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# The elements of railroad engineering (Volume 5)

. (page 9 of 15)

them will be 33 X 00 X 8 = 15,840 ft., which, divided by the

number of feet in one mile, gives ^ ' = 3 miles.

Ans.

5,280

(d) As the distance between the two bodies increases

33 ft. per sec, then, 825 divided by 33 must be the time

825
required for the bodies to be 825 ft. apart, or â€” â€” = 25

sec Ans.

(386) See Fig. 25.

(387) (a) Although not so stated, the velocity is
evidently considered with reference to a point on the shore.
10 â€” 4 = 6 miles an hour. Ans.

(^) iO + 4 = 14 miles an hour. Ans.

(c) 10 â€” 4 4- 3 = 9, and 10 + 4+ 3 = 17 miles an hour. Ans.

(388) See Fig. 26.

/
/
/

'isis

Scale l=10llK

S

<?/

[

-^46

Fig. 26.

172

ELEMENTARY MECHANICS.

(389) See Fig. 27. By rules 2 and 4, Art. 754, bc =
87 sin -23Â° = 87 X .39073 = 33.994 lb., ^ ^ = 87 cos 23Â° =
87 X .92050 = 80.084 15.

80 16,

Fig. 27.

(390) See Fig. 28. {b) By rules 2 and 4, Art. 754,
^ ^ = 325 sin 15Â° = 325 X . 25882 = 84. 12 lb. Ans.

{a) ac= 325 cos 15Â° = 325 X . 96593 = 313. 93 lb. Ans.

(391) Use formula lO.

W 125

ELEMENTARY MECHANICS. 173

W

(392) Using formula 10, m = â€”, W=mg= 63.7 X

32.16 = 1,727 lb. Ans.

(393) {a) Yes. {b) 25. {c) 25. Ans.

(394) {a) Using formula \Z, d"" \ R' :: W '. w, d =

V â€” r -^ â€” T7^ â–  = 4,749.736 miles. 4,749.736 â€”

w 100

4,000=749.736 miles. Ans.

(^) Using formula 11, R : d :: W : w, d = -jrr =

4,000 X 100 _ 2 g36,88 miles. 4,000 - 2,836.88 = 1,163.12
141

miles. Ans.

(396) (a) Use formula 18,

. ./27i ./2 X 5,280 ,.â€ž .

t=y â€” =y â€” 5^5-7^ â€” = 18.12 sec. Ans.

{b) Use formula 13, 2^=^/ = 32. 16X18. 12 = 582. 74 ft. per

sec, or, by formula 1 6, ^^ = V%gh = V^ X 32.16 X 5,280 =
582.76 ft. per sec. Ans.

The slight difference in the two velocities is caused by not
calculating the time to a sufficient number of decimal places,
the actual value for / being 18.12065 sec.

(396) Use formula 26. Kinetic energy = Wh = â€” â€” .
Wh = 160 X 5,280 = 844,800 ft. -lb.

Wif 100X582.76' oAAÂ»a(^(^ ^u a

-1^ â€” = â€” n rriTTTr- = 844, 799 ft. -lb. Ans,

2^ 2 X 32.16 '

(397) (a) Using formulas 16 and 14,

h= i- = , '7^,, = 86,592 ft. = 16.4 miles. Ans. {b)
'Zg 2 X 32.16 ' '

t = -^=1 time required to go up or fall back. Hence, total

time = â€” sec. = ,.^ .. 00 ... = 2.4461 mm. = 2 mm. 26.77
g 60 X o2.1d

sec. Ans.

174 ELEMENTARY MECHANICS.

(398) 1 hour = GO min., 1 day = 24 hours; hence, 1

5
day = 60 X 24 = 1,440 min. Using formula 7, V = -;

, ,, 8,000 X 3.1416 ,^ ,^o â–  -1 â€¢ A

whence, V= tâ€”ttt. = 17.453 + miles per mm. Ans.

1,440

(399) (a) Use formula 25.

X- f ^^' 400 X 1,875 X 1 ,875 ^^ .^^ ^_ _
Kmetic energy =-jâ€” = â–  2 X 32 IQ ^ 31,863, 339. 55

ft. -lb. Ans.

,,, 21,863,339.55 ,^ ^^, ^â€ž ^

{&) ' ' =10,931.67 ft. -tons. Ans.

{c) See Art. 961.

