Seely F.B. Analytical Mechanics for Engineers

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CONE

FRICTION

131

Graphical

Solution.

Since motion

impends,

the

angle

between the

reac-

tion

and the normal

pressure

of the

shaft

at A is

equal

to the

angle

friction;

that

is,

the

tangent

of the

angle

equal

0.2.

Hence, by

laying

off ten

spaces along

the normal

and

two

spaces

perpendicular

the

normal,

shown

Fig.

154,

the

action

line

determined.

In a

similar manner

the

action

line of R

is found.

Now

the three

forces

RI,

and

W must be

concurrent in

order

to be in

equilibrium

(Art.

48)

and

hence,

W must

pass

through

the intersection

and R

. Now

the

intersection

of these two

forces

can never be nearer

to the

shaft than the

point

since

the

angle

</>

FIG. 154.

cannot

greater

than tan-

. 2.

The

distance,

from the

shaft

found

measurement to be 24

in.

Algebraic

Solution.

The five

forces,

F\,

NI,

and

which

hold

the

lift in

equilibrium

shown

in the

free-body diagram

(Fig.

154)

form

coplanar

non-concurrent force

system

and

hence

there

are three

equations

equi-

librium

as follows

-Ni=0

(1)

=-TF

+0.2^+0.2^2=0.

(2)

which

=-Wx+lQNi

-2X0.2^=0,

....

(3)

0.2ATi

F'i

and

Q.2N

From

(1)

and

(2)

have,

fr-0.42VY-0.4tf,.

By substituting

this value of W in

(3)

the

equation

obtained

is,

whence,

z=24in.

132

FRICTION

PROBLEMS

134.

If the

weight

the

preceding

problem

is 500 Ib. what vertical

force

applied

E will

just

sufficient

to start

the lift

up?

135.

The coefficient

friction

for the

100-lb.

body

and

the table

(Fig.

155)

0.4.

What

weight,

will cause

motion

of the

three bodies to

impend?

The

pulleys

are

assumed

be frictionless

and

the cords

flexible.

Ans.

471b.

=10

lb.

FIG.

155.

136.

The ladder

shown

Fig.

156

weighs

lb.

The coefficient

friction

for

the ladder

and

vertical

wall

0.25,

and

for the

floor and

ladder

0.5.

Find

the

horizontal

force

which

will cause

motion

impend

the

right.

Ans.

51 lb.

-c

J><]

2ft.

FIG.

157.

FIG. 158.

137.

rectangular

block

wood

in.

10 in.

in.

stands on

end

horizontal

floor.

The

block is

acted on

horizontal

force

applied

through

the

center

the

top

face and

perpendicular

to one

side. The

block

ANGLE

REPOSE

133

weighs

Ib. and

the coefficient

of friction

0.20

If the

force is

gradually

increased

until motion

ensues will

the block slide

tip?

What is

the

mag-

nitude

of the force

when

the

block starts to

move?

138.

crown friction drive as

indicated

Fig.

157 is used on

screw

power presses,

motor

trucks,

etc.

The

cast

iron disk B

rotates at

1000

r.p.m.

and

drives the crown wheel

which

is faced with

leather-fiber.

The diameter of

in. and the value

is 0.35. If

turning

moment

100

ft.-lb.

is transmitted

to the

crown wheel

shaft when

slipping impends,

what

is the

normal

pressure

between

the

disk

and

the

crown

wheel?

What is

the

pressure

the

bearings

at A

and

139.

load,

of 80

Ib.

applied

to the

arm

(Fig.

158)

at a

distance

ft.

from the

axis

of the

shaft.

0.15

will

the

arm

slide on the

shaft?

140.

A cone clutch

as shown

Fig.

159

is used on

automobiles

connect

the

motor

with the

transmission.

If the

normal

pressure

between the

two

FIG. 159.

surfaces of

contact,

maintained

spring,

Ib.

per

square

inch

and

the

coefficient

of friction is

0.25,

what

the

maximum

torque

the

clutch

can transmit?

Assume

that the

frictional

force

has a

mean arm

14^

in.

What

spring

pressure,

required

produce

the

normal

pressure

Ib.

per square

inch?

Ans. T

1235

lb.-ft.;

887 Ib.

67.

Angle

Repose.

body

rests on an

inclined

plane,

as shown

Fig. 160,

and is

acted

forces

except

its

weight

and

the

reaction

of the

plane,

and if

the

angle

inclination

the

plane

to the

horizontal,

such

that

motion

the

body

134 FRICTION

impends

down

the

plane,

the

angle

a is

defined

the

angle

repose.

