Mechanics Problems For Engineering Students

MECHANICS PROBLEMS FOR ENGINEERING STUDENTS

BY

FRANK Member

of

B.

S_ANBORN

American Society

of Civil Engineers

Professor oj Civil Engineering in Tufts College

SECOND EDITION, REVISED AND ENLARGED FIRST

THOUSAND

NEW YORK JOHN WILEY & SONS London: CHAPMAN & HALL, Limited 1906

'

PREFACE There

is

an opinion among engineers that too

often students are not well grounded in the practical

problems of Mechanics

;

know more

they

that

of

theory and mathematical deductions than of practical applications.

A

prominent educator has recently

said to me, in regard to the teaching, of Mechanics,

"I in

am convinced that

to

it is

be done more thoroughly

the future than in the past

;

"

and

it

will

be done,

he believes, by sticking close to elementary principles as developed by well-chosen practical problems.

thermore, he adds, " all

it

will

Fur-

have to be recognized that

an engineering baccalaureate course can worthily

accomplish

is

to give the

raw

recruit the

'

setting-up

exercises in Mechanics." It is

now

generally recognized,

subject should cover

I

think, that this

elements and fundamental principles that form the basis of every engineer's knowledge that these necessary elements first

of all

the

;

and principles are best understood and best remembered by actually solving numerous problems that present important facts illustrative of every-day engi-

neering practice, and

arouse the

far better than abstract

student's

interest

examples which can be easily

formulated from imaginary conditions. Therefore, for the reasons indicated above, an effort

:

:

PREFACE.

IV

has been made in preparing this book to present, from

many

actual conditions,

practical problems together

with brief definitions and solutions of typical prob-

lems which should help the student in Mechanics to follow the advice once given by George Stephenson to his son "

Robert

Learn

make

for yourself, think for yourself,

yourself master of principles."

Photographs or electroplates have been furnished for certain of the illustrations as follows

Page 17 by Otto Gas Engine Works pages 20 and 32, Pelton Water Wheel Company page 24, Wellington-Wild Coal Company page 25, Harrisburg Foundry and Machine Company; page 29, Fall River Iron Works Company; page 35, Associated Factory Mutual Fire Insurance Companies page 63, Maryland Steel Company; page 64, Bucyrus Company; page 120, A. J. Lloyd & Co.; page 146, Clinton Wire Cloth Company; page 148, The Detroit page 149, The Engineering Graphite Manufacturing Company Record; pages 151 and 169, Brown Hoisting and Conveying Machinery Company; page 153, Cement Age; page 157, Fig. 84, American ;

;

;

;

;

Locomotive Company page 161, Carson Trench Machinery Company; page 164, Chicago Bridge and Iron Works; page 165, Chapman Valve Manufacturing Company. ;

FRANK Tufts College, Mass., June, 1906.

B.

SANBORN.

CONTENTS I.

WORK.

Problems

to 172.

i

FOOT-POUNDS

FAGK

Raising weights, overcoming resistances of railroad machine punch, construction of wells and trains, chimneys, operation of pumping engines. Force and distance or foat-pounds required in cases of pile. driver, horse, differential pulley, tackle, tram car .

7

HORSE-POWER; Required by Windmills, planing machines, gas engine,

—

simple, compound, triple, locomotive, steam engines Horse-power from slow speed, high speed engines. indicator cards, belts,

canals,

steam

required

crane,

streams,

force or distance

by

coal

turbines,

required

electric

lamps, driving

pumping

towers,

water-wheels. in

cases of

engine,

Efficiency, fire

pumps,

mines, bicycles, shafts, railroad trains, air brakes, the tide, electric

motors, freight cars, ships

16

ENERGY

—

Foot-pounds, horse-power, velocity Ram, hoistingengine, blacksmith, electric car, bullet, cannon, nail, pendulum. Energy resulting from motion of fly-wheel and energy required by jack-screw :

44

;

CONTENTS.

vi

FORCE.

II.

Problems 172

to 414.

FORCES ACTING AT A POINT

page

beams, derrick, cranes set as in action balloon held by rope, hammock supported wagon, trucks, picture supported forces in frames of car dumper, tripod, shear legs, dipper dredge also in triangle, square, sailing vessel, Canal boat being towed, rods,

struts,

;

;

;

rudder, foot-bridge, roof-truss.

.

.

....

.

51

MOMENTS FOR PARALLEL FORCES Beam

balanced, pressure on supports, propelling force

of oars, raising anchor force at capstan, bridge loaded

pressure on abutments, lifting one end of shaft, boat hoisted on davit, forces acting on triangle, square,

supports of loaded table and floor

72

COUPLES Brake wheel, forces acting on square

84

STRESSES Beam leaning against wall, post in truss, rope pull on chimney, connecting rod of engines, trap-door held up by chain .

.

.

.

86

CENTER OF GRAVITY Rods with

loads, metal square

and

triangle, circular

disk with circular hole punched out, box with cover open, rectangular plane with weight on one end, irregular

shapes, solid cylinder in hollow cylinder,

cone on top of hemisphere

FRICTION moved on level table, stone on ground, block on inclined plane, gun dragged up hill, cone sliding on inclined plane friction of planing machine. Weight

;

90

CONTENTS.

vii

PAGE

locomotives, trains, ladder against wall, bolt thread, rope around a post belts, pulleys and water-wheels in action heat generated in axles and bearings. ;

;

.

.

96

MOTION.

III.

Problems 414

to 527.

UNIFORM ACCELERATION Railroad

train, ice boat,

stone falling and depth of

well, balloon ascending, cable car

running wild.

.

.

.

ng

RELATIVE VELOCITY Aim

in front of deer,

rowing across

river, bullet hit-

ting balloon ascending, rain on passenger train,

wind

on steamer, two passing railroad trains

126

DISTANCE, VELOCITY, FRICTION, ANGLE OF INCLINATION Train stopped, steamer approaching dock, cannon recoil, locomotive "increasing speed, body moved on table, box-machine, motion of table, barrel of flour on elevator, man's weight on elevator, cage drawn up coal shaft

.

121

.

PROJECTILES Inclination

down

for

bullet to

strike given point,

motion

dropped from train, thrown from from hill, from bay over fortification

plane, stone

tower, projectile wall

133

PENDULUMS Simple, conical, ball in passenger car.

.

.

.

.

141

IMPACT Water suddenly shut falling

on

pile,

off,

cricket ball struck,

shot from gun, bullet from

and passenger trains

collide

hammer

rifle,

freight 142

CONTENTS.

VIU

REVIEW. Problems 528 to 600.

PRACTICAL PROBLEMS Water turbine test, suspension bridge, Niagara tower, launching data, coal-wharf incline, typical American bridge, modern locomotive tests, wood in compression, actual cableway, St. Elmo water-tower, outside-screwand-yoke valve, cast-iron pipe, retaining walls, geared drum, gas-engine test 145

EXAMINATIONS Yale, Tufts,

Harvard

i74

ANSWERS 600 problems, besides 43 under Examinations. one-half have answers given

About 1S4

DEFINITIONS Work,

force,

and motion and

their sub-divisions

.

TABLES Falling Bodies, Functions of Angles, Unit Values heights and velocities

INDEX

.

2

— 190 193

MECHANICS-PROBLEMS. INTRODUCTION. The problems and

solutions that follow have been

arranged in the order of Work, Force, and Motion.

At the beginning of each important section one problem has been solved so as to explain the method of solving similar problems

and

to serve as a guide for

An effort has been

solutions to be put in note-books.

made throughout the book have been presented

;

to simplify.

Few methods

the calculus has been used only

where necessary no discussion has been offered of the term mass many such subjects have been left for more advanced courses or extended treatises. The "gravitation system" of units the footpound-second system, or meter-kilogram-second sysknown as the engineers' system has been tem ;

—

^

—

—

used exclusively. In engineering practice one just it

what data

to use

some

;

to collect

because of

is

often puzzled to

tell

of

more data

in

this, I

of the problems,

Review, than

is

and afterward how much have

left

and especially those under

absolutely necessary for solving the

will have opportunity " to pick and choose " just as he would do in actual cases.

problem, and the student

MECHANICS- PROBLEMS.

DEFINITIONS. Mechanics

is

the science that treats of the action

of forces at rest and in motion.

Work, Force, and Motion are the three

sub-divisions

of Mechanics considered in this book.

WORK. Work

through some

is

done by the action

is

measured by the product of force times the

of force

distance.

Work

distance through which

Work = for

all

force

Work

Energy

is

x

distance,

— a formula fundamental

the amount of work that a body possesses. is

the work that a body possesses

of its position

Kinetic energy virtue of

acts.

problems.

Potential energy

by virtue

it

is

above the earth's surface.

the work that a body possesses

by

its velocity.

Horse-power

is

the rate of doing work.

power is the equivalent work done per minute.

of

One

horse-

33 000 foot-pounds of

FORCE. Force in Mechanics has both magnitude and direction,

and

in this treatise

DEFINITIONS. Force Magnitude It

may

is

3

usually expressed in

pounds.

act as pressure, a push, or as tension, a

pull..

Concurrent forces acting on a body are those that have the same point of application. Non-concurrent have different points of application.

Moment

of

a force about a point or axis

is

the

product obtained by multiplying the magnitude of the force by the shortest distance from the point or axis to the line of action of the force.

Moment =

force

dency of rotation

x perpendicular. is

Clockwise ten-

usually taken positive.

Resultant of a system of concurrent forces single force that might be substituted for

them

is

a

with-

out changing the effect. Equilibriant of a system of forces that balances them.

The

is

equilibriant

a single force is

equal and

opposite to the resultant.

Components of a single force are the forces that might be substituted for

it

without

changing the

effect.

Parallelogram of forces.

When

three forces that

are in equilibrium meet in a point they can be repre-

sented

in

magnitude and direction by the sides This parallelogram is called

parallelogram.

parallelogram of forces.

of a

the

MECHANICS-PROBLEMS.

4

2

1.

Vertical components

=

o.

When

the forces

acting in one plane upon a body are in equilibrium,

the forces can be resolved into components in any

one direction, and the algebraic sum of the components will equal

2

2.

sum

o.

Horizontal

components

components

of the

to that of

Likewise,

I

=

o.

The

algebraic

in a direction perpendicular

will equal zero

;

and

The algebraic sum of the 3. 2 Moments = o. moments of the forces taken about any point or axis in the plane will equal zero.

These three axioms can frequently be used to formulate quantities

A

three

equations

that

contain

imknown

which can then be determined.

Couple consists of two equal, opposite, parallel

forces not acting in the

Moment

of a couple

same is

straight line.

the product of one of the

equal forces by the perpendicular distance between

them. Center of gravity of a body or a system of bodies

is

a

point about which the body or system can be imagined to balance

and the forces of gravity

will cause

no

rotation.

Centroid and Center of

sometimes used Centroid

is

Mass

are terms that are

in preferenc? to center of gravity.

the point of application of a system of

parallel forces.

DEFINITIONS.

MOTION. Motion (uniform)

Motion (accelerated)

in equal times.

is

in

equal times.

Motion (uniform-accelerated)

which

is

Acceleration

is

that in which the

same amount

velocity increases the

body moves

that in which a

through unequal distances

time,

body moves

that in which a

is

through equal distances

in

each unit of

generally taken as the second. is

the gain or loss in velocity per unit

of time.

Centrifugal force.

move

wards from the center force

=

W

v"

g

r

Impact

When

curved path

in a

is

it ;

a body

is

compelled to

exerts a force directed out-

its

amount

said to take place

is

the centrifugal

when one body

strikes

against another.

A period

of compression thus occurs,

acting are Impulsive forces

and the forces

of compression.

Then

follows a period of restitution. Coefficient of restitution e for

any pair

of substances

is the ratio of the impulsive force of restitution to the

impulsive force of compression.

a.

o

I.

A

WORK.

FOOT-POUNDS.

1 00 tons weight goes 20 horizontal the reFind the amount sistances are i S pounds per ton. of work that locomotive expends per mile of travel.

1.

20th-century express of

up a grade

of

i

vertical in

Work of locomotive

1

;

MECHANICS-PROBLEMS.

8 3.

A

4.

Find what work

punch exerts a uniform pressure of 36 tons punching a hole through an iron plate of one-half inch thickness. Find the foot-pounds of work done.

in

an engine that

hour from a mine 300 If

5.

a weight of

does each 6.

A

man do

number

000 gallons

130 pounds be

i

in a minute,

per hour of

of

water an

feet deep.

by 20 men twice

feet

being done per minute by

is

raising 2

is

men

lifted

up 20

how much work

}

can each do, on the average,

495 000 foot-pounds of work per, day of 8 hours. How many such men are required to do 33 000 x 10 foot-pounds of work per minute.'' 7.

A

centrifugal

pump

delivers

water

10

feet

above the level of a lake of half a square mile area.

At the end lowered \^

of a day's

pumping, the water has been

How much

feet.

work has been done

}

" Distance" will be 10 feet plus \ oi \\ feet.

8.

Water

in a well is

the ground, and out

it is

9.

26

20

when 500

feet below.

feet

below the surface of

gallons have been

pumped

Find the work done.

Brick and mortar for a chimney 100 feet high

are raised to an average height of

35

feet.

Total

amount of material used 40 000 cubic feet or about What work was done 1 5 600 000 pounds. 10.

hangs

What work vertically,

pounds per

foot

}

is

is

done

in

winding up a chain that

130 feet long, and weighs 20

IVOI^X —FOO T-PO UNDS. !!•

feet,

A

chain of weight 300 pounds and length 1 50 with a weight of 500 pounds at the end of it, is

wound up by a 12.

g

A

What work

capstan.

stream of width 20

and mean velocity available fall of 80 feet. feet,

done

is

?

average depth

feet,

What work

is

quantity of water flowing each minute

3

hour has an

of 3 miles per

stored in the

?

Find the pounds of water flowing by observing that

=

Quantity 13.

A

horse draws

area

x

velocity.

150 pounds

of earth out of a

by means of a rope going over a fixed pulley, which moves at the rate of 2^- miles an hour. Neglecting friction, how many units of work does this

well,

horse perform a minute 14.

A

.''

cylindrical shaft 14 feet in diameter

sunk to a depth weight of which

of is

must be

10 fathoms through chalk, the

144 pounds per cubic

foot.

Find

the work done. 15.

A well

diameter.

is to be dug 20 feet deep and 4 feet in Find the work in raising the material, sup-

posing that a cubic foot of

it

weighs 140 pounds.

A horse

draws earth from a trench by means He pulls up, twice of a rope going over a pulley. every 5 minutes, a man weighing 1 30 pounds, and a 16.

barrowful of earth weighing 260 pounds. the horse goes

work done per 17.

A body

forward

40

feet.

Each time

Find the useful

hour.

weighing 50 pounds

slides a

distance

MECHANICS-PROBLEMS.

lO

down a plane

of 8 feet

a

against

Compute the

retarding

total

and the

(gravity) its weight

What work

18.

is

cord has been pulled

224 pounds

of

1

friction.

stored

in

a cross-bow whose

inches with a

5

maximum

pumped every

feet of water are

minutes from a mine 140 fathoms deep,

5

amount

work

of

is

expended per minute

In pumping

20.

force

1

25 cubic

If

19.

inclined 20° to the horizontal,

force of 4 pounds. work done upon the body by

constant

i

what

?

000 gallons from a water-cistern

with vertical sides the surface of the water

is

Ipwered

Find the work done, the discharge being 10 above the original surface.

5 feet.

feet

21.

20

A uniform

feet long,

turn freely.

it

How many

done to raise position 22.

beam weighs

i

000 pounds, and

is

hangs by one end, round which it can

from

it

foot-pounds of work must be its

lowest

to

its

highest

?

A

body

is

suspended by an

unstretched length

elastic string of

Under

a pull of 10 4 pounds the string stretches to a length of 5 feet. Required the work done on the body by the tension feet.

of the string while its length changes

4

from 6

feet to

feet.

23.

A

weight of 200 pounds

is

to be raised to a

height of 40 feet by a chord passing over a fixed

smooth pulley

;

it is

found that a constant force P,

pulling the chord at its other end for three-fourths of

1

WORK— FOO T-POUNDS. the ascent, communicates

weight to enable

Find

to

it

1

sufficient velocity to

the

reach the required height.

P.

= =

Work Work

force

X

200

X 40

P X

I of 40

distance

on weight

=

Work by pull

= Work

Work on weight

24.

by pull

200 X 40

=

P X 30

P

=

2

A horse drawing

66 J pounds

a cart along a level road at

the rate of 2 miles per hour performs 29 216 foot-

pounds of work

in 3 minutes.

What

25.

It is said that

pounds

pull in

does the horse exert in drawing the cart

.'

a horse can do about

1

3

200 000

foot-pounds of work in a day of 8 hours, walking at

the rate of

2|-

miles per hour.

What

pounds

pull in

could such a horse exert continuously during the

working-day 26.

If a

.'

horse walking once round a circle

i

o yards

across raises a ton weight 18 inches, what force does

he exert over and above that necessary to overcome friction 27.

.-'

A building

moved on

rollers

of weight 50

by a horse -that

000 pounds is

is

being

pulling on a pole

with a distance of 10 feet from the center of a capstan If the total friction is that is 1 8 inches in diameter.

200 pounds per

ton,

what force must the horse exert

"i

MECHANICS-PROBLEMS.

12 28.

The 500-pound hammer

raised to a height of 20 feet

of

a

pile-driver

is

and then allowed to

upon the head of a pile, which is driven into the ground i inch by the blow. Find the average force which the hammer exerts upon the head of the pile. fall

Work

Distance .•.

10 000 foot-pounds force

.-.

=

force

=

500

= = =

X

X

distance

20

10 000 foot-pounds -^-^

foot

force

=10

=

x tV

foot

000 x 12

120 000 pounds

A

hammer weighing i ton falls from a height 24 feet on the end of a vertical pile, and drives it half an inch deeper into the ground. Assume the 29.

of

driving force of the stant while

it

lasts,

hammer on and

find its

the pile to be con-

amount expressed

in

tons weight.

j-^

30.

Determine by the principle of work,

neglecting friction, the relation between the pull

P and the

load

W

in case of the differ-

ential wheel-and-axle of Fig.

For one revolution,

Work of

p

=

P x

2 Tra

i.

1

WORK—FO O T-PO UNDS.

A barrel

31.

pounds I

is

to

the radii

;

will

of Portland cement that weighs 396 be hoisted by a wheel and axle as in Fig. are 6, 12 and 18 inches. What force

be required

32.

13

If,

?

neglecting frictions, a power of 10 pounds,

acting on an

arm

2 feet long, produces in a screw-

press a pressure of half a ton, what would be the pitch of the screw 33.

What

is

1

the ratio of the weight to the power,

in a screw-press

working without

friction,

when the

screw makes 4 turns in the inch, and the arm to which the power is applied is 2 feet long ."

34.

18

What

inches

force

applied at

pounds upon the head turns cause the head inch

the

end

of

an arm

long will produce a pressure of

i 000 smooth screw when 1 advance two-thirds of an

of a to

}

35.

screw,

Find the mechanical advantage in a if the length of the power arm is

differential

2 feet,

and

there are 4 threads to the inch in the large screw,

and

5

36.

threads to the inch in the small screw.

In a differential pulley,

if

the radii

weight of the lower block

is

can be raised by a force of

S

37. is

7

of

the

and if the i|^ pounds, what weight pounds }

pulleys in the fixed block are as 3 to,2

;

In a wheel and axle the diameter of the wheel

feet, of

the axle 7 inches.

What

weight can be

2;

MECHANICS-PROBLEMS.

14 raised

by a force

lo pounds acting at the circum-

of

ference of the wheel

?

38. A weight of 448 pounds is raised by a cord which passes round a drum 3 feet in diameter, having on its shaft a toothed wheel also 3 feet in diameter a pinion 8 inches in diameter, and driven by a winch ;

with the

gears

handle

16

wheel.

Find the power to be applied to

inches

long,

winch handle in order to raise the

the

weight.

A tackle

two blocks, each weighing 1 5 pounds, the lower one being a single movable pulley, and the upper or fixed block having two sheaves 39.

the parts of

What

port

and what

at the

upper block

A weight

fixed to the

is

pull

on the cord

will

of

then be the pull on the staple

400 pounds

standing part of the rope

is

is

of the rope,

being raised by a

two sheaves

weighs 10 pounds.

What

whose weight may be is

;

dis-

each block

the pressure on the

point from which the upper block hangs

Two equal weights,

the

;

fixed to the upper block,

regarded, are considered to be vertical

41.

movable will sup-

?

pair of pulley blocks, each having

and the parts

and

200 pounds hung from the movable

block.'

40.

of

cord are vertical,

the

the standing end block.

formed

is

}

each 112 pounds, are joined

by a rope which runs over two pulleys

A

and

B

1

;

WORK— FO

T-PO UNDS.

I 5

apart and in the same horizontal line. weight of ten pounds is lowered on to the rope feet

A and B

way between

how

of lo-pound

A

far will the rope deflect

?

= Work

Work 42.

a

If

half-

weight

of two ir2-pound weights.

weight of 500 pounds, by falling through

by means

of a machine, a weight of 60 200 feet. How many units of work has been expended on friction, and what ratio does it bear to the whole amount of work done 1

36 pounds

feet, lifts,

to a height of

The

43.

pull

on a tram-car was registered when

the car was at the following distances along the track

0,200 pounds; 10 feet, 150 pounds; 25 feet, 160 pounds 32 feet, 156 pounds 41 feet, 163 pounds 56 feet, 170 pounds; 60 feet, 165 pounds; 73 feet, ;

;

;

What effective work was done

160 pounds.

the car through the distance of 73 feet,

in pulling

and what

constant pull would have produced the same work

In lifting an anchor of ij tons from a depth

44.

of

1

5

fathoms

power,

if

in

6 minutes, what

a man-power

per minute 45.

}

is

is

the useful man-

defined as 3 500 foot-pounds

."

Four hundred weight

of

material are

drawn

from a depth of 80 fathoms by a rope weighing 1.15

How much work is done pounds per linear foot. altogether, and how much per cent is done in lifting How many

the rope.'

units of

per minute would be required to 4|-

minutes

.''

33000

foot-pounds

raise the material in

6

MECHANICS— PROBLEMS.

1

H O R S E-P O W E R

A gas

46.

engine must hoist 3 tons of grain through 50 feet every minute. What'

a vertical height of

horse-power must be provided

?

= force X distance [minUte = (3X2 000) pounds X 50 feet per Now horse-power = 33 000 foot-pounds per minute = s X 2 000 X 2_ .-.Work engine Zi 00° = gy'y horse-power Work

of engine

I

TTT

i;o

T

of

A

47.

hod-carrier

who weighs

155 pounds carries

65 pounds of brick to the third story, a vertical height of 20 feet. How many foot-pounds of work has he done If he makes 10 such trips in an hour, at what ">.

he work

rate in horse-power does

A

48.

windmill raises by means of a

60

of water per hour to a height of it

work uniformly,

to 49.

The

22 tons

Supposing

travel of the table of a planing-machine

cutting

number

pump

feet.

calculate its horse-power.

which cuts both ways while

.'

is

be taken

9 at

feet.

If

the resistance

400 pounds, and

of revolutions or double strokes per

the

hour be

80, find the horse-power absorbed in cutting. 50.

100

A forge hammer

lifts

lift is

a minute

2 feet.

;

weighing 300 pounds makes

the perpendicular height of each

What

that operates 20 such

is

the horse-power of the engine

hammers

?

WORA'— HORSE-rO JVER.

51.

An

ilkistration.

diameter,

Otto gas engine It

is

shown

has a belt pulley that

and makes

150

17

is

revolutions

in the

above

^6 inches in per

minute.

8

;

MECHANICS— PROBLEMS.

1

What

force

is

driving,

for

shafting,

can the belt transmit

therefore,

developing

horse-power

rated

its

and machinery,

when the engine twenty-

of

one?

How many

52.

3

hundred weight

depth 53.

raise

of

is

66o

feet

horse-power would of coal a

it

take_ to raise

minute from a

pit

whose

?

Find the horse-power of an engine which is to 30 cubic feet of water per minute from a depth

440

feet.

54.

Find the horse-power required to draw a train

of 100 tons, at the rate of

30 miles an hour, along a from friction being 16

level railroad, the resistance

pounds per 55.

Each

engine

crank

ton.

is

the

of

two cylinders

i

foot.

If the driving-wheels

revolutions per minute, and the

pressure

is

horse-power 56.

mean

make 105

effective steam-

85 pounds per square inch, ^hat

is

the

.?

The weight

drawbar

in a locomotive

16 inches in diameter and the length of

is

of a train is 95.5 tons,

and the

6 pounds per ton.

Find the horsepower required to keep the train running at 25 miles pull

is

per hour.

A train,

whose weight including the engine is 100 tons, is drawn by an engine of 1 50 horse-power friction is 14 pounds per ton all other resistances neglected. Find the maximum speed which the 57.

—

engine

is

capable of maintaining on a level track.

WORK— HORSE-PO WER. In the

electrical I I

Ig

problems that follow observe that kilowatt

horse-power

Watts

= = =

1.340 horse-power

746 watts volts X amperes

58. A dynamo is driven by aq, engine that develops 230 horse-power. If the efficiency of dynamo is 0.81 what " activity " in known kilowatts is represented by

the current generated 59.

Electric lamps giving

watts (a)

lamps

?

how many

may

combined

10-

and

(b)

i6-candle

is

What

The

the candle-power available t

amperes

electrical current expressed in

be used by a 250-volt

t

dynamo, and gearing

efficiency of engine,

being 70 per cent, what

60.

how many

be. worked per electric horse-power

for every indicated horse -power

will

candle-power for 4

i

electric hoist

2 500 pounds of coal per minute from

when

raising

a ship's hold

150 feet below dump cars on trestle work, the efficiency of the whole arrangement being 50 per cent ? 61.

A

company can

prospective electric

find

a

market for 900 electrical horse-power at a city 20 Engineers estimiles from a suitable water-power. mate losses in generating machinery 10% in line ;

7%

;

in

transformers

efficiency of turbines

the river

is

at

85%.

load

end

10%

The average

2 feet per second

;

;

and the

velocity of

width available near

dam 40 feet depth 5 feet. Find (a) the waterpower that would be required (b) the net fall that proposed dam must afford. ;

MECHAiVlCS-PRORLRMS.

20

Fig-

A

3.

two jets i inch in diameter, flowing with velocity of 80 feet per second. Theoretic horse-power would be 9.9 and if efficiency of wheel is 85 per cent, and the generator which the wheel drives also 85 per cent, what power in 62.

water-motor

is

driven

l^y

;

kilowatts does the current represent 63.

What

is

}

the difference in tensions of the two

is running 4 200 and transmitting 300 horse-power

sides of a 30-inch driving belt that feet a minute,

.''

In belt problems the difference

64.

Find the speed

in

tensions represents "force."

of a driving-pulley 3.5 feet in

diameter to transmit 6 horse-power, the driving-force of the belt being

i

50 pounds.

WORK— HORSE-PO IVER.

A

65.

belt

is

designed to stand a difference

tension of lOO pounds only.

which

it

I

in

F"ind the least speed at

can be driven to transmit 20 horse-power.

A

66.

2

pulley

3 feet

6 inches in diameter, and mak-

ing 150 revolutions a minute, drives by means of a

machine which absorbs 7 horse-power. What must be the width of the belt so that its greatest tension may be 70 pounds per inch of width, it being belt, a

assumed that the tension that on the slack side 67.

An

in

the dri\dng-side

is

twice

}

endless cord stretched and running over

000

feet

Find the

dif-

grooved pulleys with a linear velocity of per minute, transmits

horse-power.

5

3

ference in tensions of the cord in pounds. 68.

A

rope drive has a grooved pulley 14 feet in

diameter that makes 30 revolutions per minute, difference in

tensions

being

The

100 pounds, find the

horse-power transmitted. 69.

power

A is

locomotive that

drawing a train

a 2 per cent grade

can develop total

500 horse-

weight 100 tons up

resistances are

10 pounds per

Find the highest speed that can be attained.

ton.

= Work

^^^ork of locomotive

500 X 5

;

of

X

of resistances

-f

Work of lifting train

33 000 = 10 X looXi/+ 100 X 10 X (/+ 33 000 = = d 50 s X 33 000 d = 3 300 feet per minute = 37a" miles per hour.

2

000 X t5o X 20 X 2 X

"' ^/

ME CIIA NICS — PR OBLEMS.

22

70.

A

lOO tons weight runs at 42 miles

train of

an liour on a level track

resistances are 8 pounds per

;

F"ind the speed of train

ton.

(i foot rise in

up a

100 feet horizontal)

i

per cent grade

the engine-power

if

kept constant.

is

71.

In 1895 a passenger engine on the Lake Shore

Railroad

made

miles an hour.

on

a run

of

Weight

level track, 15

86 miles of train,

pounds per

lo-wheeler, having drivers

and

cylinders

X 24

17

power was developed up

the average draw-bar pull

72.

A

ton.

the rate of 73 resistance

;

The engine was

5 feet 8 inches in

inches. a

at

250 tons

i

a

diameter

When 730

horse-

per cent grade what was

.''

98-horse-power automobile has by test in

Colorado drawn a special 36-ton locomotive up a 12 per cent high\\a3' grade at the rate of four miles an hour.

73.

What were

A

the frictional resistances per ton

.''

modern farming machine equipped with a

loo-horse-power automobile will plow, sow, and harrow, all at

the same time, a strip 30 feet wide at the rate

of

miles an hour, or 80 acres a day.

is

3^-

What

developed for each foot width of ground 74.