Striking force X r^^ = 21,863,339.55 ft. -lb.,

or striking force = ^^'^^^^^'^^ = 43, 726,679 lb. Ans.

(400) Using formula 18, i = i/l^ = l/lKI^ =

^ g ^ 32.16

3. 52673 sec. , when ^ = 32. 16.

/ = |/ ^ X ^QQ = 4. 47214 sec. , when g = 20.
'^ 20

4. 47214 - 3. 52673 = 0. 94541 sec. Ans.

(401) See Art. 910.

(402) See Art. 963.

(403) {a) See Art. 962.

â€ž m W 800 â€ž T^ W 500

â– ^=77= â€” â€¢ '^=Tivoo- Hence, Z/ = â€” =

V gv 1,728' ' gv 32.16Xt\V?

33.582. Ans. {b) In Art. 962, the density of water was

found to be 1. 941. {c) In Art. 963, it is stated that the

specific gravity of a body is the ratio of its density to the

33 582
density of water. Hence, -^r-^rrj- = 17.3 = specific gravity.

If the weight of water be taken as 62.5 lb. per cu. ft., the
specific gravity will be found to be 17.28. Ans.

(404) Assuming that it started from a state of rest,
formula 13 gives Â«/ =^ ^ = 32.16 X 5 = 160.8 ft. per sec.

ELEMENTARY MECHANICS. 175

(405) Use formulas 17 and 13. // = ^ ^ /' =

â€” ^ X 3' = 144.72 ft., distance fallen at the end of third

second.

t; = ^/= 32.16 X 3 = 96.48 ft. per sec, velocity at end of
third second.

96.48 X 6 = 578.88 ft., distance fallen during the remain-
ing 6 seconds.

144. 72 + 578. 88 = 723. 6 ft. = total distance. Ans.

(406) See Art. 961.

Striking force x :^ = 8 X 8 = 64. Therefore, striking

64
force = â€” = 1,536 tons. Ans.

2

12

(407) See Arts. 901 and 902.

(408) Use formula 19.

Centrifugal force = tension of string = .00034 WRN^ â€¢â€”

.00034 X (.5236 X 4' X -261) x ^ X 60' = 13.38 + lb. Ans.

(409) (80' - 70') X . 7854 X 26 X . 261 X i = 3, 997. 2933 lb. ,

weight of one-half the fly-wheel rim. Inside radius =

80- 10 .,. . ,, ,,,, ^^
â€” 3o in., or \\ = 2}| ft.

According to formula 19, I^^= .00034 W R N^ =
.00034 X 3,997.2933 X 35 X 175'

12

= 121,394+ lb. Ans.

(41 0) {a) Use formulas 1 1 and 12. Rid:; IV : zu, or
... wR 1x4,000 ,^,, A

"^= IT = -To5â€” = *" '"â€¢ ^"'-

4 000' V 40
(d) d' : R' ::IV: zu, or tv = ' ; ,,,^, = 38.072 lb. Ans.
^ ' 4,100

(411) See Art. 955.

176 ELEMENTARY MECHANICS

(412) Use formula 1 2.

_ V4,000' X 2

d^ : R':: W: zu, or d .

A

= 13,064 mi., nearly.

13,064 â€” 4,000 = 9,064 miles. Ans.

(41 3) Use formula 18. / = /^ = /1P# =
' ^32.16

1.7634 sec, nearly.

1.7634 X 140 = 246.876 ft. Ans.

(414) ^ = :5- sec. Use formula 17.
60 6

h^^gf = ^-X 32.16 X (ly = 1.78| ft. = 1 ft. 9.44 in.

Ans.

(416) See Arts. 906, 907.
(41 6) See Arts.. 908, 909.

ELEMENTARY MECHANICS.

177

(417) No. It can only be counteracted by another
equal couple which tends to revolve the body in an opposite
direction.

(418) See Art. 914.

(419) Draw the quadrilateral as shown in Fig. 29.
Divide it into two triangles by the diagonal B D. 'The
center of gravity of the triangle B C D \s found to be at a,
and the center of gravity of the triangle A B D is found to
be at d (Art. 914). Join a and d by the line a d, which, on
being measured, is found to have a length of 4. 27 inches.
From C and A drop the perpendiculars C /'' and A G on the

diagonal B D. Then, area of the triangle ABD= - (AGx

lit

B D), and area of the triangle B C D=\ {C F x B D).

