CHAPTER 8:
Frictions
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The Laws of Dry Friction. Coefficients of
Friction
Angles of Friction
Problems Involving Dry Friction
Wedges
Square-Threaded Screws
Belt Friction
1
8.1 The Laws of Dry Friction. Coefficients of
Friction
W
Keywords:
•Friction force F
•Fm, Fk
•N
• µs , µs
.
P
F
N
F
Equilibrium
Motion
Fm
Fk
P
2
W
F
Equilibrium
Motion
Fm
P
Fk
N
F
P
The forces Fm and Fk are proportional to the normal component N of the
reaction of the surface. We have
Fm = µs N
Fk = µk N
where µs is the coefficient of static friction and µk is the coefficient of kinetic
friction. These coefficients depend on the nature and the condition of the surfaces
in contact.
3
m
θ
W = mg
x
y
θ
N
W
F
N
F
θ
F = N tan θ = Ν µ
Impending motion: θ max, θ = φs
F = N tan φs = Ν µs
φs = Angle of friction
4
8.2 Angles of Friction
W
W
P
Py
Px
N
R=N=W
(a) No friction
φ < φs
R
F = Px
(b) No motion (Px < Fm)
F = Px
F < µs N
N = Py + W
5
W
W
P
P
Py
Py
Px
Impending motion
N
φ < φs
R
Fm = Px
(b) Motion impending → (Px = Fm)
Fm = Px
Fm = µs N
N = Py + W
Px
In motion
N
φ = φk
R
Fk < Px
(b) Motion → (Px > Fk)
Fk < Px
Fk = µk N
N = Py + W
6
W
Keywords:
•Friction force F
•Fm, Fk
•N
• µs , µs
• φ , φs
P
N
F = R sin φ
N = R cos φ
φ
R
F
tan φs = µs
tan φk = µk
7
8.3 Problems Involving Dry Friction
W
P
Frequired
N
The magnitude F of the friction force is equal to Fm = µsN only if the body is
about to slide. If motion is not impending, F and N should be considered as
independent unknowns to be determined from the equilibrium equations. The
value of F required to maintain equilibrium should be checked to insure that it
does not exceed Fm.
8
W
P
Fm = µsN
N
If motion is known to be impending, F has reached its maximum value Fm
= µsN , and this expression may be substituted for F in the equilibrium
equations.
9
imp
Sen
s
e of
end
ing
W
mo
tion
P
Fm = µsN
N
10
Q
P
A
F
Q
N
B
-P
Motion of A with respect to B
P
A
Motion of A with respect to B
-Q
-N
-F
B
-P
-Q
11
Example 8.1
A 100 N cabinet is mounted on the floor. The coefficient of static friction
between the floor and the cabinet is 0.30. If P = 25 N, determine:
(a) If the force P = 25 N can move the cabinet to the right?
(b) The smallest force to slide move the cabinet to the right, and
(c) When can the cabinet be moving and tipping at the same time?
500 mm
P = 25 N C
1200 mm
h
12
• Define the state of the problem
500 mm
• Construct the physical model,
mathematical model and solve
equations
(a) the force P = 25 N can move the
cabinet to the right?
P = 25 N C
500 mm
1200 mm
100 N
h
P = 25 N C
h
µs= 0.3
W = 100 N
Assuming that the cabinet can treat
as a rigid body. There are 4 forces
acting on it, P = 25 N, W = 100N, N
and F
F = 25 N
N = W = 100 N
P = 25 N < (0.3)(100) N can not
move the cabinet to the right.
13
(b) The smallest force to slide move
the cabinet to the right.
(c) the largest allowable value of h if
the cabinet is not to tip over.
500 mm
500 mm
P = 30 N
100 N
100 N
P = 30 N C
h = 833 mm
1200 mm
h
B
B
A
Fm = N µ s
= (100)(.30)
N = W = 100 N
+ ΣF = 0: P - 0.3N = 0
x
P = 0.3(100)
= 30 N
A
+ ΣMB = 0:
Fm = 30 N
N = W = 100 N
-(30)(h) + 100(250) = 0
h < 833 mm
The cabinet will move and tip at the
same time when the force P = 30 N
applies at h = 833 mm .
14
Example 8.2 (Prob 8.19)
A 61.2 kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and each
caster is 0.30. If h = 640 mm, determine the magnitude of the force P
required to move the cabinet to the right (a) if all casters are locked, (b) if
the casters at B are locked and the casters at A are free to rotate, (c) if the
casters at A are locked and the casters at B are free to rotate.