Find the

total

which are taking a train

200

horse-power of two of

force

.''

engines

250 tons down a grade

of

60 miles an hour, supposing the resistance on the level at this speed to be 35 pounds a ton. I

in

at

WORK— HORSE-PO JVER.

23

An

automobile that weighs 5 tons goes up a rough road of grade I vertical to 10 horizontal air and frictional resistances are 16 pounds per ton. What horse-power must the motor develop to maintain a speed of 20 miles an hour 75.

;

.?

76.

to

Find the horse-power

move

which

at the rate of

rises

which

is

100, the weight of the locomo-

foot in

i

of a locomotive

20 miles an hour up an incline

and load being 60 tons, and the resistance from friction 2 pounds per ton. tive

i

77.

A

steam-crane, working at

3

horse-power,

is

able to raise a weight of 10 tons to a height of 50 feet in

What

20 minutes.

friction

friction

78.

1

If

the crane

how many

hours,

work

part of the is

kept

fo
is

done against

similar

at

work

work

for 8

are wasted on

.?

The

six-master

5 500 tons of which take up

coal. i

shown on the next page carries It is unloaded by small engines

ton at each hoist

hold of ship to top of

;

average

from

lift

chutes which lead to cars,

35 feet; weight of bucket,

i

ton; 2 trips are

made

per minute, and 25 per cent of power of engine is When two towers lost in friction and transmission. are working vessel

.''

how long

will

it

take to

unload

the

;

IIORS/'.-IXIWER.

irOA'A'

The

illustration

Problem

on opposite page accompanies

six-master

of

78.

An

79.

tons.

If

average it

is

coal barge will carry

size

i

600

unloaded by two simple direct engines,

the coal being hoisted 65 feet to an elevated hopper

on the wharf, weight

of

bucket

i

ton,

and carrying

i

ton of coal, what horse-power of engines would be required to unload the

i

600 tons

The locomotive

80.

Work

=

Force

= = =

Distance

730

X

II 000

20 hours?

prolslem 71 made 360.7 What was the mean effective

of

revolutions per minute.

cylinder pressure

in

t

X distance

force \

-k

ly

y.

P

y^

2

X 360.7 x \i, (i TT 17= X T^X 2) X (24 X 360.7 X

2

ji)

Fig.

The engine shown

81.

der

1

5

inches in diameter

in ;

Fig. 4 has

steam

length of stroke,

1

5

cylin-

inches

MECHAKICS-PROBLEMS.

26

revolutions per minute, 275

76 pounds per square 82.

The

inch.

;

mean

effective pressure,

Find the horse-power.

indicator cards ilhistrated herewith

taken from an engine of the type shown 81,

diameter

of

length of stroke 300.

steam

cylinder

14

problem inches,

12 inches, revolutions per minute

Scale on cut the

mean

produced by indicator springs to an inch, and

in

being

were

ordinates,

which were

of stiffness

40 pounds

compute the indicated horse-power

of

the engine.

Fig.

83.

The

from one

5.

Full Load Indication.

indicator cards

shown below were taken

of the triple-e.xpansion

pumping-engines

at

the East Boston Station of the Metropolitan Sewerage.

The

cards were

cylinder.

from two ends of a high-pressure

Refer to the cards and compute the

indi-

(A twenty-four hours' duty trial cated horse-power. of this pumping-engine was made January 17-18, 1

90 1, by engineering students of Tufts College.)

WORK - nOKSE-rO

Fig.

6.

Headend, Card shown, one-half

inches

size areaof original, 4,69 square pounds per square inch length of stroke, revolutions per minute, 84 piston diameter, 13;, inches. ;

stiffness of spring, 50

;

30 inches

27

l! ViA".

;

;

;

Fig. 7. Crank end. Card shown, one-half size areaof original, 4.62 square stiffness of spring, 50 pounds per square inch inches length of stroke revolutions per minute 84 piston diameter, 13; inches. 30 inches ;

:

;

,

;

84.

for

The average breadth

one end

other end

it

of a piston is

crank,

85,

foot

i

is

is

1,42 inches,

pounds per square

What

;

inch.

of

an indicator diagram inches,

1,58

and

Piston,

1

The

2 inches

long; revolutions

per

the indicated horse-power

diameter of

and

for the

inch represents 32

i

diameter

minute,

;

iio.

?

cylinder of a steam-engine has an internal 3 feet

;

length of stroke, 6 feet

makes 10 strokes per minute.

Under what

pressure per scjuare inch would order that the piston

may

it

;

and

it

effective

have to work

in

develop 125 horse-power.''

MECHANICS-PROBLEMS.

28 The

illustration

of triple-expansion engines

accompanies Problem

Four

86.

on opposite page

86.

steam-engines

triple-expansion

pairs of

^

are used to drive the cotton machinery of the largest

One

Fall River corporation. in illustration

and

54.

of these engines

The steam

pressures are

pounds per square inch

;

inches.

and

low,

The mean

:

in receiver

intermediate cylinders, 40 pounds intermediate

shown

has cylinders 26\ inches diameter,

;

1

50

between high and

in receiver

between

Vacuum

pounds.

5

36|-,

In main pipe,

27

is

effective pressures in the cylin-

ders are respectively 54 pounds per square inch, 23Iand 12\. Length of stroke is 5 feet piston speed, ;

660

feet per minute.

An

87.

engine

is

Calculate the horse-power.

required to drive an overhead

traveling crane for lifting a load of 30 tons at

per minute.

The power

is

4

feet

transmitted by means of

making 160 revolutions per minute. is 250 feet the power is transmitted from the shaft through two pairs of bevel gears (efficiency 90% each, including bearings), and 2^-inch shafting,

The

length of the shafting

one worm and wheel ings).

;

(efficiency

Taking the mechanical

85%, including

bear-

efficiency of the steam-

engine at 80%, calculate the required horse-power of the engine. 88.

An

engine working

at

50

horse-power

is

driven by steam at 75 pounds pressure acting on pistons in two cylinders. If the area of each piston is 72

square inches, and the length of stroke 2 feet,

many revolutions does the

fly-wheel

how make per minute ?

MECHANICS-PROBLEMS.

30

The steam-engine in use at Weaving Mill of the Pacific Mills

the Worsted

89.

Mass.,

Lawrence,

at

a Corliss type cross-compound with steam

is

cylinders

and

19

36 inches diameter;

42

stroke,

mean effective pounds. Find how many

inches; revolutions, 100 per minute

;

60 pounds and 1 3 looms weaving worsted dress-goods

said engine will

pressures,

each loom requiring \ horse-power.

drive,

A

90.

ship laden with coal

must be unloaded

the rate of 22 tons of coal in height of hft will

150

is

be required

The

91.

coal

of

feet,

10 minutes.

what horse-power

If

at

the

of engines

.

fuel

used

in

running a steam-engine

such composition that the combustion of

is i

pound produces heat sufficient to raise the temperature of 12 000 pounds of water 1° Fahr. It is foundthat 3.1 pounds of fuel are consumed per horse-power

What

per hour. ratus

is

the efficiency of the entire appa-

.'

92.

A

that the

steam-engine uses coal of such composition

combustion of

British thermal units.

per hour, and

power 93.

is

The

pound generates 10 000

40 pounds

the efficiency

if

realized

i

If

is

of coal are

0.08,

.''

cylinder of a Corliss-type steam-engine

30 inches in diameter, stroke 48 inches, and 85 revolutions per minute.

The steam

ing 90 pounds per square inch, what

power

used

what horse-

of the engine

.''

it

is

makes

pressure beis

the horse-

WORK — J/ORSE~rO li'ER. 94.

The

diameter

;

piston of a steam-engine

stroke

its

lutions per minute

on

it is

;

is

2I feet,

the

An

is

1

inches in

5

makes 20

it

mean pressure

r

of the

revo-

steam

pounds per square inch. How many footwork are done by the steam per minute,

15

pounds of and what is the horse-power 95.

and

3

of the engine

.''

engine has a 6-foot stroke, the shaft makes

30 revolutions per minute, the average steam presis 25 pounds per sc|uare inch. Required the

sure

horse-power when the area of the piston

is

i

800

square inches, the modulus of the engine being \l. 96.

inches

The diameter ;

steam-engine cylinder

of a

the length of crank, 9 inches

revolutions per minute,

1

10

;

the

;

mean

the

number

The

51

has a

21

horse-power gas engine of

12-inch piston and

8-inch

9 (.)f

effective pres-

sure of the steam 35 pounds per square inch. the indicated horse-power.

97.

is

crank.

Find

problem

When

making 150 revolutions per minute with an explosion every 2 revolutions what will be the mean effective pressure during a cycle

98.

The

.''

area of a cross-section of the Charles River

408 square feet. The velocity of current as found by rod floats and current meter, April 17 and 22, 1902, was 1.12 feet per secat Riverside,

ond.

Massachusetts,

What would

this quantity of

gives a

fall of

is

be the theoretic horse-power of

water

12.58 feet

at .?

the

Waltham dam, which

WORK— I/ORSE-PO IVKK.

33

In water-power problems use ^^'ork.

=

(area

Force Distance

= =

X

force

X

distance

velocity

pounds

height of

62,',)

water flowing

of

dam

or available

horse-power

iiselul

water-wheel,

a

of

fall

Find the

99.

supposing the stream to be feet

wide and

loo 5

feet

and to flow

deep,

with a velocity of

per second

\ foot

;

height of the

the

feet,

and

efficiency

of

fall is

the

24

the wheel 70 per cent.

A small

100. Fig.

8.

-water-Power.

stream

velocity of 35 feet per minute,

mean

section of

5

by

feet

fall

2.

of

On

erected a water-wheel whose modulus

has

mean

13 feet and a

stream

this

is

Find

is

0.65.

at

Manchester,

the horse-power of the wheel. 101.

On

N.H., as

it

page 32

is

turing Company.

Width

feet, velocity 1.13 feet

water

is

flowing

being 27.3

shown the canal

passes the mills of the

.?

is 5 i feet,

per second.

The

feet, wdiat

Amoskeag Manufac-

is

height of

depth

What fall for

of

water 8.9

quantity of the turbines

the theoretic horse-po wer

MECHANTCS-PROBLEMS.

34

The

102.

80 per cent

an efficiency of

90 per

What

cent.

In winter,

103T

problem

reaction turbines of ;

kilowatts are available

if

2

loi have

the electric generators,

feet of

f

forms on

ice

this

canal, and the velocity drops to 0.75 feet per second,

and the available be the kilowatts

The mean

104.

before turing feet;

it

fall

becomes 25.0

what

will

Merrimac Canal

just

feet,

available.''

section of the

enters the mills of the Merrimac Manufac-

Company at Lowell, Mass., is 48.2 feet by mean velocity on Nov. 23, 1901, was 2.37

per second

the water-wheels had a net

;

fall of

10.6 feet

35-67

and gave an efficiency of about T] per cent. Find the number of broad looms weaving cotton sheetings that may be driven 2i looms requiring one horsefeet,

power. 105.

at

The

estimated discharge of the nine turbines

Niagara Falls

in

1898 was 430 cubic feet per secThe average pressure head on

ond for each turbine.

the wheels was that due to a

Compute the

bines, allowing 106.

fall

of about

The average

flow over Niagara Falls

The

height of

In round numbers what horse-power

for

feet.

all tur-

an efficiency of 82 per cent.

cubic feet per second.

107.

136

actual horse-power available from

is

270 oco

is

fall is

1

61

developed

feet. .''

Calculate the horse-power that can be obtained

one minute from an accumulator which makes

IVORA' -^ irORSE-rOWER.

one stroke

in a

minute and has a ram

35 of

20 inches

diameter, 23 feet stroke, loaded to a pressure of 750

pounds per square

Fig.

108.

g.

A

inch.

An Underwriter Fire-Pump with Standard

Fittings.

fire-pump for protection of a 50 000-spindle

cotton-mill will deliver

i

000 gallons

minute at 100 pounds pressure.

of water

Large

per

boiler capa-

MECHANICS— PROBLEMS

36 city

is

required for such a fire-pump and for the above

150 horse-power would be used. What portion would be required in actual

size

of this boiler capacity

work

of delivering

=

Work of

force

water

X

?

distance

pumping

= 000 X 8^ pounds per minute = 100 X 2.304 feet head (i pound = 2.304 ft.) Distance = 000 X 8^ X 100 X 2.304 .-.Work pumping = I 920 000 foot-pounds per minute = 58.3 horse-power Force

1

I

of

_

Portion of boiler used

=

Q

150

= An

109.

0.39, or about one-third

Underwriter fire-pump to protect an av-

erage-sized factory will deliver four streams of water

through

i|--inch

smooth nozzles with pressure

of play pipes of

at base

50 pounds per square inch.

This

would correspond to a discharge of i 060 gallons per minute. Loss of pressure through nozzle can be neglected

;

and

loss in quantity of discharge

short strokage, and so on, will be about

by

slippage,

10 per cent.

Find the work done by the pump.

A

110.

pump

of

medium

size

used for

tection of a factory will deliver three

streams, or 750 gallons per pressure. to allow

What

A boiler

minute

at

fire pro-

i^-inch

fire

80 pounds

should be provided large enough

70 per cent of

its

capacity to remain as extra.

should be the nominal horse-power of boiler

?

]V0RK~I10RSE-J'0]VIiR. 111.

A

through

a

steam

Silsby

inean velocity of

number

fire-engine

Siamese nozzle that

with a pressure of

80 pounds

is

square inch and a

j^er

feet discharged

Find

pounds pressure

;

(i)

per second

the weight of water discharged per minute

work possessed by each pound

water

delivers

inches in diameter,

2

106 feet per second.

cubic

of

37

of water

;

due

;

(3)

to

the (2)

the

80

(4) the horse-power of the engine

required to drive the i)ump, assuming the efficiency to be

70 per

112.

Station

At

cent.

the Chestnut Hill High-Service

(Boston)

month

for the

of

Pumping

October,

1904,

Engine No. 4 pumped 950 780 000 gallons of water average lift was 130.63 feet total time of pumping, What a\'erage horse-power was de744 hours. ;

;

\'eloped

}

The amount of coal burned during the month 113. was 783 148 pounds. How many foot-pounds of work were done for every 100 pounds of coal burned, that is, what was the Duty of the pumping-engine 1 114.

tion

The

burns

ordinary fire-engine

soft coal,

and

will

60 pounds per fire-stream Therefore

at

of

when

in

full

opera-

consiune in an hour about

250 gallons per minute.

the 70-million dollar

fire

in

Baltimore,

February, 1904, a 500-gallon engine that was running 30 hours, before the fire was under control, consumed

how many pounds 115.

of coal

1

Find the useful work done each second by a

fire-engine

which discharges water

at

the rate of 500

MECIIAA'ICS — PROBLEMS.

38

gallons per minute against a pressure of

lOO pounds

per square inch.

There were 6 ooo cubic

116.

mine

pump began empty

feet of

6o-fathom depth when a

of

pump

to

out.

it

Find the number

it.

It

water

a

in

50-horse-power

took

hours

5

to

of water

of cubic feet

mine during the 5 hours, supposing the work of the pump to have been

that ran into the

one-fourth of wasted.

Find the horse-power necessary to

117.

pump

out

the Saint Mary's Falls Canal Lock, Sault Ste. Marie, in

24 hours, the length of the lock being 500

width 80

feet,

and depth

being delivered

tom

at a height

The mean

Level Canal

42

above the bot-

feet

section of the branch of the First

at the

headgates of No.

Paper Co., Holyoke, ;

fall of

from

20

tail-race is

of

feet,

18 feet, the water

of the lock.

118.

deep

of water

;

Mass.,

this canal to the

feet,

i

Mill,

Whiting

78 feet wide by

is

Second Level there

but about 2 feet

is

lost in

14 a

is

penstock and

velocity of flow in canal during the daytime

0.20 feet per second, and the

driven have an efficiency of

'J']'''lo-

turbines

that

are

Find how many

96-inch Fourdrinier Paper Machines can be driven,

each machine requiring 100 horse-power. 119.

What

is

the

horse-power of a stream that

passes through a section of 6 square feet at the rate of 2\ miles

an hour, and has a water-fall of 18 feet

.?

WORK — What

120.

IfORSE-POWER.

horse-power

is

39

involved in lowering by

2 feet the level of the surface of a lake 2 square miles in

area in

300 hours, the water being

average height of

feet

5

lifted

to

an

?

Taking the average power of a man as -^^^th and the efficiency of the pump used in what time will lo men empty a tank of

121.

of a horse-power,

as 0.4,

50 feet X 30 feet x 6 feet filled with water, the being an average height of 30 feet

lift

.?

A

122. is full

shaft

560

required to

feet

deep and

How many

of water.

empty

it,

5

feet in diameter

foot-pounds of work are

and how long would

engine of 3^ horse-power to do the work

Required the number

123.

2

200 cubic

depth

is 6},

feet

take an

of horse-power to raise

water an hour from a mine whose

fathoms.

What

124.

power

feet of

it

.?

raise in

weight of coal

will

one hour from a

an engine of 4 horse-

pit

whose depth

is

200

}

A

125.

shaft

cut

is

being made on a 4-inch wrought-iron

revolving at

traverse feed

is

10

revolutions

per minute; the

0.3 inch jjer revolution; the pressure

What is the found to be 435 pounds. tool How expended at the } much metal horse-power is removed per hour per horse-power when the depth on the

tool

of cut

is

section)

?

is

.06 inch, the breadth .06 inch (triangular

MECHANICS-PROBLEMS.

40

A man

126.

rides a bicycle

up a

hill

whose slope

per minute

he doing

is

At the top

127.

is

The weight What work

20 at the rate of 4 miles an hour. of man and machine is 187I pounds. in

I

.'

of the hill the bicyclist referred" to

met by a strong head-wind, and he finds that he has to work twice as hard to keep the same rate of 4 miles an hour on the level. What force is the wind exerting against him example 126

in

is

.'

A

128.

works

bicyclist

at

the rate of one-tenth of a

horse-power, and goes 12 miles an hour on the level.

Prove that the constant resistance

of the road

is

3.125

pounds.

Prove that up an incline of the speed

tal

will

i

vertical to

50 horizon-

be reduced to about 5.8 miles per

hour, supposing that the

man and machine

together

weigh 168 pounds.

A

129.

man rows a

miles per hour uniformly.

pounds be the resistance of the water, and P

pounds of useful work are done the

number

130.

of strokes

The

R

foot-

each stroke, find

made per minute.

resistance offered

passage of a certain steamer

000 pounds.

at

If

%

at

by

still

water to the

10 knots an hour

is

power is lost in pushing by " slip aside and backward the water acted on by the screw or paddle and 8 % is lost in friction of machinery, what must be the 1

5

''

—

If

1

2

of the engine

—

total

horse-power of the engines

}

1

IVOJiA'— IIORSE-PO

The United

131.

WER.

States warship

4

Columbia has a

speed of 23 knots, with an indicated horse-power of 22 000. Find the resistance offered to her passage.

The

132. is

about

9

rise

and

fall of

the

If

feet.

the tide at Boston, Mass.,

in-coming water for one

square mile of ocean surface could be stored and

its

potential energy used during the next 6 hours with

an average available

fall of 3

what horse-power would be

feet,

.''

133. A nail 2 inches long was driven into a block by successive blows from a hammer weighing 5.01 pounds after one blow it was found that the head of ;

the nail projected 0.8 inches above the surface of the

block

the

;

1.5 feet

hammer was then

and allowed to

fall

raised to a height of

upon the head

of the nail,

which, after the blow, was found to be 0.46 inches

Find the force which the hammer

above the surface.

exerted upon the nail at this blow. 134. 5

A

500-volt motor

up a 12 miles per hour

drives a

per cent grade at a speed of

75 per cent of the

pended.

What

work

:

motor

electric current,

peres, will be required

Work of motor

135.

of the

The speed

lO-ton car

is

usefully ex-

expressed

in

am-

1

X

0.75

=

work of lifting car

of the " Exposition Flyer "

on the

Lake Shore and Michigan Southern Railroad, when running at its maximum, is 100 miles per hour. At

MECHANICS-PROBLEMS.

42

that speed what pull

one horse-power miles an hour

by the engine would represent

What

?

pull

when running

at

50.

?

An

express train of weight 250 tons covers 40 minutes. Taking the train resistances on a level track to be 20 pounds per ton at this speed find the horse-power that engine must develop. 136.

40 miles

137.

in

A

tance of

train goes mile,

I

down a grade

then runs up an equal grade with for a distance of

of

i

%

for a dis-

steam throttle being kept shut its

;

it

acquired velocity

500 yards before stopping.

Find

the total resistances, frictional or other, in pounds per ton,

which are stopping

138.

A

it.

caboose and three cars break away from a

and "coast " down a grade of 2 in 100 a distance of i mile then brakes are applied and

freight train for

;

the cars stopped in 200

feet.

Frictional resistances

over whole distance being 15 pounds per ton, what are the brake resistances per ton

Work

= Work

down grade 139.

.'

-(-

Work

of friction'

of brakes

The Baltimore and Ohio Railroad has nowin

service (1906) six electric locomotives.

its

Two of recent

construction are used in handling eastbound freight trains with

steam locomotives through the city of

a distance of about two These locomotives can start and accelerate oh a level track a train of 3 000 tons weight with a current consumption of 2 200 amperes, which

Baltimore, which

miles of tunnel.

includes

.

WORK— HORSE-POWER is

supplied from

reaches

tire

a

power

volts,

but

locomotives through booster stations and

they thus develop

These

What

horse-power

d(j

?

electric locomotives will

grade a freight train an hour.

560

station at

a storage battery at 625 volts.

140.

43

of

i

draw on

400 tons weight

Frictional resistances being 20

at

10

a

1%

miks

pounds per

ton what amperes are necessary with x'oltage of 625

?

ME CI/A A^/CS-PK OBLEMS.

44

ENERGY.

A

141.

train of 150 tons

What brake

hour.

force

running

is

is

50 miles an

at

required to stop

quarter of a mile on a clown grade of 2%, resistance being 15 pounds per ton 1

+

Work

= Work

^^'ork gained

possessed by train

= force

Work

+

in a

Work of resistances

of braltes

in \ mile

it

frictional

X distance

possessed by train

Force

=150

Distance

=

The

distance

is

body would have

found by determining the

vertical height that a

to fall in order to acquire a velocity of 50 miles an

30 miles an hour

hour.

tons

?

=

per second.

44 feet

Therefore

the

=

Now to acquire a -f- feet per second. velocity of A?,^ feet per second a body would fall a vertical distance that can be found from a fundamental formula of falling bodies, velocity

= = h =

\' 2

V

is,

x'A, in

which

^*

varies for different localities

^\'J

-'fo-

That

X 44

'^

is

84. 2 feet.

the train

if

had

vertical height of 84. 2 feet

by the action of gravity from a would have a velocity of '^\^ feet per

fallen

it

Work

second, or 50 miles an hour.

can now be analyzed as in

previous problems.

=

Work

150 X 84. 2 foot-tons

possessed by train

=

Work gained *

The value

isco, 32.15

;

of

^

for

London

for Chicago, 32.16

150

X

(iJjt

X ^'Y-

)

mile

in \

;

is

32.19 feet per second per second

for Boston, 32.16,

;

for

San Fran-

Practical limiting values for the

United States are 32.1S6 at sea level for latitude 49*^ and 32.089, latitude 25"^ and 10000 feet above sea level. In this hook the value 32 is used for ^ so that comIn many cases the table on page igo will be of putations may be shortened. ;

assistance.

The

values of

x^igh

there given are based

on

g

as 32.16.

JFOA'/r—

ENKRG Y.

45

In the Westinghou.se brake tests

142.

(Jan., 1887),

Weehawken, a passenger-train mining 22 miles an hour on a down grade of 1% was stopped in 91 feet. There was 94% of the train braked. Taking the at

frictional resistance as 8

pounds per

brake resistance per ton

(jn

ton, find the net

the part of the train that

was braked, and the grade to which

equivalent.

tliis is

A freight-car weighing 20 000 pounds requires

143.

a net pull of 10 resistance.

If

10 miles per

velocity of

before stopping

A

144.

down

pounds per ton " kicked "

to

overcome

frictional

to a level side track with

hour,

how

far will

run

it

1

cake of ice weighing

a chute the height

1

50 pounds slides

which

of

is

feet;

25

reaches the foot of the shute with a velocity of

During the motion how many

feet per second.

pounds of energy must have been

A

145.

slides

zontal

ship

and

down ways ;

its

i

frictional resistances

equivalent height the

velocity as

of fall

that

If

amount

to

hori-

constant

a

What would

be

the

ship possesses

when she

takes the

.''

the resistance of the water, anchors, and

stop ropes

how

000 tons

20 to the

would produce the same

water 150 feet d(jwn the ways 146.

5

foot in

100 tons.

foot-

.''

cradle that weigh

that sl<jpe

retarding force of

lost

it

30

far will

amount

to a constant force of

50 tons,

the ship of the preceding problem run

after she takes the water

.?

MECHANICS-PROBLEMS.

46 147. tiles

A

six-inch rapid-fire

gun discharges

horse-power expended Consider from what

148.

A railway

is

the

.'

a body would

vertical heiglit

velocity of 2 8oo feet per second,

car of

—

{v

4

fall to

have a

\/ 2.gli).

tons,

moving

at the rate

miles an hour, strikes a pair of buffers which

5

Find the average

yield to the extent of 6 inches.

force exerted

upon them.

=

Work

work

possessed by train

What

149.

car

projec-

What

a velocity of 2 8oo feet per second.

of

S

per minute, each of weight lOO pounds, with

moving

at

sengers, each

is

6 miles per hour, of

What

is

Iriiided

average weight

stopped in 2 seconds, what 150.

by buffers

the kinetic energy of a 2|--ton cable with 36 pas-

pounds.'

154 the average force

is

If

?

the kinetic energy of an electric car

weighing 2\ tons, moving at 10 miles an hour, and loaded with 50 passengers, of average weight 150 pounds .''

151.

The weight

the end of a blow second. 152.

it

What work

ram

600 pounds, and at has a velocity of 40 feet per

of a

is

done

is

in raising

Find the horse-power

of a

it

.''

man who strikes hammer of the hammer on

25 blows per minute on an anvil with a

weight 14 pounds, the velocity of striking being 32 feet per second. 153.

A

blacksmith's

helper

using

a

16-pound

sledge strikes 20 times a minute and with a velocity of

30 feet per second.

Find his rate of work.

WORK— ENERGY.

A

154. I

000

weighing

ball

47

lost in piercing the shield

I

600

A

155.

shot of

I

at

and moves on

What energy

with a velocity of 400 feet per second. is

moving

ounces, and

5

feet per second, pierces a shield,

t

000 pounds moving at the

rate of

How far

feet per second strikes a fixed target.

upon it an 000 tons ?

will the shot penetrate the target, exerting

average pressure

A

156.

weight of

ecjual to a

weighing

bullet

of a rifle with a velocity of

feet per second.

500

mean

the barrel be 4 feet long, calculate the of the powder, neglecting

The

157.

to

problem penetrates a sand bank feet.

What

158.

An

is

the

20

is

the powder 159.

the

in

to the

preceding

depth of

mean pressure exerted by

the sand

3 }

8-hundred weight shot leaves a 40-ton gun

with velocity of 2 the gun

If

pressure

all friction.

referred

bullet

2

ounce leaves the mouth

i

i

1

000

feet.

feet per

What

second

:

the length of

the average force of

is

1

A

2-ounce bullet leaves the barrel of a gun

Find the feet per second. i 000 bullet, and the height from the up in work stored which it must fall to accjuire that velocity. with a velocity of

160.

What

is

the kinetic energy of a 5-hundred

weight projectile fired with a per second 161.

velticity of

2

000

feet

.

An

8-inch

projectile,

strikes a sand butt going 2

000

weight

250 pounds, and

feet per second,

MECHANICS — I'ROBLEMS.

48 stopped

is

what

value

A

162.

25

in

is its

feet.

If

pounds

in

the resistance

hammer weighing

pound has

i

of 20 feet per second at the instant

163.

if

A

it

1

is its

kinetic

164.

and

How many moved

A

165.

if

the steam were cut

against a friction of

speed

the rate of

at

i

1

foot in diameter

what energy to provide

The

.-'

gas engine

50 revolutions per minute, must for an

increase or de-

crease in speed of 3 revolutions per minute 166.

off,

400 pounds exerted

fly-wheel on a 21 -horse-power

nominal

store

2 feet in radius.

turns would the above fly-wheel

on the circumference of an axle

of

head

.'

before coming to rest,

it

1

energy when moving

revolutions a minute

make

strikes the

000 pound.s, and is of mass may be treated as if concen-

its

trated on the circumference of a circle

5

a velocity

fly-wheel weighs 10

such a size that

What

uniform,

Find the force which the hammer exerts on it is driven into the wood -^^ of an inch.

of a nail.

the nail

is

?

fly-wheel

of

a

}

4-horse-power

engine

running at 75 revolutions per minute is equivalent to a heavy rim of mean diameter 2 feet 9 inches, and What is the ratio of the work weight 500 pounds. stored in the fly-wheel to the revolution 167.

work developed

in a

}

A

down once

3

in

horse-power stamping machine presses

every 2 seconds

;

its

speed fluctuates

from 80 to 120 revolutions per minute; and to provide for this fluctuation the fly wheel stores

-i^ths

of

IVORK — EXERGY.

