Seely F.B. Analytical Mechanics for Engineers

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KINETICS

OF A TRANSLATING RIGID BODY 311

assumed

to be

weightless

and frictionless.

Body

weighs

lb.,

weighs

lb.,

and C

weighs

120 lb.

The

coefficient of kinetic

friction

for A

and

the

plane

and for B and

the

plane

Find

(1)

the acceleration

of the

bodies,

(2)

the

tension,

T\,

in the

cord

between

and

(3)

the

tension,

in the cord between

B and

Solution. Each

body

has a

motion of translation.

Hence,

the

equations

of motion

for

translating

body

apply

each

body; namely,

(2)

(3)

addition,

two

equations

involving

the coefficients

of friction for

bodies

and

are

needed, namely,

FA=INA,

(4)

(5)

Equation

(3)

not

needed in

determining

the

quantities

asked

for.

If the

action

lines of

and

were

desired,

however,

all

five

equations

would be

needed.

free-body diagram

for

body

shown

Fig.

334(a).

Applying

the

equations

motion

to A

have,

(c)

(6)

120

lb.

FIG.

334.

From

(1),

From

(2),

From

(4),

^-36sin30-F

a*.

-36cos30=0.

(6)

(7)

(8)

solving

for

from

(7)

and

(8)

and

substituting

its

value in

(6),

the

following

equation

obtained,

ri-36sin30-iX36cos30

(I)

312

FORCE, MASS,

AND ACCELERATION

manner,

by applying

the

equations

motion to

body

the

following

equations

are obtained

(see

Fig. 334(6)

for

free-body

diagram):

Tt-Ti-FB-W

sin

30=

a*,

From the

last three

equations

obtain,

-7\-iX60

cos 30 -60

sin 30

(

32.2

(ID

For the

body

C the

equation

motion

(see Fig.

334c),

32.2

(HI)

Equations

I, II,

and

III

contain

three

unknowns, 7\,

and a

(the

x-axis

is chosen

in each

case in the direction of

the

total

acceleration,

a).

solving

the three

equations

for the three unknowns

the

following

results are

obtained;

8.02ft./sec.

7*1

34.76

lb.,

90.1 lb.

353.

An elevator starts

from

rest

and

moves

upwards,

acquiring

velocity

of 800

ft.

per

minute in a distance of 18 ft. If the

acceleration

constant

what

the

pressure

of a man

on the floor of the elevator if

the man

weighs

161 lb.?

Solution.

The

man

forms

the

body

(assumed

rigid),

the

motion

which

considered. Since

the

body

has

motion

translation,

the

three

equations

of motion

for a

translating rigid

body apply,

but

since

the forces

acting

the

body pass through

the mass-center

and

since one of

the

axes is

chosen

the direction

of the

acceleration,

only

one of

the

equations

needed;

namely,

161 lb.

In addition

the

equation

motion one

kinematics

equation

needed;

namely,

(2)

FIG.

335.

Fig.

335

shows the

free-body diagram,

the

two

forces

being

collinear.

From

(1),

P-161--

161

From

(2),

KINETICS

OF A

TRANSLATING

RIGID BODY

313

solving

the last

two

equations

the

following

results are

found,

4.93ft./sec.

185.7

Ib.

In the

above

problems

the

bodies have

rectilinear

translation

only.

The

parallel

or side

rod

a locomotive

when

running

on a

straight

track

has

motion

curvilinear

translation

with

reference

to the

engine

frame. Its

motion is considered

the

following

problem.

354. The

parallel

rod of a

locomotive

(Fig.

336a)

weighs

400

Ib.

The

crank

length,

is 15 in.

and the

radius,

the drivers is

ft.

the

speed

the

engine

miles

per

hour,

what

is the

reaction of

the

each

end

of the

rod when

the

rod is

in its

lowest

position?

9270

Ib.

314

FORCE, MASS,

AND ACCELERATION

forces,

as shown

Fig.

336(6),

the

forces

will be in

equilibrium

(D'Alembert's.

principle)

will be observed

that the

forces form a

parallel

force

system.