Since

the

body

is in

equilibrium

under

the

action

of the

two

forces

the

reaction

of the

plane,

and

its

weight,

these

forces

must be

equal,

opposite,

and

collinear.

Hence the

reaction

vertical.

Further-

more,

the

angle

which

R makes with the

normal

to the

plane

the

angle

of fric-

tion. It is evident from

the

figure

that

the

angles

and

</>

are

equal.

The

angle

repose

for two surfaces

can be found eas-

ily by experiment,

after

which the coeffi-

cient friction

for

the

surfaces

may

be found

from the relation

IJL

tan

68. The

Laws of Friction.

One

the earliest contributions

to our

knowledge

of the laws of friction

was

made

by Coulomb,

who

published,

1781,

the results of

experiments

on the friction of

plane

dry

surfaces.

Later

experiments

Morin

confirmed,

the

main,

the results obtained

Coulomb. The

results

of the

experiments

of Morin on

dry

surfaces,

published

1831,

may

stated as

follows :

1. The friction between

two

bodies when

motion

impending

(limiting

friction)

proportional

the

normal

pressure;

that

is,

the coefficient of friction

independent

of the normal

pressure.

2. The coefficient of static

friction

independent

of the area

of contact.

The coefficient of kinetic friction is

less than

the coefficient

of static friction

and is

independent

of the

relative

velocity

the

rubbing

surfaces.

Although

these laws are

probably

correct for the

conditions

under which the tests were

made,

they

must be modified

order

apply

to friction

which

developed

under conditions

quite

dif-

ferent

from those found in the

experiments.

The

pressures

used

in the

experiments

of Morin varied from

Ib.

per square

inch to

100 Ib.

per square

inch.

It has been found

later

experiments

that for

pressures

less than

Ib.

per

square

inch the value

of the

coefficient

of static friction increases

somewhat.

For

very

great

pressures

the coefficient also

increases. The

highest

velocity

used

Morin's

experiments

was 10

ft.

per

sec. For

greater

velocities than

this it has

been

found

in later

experiments

that

THE INCLINED PLANE

135

the coefficient

kinetic friction decreases with the

velocity.

The

experiments

Jenkin

show

that

for

extremely

low

velocities

(the

lowest

velocity

measured was

.0002

ft.

per sec.)

there is an

increase

in the coefficient

of kinetic

friction.

These

experiments

indicate

that the value

the

coefficient of kinetic

friction

grad-

ually

increases as the

velocity

decreases

and

passes

without dis-

continuity

into that

of static

friction.

From

experiments

made

Tower,

Goodman, Thurston,

and

others,

on lubricated

surfaces,

has been

found that the

laws of

friction

for lubricated surfaces are almost the

reverse of those

stated

for

dry

surfaces. For

example,

it is found that

the friction

two

surfaces is almost

independent

of the

nature of the surfaces

and of the

normal

pressure

long

as there is

film

of lubricant

between

the surfaces.

Again,

for lubricated

surfaces,

it is found

that the friction is

materially

affected

the

temperature,

which

not true

the case of

dry

surfaces.

69.

The Inclined Plane.

Applications

of the

inclined

plane

are found

various machines such

the

jack-screw,

the screw

type

testing machine,

wedges,

etc.

Three

variations of the

problem

which arise

connection with the

use of the inclined

plane

will now

considered.

CASE

I. To Determine

the Force

Required

Start a

Body

the

Plane. Let

Fig.

161

represent

body

resting

on a

plane,

the

angle

of inclination

of which

It is

required

find the

value of a

force,

which will

make motion of the

body

im-

pend

the

plane

when

the

action

line of

makes an

angle

with

the

plane.

FIG

161

Although

motion

of the

body

impends,

will not

occur unless the value of

increased

slightly beyond

the

critical

value here

considered.

And

since

the

body

still

equilibrium

the

equations

equilibrium

may

applied

to the

system

of forces

acting

it.

addition

the

required

force

the

forces

acting

on the

body

are

the

weight, W,

the

body,

and

the

reaction

of the

plane.

The

latter force

will

resolved

into

components

N, perpendicular

to the

plane,

and

(equal

uN), parallel

the

plane,

shown

the

136 FRICTION

free-body

diagram.