49

What energy

the energy supply for 2 seconds.

all

thus stored per revolution

A

168.

nozzle discharges a stream

i

inch in diam-

eter with a velocity of 80 feet per second,

much work

is

each minute

by

If this energy could all be what would be its power

[b)

.''

Suppose that the above nozzle drives If

An

170.

lifted

per minute

pressure

is

is

is

8590

60 pounds per square inch

nozzle should be used —

;

0.48,

.''

impulse water-wheel must provide

horse-power; efficiency of wheel

inch

a water-

water 20

lifts

the efficiency of the whole apparatus

how much water would be

ful

utilized

.'

wheel connected with a pump which feet.

How

(a)

possessed by the water that flows out

a water-wheel, 169.,

is

?

3,^ ;

use-

water-

what

size

to the nearest eighth of an

.''

Force

= =

Distance

Work

171.

The

area

x distance X velocity x 62^

=

area

x

=

60

x 2.304

five

force

8

V60 x

2.304 x 62

J-

feet.

streams shown on the next page are

being delivered through 100 feet of cotton rubberlined hose- with nozzles full

1

1

inches in diameter.

pressure at end of nozzle

What

inch.

delivering

horse-power

is

is

The

50 pounds per square the

fire-pump

thus

.

Through the 100 foot lines of hose there is a At hydrant the full pressure loss of pressure.

172.

large

75 pounds power is thus is

;

at the nozzle lost

.''

50 pounds.

What

horse-

Force illustrated by

two

fire streams being delivered by the pump service B.B. & R. Knight at Natick, Rhode Island. One stream is being held by men in correct position, the other by men who have been crowded into an awkward and dangerous position. Pressure shown on the gauge at the hydrant was 75 pounds per square inch.

of the large cotton mills of

J'VA'C/'S

II.

— AT

A

J'OJA"T.

51

FORCES.

FORCES ACTING AT A POINT. 173.

An

acrobat weighing

150 pounds stands

in

the middle of a tight rope 40 feet Imig and depresses it

5

feet.

Find the tension

in

the rope caused by

his weiirht.

r>iagr;iin

Fig

Draw

tlie

II.

constvuctior. diagram

sluiwiiig

the rope, the knnwii

force of 130 pounds and an arrow to indicate

it.s

direction.

Tlien

draw the stress diagram. I, ay off 150 jjounds parallel to the known force and at a scale of i inch = do pounds; from one end of this line AB draw a full line parallel with one of the forces from the other end a line parallel with the other force. Complete the parallelogram by drawing free hand the dotted parallel lines AT) and ED. ;

Put arrows on forces beginning with the known force of

— this positivelv in order

acts

around the

downward

— then

full-line triangle

B

to

C and C

has become a diagonal of the parallelogram and force, or

what

is

more

pro]>erly

known

i

^o

pounds

the other arrows will follow to A. is

Thus .\B

the balancing

as the equilibriant.

If this

MECHANICS-PROBLEMS.

52 force

was acting two

of the other

The is

in the

opposite direction

it

would be the

resultant

forces.

full-line triangle

ABC

constitutes the triangle of forces.

many problems in

the keynote to the solution of

It

The mag-

Forces.

and directions of the forces can be found by scaling from or by computations involving similar triangles, thus,

nitudes

this triangle,

geometry or trigonometry. Forces-At-A-Point problems therefore can be solved as follows

Draw a construction diagram showing dimensions and Draw a stress diagram. First, the known force.

:

loads.

Complete the parallelogram. Put arrows on full-hne triangle. Scale or compute the stresses.

A speed-buoy

174.

is

thrown

into the water

behind

a ship, and the pull on the buoy by the water

the ship

The two ropes make an angle of

ment.

Find the stresses on the ropes.

pounds.

Two men

175.

of ropes.

One

pull a

that connect the

60 buoy with is

15° at their point of attach-

body horizontally by means

exerts a force of 28 pounds directly

pounds in direction N. What single force would be equivalent to

north, the other a force of 42

42°E. the two

}

Three cords arc knotted together

176.

these

;

one of

pulled to the north with a force of 6 pounds,

is

With another to the east with a force of 8 pounds. whole the what force must the third be pulled to keep at rest 177.

?

Two

persons

lifting a

body exert forces

pounds and 60 pounds on opposite

of

44

sides of the ver-

FORCES— AT A tical,

POINT.

53

What

but each with an inchnation of 28°.

force would produce the

same

effect

single

?

A force of 50 units acts along a line inclined an angle of 30° to the horizon. Find, by construc-

178. at

tion

or otherwise,

and

horizontal

its

vertical

com-

ponents. 179.

Explain the boatman's meaning when he says

that greater force

is

developed when a mule hauls a

canal

boat with a long rope than with a short one.

Is the

same

master 180.

true of a steam-tug

when towing

a four-

1

Two

strings,

one of which

horizontal, and

is

the other inclined to the vertical at an angle of 30°,

Find the tension

support a weight of 10 pounds.

in

each string. 181.

Two

at a point. is

forces of 20 pounds, and one of 21 act

The

120°, and

angle between the

between

first

and second

second and

the

third,

30°.

bind the resultant. 182.

point

Forces of 9 pounds, so

that

the

A

12, 13, and 26, act at a between the successive

Find their resultant.

forces are equal. 183.

angles

weightless rod,

3

feet

long,

is

supported

horizontally, one end being hinged to a vertical

and the other attached by above the hinge

;

a string to a point 4

a weight of

1

20 pounds

is

wall,

feet

hung

Calculate the from the end supported by the string. the rod. pressure along the and string, tension in the

MECHANICS-PROBLEMS.

54

A

184.

weight of loo pounds

a weightless rod or strut rests

in

between a

a corner

upper end

wall, while its

horizontal wire

4

is

fixed to the top of

5 feet long

whose lower end and a

floor

vertical

attached to the wall by a

is

Calculate the tension in

feet long.

the wire, and the thrust in the rod.

A

185.

a

AB

rod

horizontal

hinged

is

position

at

by a

angle of 45° with the rod

A

string

and supported in BC making an

the rod has a weight of

;

10 pounds suspended from B.

Find the tension

the string and the force at the hinge.

can be neglected.)

of the rod

A

186.

in

(The weight

simple triangular truss of 30 feet span and

depth supports a load of 4 tons Find the forces acting on

feet

5

at the apex.

rafters

and

A

187.

tic rod.

derrick

is

set

as

sketch, the load being 8 tons. Fig. la.

188.

high,

as

A

stress in the stiff-leg

boom

shown

in

tackle.

189.

cut,

derrick, with

is

raising

two boilers

Find stresses

mast

in

in

Fig. 13.

feet

of

boom

stress

boom

in

tackle

and

of towers for six-

master .shown on page 24 when bucket.

weighing with

by

55

(Sec illustration on page 55.)

Find the

compression

the tackle.

85 feet long, set with tackle 40 feet long,

50 tons weight.

and

steel

boom and

shown in Find the

its

load 2 tons,

is

Fig.

set in position

13.

shown

FOA'C£S—AT A rO/XT.

s-

^

ss

ArF.CI/A.V/CS-PRO/iLilJ/S.

56

moves

the boat

Find the

canal.

when the 193.

on each rope

pull

A

angle of

down

straight

boat

is

is

800 pounds.

being towed by a rope making an

30° with the boat's length

pressure of the water and rudder

the resultant

;

inclined at 60°

is

and the tension

to the length of the boat, is

middle of the

the

total effective pull in that direction,

in the

rope

Find the

equal to the weight of half a ton.

re-

sultant force in the direction of the boat's length. 194.

In a

pressure a

22 500 pounds

is

maximum

piston. 195.

steam-engine the piston-

direct-acting ;

the connecting-rod makes

angle of 15° with the line of action of the

Find the pressure on the guides.

A man

weighing 160 pounds

sits

in a

loop at

the end of a rope 10 feet 3 inches long, the other

What

end being fastened to a point above. tal force will pull cal,

and what

will

him

2 feet 3

then be the pull on the rope

196 A man weighing 160 pounds mock suspended by ropes .which are

and 45° to

horizon-

inches from the verti-

Two

attached

to

in

flexible string

ham30°

each rope.

equal weights,

the

a

inclined at

Find the pull

vertical posts. 197.

sits in

.''

W,

e.xtremities

which passes

are

of

a

o\'er

three of

tacks arranged in the form an isosceles triangle with the

base horizontal, the vertical angle at the upper tack

being 120°.

Find the pressure on each tack.

FORCES— AT A

A

198.

hung from tached to 3 feet

2

from

a point

5

feet

C by two

and the string

strings,

hung from

is

A

BC

$7

without

long,

ends and to the point

its

long,

pounds

AB

rod

PO/jVT.

which are

the string

;

feet

2

weight,

atis

a weight of

;

and a weight

AC

is

pounds

of 3

Find the tension of the strings and the may be in equilibrium.

B.

condition that these

A

199.

strings,

7

weight of 10 pounds

suspended by two

is

and 24 inches long, the other ends

of

which are fastened to the extremities of a rod 25 Find the tension of the strings inches in length. when the weight hangs immediately below the middle point of the rod. 200.

AB

is

a wall, and

perpendicular gi\'en length If

all

AB A

The end

is

is

a fixed point at a given it

;

uniform

a

C with one end

a uniform

end B

C beam

Point

against

is

is

a

smooth

verti-

attached to a rope

vertically

above

[

A, IZ

feet.

1 ~

is

4

feet,

rope

7

I

the rope. 202.

slope

AB.

beam weighing 300 pounds.

Represent the forces acting, and find the pressure against the wall and the tension

length of

in

rod of

in ecpiilibrium.

rests against

cal wall, the

CB.

placed on

is

C

from

the surfaces are smooth, find the position in

which the rod 201.

distance

A of

wagon weighing inclination 30°,

Fig.

2

16.

200 pounds rests on a

What

are the

equivalent

forces parallel and perpendicular to the plane

?

58-

MECnA.YICS-riiOBLEMS.

,

AB

203.

A

end

round one

a rod that can turn freely

is

B

the other end

;

rests against a

smooth

in-

In what direction does the plane react

clined plane.

upon the rod

your answer by a diagram

Illustrate

?

showing the rod, the plane, and the reaction.

A

204.

a

wagon weighing

smooth road which

rises

2

tons

4

feet

tance of 32 feet horizontally road.

order

What must the that it may move What

205.

Ijy

the

parallel with the plane

zontal

of

A is

horse

is

of

is

a smooth plane (a)

when

when it

is

The

207.

is

45°

and a force

45° to the

the

hori-

rails

;

What

672 pounds.

the effective force along the rails

plane

in

attached to a dump-car by a chain,

inclined at an angle

the force exerted by the horse is

{b)

}

exceed

1

206.

which

pull

uj)

200 pounds

pull

is

rope

1

weight can be drawn

by a

drawn up

a rope parallel to the

wagon

i

5

to be

vertically in a dis-

pull of the

rising

in

is

.?

angle of inclination of a smooth inclined :

of

a force of 3 pounds acts horizontally, 4 pounds acts parallel to the plane.

Find the weight which

they

will

be just

able to

support. 208.

A body rests

on a plane of height

3 feet,

length

body weighs 14 pounds, what force actting along the plane could support it, and what would 5

feet.

If the

be the pressure on the plane

.''

J'OKCES^AT A

A

209.

one

on a given part

way, where the inchnation

is

Two

45".

smooth tram-

of a

support an ecjual

30°,

is

empty trucks on another

of

inchnation 210.

59

numlicr of loaded trucks each containing

ton, standing

number

PO/.V'J:

where the

part,

Find the weight of a truck,

planks of lengths 7 )-ards and 6 )'ards of eacli on a horizontal plane, the

one end

rest with

other ends

in

contact abo\'e that })lane

;

twri

weights

are supported one on each plank, and are connected

by a string passing over a pulley the planks

What

the weight on

;

considered

diameter of wheel

of a 5

reaction.

it

Find o\-er a

its

tlie

It is

A

;

P,

W, and R

required to find

rectangular

bo.x,

2 tons,

is

least horizontal

the I-'.)

-^^

contain-

horizontal table, and

is

tilted

'^^^'

about one of

edges through an angle of 30.°

tween the

An

ball

and the

is

''

its

lower

P"ind the pressure be-

bo.\.

iron sphere weighing 50

against a smooth vertical wall

which

load

stone 4 inches high.

a 200-pouncl ball, stands on a

213.

21 pounds.

the wheel begins to rise three

forces are acting

ing

wheel with

feet.

force necessary to pull

212.

tlie first i>lank is

.''

The weight

(When

the junction of

the weight on the other, friction not being

is

211.

at

pounds

inchned 60° to the horizon.

sure on the wall and plane.

is

resting

and a smooth plane

Find the pres-

MECHANICS-PROBLEMS.

6o

A

214.

beam weighing 400 pounds

rests with its

ends on two inclined planes whose angles of inclination to

Und

the

A thread 14 feet long fastened to A and B which are in the same horizontal

two

the horizontal are 20° and 30°.

pressures on the planes. 215.

is

points

and 10

feet apart

;

a weight of 25

P

the thread at a point therefore

BP

AP

so chosen that

is

line

tied to

—

6 feet

is

The weight being

8 feet long.

is

pounds

thus

suspended, find by means of construction or otherwise,

what are the tensions

AP

and

BP

two threads 4

feet

and

of the parts

of the

thread.

AC

216.

BC

and

are

long, respectively, fastened to fixed points

which are weight of

means it

same horizontal

in the

50 pounds

and

What

fastened

BC

is,

of

A and

A

boiler

weighing

3

by tackles from the fore tackles

make

in

Kach

A

apart

;

a

Find, by

to C.

a

5 feet

and B,

scale, the pull

of the threads

state of

tension.

1

000 pounds is supported and main yards. If the

angles of 25° and 35° respecti\'ely with

the vertical, what

is

the tension of each

}

piece of wire 26 inches long,

enough to support is

B.

covu'se,

feet

are the forces producing the tension

217.

218.

6

drawn to

of a line construction

causes at the points

AC

is

line

A

directly

a

load of

attached to two points 24 inches apart

horizontal line.

and strong 100 pounds, in

the same

Find the maximum load that can be

.7 7'

foj?cj:.s-

suspended breaking

tically

6i

middle of the piece of wire without

at the

it.

A

219.

A rO/iVT.

picture

50 pounds weight hanging ver-

of

against a smooth wall

passing over a smootli hook

two

are fastened to

the ends of the string

in

p<.)ints

supported by a string

is ;

the upper rim of the

frame, which are equidistant from the center of the rim,

and

angle at the peg

tlie

is

ViwA the tension

60°.

in the string.

A

220.

W

weight

by two connecting points A and B,

attached

cords of lengths a and b to two

fi.xed

and separated by a horizontal interval rium under the action

V and

stresses 221.

Two

O

of

are in equilib-

Required the

the cords.

equal rods

C and

together at B. in

in

c,

gravity.

AB and BC are loosely jointed A rest on tw(j fi.xed supports

the same horizontal

line,

cord equal in length to AB.

and are connected by a weight of 12 pounds

If a

be suspended from B, what is the pressure produced along AB and BC, and the tension in the cord .?

Two

222.

a pair

of

spars are lashed together so as to form

shears as shown in

sketch.

their " heels "

20 feet

They stand with apart,

and would be 40

What

vertical.

guy and thrust of

30 tons

is

is

feet high

the tension

in the legs

being

lifted

when

when

in

the

a load

f

VC Suppose

that a single leg

shoiiUl replace tht

Fig. 18

MECHANICS-PROBLEMS.

62 two

The

spars.

stress

can easily be found in

imaginary leg by

tliis

considering that at A, in this plane, three forces meet,

nary

leg, the

back guy, and the

A

consider the three forces at find the stresses in the

223.

When

will e.\ist for

two

— the

imagi-

Then

30 tons.

and thus

in the plane of the legs,

ecjual spars.

the spars

the load

vertical load of

become

(jf

1

Figs.

224.

pair

what stresses

vertical

30 tons

19-20 show a

Sparrow's

Steel

The two

pany.

Md.,

Point,

Mar}'land

the

erected at

shears

of

front legs

are hollow steel tubes feet long, Fig.

The back

19.

leg

is

feet J

hydraulic machines

f(jr

They is

75

]"3ach leg of a pair of

in is

feet

long.

I-lnd

the

of

vertical.

connected to

is

How

the

when

these legs

being

shears

are spread 20 feet at the foot. forces

ii6

and inclined 35

operating the shears.

much are the forces acting Krupp gun weighing 122 tons 225.

out

26 feet long, and

for

Com-

a

lifted.-'

50 feet long.

is

The back

stay

on each

acting

member when

lifting a load of 20 tons at a distance of 20 feet from the foot of the shear legs, neglecting the

weight of structure. 226.

Shear legs each 50 feet long, 30 feet apart on

horizontal ground, vertically

meet

at point C,

above the ground

;

which

stay from

C

is is

45 feet inclined

FORCES^ AT

A POEVT.

63

4

^ 0»^^'-

iW

^ ^'-

;

FORCES — AT A POIXT.

A

west frum the post.

Find the forces

C.

C

weight of

the

when C

tons hangs from

when C

— is

ist,

when

due east

due south.

is

The view on

228.

5

and stays

jib

southeast of the post; 2d,

is

3d,

in

65

opposite page shows one of the

Pan American," use on the Great which are 57 feet

largest dipper dredges ever built, the "

constructed

Buffalo in 1899 for

at

An

I^akes.

A-frame, the legs of

long and 40 feet apart at the bottom,

is

held at the

apex by four cables which are 100 feet long.

boom

is

feet

53

long

and weighs

handle, which weighs about 4 tons,

60 feet long, end a dipper weighing 16 tons, dredge up 8| cubic yards, or about 12

and carries on

which

will

dipper

is

is

its

tons, of material at

The

30 tons.

one load.

operated by a wire rope passing over

a pulley on the outward end of the boom. position represented is

The The

In the

by the outline sketch, the boom

inclined to the

water

surface at

an angle of

30",

the dipper

car-

rying

the

full

and the han-

load,

dle

is

in

is

a horiFig. 21.

position

zontal

with

its

boom of the

boom.

middle point supported

23 feet

at

a point

from the foot of the boom.

A-frame

is

vertically

Compute the

on the

The apex

above the foot of the

forces acting in the

100-foot

MECHANICS-PROBLEMS.

66

back-stays (considering

them

to be one rope, in posi-

tion as per sketch), in the legs of the A-frame, in the

boom, and

A

229-

whose vertex

tripod

AB, AC, AD,

are

CD

6

A, and whose legs

is

and 9

of lengths 8 feet, 8.5,

sustains a

spectively,

B, C, D,

rope which raises the dipper.

in the wire

load

2

of

form a triangle whose sides are BC 7 BD 8 feet. Find the stress in each

feet,

Sketch the figure and put on the dimensions. the base

BCD, and

re-

The ends

tons.

Then draw

feet, leg.

to scale

in this horizontal plane locate the vertices A',

A", and K'" ai the three faces of the pyramidal-shaped figure that formed by the legs of the tripod. Perpendiculars drawn from A',

is

A"

and A'"

to their respective sides of the triangle

at their intersection the projection of vertex

plane, for example, through tersection leg,

E

and the

with fine stress in

fore, also the stress in

"2-30.

A

CD.

can be graphically determined as hereto-

it

AC

and AD.

bottoms of legs be spread,

A

i

if

ton stress

232.

far apart can the

an equilateral

in

will

chandelier of weight 500 pounds 1

exist in chains

ABCD

is

2 feet

X

8

is

X

leg

to 8

?

hang

Two

.

What should What stresses

be 20 feet long. .'

1

a square

9 act in directions

triangle,

come on each

be the length of the third chain

would

to be used for

is

How

under the middle of a triangle of the chains are to

will locate

pass a vertical ;

tripod with 8-foot legs

so that not over

BCD

Now

AB and the load of 2 tons note the inAE can be considered as an imaginary

lowering a 2-ton water-pipe.

231.

A.

;

forces, of

AB, AC, and

i

AD

Find the magnitude of their resultant.

pound,

6,

and

respectively.

FORCES— AT A

A, B, C, D, are the angular points

233.

taken

PO/iVT.

in

order

6'J

of a square

three forces act on a particle at A,

;

one of 7 units from A to B, a second of 10 units from D to A, and a third of 5 V2 units along the viz.

A

diagonal from

to

Find, by

C.

construction

or

otherwise, the resultant of these three forces.

Forces

234. of

P, 2P, 3]',

a square A, B,

and 4P act along the sides

C, U, taken

in

Plnd the

order.

magnitude, direction, and line of action of the resultant.

A

235.

sinker

is

attached to a fishing-line which

is

Show by means

of

then thrown into running water. a

diagram the forces which

on the sinker so as to

act

maintain equilibrium.

A uniform rod 6 feet long, weighing

236. is

long which

is

end and to a the rod will

A

237.

A

attached to the rod

nail verticall}'

Show by

tant.

at

10 pounds,

supported by a smooth pin and by a string 6 feet

;

it

keep the rod

4 feet in

dis-

which

AB

can turn freely round a hinge

an inclined position against a smooth

peg near the end B of the rod.

pin,

from one

to rest.

light rod

rests in

foot

the position

construction

come

i

above the

;

Show at rest,

a weight in

is

hung from the middle

a diagram the forces

and name them.

which

MECHANICS-PROBLEMS.

68

A

238.

weight

W on

a plane

inclined 30° to the horizontal

supported as shown in

cut.

is

The

Find the power to the weight.

angles & being equal. ratio of the

Discuss the action of the

239. Fig. 21.

wind Let

AB

be the

keel,

CD

the

in propellinga sailing-vessel. Let the

sail.

force of the wind be represented in magnitude ^

"v^, 1^^

«''»'

and direction by EF. The component GF of EF, perpendicular to the sail, is the effecthe tive component in propelling the ship other component EG, parallel to the sail, is ;

useless

GF

but

;

The component

sidewise.

E Fig. 33.

drives the ship forward and*

GH

of

GF, perpendicular

to

AB,

pro-

and the other component HF, along produces forward motion, or headway.

duces side motion, or leeway the keel,

;

A

240.

sailing-boat

is

being

driven forward by a force of 300

pounds

What of

Show

leeway.

rudder 242.

force

motion of the boat 241.

Fig. 24.

shown in Fig. 24. is P acting in direction

as

.

Discuss the action of the

rudder of a vessel in counteracting that one effect of the action of the

to diminish the vessel's motion.

is

A thread

two points

of length / has its ends fastened to

in a line of length

vertical with angle d

by means

of a

;

a weight

smooth hook.

c,

and inclined to the

W hangs on the thread Find the position

in

FOKCF.S.

— AT

which the weight comes to

A PO/XT.

rest

69

and the tension

in

the

thread.

A

243.

smooth

along a cord that horizontal

ring weighing

is

The

line.

40 pounds shdes

attached to two fixed points distance

in a

between the points

being one-half length of cord, find position weight will come to rest and the tension

in

in

which

the string

near the points of attachment.

A

244.

a

smooth

a string

Show hoop

small heavy ring A, which can slide upon is

kept in a given position by

AB, B being the

highest point of the hoop.

vertical hoop,

that

the

between the ring and the

pressure

equal to the weight of the ring.

is

Draw

245.

a figure showing" the mechanical con-

ditions of equilibrium

when

a uniform

one extremity against a smooth

beam

rests with

and the

vertical wall,

other inside a smooth hemispherical bowl.

246.

A

ball

8

inches in diameter, weighing 100

pounds, rests on a plane inclined 30° to the horizon,

and

is

held in eciuilibrium by a string 4 inches long

attached to a sphere and to an inclined plane.

Rep-

resent the forces acting, and find their values.

247.

A

plane,

and

uniform sphere rests on a smooth inclined is

held by a horizontal string.

To what

point on the surface of the sphere must the string be

attached

.'

Draw

a figure

showing the forces

in action,

MECHANICS-PROBLEMS.

7°

A

248.

uniform bar of weight 20 pounds, length 12

with one end inside a smooth hemispherical bowl, and is supported by the edge of the bowl Draw the force's with 2 feet of the bar outside of it.

feet, rests

producing equilibrium, and find their values. The

/ stresses in a roof or bridge truss that carries a uniform load

are best determined by finding in place of the uniform loads equivalent apex loads. And a fact that is often obscure to students is, that a part of this uniform load

is

not included in

our computation of stresses. In the truss of Fig. 25 the portions a of uniform load are not

A C and be further understood by

included in the compressive stresses of

C

B.

This fact

will

solving the problems that follow.

249.

Two

floor

a post. Fig. 26.

of

16 feet length meet at

load,

10 feet width of bay for

beams

The

150 pounds per square foot. What will If it is found that be the load carried by the post the post must be removed so as to give better floor What space the plan of Fig. 27 could be used. each beam,

is

.?

would then be the stress

in the short post

(3

feet

long), and in the two rods, and in the floor beams

Fig. 26.

250.

Now

if

.'

Fig. 27.

the same conditions exist as in the

preceding problem, except that the rods, instead of

FORCES — AT A

POIA'T.

71

being fastened to the ends of the beam are fastened to straps on the outside of the wall,

what

will

be the stresses

beam

?

251.

The

in post, rods,

and

floor

then

slopes of a simple triangular roof-truss

are 30^ and 45°, and the span

is

50

feet.

The

trusses

are set 10 feet apart, and the weight of the roof covering and

snow

is

Find the stresses

50 pounds per square foot of in tie-rod

and

rafters.

roof.

MECHANICS-PROBLEMS.

72

253.

Find the stresses

in the

Dis-

king-post truss of Fig. 31.

tance between

trusses

12 feet.

^'^' ^''

is

There is a uniform load pounds per square foot of roof surface and pounds

of

100

i

000

at the foot of the post.

A

254.

king-post truss has a span of 18 feet and a

Compute the stresses due to a load of 9 feet. 14 000 pounds at the middle.

rise of

A floor

beam 16 feet long and carrying a uni200 pounds per linear foot is trussed by Confeet below middle of beam. rods that are sider a joint at the middle and find stress in rod. 255.

form load

of

H

MOMENTS The all

principles of

Work

can be used

to solve

nearly

problems that belong to the subject of Mechanics,

but

certain classes of problems shorter

in

are possible.

Moments can be used

ples of

Definition.

methods

In the following problems the prmci-

— The

Moment

to advantage.

of a force about a point or axis

is

the product of the force times the perpendicular distance from the the line of

point to force

action of the force

;

or,

briefly;

Moment

is

X perpendicular.

Clockwise motion

will

be taken positive; the opposite direction,

negative.

In beginning the solution of problems always state which point

moments are taken about; Moments about axis B."

or axis the B," or "

thus,

"Take moments about

^

FORCES— MOMENTS. 256.

A

piece of shafting 10 feet long, and weighing

100 pounds, rests horizontally on

two horses placed

A

at

keyed 2\

feet

How many

pounds

from

to just raise

one a

will

f

ends.

its

pulley weighing 75 pounds

end

73

^' \'^*

it c

'^

i

is

^\

Fig- 32-

end.

man have

to

at

lift

the other

it.?

100 pounds, the weight of shaft, acts

downward

at the

middle

point; 75 pounds, the weight of pulley, acts downward at D, 23 feet from B. Find the required force acting upward.

Take moments about B, + P X 10 — 100 X 5 — 75 X 2\ P = 67.5 pounds.

=

o.

.'.

257.

A

uniform lever

inch in length weighs

is

18

ounce.

i

inches long, and each

Find the place

fulcrum when a weight of 27 ounces

at

of the

one end of the

lever balances a weight of 9 ounces at the other end. 258.

4 feet

A

i6 feet long balances about a point

lever

from one end;

if

a weight of 120 pounds be

attached to the other end,

6

feet 259.

I

5

from that end.

A

balances about a point

light rod of length 3

pounds and

end

it

h'ind the weight of the lever.

respecti\'e]y

yards has weights of

pounds suspended

3 ;

at

the middle and

Find the

balances on a fulcrum.

it

position of the fulcrum, and the pressure on 260. tally

A

pole 12 feet long sticks out horizon-

stiff

from a vertical

of 28

wall.

pounds were hung

along the pole

it.

may

ture with safety

.''

a

It

at

would break

1

a weight

How

the end.

boy of weight

if

12

far out

pounds ven-

MECHANICS-PROBLEMS.

74

A man pulls

261.

loo pounds on the end of a 7-foot

What

oar that has 2\ feet inside the rowlock.

is

the

pressure on the rowlock, and resultant pressure causing the boat to 262shell,

if

move

?

Find the propelling force on an eight-oared each man pulls his oar with a force of 56

pounds, and the length of the oar outside the rowlock

is

263.

three times the length inside.

A

light bar,

pounds and

5

pounds from 264.

A

5

feet long, has

pounds suspended from its

middle point.

weightless lever

its

Where

AB

weights of 9 ends, and 10

will

it

balance

.

of the first order, 8

from B, has a weight pounds hung from A, and one of 17 pounds from B. From what point must a weight of 2.5 pounds be hung to keep the lever horizontal 1 feet long, with its fulcrum 2 feet

of

S

265.

A weight

of

100 pounds

is

supported by a

rope which passes over a fixed pulley and to a

1

2-foot lever at a point 2 feet

which

pended 266.

is

at the

end.

at the other

Eight

What

is

attached

from the fulcrum

weight must be sus-

end to keep the lever horizontal

sailors raise

?

an anchor, of weight 2 688

pounds, by pulling on the spokes of a capstain which

has a radius of 14 inches.

If

they

all

pull at equal

distances from the center and exert a force of

pounds each, what 267.

Is there

is

the distance

56

t

any reason why a man should put spoke of the wheel rather than to

his shoulder to the

the body of the wagon in helping

it

up

hill ?

