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DUBLIK: W. B« SELLT, 8, GBATTOK-STBEET. 

London: Simpkin, Marshall and Co. Edinburgh: Oliver and Boyd. 
Glasgow : Griffin and Co. And all Booksellers. 



A KEY 



TO THE NEW AND PREVIOUS EDITIONS OF 



A MANUAL OF MECHANICS. 



BY 



THE REV. JOSEPH A. GALBRAITH, A.M., 

FBLLOW AKD TUTOB OF TBINITT COLLSQE, DUBUK, 



AND 



THE REV. SAMUEL HAUGHTON, A.M., 

FELLOW AND TUTOB OF T&INITT COLLEGE, DUBUK. 



BT 



JAMES MCDOWELL, A.B., 

TBINITY COLLEGE, DUBLIN. 



DUBLIN : 
WILLIAM B. KELLY, 8 GRAFTON STREET. 

LONDON : W. ALLAN, 13 PATERNOSTER ROW 

1860. 



/M. C, Zi. 



By same Author: 

Key to Arithmetic, 
Key to Trigonometrt, . 



08. 

5s. 




J. F. FowLKB, Printer, 3 Crow Street, Dame Street, Dublin. 



NOTICE. 



The Reader is requested to omit the following exam- 
ples in the Key (which are solutions to Problems in an 
earlier Edition), as unnecessary to be read. 

Page 4 of Key. Examples I, 2. 



„ 5 


>» 


11 


3- 


11 7 
11 8 


11 
11 


11 

19 


lO. 

13, 14, 16. 


11 9 


11 


11 


17, 19. 


11 10 
11 II 


1$ 

11 


11 
»1 


20, 21, 22. 

2 at foot of page. 


11 12 


11 


11 


3- 


11 14 
11 16 


11 
11 


11 
11 


4. 
2,-3- 


11 17 
11 18 


It 
11 


It 
11 


4,5- 
6. 


„ 20 


11 


11 


II. 



When the word " Key" does not follow the word 
" Fig.," it will be found in Platq II. at the end of the 
Key. 

The solutions of the new questions which have been 
introduced into the Second and Third Editions of the 
Manual are inserted at the end of this New Edition of 
the Key. 



INDEX. 



N. B. — The nmncfralfl in the first t^o ctolatains indicate the ptge and 
mimber of the Example in the fourth edition of the '* Manual ;" and 
those in the third column the p&^ of the *' Kef* where the solution 
of the corresponding Example may be found. The fourth oolunm 
ghres the number of the Example in th6 '* Key'* only when it difRers 
from the number of the corresponding Exampreln the ** Manual." 



Mahval. 



Page. 
6 



12 

'3 
'3 

'5 



'5 

i6 



iS 

33 
30 



Exankple. 

I ... 

J» 4, 5 • • 
o ... 

9 . . . 

10 . . . 
I, 2 . . . 
3 . . . 

4,5,6,7 . 

8, 9, 10 

I, 2 . . . 

3 . . . 

4 . . . 

S . . . 

6 . . . 

7 . . . 

8 . . . 

9, ro, II . 

12' . . . 

13 • . 

14 . . . 
15, 16 . . 
17, 18 . . 
1,2 . . . 

*, 3,5 • • 

2, 3 . . . 



Kit. 



Page. 

5 
6 

7 

7 

7 

55 

57 

67 

8 

9 

9 

10 

10 
68 

69 

70 

71 

73 

73 
II 

15 
16 



Example. 



II 



16 

17 

19 
21 

22 



w 



\ 



\ 



11 



INDEX. 





Makual. 




Est. 


Page. 


Examine. 




Page. 


Example. 




4t 5 


17 






6. . . . 








18 




3* 


I, 2, 3i 4t 5 








18 




35 


i| », 3 • . 








19 




36 


4,5»^, 7 . 








19 




* • 


8, 9» 10 








20 






II . . . 








21 


12 


40 


2 . . . . 

3, 4» 5i 7, 8 

9 . . . . 

10 ... . 








21 
22 

23 

24 


12 
• 13 


44 


«. 2, 3, 4, 5, 6 








74 T. 




50 


I, 2 . . . 










SI 


4. . . . 

5. . . . 








75 :' 

76 ;• 




5* 


6. 7 . . . 








77 -'. 




52 


8 








79 ' ' 




59 


I1 2, 3, 4, 5 . 








25 • 




60 


6 

7 








35 
26 




61 


I, 2, 3 . . . 
4t5 . . . . 








26 

27 




t' 


6, 7 ... . 








27 




63 


I, 2 ... . 
3i4,5, 6 . . 

7, 8 . . . 








37 
28 

29 




64 


9, 10 ... 
II . . . 
12, 13 . . 








29 
30 

i 31 




• 


14 ... . 








32 




68 


I, 2 ... . 
3 








32 
33 


«•' 


69 


4,5 ... . 

6,7. . . . 
8 








33 
34 
35 




72 


I 








35 




73 


2, 3 ., . . . 








35 




fc 


4,5, 6 . . . 








36 


¥ 


h 2, 3, 4, 5, 6 . 






3<5 


^ / 


'»^iJ, 4,5, 6 . . ^ 


1 ^1 \ \ 





INDEX. 






HI 




Manual. | 

1 


Kfy. 


Page. 


Example. 


Page. 


Example. 


78 


!••• . . . . 


38 




79 


2, 3. 4 . . . 






38 




81 


4»5 . • . . 






38 






6,7, 8, 9, 10, II, 


la 




39 




84 


4,5.6,7 . . 
8, 9, lo, II, 12 






40 
41 




89 


6 

7 

8 




1 


79 
80 

81 




90 


9 

10, II, 12 . . 

13 .... . 

147 I5i 16 . . . 

17 

18 






81 
82 

83 

84 

87 




91 


19, 20 ... . 
21, 22 . . . , 
23, 24 ... . 
25 ... . 






87 
88 

89 

90 




94 


I, 2 ... . 






42 




95 


4 






42 






5,^. . . . 






43 






7,8. .. . 






44 


\ 




9, 10, II . . 






45 




99 


3»4,5, 6 . . 
7 






46 
47 




100 


8, 9, lo . . 

II 

12 

13, 14 . . . 






47 
48 

57 
58 




103 


I, 2 . . . . 






|9 




107 


h ^f 3,4* S • 






60 






6 






61 




108 


I, a, 3 . • . 






61 




109 


I, 2 

3 




1 


62 
^3 




III 


I, 2 .... . 






63 




112 


3, 4 




1 


64 


^ 




5, 6, 7 . . . 


• '. W <>v \ 


/ 


8, 9, 10 .... ^ ^^ \ 


' 1 










■^ 




IV 




INDEX. 




• 


Maitual. 


KST. 


Fage^ 


Example. 




Page. 


Exampla. 


Ii6 


I, », 3 • • • 


• • 


91 




ii8 


1,2,3 . . 








92 




121 


I, a, 3 • . 

tV. : : 

8 . . . 








93 

96 

97 




127 


I, a, 3 • 








97 




131 


I ... 
a, 3 . . 

4.5 . . 








98 

99 

100 




134 


I, 3^ 3, (to 
1, a . . 

3 . . 
4,5 • • 


3t a 

• 
• 
• 


»et) 




lOI 

101 
Eoa 
103 




135 


6, 7 . . 


• 






.104 






•••... 


IQ5 

a . -~. 



A KEY TO THE MECHANICS. 



t 
EXPLANATION OP ABBREVIATIONS EMPLOYED IN THE 

FOLLOWING WOBK. 