Measuring these distances, B Z> = 11', C F=5.l\ and
A G= 7.7'.

Area ABD=\ x7.7xll = 42.35 sq. in.

Area BCD=\ X 5.1x11 = 28.05 sq. in.
According to formula 20, the distance of O, the center of

28.05 X 4.27

= 1. 7. Therefore, the cen-

gravity, from ^ is ^g^^^^^^ 3^

ter of gravity is on the line ^ ^ at a distance of 1.7' from b.

(420) See Fig. 30. The
center of gravity lies at the
geometrical center of the penta-
gon, which may be found as
follows : From any vertex draw
a line to the middle point of the
opposite side. Repeat the
operation for any other vertex,
and the intersection of the two
lines will be the desired center
of gravity.

Fig. ao.

178

ELEMENTARY MECHANICS.

(421) See Fig, 31. Since any number of quadrilaterals
can be drawn with the sides given, any number of answers
can be obtained.

Draw a quadrilateral, the lengths of whose sides are equal
to the distances between the weights, and locate a weight on
each corner. Apply formula 20 to find the distance C, IV^ ;
9 X 18

thus. C W. =

9 + 21

5.4". Measure the distance C, W^

Fig. 31.

suppose it equals say 36". Apply the formula again.
15 X 36

31.7'.

15 +(9 + 21)

= 12". Measure C, W^ ; it equals say

Apply the formula again. C^C ^

17 X 31.7

= 8.7".

17 + 15 + 9 + 21
C is center of gravity of the combination.

(422) het A B C D E, Fig. 32, be the outline, the
right-angled triangle cut-off being E S D. Divide the
figure into two parts by the line ;Â« Â«, which is so drawn
that it cuts off an isosceles right-angled triangle m B n,
equal in area to E S D, from the opposite corner of the
square.

ELEMENTARY MECHANICS.

179

The center of gravity of AinnC D E is then at C, , its

geometrical center. ^;Â« = 4 in. ; angle Bmrâ€” 45^*; there-

A m, B

fore, Br = B7n X sin Bmr = 4: X .707 = 2.828 in. C,, the

2
center of gravity of Bmn, lies on Br^ and B C^ = â€”Br

o

= 1x2.828 = 1.885 in. BC^ = ABxsm BAC^=Ux
o

sin 45Â° = 14 X .707 = 9.898 in. C,C, = B C^- B C, = 9.8m

-1.885 = 8.013 in.

4X4

Area ABCDE=W-
4X4

2

= 188 sq. in. Area inBn

2

sq. in.

= 8 sq. in. Area Amn CDE = 1H8 - 8 = 180

The center of gravity of the combined area lies at C, at

180

ELEMENTARY MECHANICS.

a distance from C^, according to formula 20 (Art. 911)

, ^ 8X C^C, 8 X 8.013 â–  ,, . ^ ^ O.I â€¢
"^"^^ '^ ^80 + 8^ = 188 = -^^1 ^"- ^. ^= -341 m.
BC=BC^- C^C=d.89S - .34:1 = 0. 557 inches. Ans.

(423) (^) In one revolution the power will have moved

through a distance of 2 X 15 X 3.1416 = 94.248', and the

1"
weight will have been lifted j . The velocity ratio is then

94.248 H-^ = 376.992.
4

37G.992 X 25 = 9,424.8 lb. Ans.
{a) 9,424.8 - 5,000 = 4,424.8 lb. Ans.
{c) 4,424.8 -^ 9,424.8 = 46.95^. Ans..

(424) See Arts. 920 and 922.

(425) Construct the prism ABED, Fig. 33. From
Â£, draw the line Â£ F. Find the center of gravity of the

i

Horizontal G

Fig. 83.

rectangle, which is at C,, and that of the triangle, which is

ELEMENTARY MECHANICS. Igl

at C^. . Connect these centers of gravity by the straight line
C^ C^ and find the common center of gravity of the body by
the rule to be at C. Having found this center, draw the
line of direction C G. If this line falls within the base, the
body will stand, and if it falls without, it will fall.

(426) {a) 5 ft. 6 in. = 66'. 66 -^ 6 = 11 = velocity
ratio, Ans.

{b) 5 X 11 = 55 lb. Ans.

(427) 55 X .65 = 35.75 lb. Ans.