FORCES — MOM/iiVTS. 268.

props

A

AB,

rod

A

at

and B

of length 15 feet,

269.

a

A

supported by

at a

prop

tlie

is

point 7 feet from A.

sus-

Find

A.

at

9 feet long, to which is attached 150 pounds, at a point 3 feet from

light bar,

weight

one end,

is

a weight of 200 pounds

;

pended from the rod the pressure on

75

of

borne by two men.

is

the weight

is

Find what portion

of

borne by each man, when the bar

is

horizontal.

270.

A

The

on two pegs

light rod, 16 inches long, rests

9 inches apart, with

its

center

midway between them.

greatest weights, which can be suspended sepa-

rately from the

two ends

of the rod without

ing the ecjuilibrium, are 4 pounds and spectively.

Find that 271.

There is another weight weight and its position.

A

AB, 20

light rod

inches

5

disturb-

pounds

re-

fixed to the rod.

long, rests

upon

two pegs whose distance apart is equal to half the How must it be placed so that the length of the rod. pressure on the pegs

3W,

may be

equal

when weights 2W,

are suspended from A, B, respectively

272.

The

long and at its

its

horizontal roadway of a bridge

weight, 6 tons,

middle point, and

at its ends.

What

it

pressure

is

may be supposed

rests is

on

.?

30 feet to act

similar supports

borne by each

when a carriage weighing way across the bridge

supports of the

.-'

2 tons

is

of the

one-third

MECIIANICS-rROBLEMS.

']6

"

273.

We

have a

we have

times

to

Can we

go on them.

weighing one end weights

its

and

some-

at a

the correct weight by

get

time and then adding the two

" }

A

274.

hay -scales,

of

set

weigh wagons that are too long to

rod, i8 inches long, can turn about

ends, and a weiglit of

pounds

5

at

one of

fixed to a point

Find the force which

6 inches from the fixed end.

must be applied

is

the other end to preserve equilib-

rium.

A

275.

rests

end

weighing 10 pounds

straight uniform le\'er

on a fulcrum one-third of with a

its

length from one

weight

of 4 pounds at Find what vertical force must act at the Ifiaded

is

it

;

that end.

other end to keep the lever at rest.

276.

of a

A weight

of

56 pounds

uniform bar which

20 pounds

the fulcrum

;

which the weight

is

is

is

attached to one end

ten feet long, and weighs

is

2

What

applied at the other end to balance

277.

AR

is

a point in

is

a lever turning

it

weight must be

1

a horizontal uniform bar

F

a weight of

from the end to

feet

attached.

10 indies fmni A.

\ feet long,

and

Suppose that

AB

i

on a fulcrum luidcr F, and carr)'ing

40 pounds

at

R

;

weight of lever, 4 jiounds.

kept horizontal by a fixed pin above the rod, 7 inches from F and 3 inches from A, find the pressure If

it

is

on the fulcrum and on the fixed

pin.

;

FOR CES-iMOMENTS.

An

ununiform

77

weighing 4 pounds, balances about a point 4 feet from one end. If, 2 feet from this end, a weight of 10 pounds be 278.

rod, i6 feet

huns, what weig'ht must there be hum/ from the other

end so that the rod may balance about point

279.

men

Si.K

are to carry an iron

and weighing 90 pounds per one-sixth of the weight.

Where must

A

280.

)-ard

;

rail

each

Two men

one end and the other four

b)'

is

5

.'

hind

how

the rod must extend beyond the the pressures on

;<

A davit

is

nails,

i„,

>

supported by a foot-

collar B, placed

and a

far

the nails be

pounds.

A

from

A

the difference of

step

lift

of a cross-bar.

horizontal and 7 inches long,

281.

sustains

rod 2 feet long, with a weight of 7 pounds is placed upon two nails, and B.

the ends of if

long

feet

are to

means

the cross-bar be placed

60

man

at its middle point,

AB

middle

its

.''

5

feet

A boat weighing two tons is apart. about to be lowered, and is hanging 4 feet horizontally from vertical through the foot-step

and

Determine

collar.

the forces which must be acting

and

A

B.

282.

40

at

feet,

A

highway bridge

of

span 50

feet,

breadth

has two queen-post trusses of depth 8 feet

and each truss equal parts.

is

The

divided

bridge

is

by two posts

into three

designed to carry a load

;

MECHANICS-PROBLEMS.

78

loo pounds per square foot of

of

Find

floor surface.

the stresses developed. at the two panel points C and D methods of Moments, find the reactions R and Rj, observing, as explained for problem 251, that at each end half a panel of the load goes directly on the abutment and does not affect our

Find the loads for each truss

then, by the

computation of stresses

known makes

thus

it

in the

members)

(stresses in the

members

at

equal to load at

is

the abutments.

Likewise at foot of

C

or

find

One I),

A

— the

Fig.

35.

king-post truss of 20 feet span, as

has a uniform load of 10

member and

of the post.

Determine the reactions and

A

pipe

5-foot water-pipe

when

filled

pounds per square feet

;

285.

depth,

A

5 feet.

on

10 000 pounds at the foot

is

stresses.

carried across a gully

by two king-post trusses that are spaced 6

The

shown

X 200 pounds

the horizontal

284.

stress in

the other two can be

at a point.

Fig. 34-

283.

known,

is

— and

found by methods of three forces acting

in Fig. 35,

reactions

two unknown forces

to

posts three forces meet in a point.

post which

The

of the truss.

the

possible

feet apart.

with water makes a load of 200

Length

foot.

Find the

of

trusses

is

40

stresses.

storehouse has queen-post trusses in the

top story; 50 feet span, 10 feet depth, lower chord divided into

3

equal parts

;

trusses 8 feet apart, and

load 150 pounds per square foot.

Find the

stresses.

5

FOR CES — MOMENTS.

A

286.

ladder with 21

79

rungs a foot apart leans

against a building with incHnation of 45°.

when

pressure against the building

a

Find the

man weighing

150 pounds stands on the eleventh rung. 287.

Like parallel forces of 10 and 20

perpendicularly units acts

AB.

t(j

A

from

A

rod

downwards, and 2

;

units act

a force of

diagram how

in a

1

of the

acts.

it

of

Where must

upwards.

5

3

distance of two feet from this

at a

be applied to keep the rod

at rest

a force of

.''

i pound each act on hand one acts vertically upwards, the two others vertically downwards, at dis-

289.

Three

1^

acted on at one end by a force of

is

end by a force

and

Find the resultant

to B.

three forces, and show 288.

A

at

parallel forces of

a hc)rizontal bar.

The

tances 2 feet and

3

Draw and

feet respectively,

and

their resultant,

A rod

and B,

C and

D

12

is

state exactly its

first.

magnitude

suspended horizontally on two points,

feet apart;

240 ptiunds

Take

at

D

a point O,

with respect to

O

;

A

hung

at C,

B

points

3 feet

and a weight

the weight of the rod

is

;

a C)f

neglected.

midway between A and B, and find the algebraical sum of the moments

of the forces acting 291.

is

A and AC = BD =

between

are taken, such that

weight of 120 pounds

rests

from the

position.

290.

A

right

horizontal

on the rod on one side of O. rod without weight, 6 feet long,

on two supports

at

its

extremities

;

a weight of

MECHAiVICS-PROBLE.VS.

8o

672 pounds is suspended from the rod at a distance of Find the reaction at each 2i feet from one end. point of

support.

sure of only

from the

1 1

otlier

suspended

2

If

the greatest distance

is

support at which the weight could be

?

Three equal

292.

one support could bear a pres-

pounds, what

parallel forces act at the corners

of an equilateral triangle.

Find the point

of applica-

tion of their resultant.

Find the center

293.

poimds,

6,

and

of the three parallel forces

which act respectively

8,

4

at the cor-

ners of an equilateral triangle.

P,

294.

direction

Q, R, are parallel forces acting in the .same the angular points respectively of an

at

equilateral triangle

ARC.

position of their center

direction of the force

Show

295.

that

if

If

Q

P

= 2O =

also find

;

is

two

in order,

the

point in the base

296.

Draw

D

if

the

forces

of their

be represented in sides of a

triangle,

moments about every

the same.

a square

whose angular points

in order

D, and suppose equal forces (P) to act to A, to B, and B to C respectively, and a

are A, B,

from

is

sum

3R, find the

position

reversed.

magnitude and direction by two taken

its

C,

A

fourth force (2P) to act from

C

to D.

Find a point

FORCES—MOMENTS. such that,

moments

the

if

with respect to

it,

A BCD

is

297.

being 4

D

from

the algebraic

the forces are taken

sum

zero.

is

a square, the length of each

and four forces act as follows

feet,

to

of

8i

A,

pounds from B

3

2

:

side

pounds

A,

to

4 pounds from C to B, and 5 pounds Find the algebraical sum from D to B.

moments

of the

of the forces about C. Fig. 36.

The forces act as in the figure. Draw CM perpendicular to DB. Then, .-.

CM = DM. CD-= CM^ + MD-: CD ,-.

:CM-.

CM

CM = .".

sum

Algeliiairal

ABCD

is

S3 nearly.

moments about C

of the

= - 2 - DC + 3 / CK + = -2X4 + 3X4+0 + = -S + 12 X 14.15 = — 10.15 units. 298.

I.

.t

0-5

'

CM

5 (2.83)

a scjuare, and

AC

is

a diagonal

:

forces P, Q, R, act along parallel lines at B, C, D, respectively, direction,

O

acts in the direction

and

R

opposite

in

A

to C,

P opposite Find, and

direction.

show

in a diagram, the position of the center

Q =

SP and

299.

AB and

is

5

R =

Draw

a rectangle,

ABCD, such that BC forces

three-fourths of the side units act from

spectively.

when

7P.

B

to

A,

B

to C,

;

and

D

the side of

to

3,

g,

A

re-

Find their resultant by construction or

MECHA NICS-PR OBLEMS.

82

show

otherwise, and

in

your diagram exactly how

it

acts.

Prove

300.

that,

at the

are situated

parallel

if

distance of their center from

cumscribing

circle

forces

i,

2,

3, 4,

5, 6,

angles of a regular hexagon, the

is

the center of the

cir-

two-sevenths of the radius of

that circle.

Six

301.

forces,

represented by

the

of a

sides

regular hexagon taken in order, act along the sides to turn the

hexagon round an axis perpendicular to

Show

plane.

that the

moment

of the forces

is

its

the

same through whatever point within the hexagon the axis passes.

A

302. feet,

triangular table, sides 8

feet,

9

and 10

feet,

is

sup-

ported by legs at each corner, and

350 pounds is placed on it 3 feet from the 8-foot side, 2 feet from the

from the the legs 303.

feet

with

9-foot

What

side,

and

2.6

feet

are the pressures on

.''

A

A, with feet

lo-foot side.

triangular shaped platform right-angled at side

long, side long,

is

freight

AB AC

10

40

loaded at

|| ^

50

Fig- 38. pounds per square foot surface. Find the load carried by each of the three

corner-posts.

FORCES— MOMENTS. O, the center of gvavity,

is at

83

one-third the distance from the mid-

Load equals 10 000 pounds. thus find load carried by C. Take moments about axis AB Then take moments about sides AC and BC.

dle of any base to the opposite vertex.

—

304.

Four

vertical forces,

7,

5,

10,

and 12 potinds,

act at the corners of a square of 20-inch sides.

resultant and Let

ABCI)

To

point of application.

its

be the square,

Resultant

= =

Find

+

5

7n,.

+ 10+12

7

34 pounds.

find its point of application

:

and 10 will be a force of 17 pounds acting from point in hne CB distant {, of 20 inches from B. The Resultant of

7

and

resultant of 5

acting at a point

12 will be in line

of 20 inches from

A

The

.

17

AD

pounds

distant

y'y

resultant

of

these two resultants will be a force of 17

+

17

half

pounds, 34 pounds, acting at a point at a perpendicular distance from

way between them, and \ °f liV

305.

A

floor

'-<

-°

+

>'

fV

20 X 30

-°]

feet

is

four posts, one at each corner.

pounds per point O,

5

foot side,

scjuare foot

=

AB

of

7if inches,

supported mainly by

There

is

a load of 20

uniformly distributed, and at

from 30-foot side and 7 feet from 20there is a metal planer weighing 5 tons. feet

Find the load on each post. 306.

Weights

5, 6, 9,

and 7 respectively, are hung scjuare, 27 inches in

from the corners of a horizontal a side.

Find, by taking

moments about two

adjacent

edges of the square, the point where a single force

must be applied the corners.

to balance the effect of the forces at

MECHANICS-PROBLEMS.

84

A

weighing 400 pounds,

is

suspended by means of two chains fastened one

at

307.

uniform beam,

each end of the beam. is

When

found that the chains make

witli the

beam.

A

the

beam

is

at rest

it

angles of 100^ and 115°

Find the tensions

in

the chains.

50 pounds acts eastward and a Will there be force of 50 pounds acts westward. 308.

motion

force of

?

That depends, as -will easily be seen, upon the position of the If they act on the two ends of a rope there will be no motion. If they act one on the northerly j)art of a brake wheel and

forces.

one on the southerly part there

will

Such forces produce a Couple two equal, opposite, :

be motion,

parallel forces

same straight line. The tendency to motion by couples is not The measure of this tendency is, rotation.

— that

of rotation.

not acting in the

of translation but of

—

Moment

of a couple equals the product of

one of the two forces

X perpendicular distance between them.

What

is

and a force

A

309.

the resultant of a couple of 3

moment

15,

.?

brakeman

up a brake on a freight

sets

car by pulling 50 pounds with one hand and pushing

50 pounds with the other

;

his forces act tangentially

to the brake wheel, the diameter of

which

is

\\ feet.

Another time he produces the same brake resistance by using a lever in handvvheel and pulling 25 pounds. How far from handwheel must his hands be placed .''

310.

unlike

.''

When are couples said When will two unlike

to

be

like

and when

couples balance each

FOKCES — MOMENTS. Other?

a system of forces

If

(i)

show

order,

in

equivalent to a couple.

gram taken

If

(2)

ing upon a body, express the

311.

may

Show

must be

the sides of a parallelo-

system of forces

moment

act-

of the couple to

eciuivalent.

is

that a force and a couple in one plane

be reduced to a single force.

a force of

of a plane poly-

that the system

in order represent a

which the system of forces

represented in

is

magnitude and position by the sides

gon taken

55

Given

in position

10 pounds, and a couple consisting of two

forces of 4 pounds each, at

a distance of

inches,

2

acting with the hands of a clock, draw the equivalent single force. 312.

The length

ABCD sides AB

square the

the side of a

of

Along

inches.

12

and CD forces of 10 and along AU, CB forces Find the moment of 20 pounds.

pounds of

is

act,

Fig. 40.

the equivalent couple.

Moments about

1).

— i2Xio-|-i2X2o = 12

X

10

Forces P and

313.

represented

ABC.

by

Find a

AB thii'd

=

O

moment moment

of equivalent-couple of equivalent-couple

act at A,

and

AC,

force

R

and are completely sides

of

a

triangle

such that the three

forces together ma)' be equi\'alent to a couple

moment 314.

is

A

represented

b)- half

whose

the area of the triangle.

tradesman has a balance with arms of un-

equal length, but tries to be fair

by weighing

his

ma-

MECHANICS-PROBLEMS.

56

:erial first

5how

from one scale pan, then from the other.

that he will defraud himself.

A

315.

balance with arms in

tradesman uses a to 6

he weighs out from alternate pans

•atio

of

Afhat

appears to be 30 pounds.

5

jain or lose

;

How much

does he

'*.

The beam of a balance is 6 feet long, and it when empty a certain body placed .n one scale weighs 120 pounds, when placed in the Dther, 12 pounds. Show that the fulcrum must be 316.

ippears correct

;

1

distant

about J^ of an inch from the center of the

beam.

The weight

317.

movable weight

is

of a steelyard 3

is

1

2

between successive pound graduations, of the short

A

318.

end

arm

is

if

the length

weight of 247 pounds

is

attached to one

which

is

22 inches

and may be regarded as having no weight

the force

is

applied at the other end, and

angle of 27° with the lever

from the weight.

when

its

3 inches.

of a horizontal straight lever,

long,

pounds,

Find the distance

pounds.

it

;

the fulcrum

is

;

makes an 3

inches

the magnitude of the force

F"ind

just balances the weight. 319.

given

A

uniform beam rests

inclination,

6,

with

at a

one end

against a smooth vertical wall, and

the other end on smooth horizontal

ground

:

it is

held from slipping by

a string extending horizontally from

FOKCES—MOMEXTS.

87

the foot of the

beam

the tension

the string and the pressures

in

ground and

AB

is

beam

the

to the foot of the wall.

ImikI

the

at

wall.

AC

the beam,

acting at

its

BC

the wall,

the string,

W

the weight of

middle point G.

There are three forces supporting the beam

:

vertical reaction P,

horizontal reaction R, and tension in the string Y.

T?ke moments about forces

—

their lever

of intersection of two of the

B, the point

arms would be

R

AC =

X

zero.

^\

BC :•

2

Substitute for

AC

value

its

BC

'

R

must equal

-•'

tan

then

0,

tan e

2

both being horizontal resisting forces that maintain equilibrium; likewise P and must be equal.

but

F,

W

.-.

(2)'

F

=

^

2

(3)

320.

A

——

1'

=

and

tan e

W

uniform beam rests with a smooth end

against the junction of the horizontal ground and a vertical wall

;

it

is

supported by a string fastened to

the other end of the

beam and

to a staple in the ver-

and show

tical wall.

Find the tension

that

be half the weight of the beam

it

will

of the string,

the

if

length of the string be equal to the height of the staple above the ground. 321.

pounds,

A is

uniform

rod

8

long,

feet

weighing

18

fastened at one end to a vertical wall by a

smooth hinge, and

is

free to

perpendicular to the wall.

move It is

string 10 feet long, attached to

in a vertical plane

kept horizontal its

free

end and

b)'

a

to a

MECHANICS-PROBLEMS.

!8

Find the tension

the wall.

loint in

and

in the string,

he pressure on the hinge.

A

322.

ligh.

uniform beam, 12 feet

beam be

the

If

with

in length, rests

against the base of a wall which

me end

is

20

feet

held by a rope 13 feet long,

beam and

.ttached to the top of the

summit

to the

of

he wall, find the tension of the rope, neglecting

and assuming the weight 00 pounds.

reight,

ABC

323.

;onsidered

;

is

while the vertex

B.

If a

B

C

given weight

B and

B

A

324. A.

and C

beam

is

is

hung from A,

What

C.

fastened by a hinge to a

B

AB

rests on the

beam, and

Draw the figure. The weight acts

W

The

A

is

B

;

a string

at the

weight

is

P.

If

W

is

the

tlie rod.

middle point (\

reaction of the plane at

Tiie reaction of

BAD =

tire

V> is

R,. perpentlicular to the plane.

0.

tension of the string at

If

=

tension of the string throughoat

P. 'tliere are

at

R, upwards.

Let the angle

The

at

a the inclination of the plane,

ind P and the reactions nn

at

smooth ground

and, passing over a smooth peg at the

iveight of the

-

by the weight on the

1

:op of the plane, supports a

around

find the reac-

are the magnitudes and

and on a smooth inclined plane

"astened at

be

to

rests against the wall under

lirections of the forces exerted vail at

beam

a rigid equilateral triangle, weight not

the verte.x

vail,

ions at

of the

its

four forces acting on the beam,

Resolve vertically and horizontally.

W,

R, Ri, P,

;

}-0K CES

A

325.

pole 12

— MOMENTS.

feet

8Q

weighing

long,

one end against the foot

rests with

from a point 2

feet

25

pounds,

of a wall,

and

from the other end a cord runs

horizontally to a point in the wall 8 feet from the

ground.

Find the tension

of the cord

and the pres-

sure of the lower end of the pole.

A

326.

light

smooth

stick 3 feet long

one end with 8 ounces against a is

I

foot

smooth vertical from the wall.

of lead wall,

loaded at

is

the other end rests

;

and across a

nail

which

Find the position of

ec|ui-

librium and the pressure on the nail and on the wall.

A

327.

trapezoidal wall has a vertical back and a

sloping front face; width of base, 10 feet; width of top, 7

feet

;

at a point

overturn

order to

it

weight of masonry 328.

Si-x

men

horizontal force

20 feet from the top

Thickness

of wall,

to attach the rope

130 pounds per cubic

150000 pounds

to the crank axis at

the thrust along the

is

150°,

in

example

3

86, 2I is

feet

inclined

show that the moment

the thrust about the crank-pin

330.

foot.

.?

the crank radius, and the connecting-rod

and

in

foot

chimney 75 feet high. How far of the chimney was it advisable

connecting rod of the engine,

possible

i

using a rope 50 feet long were just

up from the bottom

If

.-'

in wall,

able to pull over a

329.

What

height, 30 feet.

must be applied

is

of

one-half the greatest

moment.

A

trap-door of uniform thickness,

feet wide,

and weighing

5

5

feet long

hundred weight,

is

MECHANICS-PROBLEMS.

JO

open

leld

neans

angle of

at

with the horizontal by

35°

One end

of a chain.

of chain

is

hooked

at

fastened

middle of top edge of door, and the other is Find the force It wall 4 feet above hinges.

the

in

:hain and the force at each hinge.

The

331.

sketch represents a coal wagon weighing

with

r^-

r

------.

x^—^-—-Si^'i

*

\

p//^ D

E

AE

a rod

is

''

in

its

many

load

4,i

pounds

applied

at

\ P by usual methods of hand ;B power will just lift the wagon when in the position shown in the sketch ?

tension.

CD

a connecting-bar.

is

Divide the problem into three parts

:

(rt)

Draw

{b

Find horizontal distance from C to the

)

How

tons.

the forces acting. verti-

through the center of gravity.

cal (

c)

Find force to apply

at

C

parallel to

I^

;

then

f^nd P.

CENTER OF GRAVITY

A

332.

rod of uniform section and density, weigh-

ing 3 pounds, rests on two points, one under each end a movable weight of 4 pounds is placed on ;

Where must it be placed so that one of may sustain a pressure of 3 pounds, and

the rod.

the

points

the

other a pressure of 4 pounds

1

FORCES — CEXTRR OF GRAVITY.

Two

333.

rods

pounds and

weighing 2

spectively are put

3-pound

uniform

of

one

re-

together so that the

on

stands

middle

the

Find the center

the other.

of

of gravity of ^^^-

the whole.

Take moments about AB, / — -f 3 X ?,

A

334.

I

density

pounds

3

9

5

X

metal

thin plate of

.V

is

=

«'

o

in

the shape of a

square and equilateral triangle, having one side ^G,^

C

common

the side of the square

;

Find the center

inches long.

is

12

of gravity of

the plate. Let Gi be the center of gravity of the triangle, Gj of the square,

G

of the whole plate.

From symmetry EG, GG„0 plate,

will

be a straight

line bisecting

the

and

= = w=

OGo OGi Let

Area

of triangle

Weight

Area

of square

6 inches 15.5 inches

weight of metal per square inch

= X 12 X sjiz'-— 6^ = 62.4 square inches = 62.4 pounds X w pounds =144 square inches .V

Weight = 144 X K' pounds Take moments about the axis AB, Weight of triangle X OGj-f- weight of square X 0G„ — total weight X OG = o 62.4a:'X 15.5 -f-i44WX6 — (62.4W'-f-i447f') X OG = o .-.

OG =

8.86 inches.

MECHANICS-PROBLEMS.

92

A

335.

X

bridge

member has two web

plates

i8

x f top angles

3x3

and

I inches, top plate 21

,

4x3

and I inches thick, bottom angles thick. Find " eccentricity " the distance

—

-^-^

inches

from AB,

the neutral axis through the center of gravity to C,

the middle of the section.

n

r

B

A

L Fig. 45-

Web

336.

plate

1

2

tricity."

337.

X

\,

the main

Draw-Bridge of

-

(Given in Osborn's Tables (1905) page 24.) Fig. 47 shows a cross-section of the top chord

of one of

News

Fig. 46.

46 is 10 x |^ inches, top two angles 4 x 3 X f Find " eccenplate of Fig.

June

at

15,

of this built-up

trusses in

the

Houghton, Mich. 1905.

member,

position of the axis

Portage Canal

See Engineering

In computing the strength it

AB that

is

required to find

the

passes through the center

of gravity of the section.

Fig. 48.

FORCES— CENTER OF GRA VITY. The

338.

strength of steel

puted by embodying,

from neutral loo-poand section

extreme

its

extreme

Draw

in Fig. 48.

3

;

of

A

section.

Company, has a

the section carefully

then cut

out and

it

the distance from center of gravity

is

fibres

ABC

339.

AB =

com-

center of gravity by balancing on a knife

What

edge. to

usually

the

of

iibres

to full scale on bristol board

locate

is

of the Lorain Steel

rail,

shown

rails

other factors, the distance

which passes through the center

axis,

the

gravity, to

among

93

is

.''

a triangle with a right angle at A.

AC =

inches;

ounces, 3 and

inches;

4

weights

of

4,

2

Find

are placed at A, B, and C.

the position of their center of gravity.

340.

A

ABC

uniform triangle

lying on a horizontal table,

is

of

weight

just raised

Find the magnitude

force applied at A.

by

W, and

a vertical

of this force,

and that of the resultant pressure between the base BC and the table. 341.

A

uniform circular disk has a circular hole

punched out half

way

of

it,

to the

extending from the circumference

center.

P^ind the center

of

gravity

of the remainder.

342.

A

box, including

when

its lid is

its

cover,

is

made

of six

where is its center of gravity turned back through an angle of 180° 1

equal square boards

;

ME CHA NICS-PROBLEMS.

94

ABCD

343.

thin

rectangular

weighing

plate

AB

pounds, feet,

a

is

BC

plate

50

2 feet

Ex-

10 N^-

is

the

;

.

suspended

is

by the middle point of its upper edge

AB, and course, is

then,

AB

is

horizontal, but

AB

placed at A,

"'

^'^-

of if

become

will

a weight of 5

zon.

Show how

either

by calculation or by construction.

344-

A

to

find

angle

of

inclination

circular disk, 8 inches in diameter, has a

hole 2 inches in diameter of

the

pounds

inclined to the hori-

punched out

of

it,

the center

the hole being 3 inches from the circumference of gravity of the remain-

Find the center

of the disk.

ing portion.

Find the centers

345.

of area of the

tions

of

above sec-

uniform

plate

metals. Fig. 50.

346.

Into

hollow

a

cylindrical

ves.sel

1 1

inches

high and weighing 10 pounds, the center of gravity of

which

is

solid cylinder is

just fitted.

5

inches

from

the

base,

a

uniform

6 inches long and weighing 20 pounds

Find the

common

center of gravity.

FORCES— CENTER OF GRAVITY.

95

Gj center of gravity of hollow cylinder

G2 center

of gravity of solid cylinder.

Moments about AB, + 10 X 5 + 20 X

+

+

50

=

.V

60

^

O

3

5

A'

A

30 X

30 I I

.V

=

=

A-

o

o

O ^'S-

inches.

Give examples

347.

rium.

—

3

— =

of

stable

and unstable

51-

equilib-

cone and a hemisphere of the same material

cemented together at the common circular base. they are on a horizontal plane, and the hemisphere

are If

in contact in

with the plane, find the height of the cone

order that the equilibrium

may be

(The

neutral.

center of gravity of a hemisphere divides a radius in the ratio of 3 to 5.)

A

348.

the ends in

thread 9 feet long has

of a rod 6 feet

ends fastened to is

supported

such a manner as to be capable of turning freely

roimd a point

2 feet

from one end

thread, like a bead on

on the

position in which the rod will

supposed that the rod is

its

long; the rod

no

friction

A

349.

;

a weight

a string.

come

is

placed

Find the

to rest,

it

being

without weight, and that there

is

between the weight and the thread.

circular

disk

weighs

9

ounces

;

a thin

straight wire as long as the radius of the circle weighs

7 ounces a chord

whole

;

if

the wire

of the

will

be

is

circle,

placed on the disk so as to be the

at a distance

center of gravity of the

from the center of the

equal to some fractional part of the radius. Find that fraction by construction or calculation.

circle

;

MECHANICS-PROBLEMS.

96

A

350.

base.

cone and a hemisphere are on the same

What

height must the cone be in order that the

center of gravity of the whole solid shall be at the

common

center of the

r h

= =

base

radius

.''

common

base.

height of cone.

FRICTION The

coeiKicients of friction for various pairs of sub-

stances have been found experimentally by Morin

these results however can be used only for approxi-

mate computation

actual trial should be

;

Average values

specific cases.

made

Stone on stone

0.40 to 0.65

Wood

0.25 to 0.40

on wood

Metal on metal, dry

0.15 to 0.30

well oiled

.

.

.

o.oi to o.io

proportional to normal reaction, R.

1.

Friction

2.

Is independent of area of contact.

3.

Is

is

dependent very much on the roughness of surfaces.

Define " coefficient

351.

for

are:

"

and "angle

of friction,"

and "resultant reaction." R,

352.

I

oo«..

Fig.

The

A

weight of 56 pounds

is

moved

along a horizontal table by a force of

F.,^E^->8it,. 1

How much

8 pounds.

52.

pull of 8

of friction

pounds

is

is

the coefficient

>

required to overcome friction, and

is

equal to the friction. Friction

= coefficient x

Reaction (perpendicular to plane of table.

FORCJiS

MECHAXICS-PROBLEMS.

g8

when

efficient of friction

after

it

A

359.

slide,

of

20

in

i

pounds per

when they

vertical to

the resistance on the level

Find the

ton.