L signifies angle. 

A „ triangle. 

11°^ „ parallelogram. 

The pages and figures (unless " Key" follow) refer to 
the Hey. Messrs. Galbraith and Haughton's work ; and a 
numeral in parentheses refers to the number of the 
Example. 



STATICS. 







Page 3. 




(I 


.) 


We have only to reduce to lbs. 








I ton, II cwt 2 qrs. 15 


lbs 






20 








31 cwt 








4 . 








126 qrs. 
28 








1013 








253 





3543 lbs. Ans. 

(2,) We have only to divide 3543 lbs. by the number 
of lbs, in I ton, viz. by 2240, thus: 

2240) 3543. (1.58169 tons. Ans, 
2240 



13030 

II2CX> 

18300 
17920 



3800 
2240 

15600 
13440 

21600 
20160 

1440 



A KSY TO THE MECHANICS. 5 

(3*) We most reduce both to the same denomination, 
viz. ounces, and then divide the former by the latter^ 
thus: 



25 lbs. 3 oz. 
16 


4 stones. 
14 


153 
^5 


56 pounds. 
16 


403 oz. 


56 




896 ounces. 


896) 403.0 (-44977 
3584 


Ana. 


4460 
3584 




8760 
8064 




6960 
6272 




6880 




6272 




608 





Page 6. 

(i.) In fig. 9, suppose OA, OB th^ V«c> ^Q^M!i^Vst5iR&^ 

making the L AOB = 120"^. ^oyr, «\xvRfc K^ ^ ^"^^ 

AO, tlie z AGO = AOG, \>\xt M^O =^^^> ''^ ^ 



A KBY TO THE MECHANICS. 



is II to OB, .-. AOG = BOG = 60**; and since the 3 1} of 
A AGO = 180^, it is equilateral, i. e. the resultant OG 
is = either component OA or OB. 



Page 7. 

(2.) Fig. I (Key). OB is the less component, OA the 
greater, and M the middle point of the resultant OG. 
Since OAG is a r* Z, the semicircle on OG will pass 
through A, and .-. AM = MG = GA, .-. L GOB, which is 
equal to AGM, = 60°, and .-. AOG = 30° 



(4.) 



(5) 



30' = 900 1 add, and then extract the 
40' = 1600/ square root of the sum. 

25'oo(5o Ans* 
25 • 



* 

CX) 



43' = 1849 1 add, and extract the square 
22* = 484 J root 



23'33(48.3oi I lbs. Ans. 
16* 



88 
8 



733 
704 



9^1 
3 



290x5 
2889 



96601 



I I 0000 
96601 



96602 I 



1339900 
966021 



A KEY TO THE MECHANICS. 



(6.) ' 56^ = 3 1 36 1 eubtra«jb and extract the square 
31^= 961 J root of the difference. 

2 1'75(46.6368 lbs. Ans, 
16 • • 



86 
6 



575 
516 



926 
6 



5900 



9323134400 
3127969 



93266 



643icx> 
559596 



932728 



8350400 
7561824 



(9.) 



cos 12° =.97815 
142 



195630 
391260 
97815 



138.89730 lbs. Ans, 

(10.) I cwt. 2 qrs. 14 lbs. = 182 lbs. 

182 X .97815 = 178.0233 lbs. 

= 1 cwt. 2 qrs. 10.0233 lbs. Ans, 

(II.) cos 17° = .9563. DiVidiii^ \^o>^^*\il '^i^^^^^^ 
obudn izg.^S^ lbs. ilns. 



8 



A KEY TO THE MECHANICS. 



(«.) COS 37** = •79863 

cos 530 = .60181 
Multiplying these cosines by 471, we obtain 376.15473 
and 283.452 J I. 

(13.) 3 cwt. 2 qrs. 17 lbs. = 409 lbs. 

cos 1 7° = .9563 I Multiplying these by 409, 
cos 73° = .29237 J we obtain respectively — 

391.1267 lbs. and 119.J7933. ^ns. 

(14.) We have only to reduce the answer of the last 
example to cwts. 



Page 8. 



(16.) 172= 289 
36*= 1296 



2 X 



'^xle Z Tzll^ } product of these = 1 134.8683; 



I 



1 206 t *^^ *°^ extract the square root 

iii4.86832^ of the sum. 



27'i9.86'83'zo(52.i52 Atu. 
25 



• • • • 



102 

2 



219 
204 



1 041 
I 



1586 
1 041 



10425 

5 



54583 
52125 



J04J02 24^820 
'20860^ 



A KEY TO THE MECHANICS. 

17.) 26« = 676 

127* = I6I29 
00576° = .24192 

2 X 26 X 127 X .24192 = 1597,63968 ) 

676. > add 

1 61 29. ) 



18402.63968 
The square root of this is 135*656 Ans. 

(19.) Fig. 5. OA and OB the components, and OC 
the resultant. 

74'= 5476 
123^= 15129 

cos 65° = .42262 
2 X 74 X 123 X .42262 = 7693.37444 

7<^93-37444 ) 
15129. >add 

5476- ) 

28298.37444 

The square root of this is 168.221 = OC. 

Now in A AOC we have 

OC : CA : : sin CAO : sin AOC 

but sin CAO -sin AOB, since they are supplemental z.'. 

.-. 168.221 : 74 : : sin 6$° : sin AOC 
or 168.221 : 74 : : .90631 : sin AOC 

multiplying the second and third terms of this propor- 
tion together, and dividing by the first, we obtain 

sin AOC = .398689 = sia x'^ y:! 
and subtracting this L from 6^^ ^vre^ \\^ "l^^ - 



10 A KEY TO THE MECHANICS. 

(20.) Fig. 5. Here cos 132^ = - cos, its supplement, 
= - cos 48° = - .66913. 

1 1 owt. 2 qrs. 13 lbs. = 1301 lbs. = OA 
5 »> 3 »> 17 n =661 „ =0B 



1301^ = 1602601 1 jj 

661'= 436921)'^'' 



2129522 
2 X 1301 X 661 X .66913 = 1 1 50851.40786 

This must be subtracted from the above sum since the 

cosine is negative. 

2129522.00000 1, 
1150851.40786/®"^^'^^ 

978670.59214 

The square root of this, viz. 989.277 = OC, the resultant. 

Now in A AOC we have 

OC : CA : : sin OAC : sin COA 
or 989.277 : 661 : : sin 48° : sin COA 
or 989.277 : 661 : : .74314 : sin COA 

Multiplying the second and third terms together, and 
dividing by the first, we obtain, 

sin COA = .49653 = sin 29*^ 46^ 

which L being taken from I32<> leaves 102° 14', the other 
required angle. 

(21.) The component forces are plainly (see fig. 5, 
where in A AOC, OA : AC or OB : : sin ACO or sin 
BOC : sin AOC) in the same ratio as sin 6*^ : sin 31*^; 
hence we have only to divide sin 6° = .1045 3 by sin 
31° = .51504 The result true to six places is .202955 

(22.) Fig. 5. Here OA = 22, OC = 56, and the L 
AOC = 150 ; we have .*. only to find OB or AC. 

Now AC* = A0« + CO* - 2 X AO X CO X cos AOC. 

22*= 4841 add 

3620 



A KEY TO THE MECHANICS. 11 

COS 150 = .96592 

2 X 22 X 56 X .96592 = 2380.02688 



3620.00000 1, 
2380.02688/^'^^*^*^^ 



The square root of this, viz. 35.213, is the required com- 
ponent. 