35 X 60

(428) Apply formula 20. 5 ft. = 60'.

180 + 35
9.7674 in., nearly, = distance from the large weight. Ans.

(429) {a) 1,000^50 = 20, velocity ratio. Ans. See
Art. 945. (<^) 10 fixed and 10 movable. â–  Ans. (<:) 50 h- 95
= 52.63j^. Ans.

(430) Py. circumference = PTx 3-, or 60 X 40 X 3.1416

o

-Wy\,ox W= 60 X 40 X 3.1416 X 8 = 60,318.72 lb. Since

o

the efficiency of combination is 40^, the tension on the stud
would be . 40 X 60, 318. 72 = 24, 127. 488 lb. Ans.

(431) (a) |/20' + 5' = 20.616 ft. = length of inclined
plane.

PX length of plane = Wx height, or Px 20.616 =
1,580 X 5.

/Â»-ii||221^ = 383.2 lb. Ans. (/?) In the second case,
20.616

P X length of base = IVx height, or Px 20 = 1,5S0 X 5;

hence, P=i^^^r^ = 395 lb. Ans.
^0

(432) ^Fx 2 = 42X6, or IV=^^^=126 lb.

1 68
126 + 42 = 168 lb. lGSxl=lV' X 12, otW' = ^= 141b.

Ans.

182

ELEMENTARY MECHANICS.

(433) See Fig. 34. /'X 14 X 21 X 19 = 2^ X 3^ X 2^

/* 4: O

X 725, or

14 X 21 X 19

W

ft

â€”24^â€”f6'^' 18^-4-6'^ -30

W

P

Fig. 34.

^

-IS"^

F

o

,r1

(434) See Fig. 35. {a) 35 X 15 X 12 X 20 = 5 X 3 ^ X

3 X fT, or

â€ž. 35 X 15 X 12 X 20 . ... ., .

fr = â€” = 2,400 lb. Ans.

5 X 3^ X 3 '

FIO.86.

ELEMENTARY MECHANICS.

183

(b) 2,400 -H 35 = 68y = velocity ratio. Ans.

(c) 1,932 -^ 2,400 = .805 = 80.5?^. Ans.

(435) In Fig. 36, let the 12-lb. weight be placed at A,
the 18-lb. weight at B, and the 15-lb. weight at D.
Use formula 20.

= 6" = distance C^B = distance of center of

18 + 12

Fig. 36.

gravity of the 12 and 18-lb. weights from B. Drawing C\ B,

C^C= . ' = - C^D. Measuring the distances

\i.<i -\~ loj -f- lo o

of C from BD, DA, and A B, it is found that Ca = 3.45",

Cb = 5.25", and Cd= 4.4". Ans.

(436) {a) Potential energy equals the work which the
body would do in falling to the ground = 500 x 75 =
37,500 ft. -lb. Ans.

184

ELEMENTARY MECHANICS.

(d) Using formula 18,/

sec, = -035995 min., the time of falling
37,500

^=i/'t

2X 75
32.16

= 2.1597

33,000 X .035995
(437) 127 ^62.5 = 2.032 = specific gravity. Ans.

(438)

62.5

1,728

Use i

2 X 60X 6.5

or

6.5-5.75

Fig. 37.

X 9. 823 = .35529 lb. Ans.

^

(439) Use formula 21. W=(^^\

X .48 = 499.2 lb. Ans. See ^'

(440) See Art. 961.

77., /3 io\ ^^' 1.5X25'

^x(8^^^) = -2-7=^^r32:i6'"^

1.5 X 25'

/r=!^ii!dr' = 406.42 lb. Ans.
f H- 12

(441 ) (a) 2,000 -^ 4 = 500 = wt. of cu.
ft. 500 -j- 62.5 = 8 = specific gravity. Ans.

J7

m'

500

= .28935 lb. Ans.

2>

Fig. 37.

Fig. 38.

(442) See Fig. 38. 14.5 X 2

= 29. 30X 29= J>rx 5, or W^ =

30X29 _ . ., .

â€¢ = 174 lb. Ans.

5

(443) 75 X .21 = 15.75 lb.

Ans.

(444) (a) 900 X 150 = 135,000 ft. -lb. Ans,
135,000

15

(^)A2!E = iH.p

33,000 11

= 9,000 ft. -lb. per min. Ans.
Ans<

ELEMENTARY MECHANICS.