The

equa-

tions of

equilibrium

for a

parallel

force

system

(Art. 49)

are,

(1)

(2)

Using

(1),

Using

(2),

Whence,

^1+^2-9270-400

XJ-(9270+400)X-=0.

PROBLEMS

365.

man

who is

just

strong enough

to lift

a 150-lb.

weight

when

standing

the

ground

can

lift a 200-lb.

weight

from

the floor

an elevator when the

elevator

going

down with a certain

acceleration. What

the acceleration?

What

weight

can the

man

lift from

the

floor

the elevator

when

going up

with the same

acceleration?

Ans. a

8.05

ft.

/sec.

;

l20\b.

366. A 3-ton

cage

descending

shaft

with

speed

of 9

yd.

per

sec.

brought

rest

with a uniform

acceleration

distance

ft. What

is the

tension

in the cable

while the

cage

coming

to rest?

357.

The

dimensions

body

(Fig.

337)

are

ft.

4 ft.

and

its

weight

is 1000

Ib.

Assuming

that

the

body

will not

slip

the

carriage,

what

the

maximum

weight

that B

may

have without

causing

A to

tip

over

when

the acceleration

of the

carriage

ft.

per

sec.

The

pulley

is assumed

to be frictionless

and

weightless.

Ans.

201 Ib.

-D

-2^^

FIG. 337.

FIG.

338.

358.

A man

weighing

150

Ib.

leaves

his

room

by way

of a window which

ft. above

the

ground.

has

rope

that

long

enough

to reach

the

ground

but

can

support

force

only

125

Ib. What is

the

least

velocity

with

which

can reach the

ground?

KINETICS

A ROTATING

RIGID BODY

315

369. Two

strings pass

over

a smooth

drum. On one side of

the

drum

the

strings

are attached

to a 50-lb.

weight;

the other side

one

string

attached

to a 40-lb.

weight

and

the other

string

to a

30-lb.

weight.

Find the accelera-

tion of the

weights

and the tension

in each of the

strings

during

motion.

Ans.

^30

lb.;

33.3

lb.;

5.37

ft./sec.

360. Steam is

shut off when

a train

running

at a

speed

of 30 miles

per

hour

reaches

0.4

per

cent

down-grade.

What

will be the

velocity

of the

train

after

100 sec.

if the train

resistance is 10

lb.

per

ton?

361. A small

car

(Fig.

338)

with its load

weighs

800 lb. and the center of

gravity, G,

the total

weight

5 ft.

from

the

track.

force

P of

120 lb. is

applied

to the

car

as shown.

Neglecting

the

friction

on the

track,

find the acceleration of

the car and the reactions

of the

track

each

pair

of wheels.

What

would be the reactions

the wheels if the force

acted

through

the

point

(7?

362.

door is

hung

track as shown

Fig.

339.

The

coefficient of friction

for

each

the shoes

and

the

track is

The door

weighs

FlG-

339

300

lb. What

force P is

required

give

the door an acceleration

4 ft.

per

sec.

? Find the

reactions of

the

shoes

the

track. How

far

will

the door

travel

sec.?

Ans.

112

lb.

ROTATION

146. Kinetics of a

Rotating

Rigid

Body.

rigid

body

acted

an unbalanced

force

system

that

produces

motion

rotation of the

body,

the

resulting

linear

accelerations

the

various

particles

of the

body

are

not

the

same,

but

vary directly

the distances of the

particles

from

the

axis of rotation

(Art. 133).

The

angular

velocities and accelerations of all the

particles,

how-

ever,

are the

same,

any

instant,

and

the

linear

acceleration

any

(and

every) particle

the

body

may

expressed

terms

of the common

angular

velocity,

and the

angular

acceleration,

Thus,

Fig.