N.B. — This question should more properly have been 
classed under 15. 



Page 10. 

(i.) 220 

126 

346 : 220 : : 26 inches. 

This proportion gives 16.53 inches, the length of the 
segment remote from the weight 220 lbs. 

(2.) 1 7 St. 6 lbs. = 244 lbs. 

2 cwt, I qr. 22 lbs. = 274 lbs. 

518 : 244 : : 32 feet 

This proportion gives 15.073, the length of the segment 
next the weight 2 cwt i qr. 22 lbs. 



,Page 13. 



(2.) Fig. 1 6. 0A= 31 

0B= 16 
OC = a9 
cos 1^^ = -95^^ 



12 A KEY TO THE MECHANXCB. 

3I» = 96l. ) 

16^ = 256. >-add 

2 X 16 X 31 X .9563 = 948.6496 ) 



2165.6496 

The square root of this is 46.536 = OG. 

Now, OG : GB : : sin GBO : sin GOB; but sin 
GBO = sin AOB, since these 1} are supplemental; 
hence the proportion becomes \^»SZ^ 131 : : sin 17^ : 
sin GOB; or, 46.536 : 31 : : .29237 : sin GOB .•. sin 
GOB = .19476 = sin 1 1® 14'. 

1 1° 14' 
52 o 



.-. 63° 14' = GOG, 
and cos 63° 14' = .45036 
sin 63° 14' = .89285 

Again, OG^ = 2165.6496 (found before) 

00^ = 292= 841. 

2 X 29 X 46536 X .45036 = 12 1 5.561 27 

4222.21087 

The square root of this is very nearly 64.98 = OH. 

Again, from A HOC, we have 

OH : HC (= OG) : : sin HCO (= sin GOC) : sin HOC. 

That is, 64.98 : 46.536 : : sin 63° 14' : sin HOC. 

Or, 64.98 : 46.536 : : .89285 : sin HOC. 

.'. sin HOC = .63942= sin 39° 45'. 

(3.) Fig. 16. OA = 217 lbs. 

OB = 300 

OC = 167 

cos 17° = .9563 



A KEY TO THB MECHANICS. 13 

217*= 47089. ) 

300* = 90cxx>. > add 
2 X 217 X 300 X .9563 = I245ia26) 

261599.26 

The square root of this is 5 1 1.467 = OG, 

Now, OG : GB : : (sin OBG =) sin AOB : sin GOB. 
That is, 511.467 : 217 : : sin 17° : sin GOB. 
Or, 511.467 : 217 : : .29237 : sin GOB. 
.-. sin GOB = .12404 = sin 7° 8'. 



22° 8' = GOC. 

cos 22° 8' = .92631 

sin 22° 8' = .37676 
Again, 
2 X 167 X 511.467 X 



.92631 = 158241.51692118 ) 
167*= 27889. >- 

OG' = 261599.26 3 



add 



OH'* = 447729.776921 18 

.-. OH = 669. 1 26 lbs. 
Again, 

HO : OC : : (sin HCO =) sin GOC : (sin CHO =) sin HOG. 

That is, 669.126 : 167 : : sin 22° 8' : sin HOG. 

Or, 669.126 : 167 : : .37676 : sin HOG. 

.*. sin HOG = .09403 = sin 5° 24'. 

GOC = 22° 8M , ^ ^ 
HOG= 50^^,) subtract 

.-. HOC = 16° 44' 

N. B.— Since HOC = 15^, wid^O^ = \^^ "^JU^X^ 
clear that BO ought to be on t\i^ «vd^ o^ ^Vl -aessx V3 



14 A KBT TO THB MHC»AiriC9. 

(4.) Fig. 16. OA 7 22 lbs. 

OBo 16 

AOB = 370 

BOG = 690 
cos 37'' = .79863 5 5 
sin 37° = .6018150 

2 X 22 X 16 X .7986355 = 562.239392 1 

22* =484. [add 

i6» = 256. ) 

OGi^r^ 1302.239392 

.-. OG = 36.0865 

Again, 

OG : GB : : (sia GBO =) sin AOB : sin BOG. 

That is, 36.0865 : 22 : : .601815 • sin BOG. 

.•. sin BOG = .3668942 = sin 21° 31' 27". 



BOG=2i°3i'27'M ,, 
BOC = 69° P^^ 



.-. GOC = 90° 31' 27'', and .•.HGO=89° 28'. 

Now, sin 90° 31' 27''= sin 89° 28' 33'' (its supplement) 

= -9999581 
cos 90° 31' 27" = - COS 89° 28' 33" = -.0091483 

Again, 0G*= 1302.239392 > ,, 

100'= lOOOO. J 

I 1302.239392 \ Uf ^ 

2 X 100 X 36.0865 X .0091483 = 66.02602559 /snDtract 

OH* = 1 1236.2 1 236641 
.'. OH = 106.001 

^a/n, OH : EG : : sin HGO : bVhUOGc. 

Thatisf, 106,001 : 100 : : .99995^1 ' sm'ftOCj. 



A KEY TO THE MECHANICS. 15 

.'. sin HOG = .9433478 = sin yo*' 37' 18''- 

Now, GOC = 90° 3 1' 27'M ^ 

HOG = 70° 3 7' 1 8'' / 8^^^ra<^* 

.-. HOC = 19° 54' 9'' 

N. B. — I may here remark, once/or aZZ, that I have 
used Chambers' Tables in the work of the above ques- 
tion, and occasionally elsewhere, when greater accuracy 
was aimed at than could well be secured by the too 
limited Tables at the end of the Manual. 



Page 15. 

(2.) Fig. 18. Since CD : AB : : PO : OQ 

.-. CD + AB : AB : : PO + OQ : OQ 
That is, 6^ lbs. : 23 lbs, : : 14 in. : OQ. 
/. OQ = 4.9538 in. Ana. 

(3.) Fig. 18. 

3 cwt. 2 qrs. 15 lbs. = 407 pounds. 
I „ 3 » 25 „ =221 

628 : 221 : : 3ft 7in. 
12 

43 
.". OP = 15.132 in. = 1.261 fip^. 



Page 16. 

(5.) ** It is not difficult to prove that the centre of 
gravity is the centre of the civile \Tis^TiiJck<iWsi'^^\xv- 
angle OPQJ' (Fig. 2, Key.) 

Since AQ = QB and BO =^ OC., ^so^ 



16 k KEY TO THE MECHANICS. 

.-. AQ : QB : : CO : OB, and .•. (Euclid, B. vL p. 2.) 
QO is parallel to AC, L e. the line joining middle points 
of two sides of a A is parallel to the third side, .-. AO, 
BP, CQ are ||»', and .-. QO = AP = PC = i AC, &c. 
Now, cut QO in X so that QX : XO : : BC : BA, i. e. 

. QX : XO : : PQ (= i BC) : PO (« i BA) 

.*. PX bisects the z. OPQ (Euclid, B, vi. p. 3), but centre 
of gravity lies in PX. Similarly it can be shown to lie in 
OZ, which bisects the z. POQ; .*. G is the centre of gra- 
vity, but the bisectors of the z.' of a A meet in centre 
of inscribed circle. (Euclid, B. iv. p. 4). 



Page 20. 

(2.) 13 in. : 2 in. : : 2cwt.. 

112 

224 

2 

13)448(34.4615 lbs. Ans. 

