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Friction [compatibility mode]
1. Friction
Cause of dry friction
Contact between two surfaces.
Hence first task in a friction problem is correct
identification of contact surfaces
Identify the surface, the normal and the tangential
vectors.
Also important is to get an idea of probable direction
of relative motion
The contact force acts along the normal.
Gravity is the most common cause of normal force.
Friction acts along the tangent plane opposite to the
direction of relative motion
Normal
Relative velocity
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2. Friction
Normal
Relative velocity
Friction problems are
essentially equilibrium
problems with one f the
forces being functions of
another
N
Fr=f(N,V)
Fr=f(N,V)
N
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3. The correct way of writing the dry friction force
V ˆ
Fr N N V
V
N=Normal force vector
V=Relative velocity vector of the body
m= coefficient of dry friction or Coulomb
friction
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4. Problem 1
Knowing that the coefficient of friction between the 13.5
kg block and the incline is ms = 0.25, determine
a) the smallest value of P required to maintain the block in
equilibrium,
b) the corresponding value of b.
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5. Problem 1 y
N cos60 mg f sin60 P sin 0
N sin60 f cos 60 P cos 0
f N x
mg sin60 f
P
sin sin60 cos cos 60
N cos60 mg N sin60 P sin 0
mg P sin
N
cos 60 sin60
N sin60 N cos60 P cos 0
mg P sin
sin60 cos 60 P cos 0
cos 60 sin60
mg P sin sin60 cos60 P cos cos 60 sin60 0
mg sin60 cos60 P sin sin60 cos 60 P cos cos60 sin60 0
P sin60 sin cos60 sin cos60 cos sin60 cos mg sin60 cos60
sin60 cos 60
P mg
cos 60 sin 60
1 cos 60 sin 60
mg
P sin60 cos 60
d 1 d
P 0 d cos 60 sin 60 0
d
d
sin60 sin cos 60 sin cos60 cos sin60 cos 0
d
sin60 cos cos60 cos cos 60 sin sin60 sin 0
sin60 cos 60 o
tan 2.614 69
cos60 sin60
P mg
sin 60 cos60 mg
0.866 0.125 0.72mg
cos 60 sin 60 cos 9 sin 9
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6. Problem 2
Knowing that P = 110 N, determine the range of values value of q
for which equilibrium of the 8 kg block is maintained.
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7. Problem 2
y y
x x
mg mg
P cos N 0
P cos N 0
P sin mg f 0
P sin mg f 0
f N
f N
N P cos
N P cos
P sin mg N 0
P sin mg N 0
P sin mg P cos 0
P sin mg P cos 0
P sin cos mg
P sin cos mg
mg
mg P
P sin cos
sin cos
Hence
Hence
upward movement will not start before
downward movement will not start before
mg
mg P
P sin s cos
sin k cos
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8. Problem 3
The coefficients of friction are ms = 0.40 and mk = 0.30 between all
the surfaces of contact. Determine the force P for which
motion of the 27 kg block is impending if cable
a) is attached as shown,
b) is removed
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9. Problem 3
v
T N1
f1
m1g f1
N1 T
m2g f2
N2
T N1 0
N 1 m1 g 0
T N 1 , N 1 m1 g
T m1 g
T N1 N 2 P 0
N 2 N 1 m2 g 0
N 2 N 1 m 2 g 0 N 2 m1 m 2 g
T N1 N 2 P 0
m 1 g m1 g m1 m 2 g P 0
P 3 m1 g m 2 g
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10. Problem 3
v
0 N1
f1
m1g f1
N1 0
m2g f2
N2
T N 1 m1 a
Now T 0 N 1 m1 a
N 1 m1 g 0
N 1 m1 g
N1 N2 P 0
N 2 N 1 m2 g 0
N 2 N 1 m 2 g 0 N 2 m1 m2 g
N1 N2 P 0
m1 g m 1 m 2 g P 0
P 2 m1 g m2 g
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11. Additional Problems
The 8 kg block A and the 16 kg block B are at rest on an
incline as shown. Knowing that the coefficient of static
friction is 0.25 between all surfaces of contact, determine the
value of q for which motion is impending.
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12. Toppling
The magnitude of the force P is slowly increased.
Does the homogeneous box of mass m slip or tip
first? State the value of P which would cause each
occurrence.
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13. Slip or topple?
mg
f1 f2
N1 N2
P cos 30 N 1 N 2 0
P sin 30 N 1 N 2 mg 0
P cos 30 d mg d N 2 2d 0
P cos 30 N 1 N 2 N 1 N 2
P sin 30 N 1 N 2 mg
P cos 30 mg 2 N 2
P sin 30 N 1 N 2 mg P cos 30 mg
P sin 30 P cos 30 mg P sin 30 cos 30 mg
mg
P
sin 30 cos 30
0.5mg
P 0.448mg
0.25 0.866
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14. Slip or topple?
