Transcript lect wk9 friction

WEEK 8: FRICTION
THE BEST APPLICATION OF
FRICTION
Copyright © 2010 Pearson Education South Asia Pte Ltd
FRICTION - Introduction
• 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.
• There are two types of friction:
i. dry or Coulomb friction (C. A Coulomb 1781) and – OUR
FOCUS
ii. fluid friction - fluid friction applies to lubricated mechanisms.
Characteristics of Dry Friction
• Coulomb friction occurs between contacting
surfaces of bodies in the absence of a
lubricating fluid
• Fluid friction exist when the contacting
surface are separated by a film of fluid (gas or
liquid)
Depends on velocity of the fluid and its ability
to resist shear force
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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 – NOT MOVING.
x
• 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.
N, a distance x to the right of W to balance the “tipping
effect” of P
The Laws of Dry Friction. Coefficients of Friction
x
• As P increases, the static-friction force F increases as
well until it reaches a maximum value Fm proportional to
N. Fm is called limiting static frictional force
F N
F N

m
m
No motion
s
s
Coefficient of static friction
• Further increase in P causes the block to begin to
move, & F drops to a smaller kinetic-friction force Fk.
Fm
F  N
Fk


k
k
s
k
Coefficient of kinetic friction
>

k
  0.75
k
s
Characteristics of Dry Friction
Important relationship –
you can assume but
must be technically
valid/sound
Theory of Dry Friction
Typical Values of μs
- you can used/assumed
  0.75
k
s
Contact
Materials
Coefficient of
Static Friction μs
Metal on ice
0.03 – 0.05
Wood on wood
0.30 – 0.70
Leather on wood
0.20 – 0.50
Leather on metal
0.30 – 0.60
Aluminum on
aluminum
1.10 – 1.70
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Statement: for assumption, but must be
technically sound/valid
• “Assume that the coefficient of friction between wood
and wood is µs = 0.7”
or any value you put in, but must be
realistic based on sound
Or
understanding of friction
• “Assuming that the µk = 0.7µs”
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example
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 μ = 0.3.
Mass = 20 kg
Weight = 20 x 9.81N
= 196.2 N
W = 196.2 N
Solution
Resultant normal force NC act a distance x from the
crate’s center line in order to counteract the tipping effect
caused by P.
3 unknowns to be determined by 3 equations of
equilibrium.
FBD
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Solution
∑ Fx = 0; 80 cos 30o N – F = 0
F = 69.3 N
∑ Fy = 0; -80 sin 30o N + Nc – 196.2 N = O
Nc = 236 N
Taking moment:
∑ Mo = 0; 80 sin 30o (0.4 m) - 80 cos 30o (0.2 m) + Nc (x) = O
Substituting Nc = 236 N, x = - 9.08 mm
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Solution
G
x
Since x is negative, the resultant force Nc
acts 9.08 mm (slightly) to the left of the crate’s
center line.
No tipping will occur since x ≤ 0.4m
(location of center of gravity, G is:
x = 0.4 m; y = 0.2m)
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Solution
To find the max frictional force which can
be developed at the surface of contact:
Fmax = μsNC = 0.3(236N) = 70.8N
G
x
F=69.3 N
Since F = 69.3N < 70.8N, the crate will not slip
BUT it is close to doing so.
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BACK TO BASIC
ax = y, then logay = x
102 = 100, then log10100 = 2
Logaxy = Logax + Logay
Loga(x/y) = Logax - Logay
Logaxn = nLogax
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BACK TO BASIC
ʃ4x7 dx = _4x8_ + c
8
ʃ 2 dx = 2_ ʃ _x-5_ = _2 x-4_ + c
3x5
3
3 * -4
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Frictional Forces on Flat Belts
• In belt drive and band brake design it is necessary to determine the
frictional forces developed
between the belt and
contacting surfaces
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Frictional Forces on Flat Belts
• Consider the flat belt which passes over a fixed curved
surface – to find tension T2 to pull the belt
• Thefefore T2 > T1
• Total angle of contact β, coef of friction = 
• Consider FBD of the belt
segment in contact with the surface
• N and F vary both in
magnitude and direction
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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 = μ dN
Applying equilibrium equations
 F  0;
x
 d 
 d 
T cos
  dN  (T  dT ) cos
0
 2 
 2 
 d 
T *1  dN  T *1  dT cos
0
 2 
dN  dT  0
When d small
Cos(d /2) = 1
dN  dT
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 F  0;
y
 d 
 d 
When d small
dN  (T  dT ) sin 

T
sin

0



Sin(d /2) =( d /2)
 2 
 2 
d
d
d
dN  T
 dT
T
0
2
2
2
d
dN  T
*2  0  0
The product of infinitesimal
2
size dT and d - neglected
dN  Td
2
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8.5 Frictional Forces on Flat Belts
• We have
dN  dT
dN  Td
…….1
……..2
dT
 d
T
T  T1 ,   0, T  T2 ,   

dT
T1 T   0 d
T
In 2  
T1
T2
T2  T1e 
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Example
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 μs =
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.
=F
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 8.8
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 B
180° = π rad, angle of contact between drum and cord
β = (135°/180°)π = 3/4π rad
T2  T1e  s  ;
500 N  T1e 0.253 / 4  
Pulling F=
T1 
500 N
e 0.253 / 4  
500 N

 277.4 N
1.80
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Example 8.8
For section of the cord passing over the drum at C
W < 277.4N
T2  T1e  s  ;
277.4  We 0.253 / 4  
W  153.9 N
W
153.9 N
m

 15.7kg
2
g 9.81m / s
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Sample Problem
SOLUTION:
• Since angle of contact is smaller,
slippage will occur on pulley B first.
Determine belt tensions based on
pulley B.
A flat belt connects pulley A to pulley B.
The coefficients of friction are s = 0.25
and k = 0.20 between both pulleys and
the belt.
Knowing that the maximum allowable
tension in the belt is 2.7 kN, determine
the largest torque which can be exerted
by the belt on pulley A.
8 - 23
• Taking pulley A as a free-body, sum
moments about pulley center to
determine torque.
Sample Problem 8.8
30o
8 - 24
Sample Problem
SOLUTION:
• Since angle of contact is smaller, slippage
will occur on pulley B first. Determine belt
tensions based on pulley B.
T2
 e s
T1
T1 
2700 N
 e 0.252 3  1.688
T1
2700 N
 1600 N
1.688
• Taking pulley A as free-body, sum moments about
pulley center to determine torque.
MA  0:
M A  200 mm 1600 N  2700 N   0
M A  220 N  m
8 - 25
THANK YOU
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