185

(446) 900 X .18 X 2 = 324 lb. = force required to over-
come the friction. 900 + 324 = 1,224 lb. = total force.

IS;^ = - Â«-â€¢ ^- â– .

(446) 18 -T- 88 = .2045. Ans.

(447) See Art. 962. D = f^= ^^^^

(448) See Fig. 39. 125
-47.5 = 77.5 lb. = down-
ward pressure.

77.5^4 = 19.375 lb.
= pressure on each support.

Ans.

(449) See Fig. 40.

= 12.438.
Ans.

Fig. 40.

(450) See Fig. 41. 4.5 ^ 2 = 2.25.
12

2.25

X 6 X 30 = 9G0 lb. Ans.

(451) (a) 960^30 = 32. Ans.

{d) 790 -^ 960 = .8229 = 82. 29^. AnÂ§,

186

ELEMENTARY MECHANICS.

(452) (a) See Fig. 42. 475 + (475 X .24) = 589 lb.

Ans.
(d) 475 -7-589 = . 8064 = 80. 64^. Ans.

m.

Fig. 41.

Fig. 42.

(453) (a) By formula 23, U = FS = 6 X 25 = 150

foct-pounds. Ans.

(^) 4^"^-=S-=^"^^"-

Using formula 24, Power = -^ = ^= 3,600 ft. -lb. per
min. Ans.

HYDROMECHANICS.

(QUESTIONS 454-503.)

(454) The area of the surface of the sphere is 20 X 20
X 3.1416= 1,256.64 sq. in. (See rule, Art. 817.)

The specific gravity of sea water is 1.026. (See tables
of Specific Gravity, ) The pressure on the sphere per square
inch is the weight of a column of wat^r 1 sq. in. in cross-
section and 2 miles long. The total pressure is, therefore,

1,256.64 X 5,280 X 2 X .434 X 1.026 =-5,908,971 lb. Ans.

(455) 125 â€” 83.5 = 41.5 lb. = loss of weight in water

= weight of a volume of water equal to the volume of the

sphere. (See Art. 987.) 1 cu. in. of water weighs .03617

lb.; hence, 41.5 lb. of water must contain 41.5 h- .03617 =

1,147.4 cu. in. = volume of the sphere. Ans.

Note. â€” It is evident that the specific gravity of the sphere need not
be taken into account.

/^=r^\ n 225,000 ^c, k c.

(456) Q = gQ^gQ = 62.5 cu. ft. per min.

Substituting the values given in formula 51,

^ = 1.229 / ^â– '^Â»Â°^"^-^' = 10.38'+.

Substituting this value of d in formula 49,
24.51X62.5 ^ ^ - _
^Â»'= 16.38- ^^â€¢^^^^-
The value of f (from the table) corresponding to v^ =
5.7095 is .0216, using but four decimal places. Hence, ap-
plying formula 52,

j^ 2.57 < /^^^ X 2,800 + i X 16.38)62.5- ^ ^^, ^^^_

26

188 HYDROMECHANICS.

(457) (a) Area of piston = Q\ X .7854 = .6013 sq. in.

50
Pressure per square inch exerted by piston =

.6013
83.15 1b.

A column of water 1 foot high and of 1 sq. in. cross-
section weighs .434 pound, and therefore exerts a pressure of
.434 lb. per sq. in. The height of a column of water to exert

83 15
a pressure of 83.15 lb. per sq. in. must be ^ = 191.6 ft.

Consequently, the water will rise 191.6 ft. Ans.

The diameter of the hole in the squirt gun has nothing to
do with the height of the water, since the pressure per
square inch will remain the same, no matter what the
diameter may be.

{d) Using formula 34, range = ^4: Ay, we have

/Tky = 4/4 X 10 X 191.6 = 87. 54 ft. Ans.

(458) Use formulas 44 and 43.

(d) Q^= All;i/2i[i/ T' - / !;] =

.4lx|x/^^r3^[|/^-/(4y]^

62.21 cu. ft. per sec. Ans.

(a) Area of weir = dd= 2.5 X 2 = 5 sq. ft.
Using formula 43,

v^ = yj = â€”^ â€” = 10.44 ft. per sec. Ans.