C
P
h
A
B
480 mm
15
• Define the state of the problem
P
(a) The force P required to move the
cabinet to the right.
W = 61.2(9.81) = 600 N
µ = 0.30
C
P
C
640 mm
A
B
480 mm
Assuming that the cabinet can treat
as a rigid body. There are 2 forces,
N and F are acting on caster if it is
locked and there is only N if it is
free to rotate.
• Construct the physical model,
mathematical model and solve
equations
640 mm
A
0.3NA
NA 480 mm
B 0.3NB
NB
+ ΣFy = 0: NA + NB = W = 600 N
+ ΣF = 0: P - 0.3N - 0.3N = 0
x
A
B
P = 0.3(NA+NB)
= 0.3W = 180 N
16
(b) If the casters at B are locked and the
casters at A are free to rotate.
(c) if the casters at A are locked and the
casters at B are free to rotate.
600 N
P
W = 61.2(9.81) = 600 kN
C
P
640 mm
C
640 mm
A
NA 480 mm
B 0.3NB
NB
A
0.3NA
NA 480 mm
B
NB
+ ΣF = 0: P = 0.3N
x
B
+ ΣF = 0: P = 0.3N
x
A
+ ΣMA = 0:
+ ΣMB = 0:
-(0.3NB)(640) - 600.37(240) + NB(480) = 0
NB = 500 N
P = 0.3(500) = 150 N
-(0.3NA)(640) + 600(240) - NA(480) = 0
NA = 214 N
P = 0.3(214) = 64.3 N
17
Example 8.3 (Prob 8.20)
A 61.2 kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefficient of static friction between the floor and each
caster is 0.30. Assuming that the casters at both A and B are locked,
determine (a) the force P required to move the cabinet to the right, (b) the
largest allowable value of h if the cabinet is not to tip over.
C
P
h
A
B
480 mm
18
• Define the state of the problem
W = 61.2(9.81) = 600 N
µ = 0.30
C
P
(a) The force P required to move the
cabinet to the right.
C
P
h
A
B
480 mm
Assuming that the cabinet can treat
as a rigid body. There are 2 forces,
N and F are acting on caster if it is
locked.
• Construct the physical model,
mathematical model and solve
equations
h
A
0.3NA
NA 480 mm
B 0.3NB
NB
+ ΣFy = 0: NA + NB = W = 600 N
+ ΣF = 0: P - 0.3N - 0.3N = 0
x
A
B
P = 0.3(NA+NB)
= 0.3W = 180 N
19
(b) the largest allowable value of h if the
cabinet is not to tip over.
• Conclusion
600 N
W = 61.2(9.81) = 600 N
P = 180 N
P = 180 N
C
C
h = 800 mm
h
A
A
B 0.3NB = 180 N
480 mm
B 180 N
480 mm
NB = 600 N
NB = 600 N
+ ΣMA = 0:
-(180.1)(h) - 600.37(240) + 600.37(480) = 0
h = 800 mm
P = 180 N can move the cabinet to
the right and if h > 800 mm it will tip
over before move.
20
Example 8.4
A 100 N force acts as shown on a 30.6 kg block placed on an inclined
plane. The coefficients of friction between the block and the plane are µs =
0.25 and µk = 0.20. Determine whether the block is in equilibrium, and
find the value of the friction force.
30.6 kg
100 N
5
3
4
21
• Define the state of the problem
• Construct the mathematical model
+
30.6 kg
5
P
+
ΣFy = 0:
ΣFx = 0:
3
4
N - (4/5)(300) = 0
N = 240 N
P - (3/5)300 - 0.25(240) = 0
P = 120 N
Note: Fm = µsN
Assuming that block can treat as a
rigid body. There are 4 forces acting
on it, 100 N, W = 30.6(9.81) = 300 N,
F and N.
• Construct the physical model
Assuming the block is in equilibrium, the
force P prevents it from sliding down
300 N
3
x
y
4 5
If force P < 120 N is applied, the block
will slide down.