How much

360.

;

on the traces

pull

40

work has a man, weighing 224

walking twenty miles up a slope of

in

horizontal

What

.''

be negligible,

be

friction

hill {a) if

of

-|;

the

.''

Three

361.

{b) if

i

force could drag a

dead load of the same weight up the same friction

is

are parallel with the incline.

pounds, clone

weight

and

horse draws a load weighing 2 000 pounds

up a grade 100

the block starts to

has started.

artillerymen I

drag

a

gun

700 pounds up a

weighing rising 2

hill

Sup-

vertically in 17 horizontally.

pose the resistance to the wheels Fig- 53.

g"ing

i-ip

the

hill

be

each exert to move

When

the gun

up the plane, and

is

it

pounds

16

per hundred weight, what pull parallel to the

hill

must

.?

about to move fonvai-d the pull P

parallel to

it

;

the friction

F down

-will

be acting

the plane, hold-

ing back; the force

R

perpendicular to inclined plane, partly sup-

porting the gun, and

W

the weight of the gun acting vertically down-

Weight of gun

—

given i 700 pounds. ponents perpendicular and parallel to the plane.

ward.

component I-i

;

will

the parallel

is

Resolve into com-

The

component

\\'ill

perpendicular

—

be the supporting force of the plane be the part of the pull

I'

its

reaction

required by

weight of the gun.

362.

Find the force which, acting

tion, will just

in a

given direc-

support a body of given weight on a

FOR CES ~ FRIC TION.

The

rough inclined plane. the plane as 3 to just supported

and

4,

on

it

to the base of

Find the

is

to half

coefficient of fric-

between the body and the plane.

The

363.

weighs

1

1

2

table of a small planing-machine

pounds makes

feet each per minute.

tween the

which

single strokes of

six

4^,-

Tiie coefficient of friction be-

sliding surfaces

in foot-pounds

table

is

found that the body

by a horizontal force equal

the weight of the body. tion

height

is

it

99

is

What

.07.

per minute performed

in

work moving the is

the

'>.

364.

A

double

its

floor,

rectangular block

ABCD

base, stands with its base

coefficient of friction J.

horizontal

force at

whether

will slip

it

C

till

on the

If

it

whose height is on a rough

AD

be pulled by a

motion ensues, determine floor,

or begin to turn over

round D. 365. its

A

the plank it

cubical block rests on a rough plank with

edges parallel to the edges of the plank. is

before

slipping,

how much

coefficient of friction 366.

A

If,

as

gradually raised, the block turns over on

weight of

at

least

must be the

just

be supported

.?

5

pounds can

on a rough inclined plane by a weight of 2 pounds, or can just support a weight of 4 pounds suspended by a string passing over a smooth pulley at the vertex.

Find the the plane.

coefficient of fiiction,

and the inclination

of

:

lOO

MECHAIVICS-FROBLEMS. Find the

367.

force

least

that will

weighing 200 pounds along a concrete efficient of

The in

some

drag a box floor,

the co-

friction being 0.50.

required force will of course not act horizontally, but instead

To

direction as P.

find the angle b

Resolve vertically

—P

sin ^

— R+

200

=o

Resolve horizontally

+ From

P cos

b

— nR =

_

200>t \i.

When

will

P be

by

trial

The

tangent

will

+

cos b

? When m sin b + cos b is student not familiar with the calculus can

maximum

that the

of the pull P,

sin b

as small as possiijle

as large as possible. find

o

these two equations

value of denominator, or least value

=

occur when b

tan—'

/u,

that

is,

the angle whose

is n.

By the method

of calculus,

^ sin b Differentiate, noting that m

+

cos b

= x

a constant

is

value, which in this case will be a

maximum

;

and, to find a critical

value, place

tlie first dif-

ferentia! equal to zero.

-

,

=

fi

~

cos

sni b

=

o

db f b

= ^ =

tan

/'

tan

~

'

ja

=6° 34'

p = i

X .447

+

-894

149 pounds.

368. By experiment it was found that a box of sand weighing 204 pounds required a least pull of no pounds (at angle a) to move it on a concrete floor.

What was

the value of

/u.

.?

FORCES— FRICTION. The roughness

369.

inclination 30°

is

lOI

of a plane of

such that a body of

weight 500 pounds can just rest on

What

it.

the least force required

is

Fig- 55-

draw the body up the plane

to

As

in

problem 367 a

A

370.

up

will

equal the angle of friction, or tan"'

sled of total weight 3 tons

a grade of

What angle should horizontal What pull will .'

The problems that pertain methods that have been used :

Show

Fig.

(2)

be drawn

to

The

coeffi-

to the

the traces

make

the horse exert

with the

}

wedge can be solved by the same

for the inclined plane.

The

essential

the conditions by a sketch, indicating carefully

the position and direction of plane,

p..

between the sled shoes and the snow

10.

principles are

is

vertical to 8 horizontal.

i

cient of friction is o.

?

all

forces

;

then, (i) resolve parallel to

resolve perpendicular to plane.

Thus

for

problem 371

56 shows the conditions for one form of wedge, Fig. 57 for

another.

Observe the directions of \

P should decrease

Resolve

-

||

;

to top

0.20

and

W.

As can be seen

in Fig. 56,

R

act in the

must

In previous problems the weight has

direction indicated.

the inclined plane

R

the value of R, therefore

here the plane moves. plane (Fig. 56),

R+

300 cos 3° 35'

-

W

/:

sin 3° 35'

= o.

moved on

MECHANICS-PIiOBLEMS.

I02 Resolve

J_ to plane,

+ R-300

sin 3° 35'

-

Solve these two equations for

To figure

find the pull necessary to

showing

/".R

and

\

P

WX

f 35' = o.

cos

W. withdraw the wedge, sketch another

in their

new

Then

positions.

solve as

indicated above.

A

371. is

cotter, or

wedge, having a taper' of

sure of

600 pounds.

Taking the

between the two surfaces as

wedge

the

exerts at the

pressure of 600 pounds to

in

i

8,

driven into a cottered joint with an estimated pres-

;

coefficient of friction

which

0.2, find tlie force

perpendicular to the

j<jint

also find the pull necessary

withdraw the wedge.

A

372.

floor-column with

its

load of

5

tons

is

to be

by two wedges driven towards each other. Thickness of each wedge is 2 inches, length, 12 Find the force inches; coefficient of friction, 0.15.

lifted

must be equivalent

that

to

P

in

order to drive the

wedge.

A casting of weight 5 000 pounds is to be by an iron wedge that is forced ahead by a screw and mechanism that can give an equivalent 373.

lifted

force of 3

000 pounds.

what should be

A

374.

and

its

If a

thickness

1

2-inch

wedge

used,

rough wedge has been inserted into a block

only acted on by the reactions.

is

is

.-'

If

it

is

on the

point of slipping out, and the coefficient of friction is

—p V3

,

what

is

the angle of the wedge

?

FORCES— FJi/criOiV.

A

375.

steel

wedge

103

12 inches long, 2 inches thick,

tapering on both sides to

pump

plunger weighing

3

o, is used to wedge up a 000 pounds by means of a

maul weighing 5 pounds. The is 0.15 and the striking velocity

How

per second.

A

376.

far will each

wheel

weight

of

of friction

coefficient

of the

maul

is

25 feet

blow drive the wedge

W

rests

between two

planes, each inclined to the vertical at angle a

plane of the wheel tersection of the If

fi.

is

;

perpendicular to the line of

two planes, which

be the coefficient of

is itself

friction, iind

?

the in-

horizontal.

the least couple

necessary to turn the wheel.

A

377.

ladder inclined at an angle of 60° to the

horizon rests with one end on rough pavement, and

the other end against a smooth vertical wall

down when a weight Show that the coefficient

ladder begins to slide its

•

IS

middle point.

—v^

is

;

the

put at

of friction

-'

.

6 "When the ladder begins to would be ^R.

slide

down, the

limiting friction

Resolve

vertically,

W

= R

Resolve horizontally, R'

=

/iR

Take moments about for

B,

and then soh*e

11.

If the wall

acting at in the

B

should be rough there would be

an upward force of ^'R' that woidd have to be embodied

above equations.

1

MECHANICS-PROBLEMS.

04

A

378.

uniform ladder weighing 100 pounds and

52 feet long is inclined at an angle of 45° with a rough vertical wall and a rough horizontal plane. If is at each end |, how far up man weighing 200 pounds ascend

the coefficient of friction

the ladder can a

before the ladder begins to slip

A

379.

t

uniform ladder 30 feet long

is

equally

in-

and the horizontal ground, weight 224 pounds both rough a man with a hod ascends the ladder which weighs 200 pounds. clined to a vertical wall

—

;

—

How

far

slips,

the tangent of the angle of resistance for the

up the ladder can the man ascend before

wall being 1 and for the ground i

A

380.

it

?

uniform beam rests with one end on a

rough horizontal plane, and the other against a rough vertical wall, and when inclined to the horizon at an angle of 30°

is

on the point of slipping down

pose the surfaces ecjually rough, find

A

381.

inch

;

bolt for a cylinder

mean diameter

;

sup-

^a.

head has 8 threads per

of threads

i

average

\ inches,

outside diameter of nut 2| inches, inside diameter of

The

be

tight-

ened by a pull on the end of a 3-foot wrench.

The

bearing surface, 1.6 inches.

coefficient of friction for threads

nut being

o.

i

5,

what

nut

is

to

and underneath the

pull should be exerted in order

that the stress in the bolt shall not exceed

pounds

50000

.'

Problems pertaining to bolt and nut applying the combined principles of

friction

Work and

can be solved by

Friction.

Thijs for

J-'OR

CES — FRIC TION.

105

the above prol:)lem suppose that the specified conditions should exist for one revolution.

no approximation, simply a con-

Tliis involves

venience in numerical fiyures which otherwise would have to be divided by perhaps a hundred or thousand to apply to a fractional

Then,

part of a revolution.

=

Work The

+

Work

on wrench

under nut

values of work on threads and

W

W

of lilting

work under nut can be

mined near enough for ordinary cases by As shown by Fig. 59, = .99 R, or usually to

+ Work

^^'ork

on threads

deter-

slight approximations.

R may

be taken equal

also the length of thread developed (for one revolution)

tt x and in Fig. 60 the circumference C is one that can be determined by the condition that the work done by the friction of all the particles ontside is the same as that done by all the particles inside. § (R'

i.i";

Its radius for

X ^

case,

1.

a section like inches,

1 1

As

acts like a w

Fig. result

and

which

60

is

x

or,

is

<_h1|J(-!.

Fig. 60.

59.

by takinp; a mean circnmference between

2. 7:;

— or

1

.01)

for

t.6

inches diameter

Therefore ordinarily use the mean

inches.

cumfeience for the position nf

The equation

for this

approximately the same as would

thread advauoes

tlic

it

Fi(>.

Work

friction

cir-

under nut.

thus becomes

:

For one revolution, 2

Nut

Threads

Wrencli

P(3 X 12 X

.:

3I)

= 50 000 :-,o.i5

X

4.71

+ 50 000 xo.15 x

r.09

^3

::

3I

Lifting

+ From which wrench

50 000

,\'

0.1 25.

equation find P, the pull that should be exerted on

to produce 50 000

pounds

stress in the bolt.

MECHANICS-PROBLEMS.

I06

382. By trial in a 60 000-pouncl testing machine we have obtained with a builder's lifting-jack a stress

on the machine the end of an

6 000 pounds for a certain pull on

What was that pull.? was 1.50 inches, there were and diameter of bearing that

18-inch bar.

Mean diameter 3

of

of threads

threads to the inch,

corresponds to the

mean circumference was

scribed under problem 381 cient of friction for threads

of

nut de-

1.78 inches.

Coeffi-

was 0.15, and for bearing

0.15.

A locomotive

383.

mean diameter 2

bolt has 10 threads to the inch;

inches, average outside diameter of

nut 41 inches, diameter of hole in washer on which

nut turns 2.2 feet, pull

inches. If length of vi'rench was 5 367 pounds, and stress 40 000 pounds, what

was the value

A

384.

test of

Tufts

oratory at result (The

of the coefficient of friction

rope friction in our engineering labCollege has given the following

:

iveight

motion.

Tj just moving, and pull T, resisting any increased

See Fig.

6i.J

For Weight of Pull

Number

}

of

Laps

T^

Ratio T,

Lbs.

T^

Tj

=

100 Pounds.

Number Laps

i

81

1.23

1

65 45

1.

54

2

I I I

32

li

25 19

li

J

of

Pull

Ratio

Tj

T, T.

Lbs.

14 II

2} -\

8

2-j-

4i

3

3-1

5

Compute scale of

I

the ratios of T, and T„, then plot the results, usin^ ig a inch = i lap for vertical ordinates, and i inch ratio of

=

FORCES— FRICTION. 10 for horizontal. points, observing point, origin.

n^

107

Sketch the most probaljle curve fur the plotted it does not necessaiily pass thruugh the last

that

and determine whether or not

it

should pass through the

!q= -

-

=S=

-Hz --:^

FORCES.—FKICTIOU. Determine how

this

form

109

ruling

of

Plot the points for laps and ratios, and

is

constructed.

draw the most

probable straight line through the points

Should the

line pass

as before.

How

?

do

and 2f compare on this straight with those on the curved line previously plotted

the points for line

through the origin

i

V

laps

.''

The

386.

lines

plotted

the

for

preceding prob-

lems are sufficient to answer directly many questions pertaining to that particular r
For the same conditions, how many laps are needed to hold a weight of 300 pounds with a pull of 40 pounds

.''

387. pull of

For the same conditions, with 2 laps and a 100 pounds, what weight could be lowered

into the hold of a vessel

evident that the plotted lines of the pre-

It is

388.

.''

ceding problems would not apply to other cases of

The

friction. is

contained

in

value of

/x

the coefficient of friction

the equation

of the curves, but can-

not yet be specified. For the purpose of finding the value of the coelTficietit, as will be done later on, it is advisable to determine the slope and position of the plotted line of laps)

=

;

that

is, its

equation.

Notice that

11

(the

number

T

c (a

constant depending on the slope of the line) X ^^

two tensions). The value of c (the slope of the would depend upon the stiffness of rope and roughness of the rubbing surfaces. For this particular rope and piece of wood, the (the ratio of the

line)

value of

c,

What periment

according to one plotting that

I

have,

is

2,oS.

does your plotting indicate for the above ex?

I

MECIIAA'ICS^ PROBLEMS.

lO 389.

comes

Frtim the above the equation of the line beT.

Write your equation and

2.0S

//

transpose so as to write the value of log T,, which

=

find to be log T^

+

log T^

.437

X

n.

Now

I

this

equation derived from the laboratory experiment will

be seen to bear a close relation to the general

mula

and

for rope

which

belt friction

will

for-

now be

developed. The tliat

authfjr's

he belie\'es

method it

to

usually preseiited in

of analysis

is

introduced here for the reason

be more easily understood than the methods

te.\t

books.

At B there would be a tension of Tj, at A, 1\, and at any point, C, there would be two tensions, T and T dl Now if we knew the reaction at the point C we could multiply it by ^ and obtain the

+

Fig.

62.

Fig. 63.

friction.

To

Fig. 63.

Let an arc of the angle

find this reaction

the center be arc

and arc

d B, then

for a very small angle

.

\<

draw the parallelogram of

d6 measured

T

would equal R.

R = T arc dd = ^T arc do.

^ufi

force.

from

the arc Avould be T xarc dS, a differential angle this value of the

at distance

—

at unit distance

—

.

FOK CES — FK/C TION Now

«R =

the friction

the difference in tensions d'Y

dl = That

arc d0.

IJ.T

we take

If

a very large

number

;

sum

add up and get the summing; up by the calculus,

or,

X

of small arcs

find the friction at each point,

of friction

= «T

for an infinitesimal arc, the difference in tension

is,

the infinitesimal arc.

we can

III

total

r-T.,farc.. 8

^

2 Tvn in

which

log can be changed to the

•

login T]

is

is

number

common

-

logio

^

log,T,

the

Any two

:

the tension

To

;

and the Naperian

= -4343 + 2.72SSf.«

T=

f'

••

-

^"

Tp

=

in

+

logT,

find this value solve the

of

M this

problem 389

w'as

jj.

formula

is

e\'itient.

The

the coefficient of friction.

two equations, and we have

2.72SSM" =

if

contains

number

.437 «.

similarity of this with the general

term must contain the value of

Then

It

tension T., and the

of these being given the third can be found.

log T,

The

of laps

system by multiplying by .4343.

The formula deduced by experiment

last

,10

the general formula for rope and belt friction.

variable quantities

laps n.

it

logT, =IogT,

,

Avbich

-

T,

log,.

Now

=

-437 " o.

15

be taken as the value of

/i,

how many

would be necessary, according to the general for lOO pounds to just move 14 pounds?

laps

formula,

(Check 390.

result with data of

A

weight

of

problem 384.)

100

pounds

just

moves

37

pounds, both being connected by a plain leather belt that encircles one-half of a

does not turn.

108

;

draw the

of the line.

14-inch iron pulley that

Plot the poiitt on the paper of page line of friction,

and write the equation the general formula

Then compare with

MECirANICS-PROBLEMS.

112

and determine the value of the coefficient for the plain belt and iron pulley.

of friction

Fig. 64.

391.

belt

In the same

way

as

had been treated with

55 pounds

just

moved

13.

by problem 390,

after the

" cling-fa.st " belt

dressing

Plot the line and find the

coefficient of friction. Further application of the general belt and rope friction fomiula is

seen in the problems that follow.

392.

According

to conditions of friction as in prob-

lem 390, how many turns would have to be taken around a capstan in order to lower a barrel of salt, 150 pounds, into a dory

pounds

.?

without

pulling

over

50

FOR C£S — FRIC TIOA\

A

393.

weight of

5

tons

is

I I

to be raised

from

3

tlie

hold of a steamer by means of a rope which takes

/u.

drum

turns around the

3.1

=

end

of a steam-windlass.

0.234, what force must a

man

If

exert on the other

of the rope.'

A

394.

man by

taking 2\ turns around a post with

a rope, and holding back with a hirce of 200 pounds, just 0.

16S, find the tension at the other

A

395.

pulley that

is

4

one-half feet

the coefficient

pulley

is o.

396.

ing

A

i

the

What power

friction

of

circumference

a

of

diameter and making 150

in

revolutions per minute. if

=

fx

end of the rope.

leather belt will stand a pull of 200 pounds.

passes around

It

Supposing

keeps the rope from surging.

will

it

between the

transmit belt

and

.''

belt laps 150°

around a 3-foot pulley, mak-

per minute; the coefficient of

130 revolutions

What is the ma.ximum pull on the when 20 horse-power is being transmitted and

friction is 0.35.

belt

the belt 397.

is

A

into the

just

on the point of

slipping.''

weight of 2 000 pounds

is

to

be lowered

hold of a ship by means of a rope which

passes over and around a spar lashed across the hatchcoamings so as to ha\e an arc of contact of \\ cir-

cumferences.

If

=

^

.,-V,

what force must

a

man

exert at the end of the rope to control the weight 398.

pounds. bitts, in

A

hawser

is

subjected to a stress of 10 000

How many order that a

?

turns must be taken around the

man who

cannot pull more than

MECIIANICS-rROBLEMS.

114

250 pounds may keep /i

=

o.

399.

1

68

A

from surging, supposing

it

?

rope drive carrying 20 ropes has a pulley

600 horse-power 90 revolutions per minute. Taking

16 feet in diameter, and transmits

when running M

=

at

0.7 and the angle of contact

sions

180°, find the ten-

on the tight and slack sides of the ropes.

From

the data that

force that the belt

is

given find by the principles of AVork the

is

transmitting.

— Tj^

Tj

It is

318.S ponnds

Substitute this value of T^ in the general formula, also the value of

M ^nd

}i

;

then

'ogT,

= log

{T,~2iS.Sj

+ .9550

T, •

T,

From which 400.

A

find T,

;

--

218.8

then find Tj.

dynamo The speed

belt for a

18-inch pulley.

revolutions mitted, 100

per minute ;

9.02

;

is

to encircle half of

of pulley

to

is

an

be 1060

horse-power to be trans-

coefficient of friction, 0.2

:

thickness of

belt to be || inches, and working strength 300 pounds per square inch. What should be its width ? 401.

A

main driving

54-inch pulley.

belt

is

The speed

of

revolutions per minute

;

is

its

width

.?

is

Thickness of

to be

ing strength 450 pounds

should be

pulley

to

be 350

horse-power transmitted, 520

coefficient of friction, 0.2.

ing to specifications,

to encircle half of a

|J

belt,

inches,

;

accord-

and work-

per scjuare inch.

What

I-'OKCES

A

402.

— FKIL

plain belt without dressing encircling one-

half of a pulley,

has a tension of

i

when just on the j)oint ooo pounds nn the taut

machinery being put into belt.

"5

TIOiV.

use, rosin

is

of slipping

More

side.

thrown on the

the tension on the slack side remains the

If

same as hefore, and the belt is just on the p'jint of slipping, what horse-power will be transmitted, diameter of pulley being lo feet, and revolutions per minute,

140

'>.

A

403.

single

turns on an friction, is

lifted

the force ,

A

to P.

weight of

by means

P

that

of this

tlit

;

in

R

center,

s

it -were, on its coming neaier found tlie fric-

can

lie

by multiplying by

tion will be determined

=

S

R=

I-

\j..

To

find

R

fiK'

as will be evident by plotting a parallelogram of force.

=P + W —)> —+ W \' + M P + qOO

S "o

K —

I

P+c;oo

mR.

Now

to find

P

:

Take moments about

—P

X

C, the center 3 -f

P

500 \ 3

=

=170

-I-

/iR

porrnds.

diameter,

coefficient of

500 pounds Find pulley.

axle to cieep, as iittle off

value of

tlie

6 inches

required.

is

S moves a

When

pulley,

inches in diameter

a.xle 2

O.2.

Friction causes

beatings.

h.-ved

MBCIfANICS-PROBLEMS.

Il6

A

404.

The

makes 50 revolutions per minute.

shaft

load on the bearing

bearing

is

7

friction

is

0.05.

erated

is

8 tons, the diameter of the

and the average coefficient of At what rate is heat being gen-

inches,

."

= =

S

P

=

15

^R =

W

980

ygg pounds force X distance

=

Work

-t-

8 tons

of friction

= =

A

405.

on an is

The

a.xle

inch in radius

i

A

the rope

o.

P

acts vertically.

9

inclies in

;

406.

if

407.

I

X

2

X

-/-

;

X

50)

the weight of the pulley

is

force

P

is lifted

by

required

is

}

between axle and bearing

flexible,

and without weight, and

Find the horse-power necessary

to turn a shaft

diameter making 75 revolutions per minthe total load on it is 12 tijns and /x = .015. I.et

P and

angle of 90° axle

,'3

What

coefficient of friction I

(

weight of 500 pounds

of this pulley.

is

ute,

X

73 500 foot-pounds per minute.

single fi.xed pulley, 2 feet in radius, turns

80 pounds.

means

799

inch

;

;

the relation of

W be

inclined to each other at an

radius of pulley

is

6 inches

coefficient of friction, 0.2.

P and

W

radius of

;

Determine

in case of incipient

motion.

FORCES — FRICTION.

A

408.

I I

J

horizontal axle lo inches in diameter has a

upon

vertical load

The

of 4 tons.

of

it

20 tons, and a horizontal

coefficient of friction

pull

Find

0.02.

is

the heat generated per minute, and the horse-power

wasted

in

when making 50

friction,

revolutions per

minute. 409.

The

shaft of a

i

dynamo

ooo-kilowatt

inches in diameter, makes

is

25

100 revolutions per min-

The 45 000 pounds. coefficient of friction being 0.05, find the horse-power ute,

and carries a

lost in

heat that

total load of

generated by

is

friction.

Find the horse-power absorbed

410.

the friction of a foot-step bearing with

in

overcoming

flat

end 4 inches

number

in diameter, the total load being li tons, the

of revolutions 100 per minute,

and the average

coeffi-

cient of friction 0.07.

=

Work

force

X

^^y x

(-,

distance

x

/-o)

x

27r

x

100.

of friction

The

distance being obtained

l:)y

con.sidei'ing a

circumference as in

problem ^8i, outside of which the worlc is the same as that inside. For a Ijeaiing with a flat end that ciicuniference has a radius of twothirds of R.

411.

Calculate the horse-power absorbed by a foot-

step bearing with

flat

supporting a load of

end 8 inches

4000

in

diameter when

pounds, and making 100

revolutions per minute, coefficient of friction 0.03.

A

1

50-horse-power turbine has an oak step

6 inches

in

diameter and with conical end tapering

45°.

the

412.

If

load

on the step be

2 tons,

and the

8

MECHANICS-PROBLEMS.

1 1

between the wood and its metal the horse-power thus absorbed at

coefficient of friction

seat be 0.3, find

65 revolutions per minute. To resist the load of 2 tons would require a pressure of 2.83 tons by the 45° slope of the foot-step. The mean circumference would be as in preceding problems, distant two-thirds R from center. 413.

The

shaft of a vertical steam turbine has a

conical foot -step bearing 3.5 inches in diameter,

length 3 inches.

Total load on shaft,

speed 2 500 revolutions per minute friction, 0.07.

"

burn out

"

i ;

and

500 pounds coefficient

;

of

Find the horse-power that tends to

the foot-step.

MO r/ox.

MOTION

III. 414.

second

119

A

body moving with a velocity

is

acted on by a force which produces a con-

stant acceleration of

3 feet

per second.

velocity at the end of 20 seconds Velocity gained

=

of

feet per

5

\Vhat

is

the

.''

acceleration per second

x number

of

seconds. V

Final velocity

The

415.

second

;

initial

=a X t = 3 X 20 = 60 feet per second. = 60 + 5 = 65 feet per second.

velocity of a stone

How

of 2 feet per second.

traveled in

5

Two

416.

is

12 feet per

this velocity decreases uniformly at the rate

seconds trains

other on parallel

far will the stone

have

."

A

and B moving towards each

rails at

the rate of

30 miles and

45 miles an hour, are 5 miles apart at a gi\'en instant. How far apart will they be at the end of 6 minutes

from that the

first

417.

instant,

and

pcsition of

Two

trains,

other in 4 seconds

The

A

at

what distances are they from

.?

130 and

no

when going

feet long, pass

each

in opposite directions.

velocity of the longest train being double that of

the other, find at what speed per hour each

is

going.

—

MO TION. Two

418.

each

121

trains going" in opposite directions pass

One

otlier in 3 seconds.

train

142 feet long

is

and the other 88 feet long. "When going in the same direction one passes the other in 15 seconds. How each train going

fast is

The

419.

.''

velocity of a train

known

is

increasing uniformly; at one o'clock

per hour

tion

A then, /

at 10

A

420.

hour

;

minutes past one

What was

hour.

is

moving

train

railroad train

moves 44

it

may be

How

is

moving

was 36 miles per one.''

far did

travel

it

30 miles an

The

minutes.

retarda-

.''

30 miles an hour.

In each second,

each second during the time

represented as in Fig. 66 by lines of equal length, and the vt,

represents the distance passed

over.

an illustration of uniform motion.

is

When

a railroad train starts from a station, and by uniform gain

in speed attains a velocity of 30 miles

--»

may be

an hour, the distance passed

«

Fig.

Fig. 66.

over

to have been was 12 miles

at the rate of in 2

at

it

minutes past

Its velocity for

feet.

area of the rectangle, or

This

at 7 J

brought to rest

uniform.

is

it

it

Fig. 68.

67.

graphically represented as in Fig. 67.

The

area would

represent said distance.

421.

Similarly,

what condition

of

road train would Fig. 68 represent Note 422.

that the area of Fig. 6S

A

is

vj +

stone skimming

on

speed of the

rail-

.?

\ {y

—

?„)

t,

or

ice passes

vj

-\-

\

at-.

a certain

point with a velocity of 20 feet per second, then suf-

MECHANICS-PROBLEMS.

122

fers a retardation of

423.

when

On

Find

unit.

the stone had

New York

tlie

tlie

space passed

lo seconds, and the whole space

over in the next traversed

one

come

to rest.

Hudson River

Central and

Railroad test tracks near Schenectady, an electric loco-

motive hauled 9 Pullman cars at a running speed of 60 miles per hour. The average acceleration from start to full

The

speed was 0.5 miles per hour per second.

per second per second.

by

was 0.88

retardation on applying air brakes

carefully

These

A train

is

What was

the total time

is

turned off

it

;

then runs on a

level track for 31 miles before stopping.

be the constant

retarding

Also how

force,

A

a state of rest.

through 55 feet in

A

in a certain

the next 2 seconds.

describe in the 426.

If friction

value in

its

is

turned

off

.''

body acted on by a constant force begins

move from

jj feet

far

find

does the train run in

pounds per ton. 3 minutes from the instant steam 425.

.-'

running at the rate of 60 miles an

hour when the steam

to

feet

were obtained

timing the train at measured stations.

Total distance was 4 miles. 424.

results

first

It is

2

observed to move

seconds, and through

What

6 seconds of

its

distance did

motion

it

i"

steamer approaching a wharf with engines

reversed so as to produce a uniform retardation

observed to make 500 feet during the of the retarded

first

is

30 seconds

motion and 200 feet during the next

In how many more seconds 30 seconds. headway be completely stopped .'

will

the

MO rioiv. Two

427. at

bodies are

let

For

I

St

from the same point

fall

an interval of 2 seconds.

tween them

123

the distance be-

Y\\\(\

after the first has fallen for 6 seconds.