[c) To obtain the discharge in gallons per hour (d) multi-
ply by 60 X 60 (seconds in an hour) and by 7.48 (gallons in
a cu. ft.). Thus 52.21 X 60 X 60 X 7.48 = 1,405,910.9 gal.
per hour. Ans.

(459) First find the coefficient of friction by using
formula 46, and the table of coefficients of friction.

.â€ž = a.3U->/^ = 2.315l/IJ|J|I = 3.191 ft. per sec.

HYDROMECHANICS. 189

In the table/ = .0243 for vâ€ž,= 3 and .023 for z;â€ž= 4; the dif-
ference is . 0013. 3. 191 â€” 3 = . 191. Then, 1 : . 191 : : . 0013 : x,
or .^ = .0002. Therefore, .0243 - .0002 = .0241 = /. Use
formula 50 ; substitute in it the value of /here found, and
multiply by GO to get the discharge per minute.

(2=.09445dr'

/ 7(5 V 7 5
.09445 X y-g'r 0241 X 12,000 ^ ^^ = ^^^"^ ^^^- P^' ""'"â€¢

Note. â€” It will be noticed that the term-|</, in formula 50, has
been omitted. This was done because the length of the pipe ex-
ceeded 10,000 times its diameter.

(460) {a) Use formula 46.

z/â€ž = 2.315 X r-^ X 00 =

/ 76 V 7 5
^â– ^I'^ .Omx 12,000 X ^" = 1"' "â–  P" â– "'"â–  ^"^-
{b) 447.6 gal. per min. -^ 60 = 7.46 gal. per sec. = 1 cu,
ft. per sec, nearly. Ans.

(461) See Art. 1005.

v = \/ ^gh = -1/2 X 32. 16X10 = 25. 36 ft. per sec. Ans.

(462) Use formulas 49 and 47.

24.51 Q 24.51 X 42,000 ^ ^^^ .
Vr. = â€”^^ = e.5^ X 60 X 60 = ^-^^^ ^'- P"^ ^""â€¢

h-.f ""^m |_.0233z/â€ž' =

5.36ar
.021 X 1,500 X 6.768

5.36 X 6.5

f .0233 X 6.768' = 42.48 ft. Ans.

(463) {b') Area of top or bottom of cylinder equals
20" X .7854= 314.16 sq. in. Area of cross-section of pipe =

^1) X. 7854 =.1104 sq. in. 25 lb. 10 oz. =25.625 lb.

25.625 -^ .1104 = 232.11 lb., pressure per square inch on top
or bottom exerted by the weight and piston.

Pressure due to a head of 10 ft. = .434 X 10 = 4.34 lb.
per sq. in.

190 HYDROMECHANICS.

Pressure due to a head of 13 ft. = .434 X 13 = 5.64 lb.
per sq. in.

(Since a column of water 1 ft. high exerts a pressure of
.434 lb. per sq. in.)

Pressure on the top = pressure due to weight + pressure
due to head of 10 ft. = 232.11 +4.34 = 236.45 lb. per sq.
in. Ans.

(a) Pressure on bottom = pressure due to weight + pres-
sure due to head of 13 ft. = 232.11 + 5.64= 237.75 lb. per
sq. in. Ans.

(c) Total pressure, or equivalent weight on the bottom =
237.75 X 314.16 = 74,691.54 lb. Ans.

(464) .434 X 1-: = .651 lb., pressure due to the head of

water in the cylinder at the center of the orifice.

236.45, pressure per square inch on top,4- -651 = 237.101,
total pressure per square inch. Area of orifice = 1' x . 7854
= .7854 sq. in.

. 7854 X 237. 101 = 186. 22 lb. Ans.

(465) {a) Use formulas 28 and 27. Sp.Gr. =

w 11.25

{lV-lV')-{lV^-lV,)~{9l.2o-4:l)-{lQ X5-3iX 16) "^
.556. Ans.

(466) First find the coefficient of friction. Formula
46 gives

^ = 2.310 |/-l^Â«^*

z,. = 3.315f ^ = a.31o|/ -gjj^^^j^ = 5.0719 ft. per sec.

From the table in Art. 1033,/ = .0230 for vâ€ž = 4, and
.0214 for t'â€ž = 6. .0230 - .0214 = .0016. 5.0719 - 4 =
1.0719. 2 : 1.0719 :: .0016 : x, or x = .0009. .0230 â€” .0009
= . 0221=/ for z'â€ž = 5.0719. Hence,

^- = *-^lÂ« ^ .02^X^000 = 'â– ' '^- ^' ''''â–  ^"=-

HYDROMECHANICS.