(3.) cos 126° = - cos 54° (its supplement) 

= - -58778 

217^= 470891 ,, 

725* = s^S^^S J 

2 X 725 X 217 X .58778 = 184944.977 /S^'^*'*^*^ 

387769.023 

2^e square root of this is 622.711 Iba., 1^^ t«o^\xA 
sijv/n on the fulcrum. 



A KEY TO THE MECHANICS. 17 



Page 21. 

(4,) Fig. 3 (Key). 

Let OA (= 310 lbs.) represent the power, OB (= 
422 lbs.) the resistance, then, the parallelogram AOBC 
being completed, its diagonal OC will represent the up- 
ward strain on the fulcrum. 

Now, cos AOB = - cos POR = - cos 37°, since AOB 
+ POR= i8o^ 

cos 37** = .79863 
422*= 1 78084 i^^ 
310*= 96100/ 

2 74 1 84 1 subtract, since cos 

2x 310X 422 X. 79863= 208953.5532 } AOlMs negative, 

OC^ = 65230.4468 
.'. OC = 255.402 lbs. Ans. 

(5.) Fig. 4 (Key). 

Let OA (=217 lbs.) represent the power, OB (= 
100 lbs.) the resistance, then the diagonal OC will re- 
present the downward strain on the fulcrum. 

Now, cos AOB =- cos POR = - cos 42°= - .74314 

217'= 47089 i^d 

100*= 10000/ 

^ ^ subtract 

2 X 100 X 217 X .74314 = 32252.276 / 



OC* = 248^6.qiAt 
.'. OC = 157.596 \\is, Ans. 

B 



18 A KEY to THE MECHANICS. 

(6.) COS 79® = .1908 1 

2 X 17 X 32 X .19081 = 207.60128 ) 

32^* = 1024. >add 

172= 289. ) 

1 520.601 28 

The square root of this is 38.9948 lbs., the strain on the 
fulcrum. 



Page 23. 

(i.) 2ft. Sin. : 6in. : : 1 156 lbs. 

This statement gives for the fourth proportional 
216.75 lbs. Arts. 

(2.) 79 lbs. : 17 lbs. : : 17 in. 

Hence the fourth term is 3.65822 in. An8. 



Page 24. 

< 3,) 29 lbs. X 16 = 464 lbs. 

I ft. 2 in. : 8 f t : : 464 lbs. 

The fourth term is 3 1 8 1 f lbs. = i ton 8 cwt, i qr. 1 7^ lbs. 
Ans. 

(4,) I ft. 2 in. = 14 in. radius of axle, .*. 16 = radius of 
axle ajter the rope is wound round it. Now, the re- 
sistances capable of being overcome, the length of the 
lever being the same, are inversely as the radii of the 
axles, or as 16 : 14, i. e. if 16 represent the power of 
the capstan originally, 14 will now represent its dimi- 
nished power, and the difference 2 is ^ of 16 and 100 -h 

8=I2|. 

(5.) I ton = 2240 lbs. 

1 ft. 7 Id. : 2.5 in. : : 2240 lbs. 

7^e fourth term is 294.7368 lbs. Ans^ 



A KEY TO THE MECHANICS. 19 

Page 26. 
(i.) Denoting the inclination by t^ we have 
sin t = g*y = .0161290 = sin 55' 26".6 

(2.) sin 4** = .06975, 1 = lenffth, 

h - height. 
.'. Z : A : : I : .06975 

or Z : A : : .jss : i 



Hence, t^^s = < 4*3 3^9 '^ the required number. 



Page 27. 
(3.) 6 miles = 31 680 feet 

.'. sin » = iii^ = .0161300 = sin 55' zf^ig^g. 



Also, siilo = «TMi which shows that the fall is i in 62 
nearly. 

(4.) A force equal to -ir P^i't of the weight overcomes 
friction, and to ^, the resistance arising from the slope 
of the hill, abstracting from friction, .*. to overcome both 

together the force must be « + 5 = ^i = THn -^w*. 

(6.) 14 ft. : 3 ft. : : 964 lbs. 

The fourth term is 2o64 lbs. Ans. 

(7.) By the rule given at foot of page 20 of Manual, 
we haveP : W : : sin 6° : sin 101°, or, 

sin loi® : sin 6° : : W : P. 

Now, sin 6® = .10453, and am \o\^ =.«vsk'\<|' V!>5y^w>^- 
plement)f -.98162. 

B 2 



20 A KST TO THE MECHAITICS. 

2 cwt. I qr. 17 lbs. = 269 lbs. 
.*. .98162 : .10453 : : 269 lbs. : P, 
,\ P = 28.645 lbs. AnS' 

(8.) Sin 120® = sin 60° = .86602 sin 2® = .03490 

Now, sin 120° : sin 2° :: 31 ton : P, 
That is, .86602 : .03490 1:31 ton : P, 
.-. P = 1.249278 ton = I ton 4 cwt. 3 qrs. 26.38272 lbs. 

{9.) Sin 167° = sin 13° =.22495, 

.*. .22495 : ^ : : 22 tons : P. 

.22495 X 26 s= 5.8487; dividing 22 by this we obtain, 

P = 3.7615 tons. Ans, 

(10.) Fig. 34. In A OR'V, the Z OV'R' is right, 
OR' represents the weight of the waggon (= 2 ton 
. 14 cwt. = 54 cwt), and OV the pressure upon the road, 
also R' OV = t, whose sine = ^V = .0370370 .*. i = 2° 7' 
21", and cos i = .9993214 Hence, OV = OR' cos i = 
54 cwt. X .9993214 = 53*9633556 cwt = 2 ton 1 3 cwt. 
3 qrs. 23.89 lbs. 

(11.) Fig. 36. In Ex. 7, we have, 

OV'P'= 6°\ ,, 
FOV'=ioi°/^^^ 

107^ 

Taking this from 180° we get OP'V = 73°; also, P'V 
= 2 cwt. I qr. 17 lbs. = 269 lbs. Hence, from A OP'V 
we obtain, 

Sin P'OV : sin OP'V' : : P'V' : OV, 

That is, (sin 101° =) sin 79® : sin 73° : 269 lbs. : OV, 

Or, .98162 : '.95630 : : 269 : OV', 

/. OV = 262.06 lbs. = 2 cwt. I qr. 10.06 lbs. 

In Ex. 8. OV'?' = 2° \ , , 

P^0V'=I20°/^^^ 



122° 



A KEY TO THE MECHAKICS. 21 

Taking this from i8o°, there remains the L OP'V = 
58°;alsoP'V' = 3i tons. 

.-. sin FOV : sin OP'V : : P'V : OV, 

That is (sin 120° =) sin 60® : sin 58° : : 31 tons : OV, 

Or, .866c2 : .84.805 : : 31 : OV, 

.«. OV = 30.3567 tons. 

In Ex. 9. Sin O V'P' = ^V == -^S 846 1 5 = sin 2° 1 2' 1 5''. 



P'OV'=i67° ^ 1 ,, 

OV'P'= 2° 12' 15''/*^^ 



169° 12' 15'' 

The difference between this and 180**, viz. 10° 47' 45" 
= OP'V; also P'V = 22 tons. 

.-. sin FOV : sin OP'V : : P'V : OV, 

That is, 

(sin 167° =) sin 13° : sin lo** 47' 45" : : 22 tons : OV, 

Or, .2249511 : .1^73099 : : 22 : OV, 

.-. OV = 18.318727 tons 

= 18 tons 6cwt I qr. 13.94848 lbs. 