= gf= X 32 X 6= = 576 feet s = \gt" = ,1x32X4^ = 256 feet apart = 576-256 = 320 feet.

body,

s

]^

],

For 2d body,

.•.

A

428.

distance

stone

projected vertically upwards with

is

a velocity of 80 feet per second from the a tower 96 feet high.

In what time will

ground, and with what velocity

A

429.

hammer

summit

it

of

reach the

.''

10 tons weight falling from a

of

height of 4 feet drives a wooden pile and comes to

How far does it drive the 3V second. assuming the force is uniform find it.

rest in

And

A

430.

stone

of its striking

is

dropped into a

is

heard

the velocity of sound

What

is

in air

s

=

well,

is

i

200

depth of well. s

=

seconds. I

Time

for stone to fall ^

.-.

^-2

=

is

—

__ iT

200

found from formula -J

^^

s

2 s

;

feet per second.

1

time for sound to come up

,

and

=;

16

."

and the sound it is dropped

seconds after

the depth of the well

Let .•.

2^^,,

pile

/

=

Vj-

4

1

MECHANICS-PROBLEMS.

24

Time

for stone to fall

I

+ time 200

.-.

A

431.

feet

;

4

What

is

come up

=

2/^.

12

"s/s

stone is

tower

the

dropped from a tower of height a

is

projected upwards vertically from the ;

initial

the two start at the same moment. velocity of the second

halfway up the tower 432.

to

j-

another

foot of the

sound

= 2 100 + 300 f ± 150^ + 300 V J = 3 100 ± 150^ V^ = — 310 an inadmissible value, V^ = + 10 = 100 feet, depth of well. S

or

for

A stone

of the splash

is

is

if

they meet

?

dropped into a

well,

and the sound Find

heard ^.y seconds afterwards.

the distance to surface of the water, supposing the velocity of sound to be 433.

A bucket

is

onds the sound of

How

i

120 feet per second.

dropped into a well and in 4 secits striking the water is heard.

far did the bucket drop

434.

A balloon has

?

been ascending vertically

at a

uniform rate of 4^ seconds, and a test ball dropped from it reaches the ground in 7 seconds. Find the velocity of the balloon and the height from which the ball was dropped. 435.

From

a balloon that

is

ascending with velocity

of 32 feet per second, a ball drops

ground

in

1

7 seconds.

How

far

up

and reaches the is

the balloon

.'

4389

A/OT/OiV.

A

436.

ball is let fall to the

will

they meet

A

437.

that

is

it

one

fell.

to the

When and

where

ice

down

slides

a

smooth chute

an angle of 30° to the horizon.

Through

feet vertically will the cake of ice fall in

the fourth second of The

first

thrown

is

.^

cake of

set at

how many

ball

just sufficient velocity to carry

point from which the

25

ground from a certain

same time another

height, and at the

upwards with

1

its motion.''

acceleration for a body falling vertically is^, 32 feet per sec-

The

ond per second. 30^-plane

is

2-

acceleration

s

= — =

Therefore space along plane

A cable car

72 feet, for 3 seconds I

in

A

the 4th second

56 feet

runs wild " down a smooth track

projected up a plan

is

nation with a velocity of it

How

8 seconds after starting

first

body

long before

28 feet for 4 seconds

20 to the horizontal.

of inclination

go during the 439.

"

second per second.

\ aC-

= 438.

component measured along a

sin 30^, or i6 feet per

-"•

will

come

80

far does

from

to rest

.''

of 30° incli-

How

feet per second. .•

it

rest

How

far will

it

go up the plane.

A

440.

body

is

sliding with velocity «

clined plane whose

down an

inclination to the horizon

Find the horizontal and vertical components

is

in-

30°.

of this

velocity. 441.

A

stone was thrown with a velocity of 33 feet

per second at right angles to a train that was going

ME CHANICS-PKOBLEMS.

126

30 miles an hour. It hit a passenger who was sitting on the opposite side of the car that was 9 feet wide.

How

him should be the hole

far in front of

window 442.

in the

A

deer running at the rate of 20 miles an

hour keeps 200 yards distant from a sportsman.

many

^

}

the velocity of the bullet be

A

443. of

boat

miles

5

000

i

is

How

aim be taken

feet in front of the deer should

feet per

rowed

second

if ."

at the rate

an hour on a river that

runs 4 miles an hour. In what dimust the boat be pointed

r rection

the river perpendicularly.''

to cross

With what Let

Fig_ 6g. tlie

Draw ill

OM

OX

velocity does

be 4 units

in

it

move

.-'

length to represent

velocity of the stream.

OX.

perpendicular to

The

resultant velocity is to be

the direction O'M.

With

center

X

and radius of

5 units describe

an arc cutting

OM

in P.

Join

XP, and complete

OQ The

is

the parallelogram of velocities

OXPQ.

the required direction.

angle

QOP =

sin-'

|.

Therefore the boat must not be rowed straight across, but up

stream at an angle of 53°

To

10'.

find the resultant velocity

OP2

•••

..

:

OQ^ - QP' = 52-42 ==

= 25-16 =9 OP = 3

the boat crosses the river at the rate of 3 miles an hour.

MOTION.

A

444.

A

river flows at the rate of 2 miles per liour.

boat

is

rowed

The

river

is

one shore,

3

is

such a way that

in

velocity would be

5

445.

A

000

feet

second

feet per

its

;

}

moving upwards with a

bullet

per second, hits

velocity of

balloon

a

rising with

Find the

100 feet per second.

velocity

water

in still

in a straight line.

000 feet wide the boat starting from headed 60° up-stream. Where will it

strike the opposite shore

I

127

relative

velocity.

446.

A

train at 45 miles

moving 10 yards a second Find the

a parallel road. 447.

To

an hour, passes a carriage

in

the same direction along

relative velocity.

a passenger in a train, raindrops

be falling at an angle of 30° to the vertical really falling vertically,

What

second. 448.

a

Two

is

with

at

along one 4 miles per hour, along the ;

other a carriage goes at 8 miles per hour. the velocity of the

A

449.

steamer

6 miles per hour north

;

;

man

and

its

velocity

.-'

is

.''

going east with a velocity of

is

the steamer increases

What

What

relative to the carriage

the wind appears to blow from the

per hour, and the wind north-east.

to

1

roads cross at right angles

man walks northward

seem

they are

80 feet per

velocity

the speed of the train

;

is

its

velocity to 12 miles

now appears

to

blow from the

the true direction of the wind

MECHANICS-PROBLEMS.

128

A

450.

ship

is

sailing north-east with a velocity of

lo miles per hour, and to a passenger on board the

wind appears

Vi

of lo

to

blow from the north with a velocity

Find the true velocity of

miles per hour.

the wind.

A fly-wheel revolves

451. is

What

12 times a second.

the angular velocity of a point on the rim taken

about the center 452.

with

A

}

broken casting

initial

flies

what

coefficient of friction being \

after 3 seconds

One

of the

be

will

its

The

velocity

.'

axioms for problems in Motion

P

The

along a concrete floor

50 feet per second.

velocity of

the force

:

W

the weight

is,

=

a

that .

g.

moved

=

force producing motion

:

the total weight

eration produced by the force

:

the acceleration that gravity

the accel-

would

produce.

For the above example the force producing motion is x i and

case retardation)

W WX

J

:

W=a a

=

(or in this

32 16 feet per second per second :

After 3 seconds the velocity would be zi

453.

ing is

its

A

= 50 — 16 X 3 = 2 feet per second.

locomotive that weighs 100 tons

is

increas-

speed at the rate of 100 feet a minute.

What

the effective force acting on 454.

An

ice

boat

it

.?

that weighs

i

000 pounds

is

driven for 30 seconds from rest by a wind force of

100 pounds.

Find the velocity acquired and the

distance passed over.

MO I'/OlV.

A

455.

5-pound curling iron

ice against

a

friction

of

I

is

thrown along rough of its weight

one-fifth

Comes to rest after going a distance of 40 must have been

The

456.

and 6

motion

falling

What

feet.

table of a box-machine weighs 50

by a

feet,

neglecting

it

;

velocity at the beginning?

its

pulled back to

is

29

its

What

weight of 20 pounds.

friction,

pounds

starting position, a distance of

will

be used

thus

in

time,

return

.?

A

457.

body whose mass

is

108 pounds

is

placed

on a smooth horizontal plane, and under the action of a certain force 1 1

feet in -J

describes from rest a distance of

seconds.

5

What

is

the force acting

}

458. Two bodies A and B, that weigh 50 pounds and 10 pounds, are connected by a string B is placed on a smooth table, and A hangs over the edge. ;

When A work

has fallen 10

feet,

of the bodies jointly,

A

459.

what

the accumulated

is

and what

of

them

severally

500-volt electric motor imparts velocity to

an 8-ton car so that at the end of 20 seconds

moving on a hour

What amperes are necessary Show that to give a velocity of 20

per cent. 460.

through a height of 13.4

What

461.

move

if

60

miles an lift

it

feet.

force must be exerted

a train of

eration,

is

.''

hour to a train requires the same energy as to vertically

it is

10 miles an

level track at the rate of

the total efficiency of the motor and car

;

.''

by an engine

to

weight 100 tons with 10 units of accel-

frictional resistances are 5

pounds per ton

.'

1

MECHANICS-PROBLElMS.

30

A

462.

40 miles an hour If

weighs 60 tons has a velocity of

train that

the time

at

the resistance to motion

no brakes are applied,

when the The

how

its

power

shut

is

far will

it

have traveled

velocity has reduced to 10 miles per

retardation a

Then

44 feet per second.

the time, and lastly the space by observing that space velocity

:•;

463.

A

find

average

locomotive running on a level track brings 1

20 tons to a speed of 30 miles an

in 2 minutes.

train being

what

=

time.

a train of weight

hour

hour?

be found to be 0.16 feet per second per

will

total loss in velocity is

second; the

off.

10 pounds per ton, and

is

will

The

resistance to motion of the

uniform and equal to 8 pounds per ton,

be the required horse-power at the draw-bar

and what the distance from the starting point when the speed of 30 miles an hour 464. at

A

freight train of

is

attained

}

100 tons weight

the rate of 30 miles an hour

when

is

going

the steam

is

shut off and the brakes applied to the locomotive.

Supposing the only the weight of which of friction

20 X u

:

if

465.

A

second.

is

is

that at the locomotive,

20 tons, what

is

the coefficient

the train stops after going 2 miles

100

=

train i 7f,

of

:

32.

100 tons, exxluding the engine,

grade with an acceleration of

If the

.''

a (which can be found from the data

given in the problem)

runs up a

friction

friction

is

i

foot per

10 pounds per ton, find

the pull on the drawbar between engine and train. Total force

=

force for acceleration

4- force for friction.

-f-

force for lifting

MO TION.

A

466.

body

is

projected with

a velocity of 20 feet per second

down

whose

a plane

inclination

A-.-^i^

is

r.

25°; the coefficient of friction be-

Fig.

Determine the space traversed

ing- 0.4.

X

P W = a W W=

=

V/

:

-

(•423

The space

-3625)

:

g

:

^.

traversed, J-

467.

rt

:

70.

in 2 seconds.

A body slides

+

down

1

af.

a rough inclined plane 100

whose angle of inclination is 0.6 is \. P1nd the velocity at projected up the plane with a velocity

feet long, the sine of

;

the coefficient of friction

the bottom.

If

that just carries

it

The

down

forces acting

to the top, find that velocity. the plane

= 468.

An

\'\'

X

sin a

W

X cos

electric car at the top of a hill

uncontrollable and "runs wild" vertical to

—

down

20 horizontal a distance of

ti

a

X

Jr.

becomes

grade of

| mile.

i

The

resistance to friction being 20 pounds per ton and the total

weight of car and passengers 50 tons,

will the car hill

be going when

it

how

fast

reaches the foot of the

.?

Two

weights of 120 and 100 pounds are sus-

pended by a

fine thread passing over a fixed pulley

469.

without

friction.

What

space will either of them pass

over in the third second of their motion from rest Observe that the force producing motion and the total weight moved is 220 pounds. second per second.

is in this

Then

a

.''

case 20 pounds,

=

2.92 feet per

MECHANICS-PROBLEMS.

132 470.

A

man who

pounds can

just strong

is

enough

when going down on an

How

elevator.

velocity of elevator increasing per second 471.

A

to

150

lift

a barrel of flour of 200 pounds weight

lift

fast is the

?

cord passing over a smooth pulley carries

10 pounds at one end and 54 pounds at the other. What will be the velocity of the weight 5 seconds

from

and what

rest,

will

be the tension

After computing the acceleration

the

tiiat

in

the cord

.''

two weights would

have, find the equivalent force, or tension, that would be required to

cause said acceleration on the lo-pound "weight, which

We

that is being moved.

have a

P

:

10

P, the tension

472.

Two

is

the one

= 27, and = 27 32 = 8.4 pounds. :

strings pass over a

smooth pulley

on

;

one side both strings are attached to a weight of pounds, on the other side one string

weight of

is

5

attached to a

pounds, the other to one of 4 pounds.

3

Find the tensions during motion. 473.

Weights

by a thread the ;

of i

5

pounds and

i-pound weight

is

are connected

1 1

placed on a smooth

horizontal table, while the other hangs over the edcre. If

both are then allowed to m<3ve under the action

what

gravity, 474.

table of

A

is

the tension of the thread

lo-pound weight hangs over the edge of a

and pulls a 45-pound box along

friction

475.

An

;

the coefficient

between the table and the

Find the acceleration and the tension

pit shaft

of

.''

in

bo.x

is

0.5.

the string.

engine draws a three-ton cage up a coal-

at a

speed uniformly increasing at the rate

MO rJ ON. of

5

feet per second

tension in

A

476.

which

tlie

rope

balloon

in

133

What

each second.

is

the

?

is

moving upward with a speed

increasing at the rate of 4 feet per second per

is

Find how much the weight of a body of 10 pounds as tested by a spring balance on it, would differ from its weight under ordinary circumstances. second.

477. An elevator of 300 pounds weight is being lowered down a coal shaft with a downward accelera-

tion of

second per second.

feet per

5

Find the

ten-

sion in the rope.

An

478.

elevator, starting

ward acceleration

of

,V

g

from

for

i

rest,

has a down-

second, then

moves

uniformly for 2 seconds, then has an upward acceler-

comes to rest, {a) How far does descend {b) A person whose weight is 140 pounds experiences what i-)ressure from the elevator during ation of

-J-

g until

it

it

.''

each of the three periods of 479.

A

its

motion

weight of 10 pounds rests 6 feet from the

edge of a smooth horizontal table that

A

.?

is

3 feet

high.

string 7 feet long passes over a smooth pulley at

the edge of the table and connects with a lo-pound weight. If this second weight is allowed to fall in

what time

will

cause the

it

edge of the table 480.

A

per second

body in

the horizontal.

is

first

weight to reach the

.''

projected with a velocity of 50 feet

a direction inclined 40°

upward from

Determine the magnitude and

direc-

1

MECIJAAUCS-PROBLEMS.

34

end

tion of the velocity at the

of 2 seconds

{g being

taken equal to 32.15). Let

ACE

be the path of the projectile.

The

vertical velocity

which the body possessed when it started from A carried it to the summit C of the trajectory, where it had zero \'ertical velocity, and when it reached E it would possess its initial velocity, which would be n

would

The

480^

(In Prob.

sin a.

>:

velocity

l^e

11

y

cos

is

The

40^.)

vertical velocity acquired in falling

the horizontal

AE

would be
and the time from

A

from the highest point

to

/,

=

u

''

sin a

to highest point t

and the

constant horizontal

a.

sin a

=

total time of flight

The range

AE component of

the horizontal ity

/ the time of 2

:

21

\

//

a

sin

cos a

ir sin 2

veloc-

flight

(/

The above explanation and formulas

will

ance in solving the problems that follow.

In

be of material assist-

problems

of these

all

the resistances of the atmosphere are neglected.

481.

A

bullet

per second. in

order that

tal plane, at a

From

482. at

an angle of

second

;

is

fired with a velocity of

i

ooo

feet

What must be the angle of inclination, it may strike a point in the same horizondistance of 15 625 feet

t

the top of a tower a stone 30°,

is

thrown up

and with a velocity of 288

the height of the tower

is

160

feet per

feet.

Find

the time required for the stone to reach the ground,

and the distance

it

will

be from the tower.

MOTION.

From

483.

stone feet

moving

a train

dropped

is

135

at

60 miles per hour a

the stone starts at a height of 8

;

above the ground.

Through what

distance will the stone go while falling 484.

200

A

stone from a quarry blast has a velocity of

per second,

feet

in

a direction inclined at an

To what

angle of 60° to the horizontal plane. will

it

and how

rise,

A

away

far

will

it

height

ground

strike the

}

with a velocity of which the components are 80 and 120 per second respectively. Find the range and

485.

bullet

and

horizontal feet

horizontal

1

is

fired

vertical

greatest height. 486.

The top

the level of a

300

feet

I

000

the wall.

From

city.

is

50 feet above

a man-of-war in the

below the top of the wall and distant

zontally 3 of

of a fortification wall

000

feet, a projectile is fired

The

feet per second.

Where

will

with velocity

projectile just clears

land inside the city

it

D

Fig. 71.

I

000 X I

Eliminate

/

/

X sm a

000 X

and solve

/

— X

for a (a

i,

g/-

cos a

=

bay hori-

= 300 = 3000

8° 28').

.''

1

MECHANICS-PROBLEMS.

36

Find the greatest height

d

then

h,

known

being

enable one

ivill

to find the time for the projectile to fall that height or to pass hori-

zontally over the distance

To

/.

/

add

half the range

and thus find

the distance from man-of-war to where the projectile will land inside the city.

A

487. of

I

discharged with an

ball is

greatest possible range

A

488.

that

is

cannon

A

2

is

000

What

pounds.

490.

What

be

velocity

from a

hill

its it

from the top

sea.

Its initial

its

weight 500

second and

range, and what will be the strikes

.''

must be given

to a golf ball to

just to clear the top of a fence at 12 feet

it

higher elevation and 100 3'ards distant, struck upwards at an angle of 45° 491.

the

find the time

:

fired horizontally

is

feet per

will

is

strikes the sea.

it

energy of the blow that

enable

feet high

300 feet high to a ship at

of a hill

velocity

projectile

directly

fired

is

on the coast and 900

489.

velocity

miles

.''

ball

which elapses before

initial

How many

100 feet per second.

The

if

the ball

is

.?

explosive force of a shell

is

to be regulated

by proper charging so that a required velocity can be attained. Find what velocity will be required for wall the top of which it just to clear a fortification is

distant horizontally

492.

A

velocity of

rifle i

i

Angle

above the gun.

projects

000

mile and at elevation 300 feet of projection its

feet per

is

45°.

shot horizontally with a

second

;

the shot strikes the

MO r/OiV. ground

a distance of

at

height of the

on the

is

the

?

the pres-

is

exerted

What

000 yards.

i

above the ground

rifle

What

493.

sure

137

horizontally

an engine

rails of

20 tons weight going

of

round a

level curve of

yards radius at

an hour

600

30 miles

?

W

Centrifugal force

7"

Fig. 72.

To Let

derive the above formula

A

and B be two

which

is

and

B

at

30

:

positioiLS of the engine.

an hour, would be

mile.s

in

At

A

the velocity,

direction of tangent v,

tlie

the satne velocity would be in direction of tangent

going from

A

measure of

this

to

B

In

f.

the direction of the velocity has changed, and the

change is the centrifugal force. depends upon the rate of change of miotion. To find from A draw AD to represent the velocity of position B.

Its value

the rate,

CD

Then

represent the velocity of

will

value will be

V

AB

>;

as

-,

B

found from similar triangles

r

ABO

;

and

velocity of

vt

X

AB =

—

B

- t

=

~.

r

r

velocity (on the curve)

relative to

A =

€

494.

A

train of

:

W

What

is

X

and

time, so that CI), the

and the

-,

rate of

=

change a

the rate of change or acceleration

=

W

a

:

g

v-

60 tons weight

radius one mile, with

hour.

X

ACD

its

can be found. ^-

of

?'/

Thus knowing

a the centrifugal force

A, and

relative to

is

rounding a curve

a velocity of

20 miles an

the horizontal pressure on the rails

.'

1

MECnANICS-PROBLElVS.

38

A

495.

24-ton engine

yards radius

The rim

496.

inches It

;

What

4.84 tons.

of a pulley has a is

mean

6 inches broad and

\

inch thick.

What

the centrifugal force per inch length of rim

The mass

of the

bob

of a conical

2 pounds, the length of the string

of inclination to vertical

The

is

three forces acting on the

What

45°.

bob are

:

its

is

is

}

pendulum

3 feet,

is

is

}

radius of 20

revolves at 200 revolutions per minute.

497.

rails

the velocity of the engine

is

section

its

;

rounding a curve of 400

is

the horizontal pressure on the

is

the angle

the tension

.''

weight downward, the

tension in the string, and the centrifugal force outward.

The mass

498.

of the string

is

of the

bob

is

20 pounds, the length

2 feet, the tension of the string

is

pounds weight. How many revolutions per second is the pendulum making 5007r'^

.''

pendulum be 10 feet long, the half angle of the cone 30°, and the mass of the bob 12 pounds, find the tension of the thread and the If a

499.

conical

time of one revolution. 500.

which

A is

ball is

hung by

a string in a passenger car

rounding a curve of

a velocity of

I

40 miles an hour.

000

feet radius, with

Find the inclination

of the string to the vertical. 501. car.

A

ball

is

How much

when the

train

hanging from the roof of a railroad will is

it

be deflected from the vertical

rounding a curve of

radius at speed of 45 miles an hour

1

300 yards

MO TION.

139

Find the speed

at which a simple Watt Govwhen the arm makes an angle of 30° with the vertical. I.ength of arm from center of pin to

502.

ernor runs

center of

ball,

18 inches.

(Fig. 73.)

Fig- 73-

Find the speed of a cross-arm governor when the arms make an angle of 30° with the vertical. The 503.

length of the arms from center of pin to center of ball

is

29 inches

inches apart.

;

the points

of suspension are

7

(Fig. 74,)

504. The rotating balls on a centrifugal governor make 160 revolutions per minute the distance from ;

the center of each ball to the center of the shaft 4.5 inches. in diameter.

The

balls are of cast

iron

is

and 2\ inches

Find the centrifugal force

of the gov-

ernor. 505.

Find the speed

a cross-arm governor

in

when

revolutions per minute of

the arms

make an angle

of 30° with the vertical, the length of the

arms from

center of pin to center of ball being 24 inches, and the points of suspension being 6 inches apart.

1

MECI!AXICS-riiO B LEArs.

40

in

each spoke of a six-spoked

flywheel, 8 feet in diameter

and weighing 1344 pounds

Find the tension

506.

when making 200 mass collected

its

revolutions per minute assuming

all

and that by reason of

at its rim,

cracks in the rim, the spokes have to bear the whole of the strain.

The

507.

Company's

flywheel,

mill Jan. 21,

injuring

seriously

which burst

weighed 50

tons,

nine

Cambria Steel

in the

killing three

1904 more,

is

men and have

reported to

and to have been about 20

feet

rim weighed 35 tons and its weight was acting at a mean diameter of 19 feet what centri-

in diameter.

If

fugal force did each of the 8 spokes and

the

when

have to withstand

rim

"racing"

at

its

the

150 revolutions per minute

portion of

wheel was

?

508. A man standing on a street corner was injured by a horse-shoe that was thrown from the front wheel of a passing automobile.

The rubber

tire

the horse-shoe by a protruding nail and

around to the top of the wheel when with

full force.

Diameter

of

would the rim

horse-shoe 509.

carried

was thrown

it

it

off

wheel was 30 inches, and

speed of automobile was 20 miles an velocity

caught up

of the

hour.

What

wheel thus give to the

.''

The automobile

of the

above problem was turn-

ing the corner in a curve of 100 feet radius, the road

being

level.

were acting

Therefore what two centrifugal forces at the instant the i-pound horse-shoe

at the top of the

wheel

.''

What were

was

their values

.'

Mor/OiV. the horse-shoe was thrown with the

If

510.

141 full

velocity of the rim and horizontally from the top of

how

far

away would

As

it

it

land on level ground?

man was standing 6 feet from the track how far from him must have been wheel when the horse-shoe was thrown off, pro-

511.

the

of the automobile

the

vided

it

was thrown tangentially

to track

?

A

loconK>ti\e that weighs 35 tons runs at 40 512. miles an hour on a level grade round a curve of 3 300 feet radius (about i" 44').

What

What

centrifugal force

produced

?

rail for a

standard gage track of 4 feet

513.

In the case of problem

template

ning at

maximum

the

of

'^V/

Given

/

the

length

of a

- the time of an oscillation

rails,

if

not

?

around the head 90 times a minute. 3 feet 6 inches long what will be the 515.

ruiv

What

speed of 60 miles an hour.

weighing four ounces

stone

?

512 the railroad con-

lateral pressure will the spikes

A

inches

lo-J

locomotive and

putting on a 60-ton

changed, then have to withstand 514.

is

should be the elevation of the outer

:

If

whirled

is

the sling

pull in

it

is

.?

simple pendulum,

show how

to find

i>

approximately

the

height

of

a

mountain when a

seconds pendulum, by being taken from sea level to

is

its

24 hours. If >i = 15, what the height of the mountain, the radius of the earth

summit, loses « beats being 4 000 miles

'i

in

MECHAiXICS-PROBLEMS.

142

At

516.

What

hours.

A

517. lates in

is

At

pendulum beats seconds.

sea-level a

the top of a mountain

beats 86 360 times in 24

it

the height of the mountain

pendulum

of

length

two seconds

at

London.

.''

156.556 inches

What

oscil-

the value

is

of g>.

An

518.

800-pound shot

is

gun, with a muzzle velocity of

from an 81 -ton 400 per second a

fired i

:

steady resistance of 9 tons begins to act immediately

How

after the explosion.

An

impulsive force

is

gun move

far will the

.'

a very large force that acts on a body for so

body has practically no motion, but and this change of momentum measures the Impulse or effect produced by the Impulsive Force. In the above problem the impulsive force, or action on the shot to drive it forward, is ec|ual to the reactioir on the gun to drive it short an interval of time that the receives a change of

momentum

;

backward.

—

Action

reaction

Momentum before = momentum after Momentum of gun backward ^ momentum of shot W _ W'_ and

this

forward

simple formula, with a knowledge of the principles of work,

many problems that involve questions of momentum of two or more bodies. solve

will

For the above problem -i^-6^

X

I

:

400 V

= ^

S

X

I

''-^^

at J-

=

z'

feet

per second, velocity of gun

beginning of

average velocity

its ,<

motion.

time.

To find the time Motion has been retarded by a force of 9 tons and the weight thus retarded is 8r tons. Find a the rate of retarda:

tion

;

then since the velocity of

lastly the space.

^'^'^^

equals a

''.

/,

/

can be found and

MOTION.

A

519.

of is

I

000

A

520.

ball is projected with a velocity

Find the

second.

A

and the

latter

with

What was

per second.

when

on with a

continues

feet per second,

a velocity of 380 feet

A

two pieces that weigh 12

The former

velocity of the shell

6 inches per

velocity of the bullet.

initial

20.

700

velocity of

180 pounds, kicks the

velocity of

initial

shell bursts into

pounds and

522.

?

one-ounce bullet fired out of a 20-pound

back with an

521.

gun

velocity of recoil of the

pressed against a mass of

latter

What

second from an 8-ton gun.

feet per

maximum

the

rifle

56-pound

143

the explosion occurred

man weighing 160 pounds jumps

the .''

with a

velocity of \6\ feet per second into a boat weighing

With what

100 pounds.

A

523.

velocity will boat

200

freight train weighing

move away

tons,

and

.''

travel-

ing 20 miles per hour, runs into a passenger-train of

Find the ve50 tons stanchng on the same track. which the brt)ken cars of the passenger train

locity at will

be forced along the track, supposing

Momentum

Now The

before

= momentum

c

=

\.

after.

with this forniula combine a second law, namely: differences in velocities before

ence in velocities

after.

W 1.

II

+

ill'

2, {n

and

//'

W

,

H

=

—

W

W T'

It]

€

=

V

are velocities before impact

,

1'

-|-

?

.^

.^

some constant =

:'.


—

1'' -'

;

and

-'"

after impact.)

Solving these equations, and

_ ^

'

Wii

'

Wii

+

W'li'

-

wT

+ W'// +

eW

ill

-_i/}

W"^

Wt' in

—

ii')

the differ-

.

1

iME CHANICS-PROBLEMS.

44 524.

A

freight car of 20 tons weight

on to a siding with velocity of collides with another car of

moving

in the

same

i

is

switched

is

miles an hour, and

10 tons weight that

direction at

the coefficient of impact

5

5

miles an hour.

is

If

find the velocities of

\,

the cars after they collide.

A

mass 4 pounds and velocity 4 feet per second meets directly a ball of mass 5 pounds 525.

ball of

with opposite velocity of 2 feet per second

Find the 526.

e

;

=

1

velocities after impact.

An

bowling

8-pound

second overtakes and

ball

going

12

feet

a

strikes directly a lo-pound ball

Find their velocities after going 6 feet a second. striking when the coefficient of impact it is |. 527.

A

body

at the rate of

B

weighing

10

15 feet a second,

and moving

pounds,

strikes another

body

weighing 20 pounds, and moving at the rate of 10 the direction at right angles to that

*feet a second, in

of A's motion.

and the impact

The is

bodies are to be treated as points,

supposed to take place

tion of A's motion.