Factual = µkN = 0.20(240) = 48 N
• Conclusion
300 N
M
120 N
P
N
F
= 0.25(240) = 60 N
F = 60 N
o
oti
n
N = 240 N
22
Example 8.5
A support block is acted upon by two forces as shown. Knowing that the
coefficients of friction between the block and the incline are µs = 0.35 and
µk = 0.25, determine the force P required (a) to start the block moving up
the incline, (b) to keep it moving up, (c) to prevent it from sliding down.
P
25o
800 N
Guide : Fore each part of the problem we draw a FBD of the block and a
force triangle including the 800 N vertical force, the horizontal force P, and
the force R exerted on the block by the incline. The direction of R must be
determine in each separate case. We note that since P is perpendicular to
the 800 N force, the force triangle is a right triangle, which can easily be
solve for P. In most other problem, however, the force triangle will be an
oblique triangle and should be solved by applying he law of sines.
23
• Construct the physical model,
mathematical model and solve
equations
• Define the state of the problem
P
25o
(a) Force P to start block moving Up
800 N
P
800 N
µs = 0.35
µk = 0.25
Fore each part of the problem we
draw a FBD of the block and a
force triangle including the 800 N
vertical force, the horizontal force
P, and the force R exerted on the
block by the incline.
R
φs
P
800 N
25o
tan φs = µs = 0.35
φs = 19.29o
R
25o + 19.29o = 44.29o
P = 800 tan 44.29o = 780 N, ←
24
(b) Force P to keep block moving Up
800 N
(c) Force P to prevent block from sliding
down
800 N
P
P
φs
R
φk
25o
P
800 N
25o
tan φk = µk = 0.25
φk = 14.04o
R
25o + 14.04o = 39.04o
P = 800 tan 39.04o = 649 N, ←
P
R
φs = 19.29o
800 N R
25o - 19.29o = 5.71o
P = 800 tan 5.71o = 80.0 N, ←
25
• Conclusion
(a) Force P to start block moving Up
(b) Force P to keep block moving Up
800 N
800 N
P = 649 N
P = 780 N
R
R
φs =
19.29o
25o
φk = 14.04o
25o
(c) Force P to prevent block from sliding down
800 N
P = 80.0 N
φs = 19.29o
25o
26
R
Example 8.6
The movable bracket shown may be placed at any height on the 30 mm
diameter pipe. If the coefficient of static friction between the pipe and
bracket is 0.25, determine the minimum distance x at which the load W
can be supported. Neglect the weight of the bracket.
x
W
60 mm
30 mm
Guide : Assuming that the bracket can treat as a rigid body. There are 5
forces acting on it, W, FA, NA, FB and NB. FA = µsNA, FB = µsNB, use
equilibrium equations ΣFX = 0, ΣFy = 0 and ΣM = 0.
27
• Define the state of the problem
x
• Construct the physical model
W
FA
x
x - 15
NA
W
60 mm
60 mm
FB
NB
30 mm
30 mm
Assuming that the bracket can treat as
a rigid body. There are 5 forces acting
on it, W, FA, NA, FB and NB.
28
FA
x
x - 15
NA
60 mm
+ ΣMB = 0:
NA(60) - FA(30) - W(x - 15) = 0
60NA - 30(0.25NA) -Wx + 15W = 0
60(2W) - 7.5(2W) - Wx + 15W = 0
W
FB
Dividing through by W and solving for
x = 120 mm
NB
• Conclusion
30 mm
• Construct the mathematical model
and solve equations
FA = µsNA = 0.25 NA
FB = µsNB = 0.25 NB
+ ΣF = 0: N - N = 0
x
B
A
0.5W
120 mm
W
2W
NB = NA
60 mm0.5W
+ ΣFy = 0: FA + FB - W = 0
0.25NA + 0.25NB = W
And, since NB has been found equal to NA,
0.50NA = W
NA = 2W
2W
30 mm
29
8.4 Wedges
In the analysis of wedges,
two or more free-body
diagrams are generally used
to show each friction force
and its correct sense.
A
B
P
C
6o
6o
D
W
N2
F1 = usN1
A
N1
F2
P
C
F3 = µsN3
F2 = usN2
N2
N3
30
8.5 Square-Threaded Screws
w
a
Cap
Screw
P
p = Pitch
Base
r
(P)(a) = (Q)(r )
31
W
W
Q
Q
L = (n) (p)
θ
θ
θ φs R
R
2πr
N = Type of thread
=1, single thread
=2, double thread
φs
θ
W
Q
θ
R
θ
φs
32
W
Q
L
R
θ
θ
The analysis of squarethreaded screws (frequently used
in jacks, presses and other
mechanisms) is
reduced to the analysis of a block
sliding on an incline by unwrapping
the thread of the screw
and showing it as a straight line.