The result in the Manual is incorrect. 

(12.) I ton = 2240 lbs. 

To equilibrate a weight of 2240 lbs. on a road of gra- 
dient, I in 21 will require, abstracting from friction, a 
force of TT o^ 2240 lbs. = 106.666 lbs. If,- now, we add 
to this the forces required to overcome friction, we 
obtain, 1 16.666 lbs. \ ^ 

342.666 lbs. J '^^• 



Page 30. 

(2.) xVii^* : 24in. : : 13 lbs. : R, where R = resistance. 
.'. R = 5432 lbs. = I ton 10 v«\.. T. Qjt%, \.^'^'^. 



22 A KEY TO THE MECHANICS. 

(3.) 3. 141 59 X 24 m. = 75.39816 in. 

Hence, ^in. : 75.39816 in. : : 17 lbs. : R, 

.*. R s= 1 1 535.91 848 = 5 tons 3 cwt., very nearly. 

(4.) By the last Ex. the circumference of the circle 
described by power = 2 x 75.39816 in. = 150.79632 in. 

Hence, ^ in. : 150.79632 in. : : 32 cwt. : R, 
.*, R = 12867.95264 cwt. = 643 tons 8 cwt. nearly. 

(5.) I ft. 5 in. = 17 in., 2 cwt. i qr. 17 lbs. = 296 lbs. 
2 X 17 in. X 3.14159= 106.81406, 

the circumference of the circle described by the handle. 
i|.iii. ^ 19 = ^o^ the interval between the threads. 

Hence, V^ in. : 106.81406 in. : : 296 lbs. : R. 
.'. R = 335754.86809 lbs. = 149.890566 ton?. 



(7.) 4 in. X 3. 14 159 = 1.178 1 nearly. 

1.17812=: 1.38791961 \ .. 
WtY-At- .00826446/^^^ 



1. 396 1 8407 

The square root of this, viz. 1. 18 16 in. is length of one 
ivind of the thread, .*. 14X 1.1816= 16.5424. Ans. 

N. B. — The result in the Manual is incorrect. 

(8.) T in. X 3. 141 59 = 5.89048 in. 

5.89048* = 34-6977546304 \ add 
(i)^ = ^= .140625 J*^^ 

34-8383796304 



A KEY TO THE MECHANICS. 23 

The square root of this, viz. 5.9024 in. is required length 
of thread. 

{9.) Diameter of circle described by power = 6 in., 
diameter of cylinder f in., and the ratio of these numbers 
= 6 -7- "I = 16. Also length of thread = 1.1816 in., and 
the ratio of this to -^ is 1.1816 -f- -^^ = 12.9976^ .*. 
pressure on thread =14 lbs. x 16 x 12.9976 

= 291 1.4624 lbs. = I ton 5 cwt. 3 qrs. 27.4624 

N. B. — In the Manual, Ex. 7 has entailed its error 
upon this Example. 

(10.) Diameter of circle described by power = 28 in., 
and ratio of this to diameter of cylinder - in., is 

8 14 ' 

also, length of one wind of thread = 5.9024, the ratio 
of which to I is 5.9024 .1. f = 15.7397, .'. pressure 
on thread = 37 lbs. x ??i x 15.7397 = 8696.7089 = 3 tons 

17 cwt. 2 qrs. 16.7089 lbs. Ana, 

(11.) In Ex. 9, distance between threads = -j-'y in., 
and radius of circle described by power = 6inx3.i4i59 
= 1 8.84954 in. .*. -^in. : 18.84954 in. : : 14 lbs. : R, 

.'. R = 2902.82916 lbs. = I ton 5 cwt. 3 qrs. 18.82916 lbs. 

In Ex. 10, distance between threads = f , and r^ius 
of circle described by power 

= 28 in. X 3.14159 = 87.96452 in. 
.*. f : 87.96452 : : 37 lbs. : R, 
.-. R = 8679. 16597 lbs. = 3 tons I7'cwt. i qr. 27.16597 lbs. 

(12.) Fig. 37. OP' = 5 tons, zi A = 2° = P'R'O, and 
we have to find OR'. 

Now, sin P'R'O : sin R'P'O : : OP' : OR', 
That is, sin 2° : cos 2° : : 5 tons : OR', 
Or, I : cot 2° : : ^ * OW^ 



24 A KEY TO THE MECHANICS. 

Or, I r 28.63625 : : 5 : OR', 
.*. OR' = 143.18125 tons = 143 tons 3 cwt 2 qrs. i4lbs. 

N. B. — The result in the Manual is incorrect 

(13.) Fig. 37. Here we are to find OV = P'R'. 

Ntow, sin P'R'O : sin P'OR' : : OP' : P'R', 

That is, .0348*995 : 1 : : 5 ton : P'R' 

Since sin P'R'O = sin 2° = .0348995, and 

sin P'OR' = sin 90° = i ; 

.-. P'R' = 143 ton 5 cwt. I qr. 1 1.8 lbs. Ans. 



Page 34. 

(i.) 2* = 16 .'. 16 : I : : 17 ton 12 cwt. : P, 
.'. P = I ton 2 cwt. Ans, 

(2.) 3*= 81 .". 81 : I : : 17 ton I2cwt. : P. 
.*. P = 4 cwt. I qr. \o\\ lbs. Ans. 

(3.) 2^^ = 2048 .*. I : 2048 : : 13 lbs. : W. 
.-. W = 26624 lbs. = 11 tons 17 cwt. 2 qrs. 24 lbs. Ans. 

(4.) f - 2187 .•. 1 : 2187 : : 17 lbs. : W, 
.•. W = 37179 lbs. = 16 ton 1 1 cwt. 3 qrs. 23 lbs. Atis. 

{5.) Each of the two compound blocks in Smeaton's 
Pulley has 10 single blocks; hence, its efficiency is re- 
presented by 20 ; the efficiency of a Barton of the second 
kind, with 5 moveable pullies, is represented by 3* = 
243, and 243 : 20 : : 12.15 • ' -^^^ 

(6.) 2io= 1024 

310 = J9049 

Mnd, S904g ^ J024 = 57.665039062 j Au8. 



A KST TO THE MECHANICS. 25 



DYNAMICJS. 



Page 36. 

(i.) 754 yards = 2262 feet. 

I hour = 3600 secoDds. 
2262 feet -T- 3600 = .628333 feet. Ans, 

(2.) 40 miles = 211 200 feet. 

I hour = 3600 seconds. 
211 2cx> feet -^ 3600 = 58.666 feet Ans, 

(3.) 7 hours 31 minutes = 27060 seconds. 
200 miles = 1056000 feet. 
1056000 -T- 27060 = 39.0243902439. Ans, 

(4.) 23 hours 54 minutes = 86164 seconds. 

7925 miles = 41844000 feet. 

.\ circumference of earth = 41844000 feet x 3. 141 59, 

= 13 1456691.96 and 131456691.96 -^ 86160 

= 1525.72 feet Ans, 

(5.) 64 miles = 3379^0 feet. 

337920 -T- 1090 = 3iOx%^ sees. = 5 m. .10^^^ sees. Ans, 

(6.) 30^ miles : 261 miles : : i hour. The fourth 
term of this proportion is 8 hours 37iVt minutes. 

Again, 3^7mfles|^^^^^^^ 



1 6 minutes : 86 : : 1 \iox3LT. 
The /burth term is j hour, aa^ ^"a-**. 



