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8/11/2019 CH.6.2_Dry friction
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4/7/2013
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Dry Friction6
APPLIED Mechanics
MET152
MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
5.2 Problems Involving Dry Friction
2
Types of Friction Problems
Equilibrium
Total number of unknowns = Total number of available
equilibrium equations Frictional forces must satisfy F sN; otherwise,
slipping will occur and the body will not remain in
equilibrium
We must determine the frictional
forces at A and C to check
for equilibrium
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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5.2 Problems Involving Dry Friction
3
Equilibrium Versus Frictional Equations
Frictional force always acts so as to oppose the
relative motion or impede the motion of the body
over its contacting surface
Assume the sense of the frictional force that require F
to be an equilibrium force
Correct sense is made after solving the equilibrium
equations
If F is a negative scalar, the sense of F is the reverse of
that assumed
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Example 2
4
The uniform crate has a mass of 20kg. If a force P = 80N
is applied on to the crate, determine if it remains in
equilibrium. The coefficient of static friction is = 0.3.
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Solution
5
Resultant normal force NC act a distance x from the crates
center line in order to counteract the tipping effect caused
by P.
3 unknowns to be determined by 3 equations of equilibrium.
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Solution
6
Solving
mmmxNNF
xNmNmN
M
NNN
F
FN
F
NC
C
O
C
y
x
08.900908.0,3.69
0)()2.0(30cos80)4.0(30sin80
;0
02.19630sin80
;0
030cos80
;0
236
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Solution
7
Since x is negative, the resultant force acts (slightly) to theleft of the crates center line.
No tipping will occur since x 0.4m
Max frictional force which can be developed at the surfaceof contact
Fmax = sNC = 0.3(236N) = 70.8N
Since F = 69.3N < 70.8N, the crate will not slip though itis close to doing so.
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
AnglesofFriction
It is sometimes convenient to replace normal force
Nand friction force Fby their resultant R:
No friction Motion impending No motion
ss
sms
N
N
N
F
tan
tan
Motion
kk
kkk
N
N
N
F
tan
tan
4/7/2013 MET152 Applied Mechanics Sem.121/ by Dr. Miloud S. 8
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Angles
of
Friction Consider block of weightWresting on board with
variable inclination angle
No friction No motion Motion
impending
Motion
4/7/2013 MET152 Applied Mechanics Sem.121/ by Dr. Miloud S. 9
ProblemsInvolvingDryFriction
All applied forces known
Coefficient of static friction
is known
Determine whether body
will remain at rest or slide
All applied forces known
Motion is impending
Determine value of coefficient
of static friction.
Coefficient of static
friction is known
Motion is impending
Determine magnitude or
direction of one of the
applied forces
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6.3 Frictional Forces on Screws
11
Screws used as fasteners
Sometimes used to transmit power or motion from
one part of the machine to another
A square-ended screw is commonly used for the
latter purpose, especially when large forces are
applied along its axis
A screw is thought as an inclined plane or wedgewrapped around a cylinder
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
12
A nut initially at A on the screw will move up to B
when rotated 360 around the screw
This rotation is equivalent to translating the nut up an
inclined plane of height l and length 2r, where r is the
mean radius of the head
Applyingtheforce equationsofequilibrium, we have
srWM tan
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
6.3 Frictional Forces on Screws
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Downward Screw Motion
If the surface of the screw is very slippery, the screw
may rotate downward if the magnitude of the moment
is reduced to say M < M
This causes the effect of M to become S
M = Wr tan( )
6.3 Frictional Forces on Screws
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Example 3
14
The turnbuckle has a square thread with a mean radius of
5mm and a lead of 2mm. If the coefficient of static friction
between the screw and the turnbuckle is s = 0.25,
determine the moment M that must be applied to draw
the end screws closer together. Is the turnbuckle self-
locking?
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Solution
15
Since friction at two screws must be overcome, this
requires
Solving
When the moment is removed, the turnbuckle will be
self-locking
64.352/2tan2/tan
04.1425.0tantan,5,2000
tan2
11
11
mmmmr
mmrNW
WrM
ss
mNmmN
mmNM
.37.6.7.6374
64.304.14tan520002
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
5.4 Frictional Forces on Flat Belts
16
It is necessary to determine the frictional forces
developed between the contacting surfaces
Consider the flat belt which passes over a fixed curved
surface
Obviously T2 > T1
Consider FBD of the belt
segment in contact with the surface
N and F vary both in
magnitude and direction
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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5.4 Frictional Forces on Flat Belts
17
Consider FBD of an element having a length ds
Assuming either impending motion or motion of the
belt, the magnitude of the frictional force
dF = dN
Applying equilibrium equations
02
sin2
sin)(
;0
02
cos)(2
cos
;0
dT
ddTTdN
F
ddTTdNdT
F
y
x
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
5.4 Frictional Forces on Flat Belts
18
We have
eTT
T
TIn
dT
dT
TTTT
dT
dT
TddN
dTdN
T
T
12
1
2
0
21
2
1
,,0,
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Example 4
19
The maximum tension that can be developed In the cord
is 500N. If the pulley at A is free to rotate and the
coefficient of static friction at fixed drums B and C is s =
0.25, determine the largest mass of cylinder that can be
lifted by the cord. Assume that the force F applied at the
end of the cord is directed vertically downward.
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Example 5
20
Weight of W = mg causes the cord to move CCW over
the drums at B and C.
Max tension T2 in the cord occur at D where T2 = 500N
For section of the cord passing over the drum at B180 = rad, angle of contact between drum and cord
= (135/180) = 3/4 rad
NN
e
NT
eTN
eTT s
4.27780.1
500500
500
;
4/325.01
4/325.0
1
12
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Example 5
21
For section of the cord passing over the drum at C
W < 277.4N
kgsm
N
g
Wm
NW
We
eTT s
7.15/81.9
9.153
9.153
4.277
;
2
4/325.0
12
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Example 6
22
The 100mm diameter pulley fits loosely on a 10mm
diameter shaft for which the coefficient of static friction is
s = 0.4. Determine the minimum tension T in the belt
needed to (a) raise the 100kg block and (b) lower the
block. Assume that no slipping occurs between the beltand the pulley and neglect the weight of the pulley.
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Example 6
23
Part (a)
FBD of the pulley is shown.
As tension T is increased, the pulley will roll around the
shaft to point before motion P2 impends.
Friction circles radius, rf= r sins.
Using the simplification,
8.204.0tanand06.11063
0)48()52(981;0
2)4.0)(5(
)(tansin
1
2
s
P
sf
sss
kNNT
mmTmmNM
mmmmrr
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
Example 6
24
Part (a)
For radius of friction circle,
Therefore,
kNNT
mmmmTmmmmN
M
mmrr
P
sf
06.11057
0)86.150()86.150(981
;0
86.18.21sin5sin
2
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.
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Example 6
25
Part (b)
When the block is lowered, the resultant force R acting
on the shaft passes through the point P3.
Summing moments about this point,
NTmmTmmN
MP
9060)52()48(981
;03
4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.