191

(467) Use formulas 46 and 45.

t' =2.315

/-

120 X 4

= 7.1728 ft. per sec.

.025 X 2,000

From the table in Art. 1033,/ = .0214 for v,^ = 6, and
.0205 for Â«;â€ž. = 8. 8 - 6 = 2.
.0214 - .0205 = .0009. 7.1728 - 6 = 1.1728.
2 : 1.1728::. 0009 : ;r, or ;ir= .0005.
.0214- .0005 = .0209 =/for v^ = 7.1728.
Hence, the velocity of discharge =

Â«/â€ž, = 2.315 lA

120 X 4

.0209 X 2,000 + i X 4

(468) {a) See Fig. 43. Area of
cylinder = 19' X .7854; pressure, 90
pounds per sq. in. Hence, the total
pressure on the piston is 19' X .7854
X 90 = 25,517.6 lb. =the load that
can be lifted. Ans.

{b) The diameter of the pipe has
no effect on the load which can be
lifted, except that a larger pipe will
lift the load faster, since more water
will flow in during a given time.

(469) {a) f=. 0205 for v^ = 8.
Therefore, using formula 47,

= 7. 8 ft. per sec. Ans.

/i = -

.0205 X 5,280 X 8'

+ .0233 X

5.3G X 10
8' = 130.73 ft. Ans.

(<^) Using formula 48, Q =
.0408^''z/â€ž=.0408xl0'x8 = 32. G4 gal. per sec
60 = 117,504 gal. per hour. Ans.

32. 04 X 60 X

(470) A column of water 1 in. square and 2.304 ft. high
weighs 1 lb. ; hence, to produce a pressure of 30 lb. per sq.
in. would require a column of water 2.304 x 30 = 69.12 ft.
high = head. Using formula 36,

z;=.98'/2^//=.98v'2X 32.16 X 69.12 = 65.34 ft. per sec. Ans.

192 HYDROMECHANICS.

(47 1 ) (a) 36 in. = 3 f t. A column of water 1 in. square
and 1 ft. high weighs .434 lb. .434 X 43 = 18.662 lb. per
sq. in., pressure on the bottom of the cylinder. .434 x 40
â€” 17.36 lb. per sq. in., pressure on the top of the cylinder.
Area of base of cylinder = 20* X .7854 = 314.16 sq. in.
314.16 X 18.662 = 5,862.85 lb., total pressure on the bottom.

Ans.

(^) 314.16X17.36 = 5,453.82 lb., total pressure on the
top. Ans.

(472) 2 lb. = 32 oz. 32 â€” 10 = 22 oz. = loss of weight
of the bottle in water. 32 + 16 = 48 = weight of bottle and
sugar in air. 48 â€” 16 = 32 oz. = loss of weight of bottle
and sugar in water. 32 â€” 22 = 10 oz. = loss of weight of
sugar in water = weight of a volume of water equal to the
volume of the sugar. Then, by formula 27,

specific gravity = ^^_ ^, = â€” = 1.6. Ans.

(473) 33 = i/2Â£7i (see Art. 1005), or

33"
k = â€”r-^r^ = 16.931 ft. per sec. Ans.
64.32

(474) (a) Use formula 42, and multiply by 7.48 X 60
X 60 to reduce cu. ft. per sec. to gal. per hour.

a = .41 X ^ X 1/2 X 32.16 X (^y X 7.48 X 60 X 60 =

216,551 gal. per hr. Ans.
(6) By formula 43,

r, .41x||x|/2x32.16x(j-|y

^ - ^Â« ^ 1^ ViÂ±Z_ = 3.676

*" ^^ ^yâ€” ft. per sec.

12 ^ 12 Ans.

(475) /= .0193 for vâ€ž, = 12. Therefore, using for-
mula 47,

''=Sfd+ â– "'''-' =

.0193 X 6,000 X ly ^ ^^33 ^ ^.^. ^ ^_^^^ 3^ ^^ ^^
D.oo X o

HYDROMECHANICS.

193

(476) (a) 1 cu. in. of water weighs .03617 lb.
.03617 X 40 = 1.4468 lb. = weight of 40 cu. in. of water
= loss of weight of lead in water.