26 A KET TO THE MECHANICS. 

8 hours 37tVt minutes 1 
5 „ 22^ „ 5- add 

2 „ 27 „ ) 



16 hours 277^7 minutes. Ans, 



Page 37. 

(7.) I hour : 4 days : : 12 knots. The fourth term 
is 1 152 knots, and i hour : 3 days 7 hours : : 13 knots. 
The fourth term is 1027 knots. 

1027 knots J ^^ 

2179 

Now, since i mile = 5280 feet, we have 5280 feet : 6076 
feet : : 2179 knots. The fourth term is 2507 WW miles. 
Ana, 



Page 38. 

(1.) Cubic yard =27 cubic feet, 8.9 x 27 = 240.3. 
.'. 1 1.35 -1- 240.3 = .0472326 Ans, 

(2.) 19.35 ^ 4'^7 X .64 X .31 = 16.0087968 
10.51 X 13.22 X 1. 14 X .65 = 102.9561702 
.-. 16J0087968 -T- 102.9561702 = .155491 Ans. 

(3.) By last example, mass = 102.9561702. 

5 miles = 26400 feet, and i hour = 3600 seconds. 
.'. 26400 -i- 3600 = 7-J. feet, the velocity per second. 

Hence, quantity of motion, 

= 102.9561702 X 7i = 755.0119148 Ana. 



A KEY TO TfiE MECHANICS. 27 

(4.) 1407 yards = 4221 feet. . 

4221 -r- 60 = 70.35 feet, the velocity per second. 

Also, I cubic foot = 1728 cubic inches, .'.quantity of 
motion, 

= 8.9 X 1728 X 70.35 = 1081926.72 Ans, 

(5.) 2 yards = 6 feet 

.% momentum = 6x 2 x 1.16 x 2.716 x 13 
= 491.48736 Ans. 

(6.) Here mass = i x 11.35 = '^-SS* *^^ ^7 l*st Ex. 
momentum = 491*48 

.-. velocity 491.48 -f- 1 1«35 = 43«302 Ans, 

(7.) Mass of the bullet of lead = 1 1.35 x ,267 = 3*03045 
14 feet -7- 60 seconds = yV ^^- velocity per second. 

Momentum of the ball of copper 

= 8.9x 13.47^3^ = 27.9727 

.*. velocity of bullet of lead should be 

= 27.9727 -i- 3.03045 = 9.2305 Ans. 



Page ^i, 

(i.) 17.142=293.7796) ,, 

13.11*= 171.8721/*^^ 

R2 = 465.6517 
.*. R = 21.579 feet, very nearly. 

(2.) Fig. 7. OB =13.11 

OA = 17*14 
OC = 21.579 



\ 



28 A KBY TO THE MECHANICS. 

* 

,-^. OA 17.14 o , , 

.-.COB = 90° -37° 25' = 52° 35'. 

(3.) Let (Fig. 7) OB be the direction of north, and 

A of east 

4 knots = 4 X 6076 feet = 24304 feet. 

1 hour = 3600 seconds .*. velocity of ship per second 

= 24304 -r- 3600 = 6.751 1 feet = OB. 

6.751 1»= 45-57735121 1 add 
OA2=io«=ioo. /^^^ 



00^= 145.57735121 
.*. OC = 12.0655 ^^^^' ^^^' 

(4.) Fig. 7. CosCOB=^=-^' 
^ ° ' OC 12.0655 

= •5595375 = cos 55° 59'. An^, 

(5.) 36.14'* = 1306.0996 \ ^, , 

14,3 1 V= 204.7761/'^^*^*^* 

1101.3235 
The square root of this = 33.1851 feet. Ans. 

(6.) cos 22° = .92718. 

2 X 14.37 ^ i9'22 X .92718= 512.158284504 ^ 

14.37* = 206.4969 >add 
19.22'* = 369:4084 J 

R' = 1088.063584504 
-•. B = 52,9858 feet. Atw. 



A KEY TO THE MECHANICS. 29 

Page 42. 

(7.) Fig. 5. OA = 19.22 

OB = 14.37 
OC = 33 nearly. 
AOB = 22° and sin 22° = .37460 

Now, OC : OA : : (sin OAC =) sin AOB : (sin AGO 
=) sin BOG. 

That is, 33 : 19.22 : : .3746 : sin BOG. 

.-. sin BOG = .2181761 = sin 12° 36' one angle, 

and 22° - 12° 36' = 9° 24', the other angle. 

(8.) cos 41°= .75471 

14.62 yards = 43.86 feet. 
43.86 -^ 60 seconds = .731 ft. per second. 

2 X 13.61 X .731 X .75471 = 15.0170837322 1 

1361^=185.2321 >add 

•731'= -534361 ) 



R* = 200.7835447322 
.'. R = 14.17 feet, very nearly. Ans. 

(9.) Fig. 5. 0A= 13.61 . 

0B= .731 (See last Ex.) 
0G= 14.17 
AOB = 41° 

.'. sin AOB = sin 41° = .65606. 

Now, OG : (G A =) OB : : (sin AG =) sin AOB : sin AOC. 
That is, 14.17 : .731 : : .65606 : sin AOG. 

.*. sin AOG = .0338447 = sin 1° 56' 22'', 
and .-. BOG == 41° - 1° 56' 22" = 39° 3' 38''. 

(10.) Fig. 5. (Key.) 

Let O be the point from viVi\s^ A5JcL^\ic>'^\.%\a3t\R»^^^ 
~ 2.5 in direction of stream, O^ = 'i, «cA l^^^^"^"^ 



30 A KEY TO THE MECHANICS. 

OD being JL to the bank of the river. Complete the 
parallelogram AOBC, and OC will represent the direc- 
tion of the boat, and we are to find DF. 

Now, cos AOB = - cos BOG = - cos 87° ==-- .052336. 



30.29 
2 X 5 X 2.3 X .052336 = 1.203728 

Subtracting this from the last sum, we have, 

OC* :^ 29.086272 
/. OC = 5.393 

Again, 

OC : (AC =) OB : : (sin CAO =) sin BOG : sin AOC. 

That is, 5.393 : 5 : : sin 87° : sin AOC. 

Or, 5.393 ' S ' ' .9986295 : sin AOC. 

.-. sin AOC = .9258571 = sin 67^48', and 

. •. DOF = 90^ - 67° 48' = 22° 1 2'. 

.*. tan DOF = .4080924. 

Again, DO = i|- mile = 2200 yards. 
.-. DF = DO X tan DOF, 
= 2200 X .4080924 = 897.80328 yards. Ans. 

(II.) Fig. 6. (Key.) 

Here, OA = 2.3 

OB = 5. 

L AOB = Sf 

and we are to find DF. 

^onr, since OC' = OA^ + OB* + 2 OA x OB cos 
AOB, we have by last Bx.^ 



A KEY TO THE MECHANICS. 3t 



0A2= 5.29 
OB* = 25. Y add 

2 OA X OB cos AOB = 1.203728 



I 



0C«= 31.493728 

.*. OC = 5.612, very nearly. 

Again, 

OC : (AC =) OB : : (sin CAO =) sin AOB : sin AOC. 

That is, 5.612 : 5 : : sin 87° : sin AOC. 

Or, 5.612 15:: .9986295 : sin AOC. 

.-. sin AOC = .8897268 = sin 62"* 50^ 

.-. DOF = 27° lo', and tan 27° 10' = .5131949 

Also, DO = 2200 yards (by last Ex.) 
.*. DF = 2200 X .5131949 = 1 129.02878 yards. Ans. 

