Chap 5 Friction

8/13/2019 Chap 5 Friction

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VECTOR MECHANICS FOR ENGINEERS:

STATICSCHAPTER

8Friction


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Vector Mechanics for Engineers: Statics

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Contents

Introduction

Laws of Dry Friction. Coefficients

of Friction.

Angles of Friction

Problems Involving Dry Friction

Sample Problem 8.1

Sample Problem 8.3

Wedges

Square-Threaded Screws

Sample Problem 8.5

Journal Bearings. Axle Friction.

Thrust Bearings. Disk Friction.

Wheel Friction. Rolling Resistance.

Sample Problem 8.6

Belt Friction.

Sample Problem 8.8


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Introduction In preceding chapters, it was assumed that surfaces in contact were

eitherfrictionless(surfaces could move freely with respect to each

other) or rough(tangential forces prevent relative motion between

surfaces).

Actually, no perfectly frictionless surface exists. For two surfaces

in contact, tangential forces, calledfriction 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.

The distinction between frictionless and rough is, therefore, a matter

of degree.

There are two types of friction: dryor Coulomb frictionandfluid

friction. Fluid friction applies to lubricated mechanisms. The

present discussion is limited to dry friction between nonlubricated

surfaces.


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The Laws of Dry Friction. Coefficients of Friction

Block of weight Wplaced on horizontal

surface. Forces acting on block are its weightand reaction of surfaceN.

Small horizontal forcePapplied to block. For

block to remain stationary, in equilibrium, a

horizontal componentFof the surface reaction

is required. F is astatic-friction force.

AsPincreases, the static-friction forceF

increases as well until it reaches a maximum

valueFm.

NF sm

Further increase in P causes the block to begin

to move asFdrops to a smaller kinetic-friction

force Fk.

NF kk


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Maximum static-friction force:

NF sm

Kinetic-friction force:

sk

kk NF

75.0

Maximum static-friction force and kinetic-

friction force are:

- proportional to normal force

- dependent on type and condition ofcontact surfaces

- independent of contact area


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Four situations can occur when a rigid body is in contact with

a horizontal surface:

No friction,(Px= 0)

No motion,(PxFm)


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Angles of Friction

It is sometimes convenient to replace normal force

Nand friction forceFby 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


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Angles of Friction

Consider block of weight Wresting on board with

variable inclination angle q.

No friction No motion Motion

impending

Motion

V M h i f E i S i


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Problems Involving Dry Friction

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

V t M h i f E i St ti


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Sample Problem 8.1

A 100 lb force acts as shown on a 300 lb

block placed on an inclined plane. The

coefficients of friction between the block

and plane are s= 0.25 and k= 0.20.

Determine whether the block is in

equilibrium and find the value of the

friction force.

SOLUTION:

Determine values of friction forceand normal reaction force from plane

required to maintain equilibrium.

Calculate maximum friction force

and compare with friction force

required for equilibrium. If it isgreater, block will not slide.

If maximum friction force is less

than friction force required for

equilibrium, block will slide.Calculate kinetic-friction force.



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Sample Problem 8.1SOLUTION:

Determine values of friction force and normalreaction force from plane required to maintain

equilibrium.

:0 xF 0lb300-lb100 53 F

lb80F

:0 yF 0lb300- 54 N

lb240N

Calculate maximum friction force and compare

with friction force required for equilibrium. If it isgreater, block will not slide.

lb60lb24025.0 msm FNF

The block will slide down the plane.



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Sample Problem 8.1

If maximum friction force is less than friction

force required for equilibrium, block will slide.Calculate kinetic-friction force.

lb24020.0

NFF kkactual

lb48actualF



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Sample Problem 8.3

The moveable bracket shown may be

placed at any height on the 3-in.diameter pipe. If the coefficient of

friction between the pipe and bracket is

0.25, determine the minimum distance

xat which the load can be supported.

Neglect the weight of the bracket.

SOLUTION:

When Wis placed at minimumx, the

bracket is about to slip and friction

forces in upper and lower collars are at

maximum value.

Apply conditions for static equilibriumto find minimumx.



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Sample Problem 8.3SOLUTION:

When Wis placed at minimumx, the bracket is about toslip and friction forces in upper and lower collars are at

maximum value.

BBsB

AAsA

NNF

NNF

25.0

25.0

Apply conditions for static equilibrium to find minimumx.

:0 xF 0 AB NN AB NN

:0 yF

WN

WNN

WFF

A

BA

BA

5.0

025.025.0

0

WNN BA 2

:0 BM

05.1275.026

05.125.036

0in.5.1in.3in.6

xWWW

xWNN

xWFN

AA

AA

in.12x



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Wedges

Wedges- simple

machines used to raise

heavy loads.

Force required to lift

block is significantly

less than block weight.

Friction prevents wedge

from sliding out.

Want to find minimum

forcePto raise block.

Block as free-body

0

:0

0

:0

21

21

NNW

F

NN

F

s

y

s

x

or

021 WRR

06sin6cos

:0

0

6sin6cos

:0

32

32

s

y

ss

x

NNF

P

NN

F

Wedge as free-body

or

032 RRP

Chap 5 Friction

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