340

represented

physical

body

acted on

the

external

forces

and the reaction

(not

shown)

of the

axis at

which

cause

the

body

rotate

with

angular

acceleration,

about

an axis

through

0. Each

particle

of the

body

moves

cir-

cular

path

and the

linear

acceleration

any

particle

in the

body

316

FORCE,

MASS,

AND

ACCELERATION

has

two

components;

tangential

component,

and a

normal

component,

(Art. 118)

such

that,

ra,

directed

tangent

to the

path

the

particle,

and

directed normal to the

path

the

particle,

towards

the center

of rotation.

which

and

a are the

angular velocity

and

angular

acceleration,

respectively,

the

particle

(and

also of the

body)

at the

instant,

is the

linear

velocity

of the

particle

at the

instant,

and

r is

the distance

the

particle

from

the axis

of rotation. The

accel-

eration

components

for

several

particles

are shown

Fig.

340.

From the

components

the

acceleration of

any particle

of the

body, expressed above,

the

components

of the resultant force

(effective

force)

required

produce

these

acceleration

components

may

now

found.

Thus,

m denotes the

mass of

the

particle,

then

from

Newton's second

law,

=mra is the

tangential component

of the effective

force,

and

mru>

is the normal

component

of the effective

force,

shown

Fig.

341.

Resultant

the

Effective

Forces. The

resultant

of the effective

forces

for the

whole

body

may

now

found. As

already

noted

KINETICS

OF A ROTATING

RIGID BODY

317

(see

footnote

under

Art.

144),

the effective

forces

form

coplanar

system

the

plane

of motion

and,

as indicated

Fig. 341,

the force

system

is non-concurrent.

According

to Art.

28,

the

resultant of

such a

coplanar

system

is either

force or a

couple.

either

case

the resultant

may

determined

completely

from

the summa-

tion

the

components

the forces

along

each of two

rect-

angular

axes

and the summation

of the moments

the forces

about some

point

the

plane

the forces.

Thus,

let the two

axes be

the T- and TV-axes as

shown

Fig.

341,

where

the

TV-axis

drawn

through

the center of

rotation, 0,

and the mass-center

the

body.

Let each of the

two

components,

mra and

rarco

of the

effective

force for

any

particle

be resolved

parallel

the

N- and

T-

axes

respectively.

Let

6 be

the

angle

(measured

clockwise from

the

TV-axis)

between the

radius, r,

to the

particle

and

the TV-axis.

Then,

from

Fig.

341,

the

component

of the effective force

parallel

to the TV-axis

is,

wrco

cos 9

mra sin

and the

component

of the effective force

parallel

to the T-axis

is,

mra cos 6

mrai

sin

Hence,

The

algebraic

sum of the

components

of the effective forces

parallel

to the TV-axis

(

mrco

cos 6 mra sin

Smr

cos

da'Lmr sin

-Mrco

-0.

the above

equation

Srar

cos 6

represents

the moment

the

body

with

respect

to the T-axis

and

hence

may

replaced

Mr,

and Srar

sin

represents

the moment of the

body

with

respect

the TV-axis and hence

is zero since the

TV-axis

passes

through

the

mass-center

of the

body.

Further,

the

negative

sign

indicates

that

Mrco

is directed

from the mass-center

towards

the

center of

rotation.

manner,

The

algebraic

sum of the

components

of the

effective

forces

parallel

the

T-axis

(mra

cos 6

mru

sin

a1l

mr cos

2mr sin

318

FORCE, MASS,

AND

ACCELERATION

The moment of

the

effective

force for

any

particle

about

the

center of

rotation, 0,

mra r

since the

normal

component,

rarco

the

effective force for

each

particle

passes

through

the axis of

rotation and

has,

therefore,

moment about

the center

Hence,

taking

moments

about

The

algebraic

sum of

the

moments of

the

effective

forces with

respect

to the

axis of

rotation

=S(rara-r)

where

denotes

the

moment of

inertia of

the

body

with

respect

to the

axis

of rotation.

Since the

sum of

the

components,

any direction,

of the forces

of a

system

equal

the

component

their resultant in

the same

direction,

and since

the sum of the

moments of

the forces about

any

point

equal

to the

moment

their

resultant about the

same

point,

the

above results

may

summarized as

follows:

The

component

of the resultant

of the

effective

forces

parallel

to the

JV-axis

Mrco

The

component

the resultant of the

effective

forces

parallel

the

T-axis

Mra.