If the

z-axis be

taken

parallel

to the

plane

and the

y-axis

perpendicular

the

plane,

the

equations

equi-

librium

may

written

follows :

cos

0-nN-W

sin

N+P

sin

B-W

cos

Eliminating

N from

the two

equations

have:

p_TF

(sin

a+M

cos

6+iJ,

sin

tan

substituted

for

JJL

this

may

written,

sin

(a+0)

cos

(00)"

the values

W, a,

and are

specified,

may

regarded

function

of 0. The

value

is a minimum when

equal

since this value of makes cos

(0 0)

maximum;

the

minimum

value of

P, then,

sin

(a+0).

If the force

applied paral-

lel to the

plane

its value

becomes

cos

CASE

II. To Determine

the Force

Required

Prevent Motion

Down

the

Plane when a

this case the forces

acting

on the

body

are

the same

as shown

Fig.

161

except

that the

frictional

force is directed

the

plane.

writing

the

equations

equilibrium

before

and

solving,

the

equation

obtained

is,

cos

(0+0)

For this case

is a

minimum

when

and the minimum value

sin

If the force

applied parallel

the

plane

W sin

(</>)

the value

cos

CASE

III.

To Determine

the Force

Required

Start

the

Body

Down the Plane when

a<<t>.

In this

case

the

forces which

act on

the

body

are as

shown in

Fig.

161, except

that

and

/zTV

are

reversed.

By writing

the

equilibrium

equations

as before and

solving,

the

equation

obtained

is,

cos

(0+0)

THE

INCLINED PLANE

137

For

this

case

P is a minimum when

and the

minimum

value

P is

sin

a).

HP is

applied parallel

to the

plane,

the

sin

(<f>

value

becomes

cos

<f>

ILLUSTRATIVE

PROBLEM

141.

A small

body weighing

50 Ib.

placed

on a

rough plane

which

is in-

clined

30 with

the

horizontal

(Fig.

162).

The

body

acted on

a force

the action

line of

which

lies in

the

plane

and makes an

angle

of 30

with

the

line of

greatest

slope

in the

plane.

If the

coefficient of friction is

find

the

value of

P that will

just

start

the

block in

motion,

and find the direction

which

the

block will

begin

move.

Solution.

There are

three

forces

acting

on the

body,

namely,

the force

the earth

pull

Ib.,

and the

reaction of

the

plane.

The 50-lb. force

may

resolved

into two

components;

50 sin

the

plane (Fig.

162)

and 50

cos

(not

shown)

perpendicular

to the

plane.

The reaction of

the

plane may

resolved into two

components,

(not shown)

perpendicular

the

plane

and

lying

in the

plane.

The

frictional force N acts

opposite

to the

direction

which

the

body

will

begin

move.

Let

the

angle

which

the

frictional

force

makes

with

the

line

greatest slope.

The

two

forces which

are

perpendicular

to the

plane

are

equilibrium

and

hence

are

equal.

Thus,

50cos30

(1)

The three forces

in the

plane

are in

equilibrium; hence, by

Lami's

theorem

(Art.

47),

sin 30

sin

(0+30)

sin

sin 30'

eliminating

N from

(1)

and

(2)

the

equation

obtained

is,

in(

+30)=?^i|?!

Hence,

0+30

60.

Therefore,

Substituting

(2)

have,

FIG.

162.

=^-,

2V3

PROBLEMS

142.

body weighing

W Ib.

rests on

rough plane

inclined at an

angle

the

horizontal.

What

horizontal force

must be

applied

order

start

the

body

the

plane

the

angle

of friction

Express

the force

terms

<t>

and also in terms

ju.

cos u

sin

138

FRICTION

P-120

lb.

143.

homogeneous

rectangular prism,

the dimensions of which

are 1

ft.

ft.,

stands on end on a

flat

square

board,

the

edges

of the base

of the

prism

being

parallel

to the

edges

of the

board. The

coefficient

of friction

for the board and

prism

0.2. If one

side of the board

gradually

raised

will the

prism

slide or

tip?

W-200

lb.

FIG. 163.

from

moving

down the

the

coefficient

friction

for the

body

and

plane.

Ans.

145. Two bodies

weighing

50 lb.

and 100 lb. rest on

an inclined

plane

and are

connected

cord

which is

parallel

to the line of

greatest

slope.

The

body weighing

lb.

below

the one

weighing

100

lb.,

and the

coefficient of friction

for the 50-lb.

body

and that for the

100-lb.

body-

Find the

inclination

of the

plane

to the horizontal and the tension

the

cord,

when

motion

impends.