θs
2θr
In doing this, r denotes the mean radius of the thread, L is the lead of the screw (the
distance through which the screw advances in one turn), W is the load, and Qr is the
torque exerted on the screw.
33
Example 8.7
The position of the machine 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 the force P required (a) to raise block B, (b) to lower
block B.
400 N
B
8o
A
P
Guide : Assuming block B and A can treat as a rigid body. To raise block
B,the direction of P and impending motion of A are right to left and the
impending motion of B is up. To lower block B, the direction of P and the
impending motion of A are left to right and impending motion of B is down.
The direction of friction forces are opposite from impending motion. Draw
FBD for each block B and A, we can find unknown from block B and sent
it to block A to solve P.
34
• Define the state of the problem
400 N
B
8o
A
P
Assuming block B and A can treat as a rigid body. To raise block B,the
direction of P and impending motion of A are right to left and the
impending motion of B is up. To lower block B, the direction of P and the
impending motion of A are left to right and impending motion of B is
down. The direction of friction forces are opposite from impending
motion. Draw FBD for each block B and A.
35
• Construct the physical model, mathematical model and solve equations
400 N
R2
F2
19.3o
B
N2
F1
Impending motion of B
(a) to raise block B
φs = tan-10.35 = 19.3o
8 + 19.3 =
8o
27.3o
400 N
R1
180o - 109.3o - 27.3o = 43.4o
90o + 19.3o
= 109.3o
R2
R1 = 549 N
27.3o
R1
R1= 549 kN
P
N1
N1
27.3o
8o
27.3o
549 kN
F1
A
8o
F3
19.3o
R3
R1
400
=
sin 109.3o sin 43.4o
N3
R3
27.3o +19.3o = 46.6o
P
Impending motion of A
P
549
=
sin 46.6o sin 70.7 o
P = 423 N
36
(b) to lower block B.
400 N
N2
B
19.3o
F2
F1
R2
R2
90o - 19.3o
= 70.7o
180o - 70.7o - 11.3o
= 98o
400 N
R1
R1
400
=
sin 70.7 o sin 98.0o
11.3o
R1 = 381 N
8o
19.3o - 8o
= 11.3o
90o -19.3o = 70.7o
R1 N
1
N1
Impending motion of B
φs = tan-10.35 = 19.3o
11.3o
19.3o
R3
R1= 381 kN
11.3o
A
8o
F3
19.3o
R3
381 kN
19.3o + 11.3o
= 30.6o
F1
N3
P
P
Impending motion of A
P
381
=
sin 30.6 o sin 70.7 o
P = 206 N
37
Example 8.8
A clamp is used to hold two pieces of wood together as shown. The clamp
has a double square thread of mean diameter equal to 10 mm with a pitch
of 2 mm. The coefficient of friction between threads is µs = 0.30. If a
maximum torque of 40 N•m is applied in tightening the clamp, determine
(a) the force exerted on the pieces of wood, (b) the torque required to
loosen the clamp.
38
• Construct the physical model, mathematical model and solve equations
(a) Force Exerted by clamp.
φs = tan-1µs = tan-1 0.3 = 16.7o
W
θ = tan −1 (
Impending motion
4
) Q
10π
= 7.3o
F
L = (n)(p)
= 2 (2) = 4 mm
R
7.3o
N
2πr = 10π mm
16.7o
7.3o +16.7o = 24o
W
T = Qr
40 = Q(0.005)
Q = 8000 N = 8 kN
W
8
=
sin 66 o sin 24 o
R
90o - 24o = 66o
Q = 8 kN
W = 17.97 kN
39
(b) Torque required to loosen clamp
φs = tan-1µs = tan-1 0.3 = 16.7o
9.4o
W = 17.97 kN
Impending motion
Q
7.3o
F
L = 4 mm
16.7o - 7.3o
= 9.4o
R
7.3o
N
16.7o
2πr = 10π mm
90o -
R
9.4o =
W = 17.97 kN
80.6o
Q
Q
17.97
=
sin 9.4 o sin 80.6 o
Q = 2.97 kN
T = Qr
= 2.97(0.005)
= 0.01485 kN•m
= 14.85 N•m
40