Find the

velocities

in

the direc-

and directions

of the motions of the bodies after impact, the restitu-

tion being perfect (coefficient of elasticity I'lot

a figure to

show the conditions.

The

=

i).

velocity of each in a

unchanged

direction perpendicular to a line through their centers

is

by the impact. Note that u of formula for problem 523 becomes u' becomes u' X cos 90°; z', v cos iSo°; and v', v' cos

45'.

i/

X

cos 45°

;

RE]-IE\V.

REVIEW

IV.

Many

the

of

145

review problems that

have

follow

been prepared from actual engineering,^ conditions. They arc classified somewhat according to their subgraded order, nor

ject matter, but are not given in

with solutions, as they are students

who have

intended

pursued

already

Mechanics, or for engineers

wish to "brush up a

One

528.

in

a

practice

course

in

who may

bit."

of the largest

of the Clark

especially for

Thread Co.

chimneys

at

in

America

Newark, N.J,

is

that

Its height

33 5 feet, interior diameter ii feet, outside diameter Find the work done at base 28^ feet, at top 14 feet. is

in raising the material

from the ground

to its place in

the chimney. The volume may be determined by considering the whole cone and subtracting the top and the core. The average height to raise the average weight of matethe material is found to be 119.07 feet rial is 130 pounds per cubic foot. ;

U.S. Naval Academy a concrete pile of and tapering from 6 inches at the point to 20 inches at the head, was driven 1 inches by 2 blows from a 2 100pound hammer falling 20 feet. The same hammer and fall by 529.

By

tests at the

conical shape 19 feet long

two blows drove a wooden

pile also 19 feet long,

but

gi-

inches at the

point and 11 inches at the head, a cUstance of 5y'ij inches. port says " This shows the comparative bearing power."

The

re-

;

What were (b)

of the

the resistances

wooden

t

{a) of

the concrete pile

ME CHA NICS-rR OBLEMS.

146 '0:

.-^

'

NET/Eir.

147

horse-power was developed, that

"The

output?"

in-put"

is,

what was "the

by problem

530 being

161 horse-power, what was at that time the efficiency of the turbine with its set of bevel gears of horizontal shaftine;

and 50 feet

?

Fig.

76.

532. The strap underneath the pulley

in Fig. 76

is

tightened

by the nuts A and B on top of the lever. When the brake is in use one nut can be tightened by a pull of 50 pounds on a 3-foot wrench, the other by 25 pounds.

Which

nut should tighten

harder and what will

be the approximate tension in that end of the strap, the threads being 6 t(j the inch with mean diameter of

i;V

inches,

and the hexagonal nut having a mean Use 0.25 as the co-

outside diameter of 2; inches.? efficient of friction

and

o.

533.

I

5

between the nut and

its

washer,

for the threads.

For a bridge

like Fig.

'/'j

with lower chord a

parabola 200 feet span and 30 feet rise what would be the stresses at the abutments and middle vertical reaction at

each end

is

when the

600 000 pottnds

.''

1

MECHANICS-PROBLEMS.

48

The Driving Park Bridge at Rochester, N. Y.

Fig. 77.

The

534.

platform of a suspension foot-bridge 100

feet span, 10 feet width, its

own

supports a load, including

weight, of 150 pounds per square foot.

two suspension cables have a dip

of

20

The Find

feet.

the force acting on each cable close to the tower, and in

the middle, assuming the cable to hang in a para-

bolic curve. 535.

Find analytically the stress

in

the cable at

a horizontal distance of 30 feet from the center. 536.

The weight

of a fly-wheel

the diameter, 20 feet coefficient of

;

it

stops.

000 pounds and

0.2.

If

the wheel

14 inches is

;

discon-

when making 27 revolutions how many revolutions it will make

nected from the engine

before

8

diameter of axle,

friction,

per minute, find

is

REVIEW. 537.

149

Plxperiment shows that a weight can

three-quarters of

its

own weight by means

lift

only

of a rope

over a single pulley, this being on account of the stiffness

of

Hence show

the

rope and the friction of the axis.

that the mechanical advantage of four

such pulleys arranged in two blocks

Fig.

Height of

78.

Concrete

dam

is

about 2.05.

falling into place, Niagara, N. Y.

trestle is 20 feet, of

dam

50

feet, cross section

of

dam

and base of trestle i foot wider on the water Tower was tipped by three heavy jacks until it fell, as shown

7 feet, 4 inches square, side.

incut.

538. of

(See Eng. Record of Nov. iS, 1905.)

How many

feet

out of plumb would the top

tower move before starting to

fall

.''

MECHANICS-PROBLEMS.

150

What

539.

tower possess base

energy did the

at the instant

passed the level of the

it

falling

?

A bolt

540.

inch,

foot-pounds of

head

\\ inches in diameter, 9 threads to an

2 inches outside diameter,

by a wrench 12 inches pounds.

for threads

/%

long, is

and

is

tightened up

on end of 50 Find the nut 0.3. pull

0.2, for

stress in the bolt. 541. In a certain piece of that

tlie

bolts in the iish plates

street railway construction I

observed

were being tightened unusually hard,

and the workmen told me that such bolts often broke. By test we found that a pull of 170 pounds was being used at a distance 013.75 There were 9 threads to the inch, mean feet from center of nut. diameter 0.S6 inches, average outside diameter of nut diameter of inside bearing 0.S9

The

was about 0.2 and for the nut

for the threads

Find

inches.

stress

in

bolt,

1.4 inches,

coetticient of friction

0.3.

and then the

stress

per

square inch at root of thread which was 0.60 inches in diameter.

542.

The mean diameter

inch bolt

the

is

mean circumference

of nut 0.3S inches

coefficient of friction 0.16.

bolt

the threads of a ^

of

0.45 inches, the slope of the thread .07,

when tightened up by

and the Find the tension in the

a force of 20

pounds on

the end of a wrench 12 inches long. 543.

The

illustration

floating cantilever

is

Monitor "Florida." to the legs

is

crane shown

in

the

hoisting a 90-ton turret off the U. S.

Distance from middle of crane

45 feet;

from legs

to turret

15 feet.

Vertical height of legs 60 feet, distance apart at top

KE VIE W. 10

feet, at

case

151

bottom 50 feet. Find the stresses for this and in the inclined members at

the legs,

in

center considering a joint at that point. 544.

The

load of 90 tons

steel blocks.

is

Each block has

thus using 8 strands of

i-J

supported by two large four 4-foot steel shea\'es

inch cast-steel rope.

The

Fig. 79-

speed of vertical travel

is

50 feet per minute,

{a)

What

will be the velocity of travel of the hoisting strand (b)

What

will

be the stress per square inch

rope of each block 545.

in

.?

each

.'

Launching data

for nine of

in papers read before the Society of

our recent warships was given Naval Architects and Marine

MECHANICS-PROBLEMS.

152

New York, November, 1904, and published in abstract by Engineering News, Dec. 22, 1904. For the " Ca;lifornia," built by the Union Iron Works of San Engineers at

Francisco,

and launched April

28,

1904

Total moving weight, ship, cradle,

Mean

the above the

It

Travel to point of

maximum

mum

6 062 tons i

moved the first-

in 27.6

foot in

1 1

seconds.

Check

has

this result.

Total travel of the ship, in preceding prob-

lem, was 786 feet

to

...

initial coefficient of friction

been computed as .036. 546.

etc.

slope of upper end of ways

Ship started " very slowly."

From

:

;

total time,

maximum

velocity,

minute 16 seconds

i

;

320 feet time 41.5 seconds; amount of maxivelocity,

velocity, 22.1 feet per second.

ship exclusive of cradle

was

5

980

;

The weight of Means for

tons.

checking were rope stops (72 were broken of total Find the strength 30 tons), anchors, and mud banks. average resistance that the means of checking must

have afforded. 547.

Consult the reference of problem 545 (Eng.

News, Dec. Explain

22, 1904),

why

and copy Fig. 6

in note-book.

the velocity and distance curves start

horizontally frorri

the

In the acceleration

origirf.

curve what was the cause of the rise between 15

and 30 seconds 548.

of elapsed time

From curve

of

problem 547 what was the

velocity at time of 41.5 seconds

What was

."

the acceleration at

5S

.'

The

seconds

acceleration .'

data compute the velocity for 55 seconds.

From

?

that

RE VIE W. 549.

Coal from a barge (Figs. 80 to 82)

where

to a steeple tower

feet long in

or " stopper

"

car goes

24 seconds.

hoisted

down

strikes a

It

a grade cross-bar,

is pushed back a distance of 30 empties and for an instant comes to

which

feet while the car rest.

is

run into a car and by the

it is

action of gravity alone the

294

153

The weight

the coal 4 000.

of the car

If the car

the 30 feet, what

is

is

2

000 pounds and

of

empties uniformly during

the average force of resistance

that the cross-bar exerts

?

Fig. 82.

550. The method of stopping the car of problem 549 may be understood by referring to Pigs. So to S3. The car, going down the grade, piclcs up the cross-bar C, which is clamped to the wire cable

AB. As

the cross-bar

is

pushed along the cable moves, and the

MECHANICS-PROBLEMS.

154

pulley E, around which the cable passes, as

sketch E", goes from E,

its initial

shown

in the enlarged

position, to E', its final position.

The

travel o£ this pulley raises a triangular frame, that is partly filled with broken stone, from the position shown dotted to that shown by full lines in Fig. S2. The mass of stone is 5.5 feet X 4-8^ as shown,

and

I

The space

feet thick.

V

150 pounds per cubic foot

Show is,

is

one-third voids

;

weight of stone,

weight of wooden frame,

000 pounds.

I

that the force exerted parallel to the cable

for the initial

220

;

pounds;

pulley at E, about

position with

for the final

position E' about

1925

pounds.

When

551.

the cross-bar and wire cable of the pre-

ceding problems

move through

the travelling pulley

The diameter distance

were

E to

of pulley being \\ feet

What would be

.-"

diameter

2 feet in

552.

mass

E E'

E

a distance of 30 feet

goes from

E' as described.

what

will

the distance

if

be the pulley

t

One-fourth of the energy possessed by the

of stone

when

lost in friction,

in

the final position E', Fig. 82,

and three-fourths

of

it

is

is

utilized to

"kick" automatically the empty car up the incline C back to its starting point A. If this energy is

from

expended on the car

what

will

starts 553.

back

in a

return distance of 30 feet

maximum

be the

velocity of the car as

it

t

When

the bridge of Fig. 83 carries a crowd

of people making a load of 150 pounds per square foot what will be the reactions for the middle truss and

the stresses in the inclined and horizontal at the

abutments

.''

members

REVJEir.

155

feiS

A

modern highway bridge over

the main tracks of the Reading and Indiana Streets, Pliiladelpltia. Three with 24-feet clear roadways and two lo-fuot sidewalks

railroad at Seventeenth

Pratt trusses, outside.

Concrete abutments. The middle truss

end

from center

to center of

of chords.

Equal panels.

pins,

and iz

feet

is

i

^5 feet

from center

oJJ

inches

to center

I

ME CHA NICS-PROBLEMS.

56

A bridge of the type

shown in Fig. 83 is sued Length of bridge is 150 feet, there are 6 equal panels, and height of trusses With loading of Fig. 85 and the second* is 25 feet. 554.

for a double-track railroad.

driver of forward locomotive placed at the point from abutment,

member and

inclined

the truss which

is

what the

first

panel

first

be the stresses

will

in the

panel of lower chord of

carrying two-thirds of the loadings

.'

555. Span 1 1 of the Benwood Bridge on the Baltimore and Ohio Railroad is 347 feet long. It was reconstructed in 1904 and designed to carry the heavy loading shown in Fig. 85. What would be the

reactions for a train half-way across the bridge 556.

bridge, 557.

With the forward truck what

The

has four

pounds

;

be the reactions

will

going

just

?

off

locomotive of the Empire State Express

drivers

and a

total

weight

the weight on the drivers

is

of

124 000

84 000 pounds

the coefficient of friction between wheels and 0.18. it

Find the

total

weight of

can draw up a grade of

to motion 558.

is

the

."

1

From

2

pounds per

i

itself

in 100,

and

if

train

;

rails is

which

the resistance

ton.

the following data given by a General

Superintendent, determine what revolutions per minute the locomotive drivers were making "

The

:

driver wheels are 42 inches in diameter, each pair being

geared to a 200

HP. 625

volt-motor, with ratio of gearing 81 to 19,

providing for a total tractive effort at

full

working load on 8 motors

of 70 000 lbs. and at starting of 80 000 lbs., assmning coefficient, giving

a nominal rating of

speed of these locomotives

is

i

600 HP.

25%

The

about 20 miles per hour."

tractive

free running

RE VIE IV.

157

•7c- 25 000 CO

C

-^- 50 000

c

-<-

50 000

--y--

50 000

-X-

50 000

B

-V

m

-

-J-

32 500

--^-

32 500

-^/7-

32 000

q-

-5i:-

P C

32 500

25 000

CO

Q

•\- 50 000

c-

c c B

4-

50 000

-X-

50 000

~^-

B

tt,

6

'

I

5

MECHANICS-PROBLEMS.

8

559.

Also from the foregoing data determine what

revolutions the motors were making.

were supplied

An

560.

to the 8

motors

What amperes

}

—

enormous freight locomotive a Mallet designed as a " mountain helper "

duplex compound

—

was put

on the Baltimore and Ohio Railroad

in service

in January, 1905.

This locomotive has drawn 36 steel

cars weighing 702 tons,

a

1%

and

668 tons

i

of lading,

up

grade, and with an average speed of io\ miles

an hour

;

weight of locomotive with tender and an

average amount

of coal

and water

is

225 tons.

What

horse-power without friction was developed for hauling the above total load 561.

What

."

per cent of the work done in the preced-

ing problem would be paying 562.

The draw-bar

work

.?

pull of this Mallet

Compound

has been found to be 74 000 pounds. When running with conditions according to problem 560 what frictional resistances

would

exist

.''

563. At the testing-plant of the Pennsylvania Railroad at the Louis Exhibition in 1904 a freight locomotive of type two-cylinder cross-compound consolidation (2-8-0), size 23 435 x 32, made by St.

the American Locomotive

Company

for the Michigan Central RailDriving wheels 5 feet 3 inches in diameter, total weight 189000 pounds, on drivers 164500; maxiroad, gave the following data

:

mum

tractive effort, sand being used and locomotive acting as a compound, was 31 838 pounds.

According to the above data what would be the between drivers and rails }

coefficient of friction

REVIEW. 564.

With speed

of

15.01

IS9 miles per hour, 80.18

revolutions per minute, piston speed of 428 feet per

minute, the indicated horse-power was 734.9 dynamometer horse-power 675.7. Find the draw-bar pull ;

and the per cent

of

indicated horse-power that

was

lost in friction.

565.

A

car

with 4-inch

What

is

supported on four

axles

traction will

and

coefficient

wheels

36-inch of

friction

0.05.

be required to move the car on

a level track with a total weight of 20 ton on the axles

?

What

energy

will

be

lost

in

friction

minute with the car moving 30 miles an hour L. <

.?

per

1

MECHANICS-PROBLEMS.

60

The load on the test piece at the time of failure is 45 400 pounds. The plane of fracture is shown by the white hne on the test-peice, This plane makes an angle of 23° with the horizontal. Fig. 86. 566.

What would be

the pressure in the direction

Find the

of the

plane at the time of breakage

number

of square inches thus resisting this pressure

and then the stress per square inch in applied mechanics as the Shear. 567.

A horse

?

— which

is

known

pulling a 300-pound cake of ice up

is

a plank run which makes an angle of 40° with the

There are two single pulleys which have

horizontal.

80% each. Coefficient of friction What pull must horse exert 0.05.

efficiencies of

plank run 568.

A

on

.?

ring of weight

smooth

W

free to

is

slide

on a

circular wire that stands in a

vertical plane.

A

string attached to

the ring passes over a smooth pin at point of

the circle and

sustains a weight P.

Determine the

the highest

position of equilibrium. 569.

CD

is

a vertical wall.

port 12 feet from the wall. feet long resting

on

ED

A

A and against

the surfaces are smooth.

is

is

a point of sup-

a uniform bar 32

the wall

CD.

All

Find the position of equi-

librium of the bar. 570. Fig. 88 shows a Carson-Lidgerwooda cableway in use Hartford, Conn., lowering a

The

pipe

river

by means

is

1

part of an intercepting sewer of an inverted siphon.

span of cableway 300

feet,

at

2-foot length of 36-inch cast-iron pipe.

that passes under a

Weight of pipe

and the design of cableway

is

is

3 tons,

such that.

I^E VIE W.

i6i

Fig. 88.

as customary, the sag by a full load in the middle -.}q

The for

is

allowed to be

of the span. practical

method used by manufacturers

computing the

total stress in

sider the condition of

maximum

of trench machinery

cableways of this sort

is

to con-

loading, namely with the load in

the middle of the span, and for that condition to divide half the

—sag—

-

2

This rives a factor which mul-

^

y

1

Mli CHANICS-FK OBL EMS.

62 by the

tiplied

+

total load (in this case 3 tons

weight of cablej gives

the required stress in the main cable-

How

does the result obtained by the above pracmethod check with result obtained by the method of parallelogram of forces, considering the

tical

+

total load of 3 tons

middle of span

weight of cable as acting at

?

571. Each tower for the above cableway sists

is

10 inch legs spread 9 feet apart at the

The main

30 feet high, and con-

formed by two S X bottom and 2 feet at the top.

of a vertical timber frame, or bent, that

is

cable passes over an iron saddle at the top of the tower

and is fastened to a " dead man " anchorage that is buried 60 feet The timber frame is kept vertical by from the foot of the tower. steel guys made fast at top of tower and to the same " dead man,"

What

horizontal

stress

do these guys

when the main cable is free What vertical load do they add

have to

provide for

to slip

saddle

to the

'>.

The

572.

pull

on

additional stress that

the three vertical

steel guys, in

}

the hoisting ropes causes an

would

be, for this case, equiva-

lent to a vertical load of about 3 tons

Add

on the tower

stresses

on one tower.

due to main cable,

and hoisting ropes and then

find the stress

each leg of the tower.

A

573.

jiower

is

Pelton water-motor

12-inch

tested by a friction brake

three fijurths of a 4-inch })ulley a lever

arm

The

is

making

horse-power

is

I

I

3

scales read

5

center of

pounds when

50 revolutions per minute.

being developed

.'

horse-

encircles

the mot
that extends 22 inches from

pulley to scales.

motor

(jn

of

that

What

REI-JEIV.

163

A

6-inch water-pipe that is 600 feet long is 750 gallons of water per minute the water shut off by uniformly closing a 6-inch valve in 3

574.

.delivering is

;

How much

seconds of time.

near the valve be increased

A

575.

on

stands

t

water-works tank

uneven ground

The tank weighs 30 000

will the static pressure

is

as

on a trestle which

shown

the stress in the plane of the legs is

empty

;

when

{b)

how much water

will

the tank

DB is

diagram.

strong wind

40 pounds per square

gives a pressure of

tank

in

A

pounds.

foot.

{a)

full,

P^ind

when (c)

prevent the tank from

the

Find over-

turning.

The wind pressure

tliat

acting on

tlie

may be taken

curved surface of as 0.6

througli the middle of the tank.

of

tliat

tlie

on a

tank causes a vertical section

ME CI/A NICS-PROBLEMS.

164 576.

Name two

advantag'es

of

the hemispherical

bottom over a flat bottom as representedA tank 20 feet high on the sides and 20 in Fig. 89. with hemispherical bottom will hold diameter feet in (or similar)

how many gallons of water ? What will be the wind pressure

taken as

problem

A

577, posed

in ?

pro-

clause in

for

specifications

water valves requiies "

if

preceding

tliat

a valve shall stand with-

out injury a

pull

pounds on

wrencli

is

a

length

in

radius

of

i-^-

of

175 that

times the

the wheel.''

In

considering these specifica-

an engineer and

tions

spector

asks

this

if

unduly strains a valve following analysis

valve:

wheel Fig. 90. A modern form of water-tank, (Erected at St. Elmo, 111., by the Chicago

Bridge and Iron Works.)

is

The

can be

an S-inch

luaile relative to

outside

?

in-

test

S(rew-and-}oke

Diameter of hand 14 inches, diam-

inches, 4 eter of spindle, i threads to the inch (mean ;^

,

.

,.

bearing diameter

.... inches).

i-J

Valve seats taper from 4 inches to 2 inches in diameter of valve Bearing of hand wheel has a mean diameter of 2 inches.

— 8 inches. The

coefficient of friction for the bearing of

and the face of valve against

Find the stress of 175

its seat,

hand wheel, the threads,

may be

in the spindle

pounds as indicated above.

taken as 0.15.

caused by the pull

REvn:\v.

To

find this stress

it

i6s

only necessary to consider

is

conditions that affect the friction of the hand wheel.

The

resistances at the valve have no affect on the

stress in the spindle

which

that part of the stress that

wheel.

To compute

tion of the

this

in is

any case

is

subjected to

transmitted by the hand

stress consider

one

re\'olu-

hand wheels.

Work of pull

of lifting

Substitute and find the

Fig. 91.

+Work

Work

+ Work

on threads

unknown term W,

Fig. 92.

on bearing

the stress.

1

MECHANICS-PROBLEMS

66 578.

Find the pressure against the seat

(no water pressure being considered)

17s pounds 579.

is

When

of the valve

when a

pull of

applied as in problem 577.

a water pressure of

100 pounds per

square inch acts on one side of the 8 inch valve in

problem 578 and the test of 175 pounds pull is applied what normal pressure exists on each side of the valve

?

ji

[Li IS/t

A length of -Vfater pipe

Half Seotiqn of an

18-incli

water pipe

Fig. 93.

Dimensions and weights for cast-iron water pipe are given in the Standard Specifications for Cast-iron Pipe and Special Castings," issued Sept. lo, 1902, by the New England Water Works Asso"

ciation.

REVIEW.

167

580. The dimensions for an 18-inch pipe of class D designed for a hydrostatic pressure of 300 pounds per square inch are represented in Fig. 93. Find the

weight of portion

D

\\

and then the

total

weight of

the whole length of pipe. 581.

Weight

A

has triangular trus.ses 12 feet apart.

roof

of roof covering

and snow equals 30 pounds

per square foot, and the floor gives a load equivalent to

20 000 pounds concentrated

at the foot of a vertical

rod at the center of the truss

40 feet, height 10 and tie-rod. 582-

A

triangular

60 000 pounds, the which is vertical. 1 1

feet.

in

AB 583-

;

length

of truss

Find the stresses

feet.

jib-crane

AB =

is

rafters

at

A

being parallel to

BC

ABC

line of action

in

10 feet,

Find the amount and kind

carries

BC

8

feet,

AC

of stresses acting

and AC.

A derrick with

55 feet long, set at 60°

mast 40 feet long and boom from the horizontal, is lower-

ing into water a wrought-iron pipe 12 feet long, 60 inches internal diameter, 66 inches external diameter.

Density of wrought-iron

boom and when it is in in

584.

tackle

when

is

7.8.

the pipe

Find the stresses in air,

is

and also

water.

Is the retaining wall

shown

in

Fig.

94

safe

against overturning by the earth pressure acting as represented.''

Does the

resultant pressure

the weight of the masonry, taken at

1

between

70 pounds per

"

MECHANICS-PROBLEMS.

i68

cubic foot, and the earth pressure cut the base " within

the middle third

?

_j

L,

Fig. 94-

585.

Fig. 95

shows a retaining wall

As

built at Northfield, Vt.

lem,

is

masonry as

Where

the wall safe against overturning?

does the resultant cut the base 586.

of

in the preceding prob-

The waste gate

.'

Nashua

of the canal for the

Manufacturing Company

at

Nashua, N.H.,

about

is

When this gate is 1\ feet high and 4^ feet wide. closed there is usually a head of 10 feet of water on its

center.

The

coefficient of friction of this

wooden

gate against an ordinary metal 1

i

seat

taken as 0.40 and the

is

weight of the gate pounds. will

What

is

i

000

force in tons

be required to

lift it

.'

^

587.

be the

How wide on top should dam shown

in Fig.

96

to

withstand the reservoir pres-

AV-;

588.

that

Two

VIE \v.

169

Indians wanted to divide a birch log,

was 30

feet long

and tapering from 8 inches

diameter to 12, so that each would have one-half.

them At what

school teacher told

to balance

at that point.

point should

589.

gun

A

20 pound shot

of length 10 feet

being

i

up an

recoil

slope

200

}

feet

;

is

per second

long

through the gun

and saw it

be cut

from a

2

it

A

open

'>.

000-pound

the muzzle velocity of the shot

incline rising

How

fired

it

in

will

it

how

far will the

vertically to

i

take

the

1

5

gun

on the

shot to travel

1

Fig. 97.

590. The engine and geared drum shown

in the illustration are

used for hoisting ore to the top of a blast furnace.

The engine

cyl-

MECHANICS-PROBLEMS.

170

makes a 15 -inch stroke, and 300 revoThe mean effective pressure of the steam being 100 pounds per square inch, what horse-power is developed? The ratio of gears is 5.6. to i the diameter of drum, 4 J feet. The efficiency of engine, geared drum and rest of mechanism is about 85

inder

is

12 inches in diameter,

lutions per minute.

.

;

per cent.

Therefore, under the above conditions, what force will the

engine give to the cable for drawing the

loaded " skip-car blast-furnace

"

up the

incline to the top of the

}

The drum of a hoisting engine is 4 feet in The angle between the engine crank and diameter. 591.

60°.

connecting rod

is

necting rod

feet.

5

Length of crank i foot, conSteam pressure on the piston

W

that is 100 000 pounds which just balances a load compresthe load W, the Determine being hoisted. sion in

the connecting rod and the side pressure

against the cross-head guide. 592.

Sixteen horse-power

is

to

be transmitted by a

belt which embraces | of the circumference of a 20-inCh pulley that makes 1 20 revolutions per minute Find {a) the tension in coefficient of friction is 0.35.

;

the two sides of the belt

when

slipping

is

just pre-

the width of belt required, thickness

vented and. being | inches, and working stress 300 pounds per square inch of section. [h)

593. Find the width of a belt necessary to transmit 10 horse-power to a pulley 12 inches in diameter, so

that the greatest tension

may

not exceed 40 pounds

REVIEW. per inch of width

when the

lutions per minute,

and

171

pulley

makes

i

500 revo-

the coefficient of friction

is

0.25.

A

was made, Aug. 15, 1905, of the new steam plant of the Wolff Milling Company at New Haven, Mo. The engine had a high-pressure cylin594.

der

1

test

2 inches in

diameter

;

length of stroke, 36 inches.

low pressure, 24 inches

The

77.1 per minute; the indicated horse-power

tif

inders

in

high-

What

pressure cylinder, 85.74, low pressure, 66.19.

were the mean effective pressures

;

revolutions were

the two cyl-

1

595. The engine and boiler test of problem 594 was continued 10 hours. During that time 177.72 barrels of flour (each weighing 196 pounds) had been made, and 3 685 pounds of coal had been burned cost of coal per ton of 2 000 pounds in the boilers was $2.90. Find the cost of coal for each barrel of flour made, and the pounds of coal burned per hour ;

per indicated horse-power. 596.

A

Columbus Gas Engine tested

as

shown by

Fig. 98, Oct. 21, 1905, gave the following data

lutions during 15 minutes 3 688, explosions

load on brake

3.024 inches. test

i

:

Revo-

516, net

arm 30 pounds, length of arm 5 feet Four indicator cards taken during the

gave average areas 0.69 square inches, length

3.00 inches.

Stiffness of spring that

was used 300.

MECIIANICS-PliOBLEMS\

1/2

The diameter

engine cylinder

of

the length of stroke

1 1

inches,

is

776

inches and

t'ind the efficiency.

In computing the horse -power of a gas engine by indicator cards number of explosions corresponds to the number of revolutions

the for

an ordinary engine.

Fig. 98.

What would be

597.

the indicated horse-power of

the gas engine, shown on page

17,

and which has a

piston 12 inches in diameter and a crank 8 inches

long

The engine works

.'

there

mean

is

at

150 revolutions a minute,

an explosion ever)- 2 revolutions, and the

62 pounds

effective pressure in the cylinder is

per square inch. 598.

The speed

revolutions

per

of the

minute.

be replaced by two

governor shaft

The

20-pound

AB

lo-pound balls that

is

ball

500 is

to

revolve in

RE VIE IV. planes distant

I

from the plane

foot

and 4 feet lo-pound

of the

Take the distance

ball.

173

7 inches and find

R

and

AC

as

Rj, the

distances at which the 20-pound balls will revolve

ernor shaft forces

from the

gxjv-

when

their centrifugal

the

same moment

liaA'e

about the speed

controller

at

Fig. 99.

A

as

the

lo-pound

one alone. In the preceding problem could the distance changed and still have the moments of the two 20-pound balls balance the 10.' If so what is one such distance 599.

AC be

.''

600.

A

'The

breaking strenglh of rope

rope manufacturer's catalogue states

may be

taken as

7

:

000 X diam-

For a constant transmission tire best results are obtained when the tension on the driving side of the rope is not more than Jj- of the breaking strength and the teirsion on the driveter sijuared.

;

ing side

is

nsrrally twice the tension

Find the horse-power

on the slack side."

of a 2-inch diameter rope, of

weight per foot 0.34 X diameter squared, that runs at 3 000 feet per minute when centrifugal force is considered. Observe that the tension on slack side belt

-|-

\ of difference in tension.