16. 4 â€” 1.447 = 14.953 lb. weight of lead in water. Ans.

(d) 16.4 -^ 40 = .41 â€” weight of 1 cu. in. of the lead.
16.4 â€” 2 = 14.4 lb. = weight of lead after cutting off 2 lb.
14.4 -i- .41 = 35.122 cu. in. = volume of lead after cutting
oflf 2 lb.

(477) (a) See Fig. 44. 13.5 X 9 X .7854 = 95.4261 sq.
in., area of base.

.03617 X 20 = .7234 lb. per sq. in., pressure on the base
due to the water only.

12 + .7234 = 12. 7234 lb.,
total pressure per square
inch on base.

12.7234 X 95.4261 =
1,214.144 1b. Ans.

{d) 47 X 12 = 564- lb.,
total upward pressure.
Ans.

(478) {a) See Fig. 44.
4 sin 53Â° = 3.195', nearly.

20 â€” 3.195 =16.805'=dis-
tance^ of center of gravity
of plate below the surface.

.03617 X 16.805 + 12 =
12. 60784 lb. per sq. in. = per-
pendicular pressure against the plate,
area of plate.

12.60784 X 40 = 504.314 lb. = perpendicular pressure on
plate. Ans.

(d) 504.314 sin 53Â° = 402.76 lb. = horizontal pressure on
plate. Ans.

(c) 504.314 cos 53Â° =303.5 lb. = vertical pressure on
plate. Ans.

Fig. 44.

5 X 8 = 40 sq.

in.

194 HYDROMECHANICS.

5' X 7854
(479) T-n â€” ^^^^ ^^ P^P^ ^^ square feet. Using

formula 31,

Q = Av^ = â€”J X 7.2 = discharge in cu. ft. per sec.

5' X 7854

X 7.2 X 7.48 = discharge in gal. per sec.

144
5' X .7854

X 7.2 X 7.48 X 60 X 60 X 24 = 634,478 gal. dis-

144
charged in one day. Ans.

(480) 38,000 gallons per hour ^q'^^^q gal. per sec. = Q.

Using formula 49,

_ 24.51 (2 _ 24.51 X 38,000 _
'^"*- d' â€”5.5^X60X60 -^â€¢^^^^"' P^"^^^""- ^"^â€¢

(481) Area of 2 - in. circle = 4. 9087 sq. in.; area of a

2-in. circle = 3.1416sq. in. (4.9087 - 3.1416) X 12 = 21.2052
cu. in. of brass.

21.2052 X .03617 = .767 lb. = weight of an equal volume
of water.

6 lb. 5 oz. =6.3125 lb. 6.3125 ^ .767 = 8:23 Sp. Gr. of
brass. Ans.

(482) (^) A column of water 1 ft. high and 1 in. square
weighs .434 lb. .434 X 180 = 78.12 lb. per sq. in. Ans.

(a) Projected area of 1 foot of pipe = 6 X 12 = 72 sq. in.
(See Art. 985.) 72 x 78.12 =5,624.64 lb., nearly. Ans.

(483) Use formula 42.

(a) (2a - .41 ^i/2^ = .41 X ^ X /2 X 32.16 X (^^J =
38.44 cu. ft. per sec. Ans.

(^) (2 = -^ = ^1^ = 62. 5 cu. ft. per sec. Ans.

(484) (a) Area of pipe : area of orifice :: 6" : 1.5'; or,
area of pipe is 16 times as large as area of orifice. Hence,
using formula 35,

HYDROMECHANICS. 195

^-J '^gh _J ^X 32.10 X45 ~_

z/-r ^^-r ^^^^^^^^^^^g^_5d.9tt. persec. Ans.

' A-" ^ (6^ X. 7854)*

{b) 2.304X10=33.04 ft. = height of column of water
which will give a pressure of 10 lb. per sq. in. 45 + 23.04
= 68.04 ft.

_ V2 X 32.16 X 68.04

- =66.28 ft. per sec. Ans.

(1.5* X .7854)
â– ^ (6* X .7854)'

(485) Use formula 48.

Q = .0408 Â«^Vâ€ž. = .0408 x 6' X 7.5 = 11.016 gal. per
sec. Ans.

(486) 14' X .7854 X 27 = volume of cylinder = volume
of water displaced.

14' X .7854 X 27 X .03617 = 150 lb., nearly, = weight of

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