
Block moves w/ const. velocity  equilibrium
Three-force member: reaction at C must pass through G
tan θ = µk = x/ ( H/2 ) , ∴ x = µk H/2
4.3 Dry Friction
Ch. 4: Friction
P. 4/5
220
The three flat blocks are positioned on the 30°incline as shown,
and a force P parallel to the incline is applied to the middle block.
The upper block is prevented from moving by a wire which
attaches it to the fixed support. The coefficient of static friction
for each of the three pairs of mating surfaces is shown. Determine
the maximum value which P may have before any slipping
takes place.
4.3 Dry Friction
Ch. 4: Friction
221
P. 4/5
4.3 Dry Friction
Ch. 4: Friction
222
P. 4/5
 ∑ Fy =
0 
N1 − 30gcos30 =
0, N1 =
254.87 N
N 2 − N1 − 50gcos30
= 0, N=
679.66 N
2
N 3 − N 2 − 40gcos30
= 0, N=
1019.5 N
3
Since 30 kg-block cannot slide and 50 kg-block is pulled, 50 kg-block tends
to move and only 2 cases are possible. Either 50 kg-block alone or
50&40 kg-blocks move together.
50 kg-block tends to move alone  F1 & F2 max (either one alone will not slip)
=
F1 µ=
76.46 N,=
F2 µ=
271.86 N
s N1
s N2
F=
µ=
458.8 N
3max
s N3
block #3: F2 − F3 + 40gsin30
= 0, F=
468.06 N > F3max
3
∴ block #3 cannot stay still -- the assumption is invalid
50&40 kg-blocks tend to move together  F1 & F3 max (either one alone will not slip)
block #3: F2 − F3 + 40gsin30
= 0, F=
262.6 N < F2max
2
∴ block #2 & #3 does not slip relative to each other
block #2: P − F1 − F2 + 50gsin30
= 0, P
= 93.8 N
4.3 Dry Friction
Ch. 4: Friction
223
P. 4/6 The light bar is used to support the 50 kg block
in its vertical guides. If the coefficient of static
friction is 0.30 at the upper and 0.40 at the lower
end of the bar, find the friction force acting at
each end for x = 75 mm. Also find the maximum
value of x for which the bar will not slip.
4.3 Dry Friction
Ch. 4: Friction
N
P. 4/6
R
F
Bar is a two-force member.
Assume the system is in equilibrium.
Hence the reaction forces at both ends
act along the axial direction.
 ∑ Fy = 0 
y
F
224
N − 50g = 0, N = 490.5 N
limitation of the reaction force on each end
φA= tan −1 µ= 21.8°, φB= tan −1 µ= 16.7°
R
N
θ sin −1 ( 75 / 300
=
at x = 75 mm: =
) 14.5° < φB < φA
R inside the static friction cone, system is in equilibrium and
=
F Ntan
=
θ 126.6 N
max x before slipping when the bar angle = that of small friction cone
=
=
x/300 sin
φB , x 86.2 mm
4.3 Dry Friction
Ch. 4: Friction
P. 4/7
225
Find the tension in the cable and force P
that makes the 15 kg lower block
(a) to start sliding downward
(b) to start sliding upward
4.3 Dry Friction
Ch. 4: Friction
P. 4/7
226
=
= 73.75 N
N1 8gcos20
= 0, N=
N 2 − N1 − 15gcos20
212 N
2
=
=
F1max 0.3N
22.12 N
1
=
=
F2max 0.4N
84.81 N
2
a) pulling down, 15 kg block impends to slide downward
T
P − F1max − F2max + 15gsin20 = 0, P = 56.6 N
8g
F1max + 8gsin20 − =
T 0, =
T 49 N
F1
15g
N1
N1
F1
P
N2
F2
4.3 Dry Friction
Ch. 4: Friction
P. 4/7
227
=
= 73.75 N
N1 8gcos20
= 0, N=
N 2 − N1 − 15gcos20
212 N
2
=
=
F1max 0.3N
22.12 N
1
=
=
F2max 0.4N
84.81 N
2
b) pushing up, assume 15&8 kg blocks impends to slide upward together
the cable slacks  T=0
T
8g
8 kg block: 8gsin20 − F=
0, F=
26.84 N > F1max
1
1
∴15 kg block impends to slide upward alone
−P + F1max + F2max + 15gsin20
= 0, =
P 157.3 N
N1 F1
−T − F1max + 8gsin20
= 0, =
T 4.72 N
15g N1
F1
P
F2
N2
4.3 Dry Friction
Ch. 4: Friction
P. 4/8
228
The uniform slender rod of mass m and length L
is initially at rest in a centered horizontal position
on the fixed circular surface of radius R = 0.6L.
If a force P normal to the bar is gradually applied
to its end until the bar begins to slip at the angle
θ= 20°, determine the coefficient of static friction.
4.3 Dry Friction
Ch. 4: Friction
229
P. 4/8
no slip until θ=20°
distance on bar = length on curve
π 
=
=
[θ a/r
] a  20 =
 R π R/9
 180 
α
α
L/(2tan20)
20°
=
µs tan=
α F/N
=
( L/2 − π R/9
)
=
L/ ( 2tan20 )
0.211
4.3 Dry Friction
Ch. 4: Friction
P. 4/9
230
The three identical rollers are stacked on a
horizontal surface as shown. If the coefficient
of static friction μs is the same for all pairs
of contacting surfaces, find the minimum value
of μs for which the rollers will not slip.
4.3 Dry Friction
Ch. 4: Friction
P. 4/9
231
Lower roller tends rolling out at upper contact
while tends to slide out at lower contact
condition: one or more contacts impend to slip
FBD: lower left roller (three-force member)
mg
FA
=
 FA FB
∑ M O 0=
from the figure, N A < N B
R
α N
A
∴ FAmax < FBmax so FA reaches the limit value before FB
A
∴ slipping does occur first at contact A
FAmax and FB determined by equilibrium equation
∴ FA =
O
FB
B
NB
r
A
lower roller: three-force member
reaction force at A must pass through contact B
from geometry,
15°  F =
FAmax , ∴ tan α =
µs =
tan15 =
0.268
30°
O
r
B
4.3 Dry Friction
Ch. 4: Friction
232
P. 4/10 The industrial truck is used to move the solid
1200 kg roll of paper up the 30°incline. If the
coefficients of static and kinetic friction between
the roll and the vertical barrier of the truck and
between the roll and the incline are both 0.40,
compute the required tractive force P between
the tires of the truck and the horizontal surface.
4.3 Dry Friction
Ch. 4: Friction
P. 4/10
To move the paper roll, 3 possibilities
1) A and B both slip
2) only B slips
3) only A slips
after calculation, only case 3) is viable
slipping at A, F = 0.4N A
1200g
=
 FB 0.4N A
∑ M O 0=
 ∑ Fx =
0  N A − FB cos 30 − N B sin 30 =
0, N B =
1.307N A
0.4NA
NA
A
233
FB
O
B
 ∑ Fy =0  − 0.4N A − 1200g − FBsin30 + N B cos 30 =0
N A 22.1 kN,
N B 28.9 kN,
FB 8853 N < 0.4N B
=
=
=
NB
4.3 Dry Friction