146.

body weighing

200

lb. rests

on a

plane

inclined

to the hori-

zontal and

acted

force

120

lb. as

shown

Fig.

163. The

coefficient

friction for the

body

and

plane

is 0.3.

Find the

friction between

the

body

and

plane.

Ans. 3.92

lb.

70.

The

Wedge. Wedges

are

used

for

lifting-devices,

cotter-

pins,

keys,

etc. In the

wedge

shown

Fig.

164

(a),

it is

required

to find

the value of

the

force P

which

must

applied

the

wedge

overcome the forces

W, applied

blocks

and B.

It will

assumed that

the

angle

of friction for all

rubbing

surfaces

the same.

free-body

diagram

the block

and

of the

wedge

144.

body weighing

100

lb.

rests on

rough

plane

inclined

45 to the

horizontal.

horizontal

force of 50 lb.

just

sufficient to

prevent

the

body

plane.

Find

FIG. 164.

THE

WEDGE

139

C is shown

Fig.

164(6)

and

164

(d), respectively.

The

force P

may

be found

graphically by

constructing

the

force

polygons

for

the

forces

acting

and C. Since

the force

known com-

pletely

and since the directions of the reactions

(Ri

and

R%)

of D

and C on

are known

direction,

the

force

polygon

for

can

be drawn

as shown

Fig.

164(c).

Having

found the

reaction

R%,

now

possible

complete

the force

polygon

for

the forces

acting

the

wedge

C. The reaction of

A on C is

equal

and

oppo-

site to the reaction of C on

that

is,

it is

equal

the

force

just found,

but reversed

sense. The force

polygon

for C

shown in

Fig. 164(e)

from

which

may

found.

Fig.

164(/)

shows the two force

polygons superimposed.

From the force

polygons

it is

easy,

means of

trigonometry,

derive

the

expres-

sion

for P in terms of

<f>,

and a. It

will

be of

interest,

however,

to find

the value of

use of the

equations

equilibrium.

Thus,

the

equations

equilibrium

for the

forces

acting

on A

and

respectively,

are,

in^-lfe

cos

(+0)

....

(1)

For

cos

0-^2

sin

(a+0)

......

(2)

cos

(+</>)

-#3

cos

(+<)=

....

(3)

sin

(+</)

-P

0. ...

(4)

eliminating

from

(1)

and

(2)

the

equation

obtained

is,

cos

By eliminating

from

(3)

and

(4)

the

equation

obtained

is,

sin

(a+0)

........

(6)

eliminating

from

(5)

and

(6)

the final

equation

is,

p_2TF

cos

<f>

sin

(a+<j>)

cos

(a+20)

Another

problem

which arises in

connection with

the

wedge

the

determination

of the least

force,

which

will

prevent

the

wedge

from

being

forced

upward

the

loads

W. The

pro-

cedure in

the solution

of this

problem

similar to

that

used in

the

problem

just

discussed.

The

frictional

forces,

however,

will

reversed and hence the

reactions

will

make

angles

the other

140

FRICTION

sides

of the normals to the surfaces.

The

value of P

for

this case

given

the

equation,

p_2W

cos sin

<ft)

cos

(a 20)

When

a is less than

<J>,

is evident

that the

value

negative

and

hence the

wedge

self-locking,

that

is,

it would

stay

place

even

if the force P

were reduced

to zero.

third

problem

which

arises

connection

with the

wedge

is the

determination

the least

pull, P,

which

required

withdraw a

self-locking wedge,

as for

example,

lowering

heavy body

means of the

wedge.

By writing

the

equations

equilibrium

and

solving,

the

value

the

pull

found

to be

_2TF

cos

sin

(<J>at)

cos a

ILLUSTRATIVE

PROBLEM

147.

Fig. 165(a)

represents

a cotter

joint.

The

angle

equals

and

the

angle

friction

for all

rubbing

surfaces is

12. What is the value of the force P re-

quired

overcome the

1000-lb. forces

applied

parts

and

J5?

Solution.

Free-body

diagrams

the

block

and

the

cotter

C are shown

Fig.

165(6)

and

165(c)

respectively.

The

equations

equilibrium

for the two

blocks

may

written as

follows:

1000

lb.

(a)

1000

lb.

For

n 12

-R*

cos

-Ri

sin

ForC

cos 27-fl

cos

(1)

(2)

(3)

(4)

eliminating

from

(1)

and

(2)

the

equation

obtained

is,

1000 cos

12'

cos 39

(5)