(12.) Fig. 7 (Key). 

N. £. S. W. denote the 4 cardinal points respectiyely, 
and .-. L AOE = 22° 30^ and BOB = 1 1° 15'. 

O A = 5 knots, OB = 3. 

And completing the H"* OACB, OC will denote the ship's 
real motion in direction and magnitude. 

Now, L AOB = 22° 30' + 1 1° 15' = 33^45', 
and sin 33° 45' = .5555702 
cos 33° 45' = .83 14696 

2 X 5 X 3 X .8314696 = 24.944088 1 

5^ = 25. [ add 

»2 _ 



3 - 9- 



i 



OC* = 58.944088 
.*. 0C= 7.677 knots. Ans, 

(13.) See fig. in last Ex. 
OC :0A : :(8iuOAC=)Axik!:f^\^\v.^^^-^'^^^'^^^ 



;» 



32 A KEY TO THE MECHANICS. 

That is, 7.677 : S '' •555570^ • sin COB. 
^v •*• sin COB = .3618406 = sin 21° 12', 

From which, subtracting BOE= 11° 15', we get COE = 
9** 57', and therefore NOC = 90^ - 9° 57' = 80° 3'. Atis, 

(14.) Fig. 8 (Key). 

Let OA in direction of stream = 2.3, and OB in di- 
rection in which boat should be rowed = 5, then com- 
pleting the II™ OACB, its diagonal OC should be JL to 
OA, we are therefore to find z BOG = CAO, 

Now, since Z. COA is right, we have 

cos CAO = -7-^ = — ^ = .46 
AC 5 ^ 

= cos 62° 36' 47". Ans, 



Page 44* 

(i.) A force = weight of one grain would produce a 
velocity of 32.1948 feet per second in the quantity of 
matter contained in i grain of water, therefore it 
would produce a velocity of i foot per second in the 
quantity of matter contained in 32. 1948 grs. The unit 
of volume will therefore be the fraction of a cubic inch 
contained in 32.1948 grs. of water, which is found 
thus — 

252.458 grs. : 32.1948 grs. : : i inch. 
The fourth term is .1275 in. Ans. 

(2.) By reasoning the same as in the last it will ap- 
pear that the unit of mass will be 32.1948 tons, reduced 
to cubic yards, which will be done thus—- 



A KEY TO THE MECHANICS. 33 

252.458 grs. : 32.1948 tons : : i inch. 

2240 



Product = 72116.352 lbs. 

7000 



Product 5048 1 ^6^,000 grs. 

The last product, divided by 252.458, gives for quotient 
1999597.810 in., 1999597.810 in. -r- 1728 = 1157.174 
feet, 1 157.174 -^ 27 = 42.858 yards. Ans, 

(3.) Here we have unit of 

weight of cubic foot of water 
32.1948 ' ' 

But weight of a cubic foot of water 

= 252.458 grs. X 1728 = 436247.424 grs. 

.-. unit of F = "^^^^"^l^y = 13550.244 grs. 

And dividing this by 7000 we obtain 1.935749 lbs. 

Ans. 

(4.) A cubic inch of water = 252.458 grs., .*. a cubic 
inch of lead = 252.458 x 11.35 " 2865.3983 grs. 

.-. unit of F = ? — ^^^J = 89.001 8078 ffrs. Ans. 

32.1948 "^ ^' ^ 

(5.) The weight of the body would produce in i second 
a velocity of 32.1948 feet, 

.-. .317 ft. : 32.1948 ft. : : 3 lbs. : weight of body (F). 

.-. F = 304.68265 

Which is reduced to cubic feet of water thus — 

252.458 grs. : 304.68265 lbs. : : i inch. 

7000 



Product = 2132778.55 

C 



«. 

34 A KEY TO THE MECHANICS. 

Dividing this by 252.458, we find 8448.052 inches, and 
8448.052 in. -7- 1728 = 4.8889 feet. Ans. 

(6.) 252.458 grs. = one cubic inch of water, .*. 252.458 
X 1728 = 436247.424 grs. = one cubic foot of water. 

13.16 lbs. = 92120 grs. 

Now, 436247.424 grs. would produce in a cubic foot of 
water a velocity of 32.1948 feet in one second, heqce, we 
find the velocity which 13. 16 lbs. would generate in one 
foot of water in one second, thus — 

436247.424 grs. : 92120 grs. : : 32.1948 feet. 

The fourth term is 6.798 feet, and as the velooities ge- 
nerated in the same volume by the sam^ force in one second, 
must be irwersdy as the specific gravities, we have (the 
sp.gr. gravity of water being unity), 

4,16 ft. : 6.798 ft. : : I : sp. gr. of the matter, 
.*. sp. gr. of the matter = 1.6341 • Ans. 



Page 45. 

(7.) A cubic inch of water = 252.458 grs., .*. cubic 
inch of gold = 252.458 x 19.35 =4885.0623 grs. 

Now, 4885.0623 grs. would generate in a cubic inch 
of gold a velocity of 32.1948 feet in one second; and 
since the velocity varies as the force, we have 

j2.ip^8 ft. : 42.^1 ft. ; ; 4885.0623 grs. : required force, 

. : required force = 6419.8S7 gxs. Aus. 



A SET TO THE MECHAXIC8. 35 

(8.) ; 7 lbs. = 11 9000 grg. 

One cubic foot of water = 436247.424 grs. (See Ex. 
6, p. 44.) Hence, we find the velocity (V) which 1 7 lbs. 
would generate in a cubic foot of water in one second, 
thus — 

436247.424 : 1 19000 : : 32.1948 : V, .•. V= 8.782 ft. 

Hence, 14 ft. : 8.782 ft. : : i sp. gr. of water : sp.gr. 
of first matter, /. sp. gr. of first matter = .627. 

Again, one cubic inch of water = 252.458 grs., and 

7000 
I ounce = -^-^= 437.5 grs. Hence, we find the velocity 

(v) which I ounce would generate in a cubic inch of 
water, thus — 

252.458 grs. : 437.5 grs. : : 32.1948 feet: v, 
.'. V = 55.792 feet. 

Hence, 11 : 55.792 : : i : sp. gr. of second matter. 
.*. sp. gr. of second matter = 5.972. 



Page 47. 
(i.) 3 minutes =180 seconds. 

(2.) 4 miles = 2 1 120 ft., and i hour = 3600 seconds. 

.'. f= 7 — = .4? I ft. per second. 

•^ 13 X 3600 ^^ '^ 

(3.) 3 minutes = \%o ^%ccrcA^. 

... r =yit = 7 X ito = laSoit. '^«t ^^was^V 



36 A KBY TO THE MECHANICS. 

(4.) v=^ = 3*-i95 X 5 = 160.975 ^*- 



100 



(5O < = :? = > = 8,96 seconds. 

^"^ J II. 16 ^ 

(6.) « = •%= ^^ = 1 1.647 seconds. 

/ 3«.i9S ' 



Page 49. 

(I.) ^^i/i*. 

.*. = -r = — = I.I 1 1 feet 
f 9 

(2.) 300 yards = 900 feet 

- 28 1800 

/=-2 = =18 feet 

r 100 

(3.) 3 minutes = 180 seconds. 

II X 180' II X 32400 
s^^jP=—- 

= 178200 feet 
(4,) 8 = iff = ^^^^^^^^^ySS.6ssfeet 

( J.) 137 yards = 41 1 feet 

^ 28 822 ^ 

«« = --= =67.543138. 

