copyright © 2010 pearson education south asia pte ltd week 8: friction the best application of...
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Copyright © 2010 Pearson Education South Asia Pte Ltd
WEEK 8: FRICTION
THE BEST APPLICATION OF
FRICTION
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.
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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
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.
• 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
x
The Laws of Dry Friction. Coefficients of Friction
NFm
• 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
NFsm
• Further increase in P causes the block to begin to move, & F drops to a smaller kinetic-friction force Fk.
NFkk
No motion
Fm
Fk
x
s Coefficient of static friction
k Coefficient of kinetic friction
s
k> sk
75.0
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Characteristics of Dry Friction
Theory of Dry Friction
Typical Values of μs
- you can used/assumed
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
sk 75.0
Important relationship – you can assume but must be technically valid/sound
Statement: for assumption, but must be technically sound/valid
• “Assume that the coefficient of friction between wood and wood is µs = 0.7”
Or
• “Assuming that the µk = 0.7µs”
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or any value you put in, but must be realistic based on sound understanding of friction
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 kgWeight = 20 x 9.81N = 196.2 N
W = 196.2 N
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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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∑ Fx = 0; 80 cos 30o N – F = 0
F = 69.3 N
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
Solving
Solution
∑ Fy = 0; -80 sin 30o N + Nc – 196.2 N = O
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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)
Solution
x
G
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To find the max frictional force which can be developed at the surface of contact:
Fmax = μsNC = 0.3(236N) = 70.8N
Since F = 69.3N < 70.8N, the crate will not slip BUT it is close to doing so.
Solution
x
G
F=69.3 N
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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Copyright © 2010 Pearson Education South Asia Pte Ltd
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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dTdN
dTdN
ddTTdNT
ddTTdN
dT
Fx
0
02
cos1*1*
02
cos)(2
cos
;0=1
=1
____
____
When smallCos( /2) = 1
dd
=1
=1
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 dNApplying equilibrium equations
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=negelct, 0
TddN
dTdN
dT
ddT
dTdN
dT
ddTTdN
Fy
002*2
0222
02
sin2
sin)(
;0
When smallSin( /2) =( /2)
dd d
The product of infinitesimal size dT and - neglected
2
d
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8.5 Frictional Forces on Flat Belts
• We have
eTT
T
TIn
dT
dT
TTTT
dT
dT
TddN
dTdN
T
T
12
1
2
0
21
2
1
,,0,
…….1
……..2
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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
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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
NN
e
NT
eTN
eTT s
4.27780.1
500500
500
;
4/325.01
4/325.01
12
Pulling F=
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Example 8.8
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
8 - 23
Sample Problem
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.
SOLUTION:
• Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B.
• Taking pulley A as a free-body, sum moments about pulley center to determine torque.
8 - 24
Sample Problem 8.8
30o
8 - 25
Sample Problem
SOLUTION:
• Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B.
N16001.688
N 2700
688.1N2700
1
3225.0
11
2
T
eT
eT
Ts
• Taking pulley A as free-body, sum moments about pulley center to determine torque.
0N2700N1600mm200:0 AA MM
mN220 AM
THANK YOU
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