=

centrifugal force of

MECHANICS-PROBLEMS.

174

EXAMINATIONS. MECHANICS.* YALE UNIVERSITY, SHEFFIELD SCIENTIFIC SCHOOL. Senior Mechanical and Mining Engineers. March, 1905. 1.

Define

(a)

balloon

is

five different units of force.

ascending with a speed which

at the rate of

4

feet per

(3)

A

increasing

is

Find

second in each second.

the apparent weight of 10 pounds weighed by a spring balance in the balloon.

A

weight of 20 pounds rests 7 feet from the edge of a smooth horizontal table 4 feet high. string 2.

A

8 feet long passes over a

table

smooth pulley

at

edge of the

and connects with a lo-pound weight. is allowed to fall, in what time weight reach the edge of the -table.

second weight first

3.

(a)

A

be the tension in the cord

.'

(p)

a balance with arms in ratio of froiti

4.

(«)

A 5

does he gain or lose

;

what

will

shopkeeper uses to

He

6.

alternate pans what appears to be

How much

the

cord passing over a smooth pulley carries

10 pounds at one end and 54 at the other

out

If this

will

weighs

60 pounds.

1

Define a force couple.

Show

that a force

couple cannot be replaced by a single force,

{b)

Show

About 20 weeks of Mechanics in a three hour a week course, and the present course of 10 weeks with three hours a week preparation. * Preparatory Studies

:

EX A iV/iVA T/OiVS. how

to find the resultant of any

I

number

75

of non-con-

current forces acting on a rigid body.

Find the force of attraction of a homogeneous sphere on a particle within the sphere, {b) The {a)

5.

mass

of the sun

earth,

and

its

How

earth.

is

radius

300 000 times the mass is 100 times the radius

far will a stone fall

second at surface of sun 6.

A

{a)

pounds,

of the

rest in

one

?

uniform rod 8 feet long, weighing i8

fastened at one end to a vertical wall by a

is

smooth hinge.

It

is

kept horizontal by a string 10

feet long, attached to its free

Find the tension

the wall.

on

pressure

from

of the

the

hinge.

end and to a point

A

(^7)

in

the string and the

in

uniform rod

AB,

20 inches long weighing 20 pounds, rests horizontally upon two pegs whose distance apart is 8 inches.

How must the pegs

the rod be placed so that the pressure on

may be

equal

when weights

pounds are suspended from 7.

A

and

Find by the principle of

of

40 and 60

B, respectively

virtual

1

work the con-

dition of equilibrium for a differential screw consider-

ing friction. 8.

A

uniform ladder 70 feet long

is

equally inclined

to a vertical wall and the horizontal ground.

A

man

weighing 224 pounds ascends the ladder, which weighs 44S pounds. How far up the ladder can the man ascend before the wall

is

\

it

and

slips

if

for the

the coefficient of friction for

ground

\

?

MECI/AAVCS-PJiOBLEMS.

176

Find the work

9.

pivot, bearing 10.

A

lost

by

a shaft with a truncated

an end thrust.

belt passing

around a drum has an angle of

contact a and a coefficient of friction

Find the

/^.

horse-power which can be transmitted. 11.

Two

rough inclined planes are placed end to

A

body of 100 pounds rests on one of the planes, which has an inclination of 60°. A string end,

attached to this body passes over a smooth pulley at the apex of the planes and holds another body on the

second plane of inclination, friction for

body

each plane

is I,

to just hold the first

=

sine 30^

30°.

coefficient

If

of

find the weight of second

from sliding down the plane. cosine 30°

.5

=

.86

*

MECHANICS. TUFTS COLLEGE, DEPARTMENT OF ENGINEERING. Ex.\MINATION AT MiD-YeAR, FeB.

Answer any

In tests of cast-iron

1.

1904) record

15,

of

wheel 4

feet,

is

weight

of

I905.*

fly

wheels (Eng.

given of one as follows

stress in each

fugal force of its

6,

eight questions.

arm due

portion of the rim

same portion

of

rim

7^-

A'ercv, :

Dec.

Diameter

to the

centri-

1680 pounds, pounds.

Find

bursting speed in miles per hour. 2.

A

leather belt treated with dressing has coeffi-

cient of friction *

Preparatory studies

mechanism one-half week and two hours

year,

on an iron pulley of :

Physics lectures

and present course

of preparation for each.

erne year,

0.3.

laboratory

The one-half

belt year,

of half year with three class htiurs per

EXAMIiVA TIONS.

I "]"]

encircles 200° of a pulley 10 feet in diameter.

W'licn

running at 140 revolutions per minute the belt must

How

transmit 300 horse-power.

be

if

width

?

A

3.

wall derrick has a vertical post 9 feet high,

member

at top a horizontal

back from the load brace

i

A

4.

of

up from ground.

highway bridge 80 from one end and

feet

Uniform load on bridge

A

foot.

across 5.

;

A

feet I'jng,

15

and

what load

is

Find

the other.

300 pounds per

is

tons weight

is

then on each abutment

&

How

speed

—

{a)

heavy a train

40 miles an without wind or of

frictional resistances, [b) with resistances of

per ton acting

train of

one minute.

400 tons

starts

stiff-leg

high,

from a station and on

40 miles an hour

in

Neglecting resistances, what would be

the draw-bar pull

feet

20 pounds

?

a level track attains a speed of

55

}

H. R. R. has developed ap-

hour, up a 2 per cent grade

A

linear

half-way

large type of locomotive recently put in ser-

could this locomotive draw, at

7.

a

feet long has supports

proximately 2 000 horse-power.

A

is

stresses.

10 feet from

road roller of 10

vice on the N. Y. C.

6.

feet

3

10 tons at outward end

3 feet long connecting with the vertical post

at a point 4 feet

2

wide should belt

designed to stand 100 pounds per inch of

is

it

.''

steel

boom

derrick

85

with

feet long,

vertical set

mast

wilh tackle

1

MECHANICS-riiOBLEMS.

78

40

feet long

two

raising

is

Find stresses

weight.

in

back stay which makes an angle If

50 tons total

boilers of

boom and

ford,

would be the

\\'hat

21

1

at

top and 20

bottom what stresses must they sustain

feet apart at

8.

in

30° with vertical.

of

mast be made of two members joined

working

and

tackle

total

horse-power of

.?

pumps

hours per day to supply the City of Med-

2

100 gallons of water

600 population, with

per day (for each person) and forced against 60 pounds

pressure (equals a height of 138 feet) of engines 9.

A

and pumps

shell

per second

;

is

The

efficiency

80 per cent.

can be fired with velocity of 2 000 feet

neglecting resistances,

can a man-of-war be

of projection being 30°

how near

order to have

in

500

clear a fortification wall

10.

to be

.''

feet

above sea

to shore

shells just

its

level,

angle

}

Derive the formula for centrifugal force.

The

20th Century express attains a speed of 60 miles per

When

rounding a curve of 4 000 feet radius how much should the outer rail be elevated to avoid hour.

lateral pressure

1

(Center to center of

rails is

4 feet

10 J inches.) 11.

Define acceleration, work,

coefficient of friction.

to pull a

Find the

moment

of a force,

least force

necessary

packing case of 300 pounds weight along a

horizontal floor.

Coefficient of friction 0.58.

Total number of problems taken during the half-year has been

about 195.

l-:XAMIiVA riONS.

179

MECHy\NICS. TUFTS COLLEGE, DEPARTMENT OF ENGINEERING. Examination at Division

1\Lij-Yi;ar, Fek.

answer any S questions.

I?

Division

/'

njor..*

i,

answer No.

and

1

1

7 otlrers.

In a clircct-acting steam

1.

sure

is

23 500 pounds

maximum

angle of 15° with the

An

iron

wedge

make an angle column which of

coefficient

What 3.

the

line of action of

Find the pressure on the guides.

piston. 2.

;

the piston pres-

eni;-inc

the connecting-rod makes a

force

An

is

hax'ing faces of equal taper that

10°

of

being forced under an iron

is

supporting a load of

is

friction for the ii^on

needed

electric

to

is

push the wedge forward

car that

is

filled

with

and weighs 25 tons goes up a grade the speed of 10 miles an hour. to traction are 30

The

tons.

5

surfaces

pounds

of

The

]5er ton.

0.18. .''

passengers i

in

100 at

total resistances

What

horse-power

must be supplied when the efficiency of the mechanFor an electro motive force of ism is 60 per cent 500 volts what amperes would be necessary.'' ."

4.

A

shaper head that weighs 500 i)ounds makes

The forward stroke of 12 inches in 6 seconds. equivalent are machinery and of cutting resistances of its

to a coefficient of friction of 0.5.

work being done * Preparatory studies

page

176.

At what

rate

is

bottom

of

.-'

same

as for examination of 1905

and given

at

I

IME CHA NICS-PR BLEATS.

So Coal

5. is

in

hoisted from a barge to a tower where

is

it

run into a car that goes down a grade 294 feet long " 24 seconds. It strikes a cross-bar or " stopper

which

is

pushed back a distance

of

30 feet while the

comes to 000 pounds and

car empties and for an instant

weight of the car

4 000. feet

If the car

what

is

empties uniformly during the 30

."

A highway bridge

span 48

of

feet,

width 40

has two queen-post trusses of depth 9.2 feet

each truss

The

rest.

of the coal

the average force of resistance that the

cross-bar exerts 6.

2

is

is

feet,

and

;

divided by two posts into three equal

parts.

The bridge

load of

I

crowded with people making a

is

50 pounds per square the

tric car one-third

way

foot,

and also an

elec-

across the bridge causes an

additional load equivalent to a concentrated load of

20 tons.

Find the stresses

in

chords and posts.

The head

7.

engine

is

to

plate of a

Buckeye

be hoisted by a con-

tinuous rope

tliat

passes through eye

bolts

are

5

that

feet

and

apart,

through a chain-hoist hook that 3 feet

above the plane

bolts.

The rope

is

of

is

the eye

free to slip,

the plate weighs 500 pounds.

and

Find

the total pull that tends to break the eye bolts. 8.

The

16 pounds

center is

of

a

steel

crank-pin that

weighs

12 inches from the center of the engine

EXAMIiYA TJOA'S.

The

shaft.

Find

tlie

8

I

makes 190 revohitions per minute. centrifugal force caused by tlie pin. shaft

The San Mateo Dam

9.

1

in

Cahfornia was designed

for a height of 170 feet, width at top 25 feet, at base

176 I,

feet,

with a uniform batter on the water side 4 to side near the top 21 to i, then a

and on the back

curve of radius 258 feet to near batter of

is

i

to

The

i.

150 pounds

weight

bottom where the

tire

material throughout

per

is

concrete

Compute

foot.

cul^ic

approximately the factor of safety of such a section against overturning.

Define

10.

example.

moment

and

of a force

Also define and

illustrate

by an

illustrate " resolve parallel

and perpendicular to plane," a couple and three other important terms or equations of Mechanics.

Show

to find the least force necessary to pull a box

how

along a horizontal

W

--«H

11.

floor.

W

g

,

;/'

=

—

W'

,

v'

g

S V

W 7'H —

7''

=

£>

(u

—11)

Tell what the above formulas mean. "

Horizontal range

How

A

obtained

bullet

is

=

2 //

cos a

X

u sin a

" "

?

fired with a velocity of

i

000

feet per

What must be the angle of inclination in that it may strike a point in the same horizontal

second.

order

plane at a distance of 15 625 feet

.?

Total number of problems taken during the half-year, about i8o.

I

MECIIANICS-PKOBLEMS.

82

STATICS HARVARD UNIVERSITY First Course in Mechanics 1.

Find the components of a force of 500 pounds along it by (
lines inclined to

;

cally only. 2.

Find the moment

about {a)

(4,

6)

;

(^b)

(o,

of o)

(—

(300 pounds 48° ;

{c)

(-

4, 6)

;

{d)

6))

4,

(3,

-

7;.

Algebraically only. 3.

A

uniform body

in the

shape of an isosceles triangle

with base of 60 feet and altitude of 20 feet weighs 200 It is supported at points in its base 20 feet pounds.

Forces of 20 and 60 feet respectively from the left end. pounds and 40 pounds act vertically upward and downward respectively from points bisecting the left and right sloping sides respectively.

Determine the pressures upon the supports. 4.

A

rectangle, 10 inches

at the origin,

a corner

by 8 inches, has one corner

two sides coincident with OA^ and OY, and

at fio, 8).

Two

forces, of 20

one along the upper edge of

it

pounds each,

toward the

along the lower edge toward the

Two more

left.

forces,

40 pounds each, act respectively upward along edge and downward along the right edge. of

{a)

Is the

{l>)

If

body subject

act

right, the other

left

either to translation or rotation

any further forces be needed

state the value of the simplest

to

?

cause eciuilibrium

system that

will

do

it.

EXAMTNA riONS. 5.

Find the center

of gravity of

1S3

a plane

sides with corners at (o, oj, (5, o), (4,

5),

figure of fi\e (4, 3),

(14, 3),

(8, o).

Solve both algebraically, and graphically, using latter case the general string

the

in

polygon method.

6. A sphere weighing 1000 pounds rests between two smooth planes which are inclined to each other by 30°, the less steep of which is inclined 10° to the horizontal.

Determine the pressure on each plane algebraically. 7. A plane rectangular frame 60 wide stands on two supports, one

A

corners.

applied at

pounds

horizontal wind

t,o

force

high and 10 feet

each of the lower of

4000 pounds

is

from the ground and a load of 6000

rests at the middle of the top.

If the thrust

equally

feet

feet at

b}'

of

the

wind be assumed

to

be resisted

the supports, determine the remaining forces at

the supports. 8.

Determine graphically stresses

the given truss.

hand sketch 9.

in all

of the bars of

results

upon large

stresses in

Q, R, and

Show numerical

free-

of truss.

Determine algebraically

5

of

truss of last question without finding other stresses. 10.

Determine the reactions {Hi, H„,

supports of the given three-hinged arch. a

Fj,

/:,)

at the

:

1

MECHANICS-PROBLEMS.

84

ANSWERS TO PROBLEMS. new

In preparing this tions have all

two opposing sugges-

edition

been offered to

me

:

one that

I

should give

the answers to the problems, the other that

should give none.

and

am

I

have taken the middle ground,

I

giving about half of the answers, believing

method will serve both for engineers in pracand others who wish to know that their results are correct, and for college classes where it is often that this tice

preferred that

some

the student place too is

generally agreed,

answers be omitted

lest

much dependence on them.

It

of the

I

think, that with students the

advice frequently given by Professor Merriman in his excellent text books should be emphasized,

namely

that the answers are not the main part of a problem.

In

fact,

the student

is

urged not to consult the answer

at the beginning of a problem,

to get that numerical result. of the

and then aim merely

First an understanding

problem should be obtained, then a diagram

representing the data should be drawn, and an

mate

of the

esti-

answer based on experience should be

noted.

Furthermore, in

my own

solutions shall be carefully

classes I require that the

made in

special note books,

AA'SH'/iRS rO

rKOBLEMS.

1S5

and that the student's method of analysis shall be plain, concise, and easily understood for the ability ;

to reason soundly and to demonstrate clearly should

be leading aims

Work.

in the

study of Mechanics.

(10) (6) 320 men. (3) 3 000 foot-pounds. foot-pounds. foot-pounds. (12) ig 800 000

169 000 (15) 352

000 foot-pounds.

104 167 pounds.

foot-pounds.

(40)

522.5 pounds. 12.9

(55) 435 67.8 amperes.

pounds (81)

1

power.

(89)

horse-power.

(79)

power.

no

10 500

9^',-

i

566

1

horse-power

10.5

(70)

theoretically.

107 horse-power.

665 horse-power. looms.

(91)

(83) 97.5

(87) 13.2 horse-

0.061.

horse-power.

(106)

(loi) 5

pounds i

594

million horse-

horse-power.

132 million

12.6

(92)

(97) 62

horse-power.

Duty.

(113) (116)

machines.

(121) 21 pounds. (125) 0.14 (127) 18a pounds. (129)

(118) 3.44

(124)

1-2

kilowatts.

139

of

39600

horse-power; 97 cubic inches. 88 a R/P strokes per minute. (137)

^j

horse-

12

(100) 5.6

feet.

2. i

horse-power.

(68) 4

000 foot-pounds, or cubic

;

(48)

(53) 25

(58)

(104)

hours 18 minutes.

pounds.

horse-power.

9448100ms. (no) 157 (108) 0.39.

horse-power.

(34)

pounds.

About 80 pounds per inch

(82)

(95) 450

per square inch.

112

15

+';|

(86) 2

125

pounds per ton. (74) i 000 against friction; 23 520 000 foot-

(72)

40 horse-power.

horse-power.

132

(63)

(77)

wasted.

inches.

1

horse-power.

inches.

an hour.

horse-power.

1.5

(38)

horse-power.

power.

miles

(46)

(50) 36/]-

(66) 4

(32)

(20)

(25)

(42) 6 000 foot-pounds; ratio 3

man-power.

horse-power.

width.

foot-pounds.

pounds.

28.5

(36)

(44)

(60)

20

(28) 120 000 pounds.

pounds.

0.54

(17) 104.8 foot-pounds.

(22)

pounds

(131) 156 tons.

per

ton.

(133) 265

(140)

i

783

1

86

MECHANICS-PROBLEMS.

'

(142) 393 pounds; 19.6 per cent. (145) 4.5 (149) 12 758 foot(147) I 856 horse-power. pounds; I 450 pounds. (152) 0.17 horse-power. (156)

amperes.

feet.

pounds.

549

height.

Ratio

('58)

to

I

250

I

tons.

pounds.

(162) 750

(167) Average

1.94.

625

15

(159)

feet

turns.

(164) 44.3

(166)

28.75 foot-pounds.

of

Force. (171) 15.2 horse-power. (169) 62.5 cubic feet. (176) ID pounds. 43 units (178) (173) 300 pounds. ;

(181) 29 pounds.

25.

Rafters 6.32 tons;

pounds;

(183) 150

rod 6

tie

tons.

90.

(186)

Boom

(188)

77.3

(191) 50 pounds. (193) 580 pounds. (196) 117. 1 pounds; 82.8. (194) 6 030 pounds. (199) 2.8 pounds bja, b being distance 9.6. (200) cos"^

tons

tackle 36.4.

;

=

;

C

from

to

AB

Perpendicular

and

100.

(217)

plane.

to

(209) 2.45 tons. I

a the length of the rod.

2

(211)

(205) 1.

460 pounds;

I

990.

D being area of triangle, P = 4fD Q =W(?(6- -H r — tr)/4 c D.

W

pounds.

(218) 77 /;

+

{a"

c"

—

-h

Z'^)

(222) In guy 10.3

;

(224) Back stay, 90 tons; legs,

tons; in legs 18.6 tons.

(226) 7.8 tons; 6.5.

100.

000.

i

(213) 86.6 pounds;

15 tons.

(220)

(203)

pounds;

1020

(228)

Back

stays, 81

tons;

A-frame 36; wire rope 29; upper boom 46; lower 51; guy or tackle 49.5. (229) 1. 16 tons 0.55 0.53. (232) ;

;

(233) 13 units at tan~' ^V '^'^'ith AB. (234) parallel to CA at distance from AC of 3 \/2/2 X

14.2 pounds. 2

V2 P

AB.

(238) Ratio

2 \l/'^

—

c'

i

to

V3.

(242)

=

W ^ /

(246) 115. 5 pounds; 57.7. (252) 6250 5 000. (257) 6 inches from end. (259) 9

pounds; inches from middle;

18 pounds.

the middle.

feet.

3.5

Tension

sm9.

(266) 7

pounds; i inch from middle.

(276) 6.5 pounds.

(263)

5

inches from

(268) 1065 pounds. (272)

(278) 9.5 pounds.

4',

(270)

tons;

(280) 11

3^.

inches

1

ANSli'ERS TO PRO/iLEMS.

187

and 6. (282) Posts 40 tons; lower chord, 37,5 and 32.5 upper chord 32.5. (284) Post 6 tons; tie 12.8; chord (2S7) 33.6 pounds. 12.5. (286) 75 pounds. (290J ;

540.

(292)

On

tex.

(293)

2 "73 (7/9,

=

each side

2

line bisecting- vertical angle

y/^ir/s^

V3

(294) 6

(7.

(?/i 1,

from sides; outside the triangle

3 at

V3

from ver-

5

the sides,

fi'onr

A-'^j"

V3

2

i^/i i,

if

,7/1

6V3<7/5,

distance

-'^3"/5- (296) Any point of line parallel to passing through X which is in 1>C produced so that = 2 BC. (299) 5 units acting parallel to BD, cutting

"S^S'VS'

CD CX

BC produced

X,

at

so

that

4

CX =

BC.

(302) 124

pounds, 92, 134. (306) At point 15 and 16 inches from adjacent sides. (309) 2]- feet from rim. (313) If D be the middle point of BC, R is represented in magnitude b)' 2

DC

AD, and orDB, so

(317)

I

each. at

B

tan~^

and

DX

to vertical

from

is

and

('327)

I.

P=

15

DA,

X

being

(315) Heloses

horizontal,

nail to wall '^3

—

to

pounds.

(318) 85.9

^3/~

(331)

X parallel

=BC/8.

At C force

8 V
feet.

through

that

inch.

(^23)

of stick

acts

= :

(321) 15

and

W V7/2. pressure

(33:^)

=

in

pound.

pounds

W V3/2

;

(326) Length

= 8 V3

18900 pounds.

000 pounds.

i

ounces

(328)

f of length

35.3

from

end where pressure is 4 pounds. (335) 1.33 inches. (342) It divides (339) I inch from AC, ij from AB. in ratio of i to is hinged the cover which the face to 2.

(345)

(348)

2

From

cos^

=

left-hand edge 2.84 inches; 5.36 inches. 3 cos

(tt

= rV3.

—

6)/3,

being angle with hori-

(352) (3Ss) 373 pounds. pounds. 20° (362) (366) i; /j. (359) (357) i/^3pounds. (37 r) 1140 tan'^ inclination, (369) 433 f. W;y(i sin a, 60°. /x -y) -f pounds; 314(376) (37+) wheel. of weight the and radius, (378) being 47 r zontal.

(35^0)

//

W

J-.

;

1

MECHANICS-PROBLEMS.

88

(388) About (382) 104 pounds. pounds. pounds. 2 800 (396) (394) (393) 65 898 pounds. (398) 2,\- (400) 13 inches. (401) 37-6

(380) 1/ V3.

feet.

2.08.

inches.

(407)

(405) 504

W/P =

pounds.

0.95 or 1.05.

(410) 0.44 horse-power.

(415) 3S 30 miles

horse-power.

1,92

137

thermal

(412) 3.5 horse-power.

units.

Motion.

(417) 13/T miles per hour 27tV. (419) (422) 150 feet; 200. (424) 13

feet.

;

an hour.

pounds per 6

(406)

(408)

ton.

seconds;

per second

(425) 99 feet. (426) 5 seconds. (428) 112 feet per second. (431) '^fa~g feet

'

(433) 231 feet. (435) 4080 feet. In "sfhJTg seconds, and 3/2/4 feet from ground. (440) 2/V3/2; 11/2. (442) 17.6 feet. 3 SO feet. .

(436) (438)

(444) (446) 24 5 miles an hour. (449) Northwest 6 V2 miles an hour. (451) 24 t. (454) 65.5 miles an hour; 0.27 miles. (457) 300 pounds. (459) 24.3

About ^ mile up-stream.

amperes for maximum

velocity. (462) 1.9 miles. (464) feet 16 per second. (469) 7.25 feet. (467) Vs (470) 8 feet per second. (473) 3/^ pounds. (475) 6938 pounds. (478) {a) J/- ^ feet; {b) 70 pounds, 140

0.014.

and i86|.

i second. (481) 15° or 75°. (483) (486) 3903 feet inside of city. (487) 7.16 miles. (489) 8 600 feet 31 250 000 foot-pounds. (494) 0.3 tons. (497j 2.83 pounds. (500) Tan-i ^^\\. (503)

44 V2

(479)

feet.

;

43 revolutions per minute. tons.

(506)

5^ tons.

(514) 3.4pounds.

(510) 23.ifeet.

(507) 321

(515) 3 666§

(519) 3.1 feet per second. (521) 496.8 feet per second. i feet per second; +2. (525) (527) returns 5 feet per second B moves at 45" with its course

feet.

—

A

;

and

velocity of 10 \/2.

Review.

(531) 56-7 per cent.

(529) (a) 58 tons(534) 60 000 pounds close to tower

47000

(535)

in

middle.

52222

pounds.

(536)

17

PROBLEMS.

ANSiyE/<:S TO

1

89

revolutions.

pounds.

(538) 7.25 feet. (541) In bolt 27 500 (543) In leg 63.24 tons; in inclined members,

18.75 tons.

(545) 0.036. (549) I 250 pounds. (552) per second. (557) 472 tons. (558) 160. (559) 682 revolutions per minute; igio amperes for the 8 motors. (560) i 453 horse-power. (563) 0.19. (564) 27.8

feet

16 879 pounds; 8.1 per cent.

square inch. stress

(568)

cos ^

=

P

(566) 2

W.

2

720 pounds per

(571)

Horizontal

tons; vertical additional 0.9 tons.

(573) 2.0 horse-power. (574) 35 pounds. (575) {a) 41 800 pounds 1.8

;

(c) gallons. i 195 458 000 pounds (S77) 7 419 pounds. (578) 30 862 pounds. (579) 25 800 pounds, (580) I 780 pounds. (583) In air, boom 16.57 35 900. (/')

;

boom 14.45 tons. (586) 4^ tons. (587) (590) 9 518 (589) 33.75 feet; ^V second. pounds. (594) 542 pounds; (592) (b) 19.2 inches. 10.4. (596) 80,6 per cent, (600) 42 horse-power.

tons; in water, 15.2 feet.

I

go

Falling Bodies

^

MECHANICS-PROBLEMS. Velocity Acquired by a Criveii Heiglit*

Body Falling

a

Functions of Angles Angle

A FEW IMPORTANT UNIT VALUES TO BE USED IN SOLVING THESE PROBLEMS = 44 feet per second = 2 000 pounds = 6 feet = 6 080 feet = 62^ pounds = 7i gallons = 8J pounds = 2.304 feet head = 778 foot-pounds of energy = 32 feet per second per second,

30 miles an hour I

ton

I

fathom knot

I

cubic foot of water

I

I

gallon of water

I

pound of water pressure

I

British thermal unit

g, acceleration of gravity

unless otherwise specified e the base of Napierian

system of

= 2.7 1828 1828 = 746 watts = 1.34 horse-power

logarithms 1

horse-power

I

kilowatt

Watts

=

volts

X amperes

IMPORTANT FUNCTIONS OF ANGLES AND TRIGONOMETRIC RELATIONS.

INDEX.

193

INDEX Acceleration,

5,

1

EquiHbriant, 3

19

Examination papers, 174

Angular

velocity, 12S

Answers

to problems, 1S4

Automobile,

Axle

friction,

Falling bodies, 123

140

22,

Fire engines, 37 streams, 49

115

Belt friction, 108, 170

Floor-posts, 83, 102

Bicycles, 140

Floating cantilever, 151

Bolt Friction, 104, 150

Fly wheels, 4S, 140, 148

Bridges, 75, 77, 78,147, '54

Foot-pounds,

Center of gravity,

4,

90

Centrifugal force,

5,

137

Force problems, Forces

Friction coefficients, 96

Chimney, 145

problems, 96

Coal, unloading, 23, 153

Friction of angles, 191

wagon, 90 Coefficient of friction,

Components,

3, 4,

96

98, 100

Concurrent forces, 3 Cooper's loading, 157 4,

Gas

engines, 16, 171

Governors, 139, 172 Gravity, acceleration of, 44, 123,

190

84 Florse-power,

Dam

falling,

Impulse,

Derricks, 54, 167

Dipper dredge,

64, 65

for hoisting, 169

Electric car, 131 current, 19

motors, 41, 129 i,

16

44

5,

142

Indicator cards, 26

Definitions, 2

Energy,

2,

149

Davit, 77

Dram

51

at a point, 51

Fortification wall, 135

Centroid, 4

Couples,

7

Foot-step bearings, 117

Cast-iron pipe, 166

Ladders, 79, 103, 104

Launching data, Least

pull,

45, 151

100

Levers, 73

Locomotives,

22, 41, 122, 128, 130,

141, 156

Logarithmic-decimal paper, loS

MECJIANJCS-PROJILEMS.

194 Moments,

3,

Momentum,

68

72

Sailing vessel,

142

Shears, 6r, 62, 63

Motion problems, 119

Ship resistance,

Sound Parallel forces, 72, So

40, 151

velocity, 123

vSteam engines, 25, 56, 89, 171 turbine shaft, 118

Parallelogram of forces, 3

Steel rails, 93 Structural plates, 92

Pendulum, 141 Pile driver, 12

Trench machine, 160

resistance, 145 Plates, structural, 91

Tripod, 66

Projectiles, 46, 134

Trusses, 70, 71, 72, 77, 78

Pulleys, 13, 21

Pumps,

Unit values, 192

35, 38

Velocities,

Relative velocity,

1

19

of falling bodies, 190

Rail sections, 92 1

36

Water

gates, 97, 164, 168

Resolution of forces, 100

motor, 20, 162

Restitution, coefficient, 5

turbine, 33, 146

power, 33, 38

Resultant, 3

Review problem, 145 Roof trusses, 70

Water-works tanks, 163 Wedges, loi Wire rope, 173

Rope

Work

Retaining walls, 89, 767

friction, 106, 114, 173

proljlems, 7