The

moment of the resultant

of the

effective

forces

about the

center

of rotation

/o.

If the

resultant

the effective forces

is a

force,

all of

the

above

expressions

are

required

to determine

completely. If,

however,

the resultant

is a

couple,

the first

two

expressions

reduce

zero,

in which

case

loa

expresses

the moment

the resultant

couple,

as will

be discussed

greater

detail later in

this article.

Relation

between

Effective

Forces and

External Forces.

The

relation

between

the resultant

of the effective forces for the whole

body

and the

resultant

the external forces which act

the

body

may

now

be found.

Since

the effective force

for a

particle

is the

resultant

all the forces

(both

internal

and

external)

which

act

the

particle,

the effective

forces for all the

particles

will

include

all of

the

internal forces

(exerted by

the

particles

among

them-

selves)

and

all of the external

forces.

Hence,

the

summation

the

Af-components

the effective

forces is

the

same

the

summa-

KINETICS

OF A

ROTATING

RIGID

BODY

319

tion of

the

N-components

the internal

forces

plus

the

summation

of the

AT-components

the external

forces,

and a similar

statement

applies

to the

^-components.

Therefore,

if F

denotes

force

and

the

moment of a

force,

the

relation

between the above

expressions

and the forces

acting

on the

particles

expressed by

the

following equations:

External

(

)

Internal

(2F

)

External Internal

To)

External

+(2

TO)

Internal

/0.

But the

algebraic

sum of the

components,

any

direction,

the internal

forces

and the

algebraic

sum of the

moments,

about

the axis of

rotation,

of the internal forces

are

zero,

since the

internal

forces

occur

pairs

equal,

collinear,

and

opposite

forces

(Newton's

third

law). Therefore,

the

resultant

the

effective forces

for the whole

body

and

the

resultant of

the

exter-

nal

forces,

only,

are

identical

(D'Alernbert's

principle).

Therefore,

the

above

equations,

the

expressions

which involve

the internal

forces

drop

out since as

just

stated,

Internal

0, (^F

)

Internal

0, (27o)

Internal

Hence,

if F

now used

to denote an

external

force

only,

and T

used to

denote the

moment of an external

force,

the

equations

which

express

the

relations between

(1)

the external

forces,

(2)

the kinetic

properties

of the

body,

and

(3)

the

change

of motion

(the

equations

motion

for a

rotating rigid

body)

are,

will

noted

that

rco

and

are the

normal

and

tangential

accelerations, respectively,

the mass-center

of the

body,

and

hence the first

two

the

above

equations may

written

and

2^,

Ma*.

Further,

the

body

rotates about an axis

through

the

mass-

center,

that

is,

if the

points

and G

coincide,

then the

right-

320

FORCE,

MASS,

AND

ACCELERATION

hand

members of the first

two

equations

are

equal

zero since

equal

zero,

and therefore

the

algebraic

sums of the

components

of the

external forces

in the

two

directions are

equal

zero.

Hence,

if a

rigid

body

rotates about an axis

through

its

mass-

center,

the

equation

motion

for the

body

is,

in which

is the

algebraic

sum of the moments of

the external

forces

acting

on the

body

with

respect

to an axis

passing

through

the mass-center

(axis

rotation)

and

the moment

inertia

of the

body

with

respect

to the

same axis.

ILLUSTRATIVE

PROBLEMS

363.

The rod BCE

(Fig.

342a)

is made to oscillate

means

the

crank

A D

and

link

DC.

The

members

are

connected

smooth

pins

B, C,

and D.

The

rod BCE has a constant cross-section

and

weighs

16.1 Ib. In the

position

shown its

angular

velocity, co,

is 60

r.p.m.

and

its

angular

acceleration,

is 40

rad./sec.

(see

Art. 134

for method of

finding

and

a).

Find the

pressure,

of link

DC at

and

the

reaction,

of the

FIG.

342.

Solution.

The

rod rotates about

the

center B. The

equations

of motion

for the

rod

are,

2/^=M?o>

(1)

(2)

(3)