copyright © 2010 pearson education south asia pte ltd week 8: friction the best application of...

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.


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


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”


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


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


∑ 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


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


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


BACK TO BASIC

ʃ4x7 dx = _4x8_ + c

8

ʃ 2 dx = 2_ ʃ _x-5_ = _2 x-4_ + c

3x5 3 3 * -4



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


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


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


=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


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


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


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=


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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