mg
f1 f2
N1 N2
P cos 30 f 2 0
P sin 30 N 2 mg 0
P cos 30 d P sin 30 2d mg d 0
P cos 30 f 2
P sin 30 N 2 mg
P cos 30 2P sin 30 mg 0
P cos 30 2 sin 30 mg
mg
P
cos 30 2 sin 30
mg
P 0.536mg
0.866 1
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18. Wedge
f=mN
P
q
N mg
P f cos N sin 0
f sin N cos mg 0
f N
mg
N sin N cos mg 0 N
sin cos
P N cos N sin 0
cos sin
P cos sin N mg
sin cos
P tan
mg tan 1
P tan tan
tan
mg tan tan 1
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19. A screw thread is a wedge
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20. A screw thread is a wedge
M=Pa
Q=Pa/r
= equivalent force
W
Q
N
f=mN
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21. A screw thread is a wedge
W
Q
N
q f=mN
Q f cos N sin 0
f sin N cos W 0
f N
W
N sin N cos W 0 N
sin cos
Q N cos N sin 0
cos sin
Q cos sin N W
sin cos
Q tan
W 1 tan Pa tan
Wr 1 tan
1 P
if tan
W
Screw locks
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22. A screw thread is a wedge
The
W
P
N
q f=mN
1 Pa
tan Screw locks
Wr
1 Wr
0
tan Pa
There is a critical value of friction
coefficient beyond which the thread
does not move irrespective of the
force applied.
This happens when a screw is not
maintained properly. Because of dirt
and rust m becomes more than
critical.
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23. A screw thread is a wedge
W
N
f=mN
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24. A screw thread is a wedge
W
N
q f
For no movement
f cos N sin 0
f sin N cos W 0
Self locking
f sin
N cos
sin
tan
cos
Therefore after raising the load if we
let go of the screw the load will not
cause the screw to unscrew by itself.
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25. Terminologies
Lead (L)
2pr
Pitch (p)
Lead L np
where
n=no. of parallely running threads = starts
L
tan =
2 r
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26. Turnbuckle
T1 T2
Used to apply tension.
The sleeve is rotated to pull the
threads together.
M tan
T2 T1 r 1 tan
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27. An improved screw jack
W
q
q
W
T
T T
T
2T cos q
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28. An improved screw jack
W
f
f
M=Pa
W=2T cos f
Pa cos sin
Wr sin cos
M cos sin
2Tr cos sin cos
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29. Worm gear
MG
M0/R
R N
M f
MG
M G WR W
R
Pa tan
Wr 1 tan
M tan
MG 1 tan
r
R
MR tan
M G r 1 tan
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30. Belt drives
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31. Belt drives
F x
0 T cos F T T cos 0
2 2
Fy 0 T sin 2 N T T sin 2 0
F s N
T cos F T T cos 0
2 2
T cos F T cos T cos 0
2 2 2
F T cos 0 T cos F s N
2 2
T sin N T T sin 0
2 2
T sin N T sin T sin 0
2 2 2
2T sin N T sin 0
2 2
2T sin N 0 2T sin N
2 2
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32. Belt drives
T cos s N
2
2T sin N
2
T cos 2T s sin
2 2
sin
1 T 2
cos s
2T 2
sin
1 T 2
Lim cos Lim
0 ,T 0 2T 2 0 ,T 0 s
1 dT
1 s
2T d 2
dT d
T s
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33. Belt drives
Belt is just about to
slide to the right
dT d
T s
T2
dT
s d T2
e s T e s T
T T 10 2 1
T2
T1
lnT
T1
s 0
Torque required to drive the pulley
lnT lnT2 1
s
T2 T1 e s 1 T
ln T
2
1
T 1
s
T
2 es
T
1
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34. Belt drives : Important points
T
2 es
T
1
Angle b must be expressed in radians
Smallest mb determines which pulley slips
first
Larger tension occurs at that end of the
belt where relative motion is about to begin or
is already moving
T2 is used to denote the larger tension
A freely rotating pulley implies no friction
For a rotating pulley where slipping is
about to start friction is ms since relative
velocity between belt and pulley is zero.
Once slipping starts friction coefficient is
dynamic or kinetic i.e. mk
If pulley does not rotate at all then rope
has to slide and not slip, hence friction is mk
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35. Belt drives
A B
q
q
A B
q
q
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36. Belt drives
T2
T2
exp 2 ???
T1
A Always check
for mb value for
each pulley in a
system. The one
with the
T1
smallest mb
q
value will
determine the
Which expression is tensions.
correct???? Is T2>T1
Or T1>T2.
One pulley must slip. T2
Friction force is larger
for the larger pulley since
angle of wrap is larger.
Hence smaller pulley
B
slips and determines the
tension
T1 q
exp 2 ???
T1
T2
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37. Band brakes
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38. Band brakes
TB T1 T3
B
T1 T3
75
T2 T1 exp
T1 exp 0.25
T
75 4
T2
180 180
T2 1.39T1
135 135
T3 T4 exp T4 exp 0.25
180 180
T4 0.55T3
M 0 T3 T4 T2 T1 R
Consider the pin B
F x
0 T1 cos 45 T3 cos 45 T1 T3
F y
0 T1 cos 45 T3 cos 45 TB 2T1 cos 45 TB 2T1 TB
T2 1.39T1 ,T3 T1 ,T4 0.55T1
Thus the largest tension is T2 5.6 T1 T3 4.03,T4 2.22,TB 5.7
M 0 5.6 2.22 R 3.38 0.16 0.54 KNm 540 Nm
Taking moments about D
50TB 250 P P 0.2TB 1.14KN
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