/ 12.17 

,*. < = 8.218 seconds. 

.^x >s« 32.18 X 121 ^ ^ ^ 

(6.) 5 = iyi5» = -2 = 1946.89 feet 



A KEY TO THE MECHAKIG8. 37 

Page 50. 

I.) 100 yards = 3cx> feet. 

v^ = 2/3. 

^ V* 35* I22C ^^_ 

.•./=— = ^ = — -? = 2.04166 feet. 

•' 25 600 600 

(2.) 1 142 yards = 3426 feet. 

.*. ^^ — = S7»i ft» per second «= r. 

I mile = 5280 ft. = 8. 

^ v^ 3260.41 ^ ^ 

.-./= — = "^ — 2^ = -30875 feet. 
•^ 28 10560 ^ '^ 

(3.) t^ = 2/8=2 X 1^ X 1 100 = 28600 

.*. V = 169.115 feet. 

(4.) v* = 2^ = 2 X 32.19 X 467 

= 30065.46 
.-.©=: 173.3939 feet 

(5-) "-i^-6^8-6^8-^7-37feet. 

(6.) 3 14 yards = 942 ft, and ^^ = 1 5.7 ft per second. 



38 ▲ KEY TO THE MECHANICS 



Page 52. 

(i.) In this and the three following examples ^ = 32 
feet 

The ninth term of the series, i, 3, 5, 7, 9, 11, 13, 15, 
17, &c., is 17 (found either by actually continuing the 
series to the ninth term, or by the common rule for 
finding any term of an arithmetical progression when the 
first term and the common difference are given; we 
might also have found the term thus-^ 

9»-8'*=8i -64= 17) 
.*. 17 X 16 = 272 feet Ans, 

(2.) 8 = igt^ = 16 X 81 = 1296 feet. Ans. 

is-) 7' -4' =49 -16=33 

.*. required space =33x 16=528 feet Ans» 

(4.) 11*- 3* =s: 121 -9= 112 

/. required space =ii2x 16= 1792 feet Ans, 



Page 54. 

(4.) 5^ = 32.1412 

.'. ^8 = 16.0706 

^ = 3.151 
,\f= 9.928801 

/. 8 = igf- 16.0706 X 9.928801 
= 159.5617893506 feet ATis, 

(5.) y = 32. 1 6 feet 

5= 316 feet 
and, 8 = ^flff . 



A KEY TO THB MEGHAKICa. 39 

^ 28 632 , 

g 32.10 
.\ t = 4.43 seconds. Ans. 

(6.) ^ = 32.182 

v = ^ = 32.182 X 9 = 289.638 feet. Ans, 

(7.) g ^ 52.0927 

t/*sa 2^= 2 X 32.0927 X 216. 171 
= I3875.O22IO34 

.». V = 117.7922 feet. Ans, 

(8.) ^.L* = 43^:341. X3.4716 

^ Sf 32.0927 

.•. t = 3.67 seconds. Ans, 

(9-) 5^ = 32-1403 

t; = 7i.3 

V* 5083.69 - . ^ 

2^ 64.2806 '^ -^ 

(10.) v = ^. 

.•. f = - «= . ' '^ ' =2.218 seconds. -4w5. 
9 32.1403 

(11.) ^ = 32.1691;!;= 131. 17 

V 131. 17 J 

i = - = '^ ^ ^ = 4.077 C seconds. 

^ 32.1691 
(12.) ^= 32.0917; 5= 200 

25 . 400 J 

.-. <2 = — = -i = 12.4^0^1. 

flf 32.0911 



40 A KEY TO THE MECHANICS. 



Page 57. 
(4-) 5^=32.1691 



sm % = 



ZT 



^ = 427 

. . 32.1691 
and/s;^^ sm 1 = r- = •4949' 

.% v" = 2/» = 2 X 427 X .49491 

= 422.65314 

.'. t> = 20.558 feet Ans, 

(5.) 8=ifP 

• 25 854 ^, 

.-. <*=-?= ^-^ = 1725.5662 

/ .49491 ^ ^ ^ 

.*. ^ = 41.54 seconds. Ans, 

(6.) «=i^=ix. 49491x9 

= 2.22709 feet. Ans, 

(7-) 5' = 32-i9 

2233 31Q 

2240 lbs. - 7 lbs. = 2233, and — ^ = ^-^ 

^ ' ^^ 2240 320 

.*. the accelerating force of gravity is diminished in the 
ratio of 319 : 320 by the resistance from friction and 
the air. 

31 



sm ts 



2164 



71 7JQ 
•^ ^ ^ 2164 320 ^^^ ^ 

But it is not necessary to use this value o^ fhere since 
the velocity acquired in falling down the vertical height, 



A KEY TO THE MECHANICS. 41 

3 1 feet, will be the same as that acquired by running 
down the inclined plane, g being diminished in the ratio 

of 319 : 320, which will then become ^-^ — ^ . 

320 

310 X 32.10 
/. v' = 2 X ^-^ — - — ^ X 31 = 1989.54318 

.*. V = 44.6 feet. Ana. 

(8.) 8 = 1000 feet. 

/= •45969 (Bj last Example.) 

,\f=— = — . = 4350.7581 

/ 45969 ^^^ ^^ 

,\t= 65.96 seconds. Ans» 

(9) . Sf = 32-H35 

sin t = sin 14 = .24192 

-/=5^ sin I = 32.2435 X .24192 

= 7.80034752 

.'. 8 = iJ^ = 3.90017376 X 25 

- 97*5^4344 ^^^^ '^'i'* 

(10.) ^=32.093 

.-./= 32.093 X .24192 = 7.76393856 

.•. s = i/i?2 = 3.88196928 X 25 

= 97.049232 feet. An8. 

(11.) 5 = 100, /= 7.8 very nearly (by Ex. 9). 
tr» = 2/^ = 2 X 100 X 7,8 = 1560 
.*• v = 39.5 feet. Ans. 

(12.) v=yil = 7.8 X II = 85.8 feet. Atw. 



42 A KEY TO THB MECHANICS. 

Page 64. 

(i.) 300 yards = 900 feet 

.'. 1; = ~— = 1 5 feet per second. 

^ v* 22c ^ 

.*./= — = — /■ = 14.0625 feet Arts, 

r 16 

(2.) r = 17 yards = 51 feet, v = 361 feet 

v^ 130321 . ^ . 

.•./=-—= -^^^-=^ — = 2555.3157 feet -4n5. 
r 51 



(4.) Let W = weight of body, F = statical measure of 
centrifugal force, /= dynamical measure of centrifugal 
force, 

... W = 1000 lbs. 
F = 450 lbs. 
r = 100 feet 

Now, centrifugal force in lbs. : weight of body in 
lbs. : : centrifugal force in feet : gravity in feet. 

That is, F: W ::f:g 
Or, 450 lbs. : 1000 lbs. : : — : 32.1948 feet 

T 

Or, 45 : 100 : : — : 32.1948 

r 

Oii 45 : lOo : : — : 32.1948 

.*. t;^= 45 X 32.1948 = 1448.766 
and .'i V ^ 38.06 feet Ans. 



A KEY TO THE MECHANICS. 48 

(5.) 30 miles = 1 58400 feet. 

I hour = 3606 seconds. 

... v = -^-r- — = 4.4 feet. 
3600 

r = 460 yards =1380 feet. 

^ v* 44* 1036 . ^ , 

Also, W = 7.21 tons. 

That is, 32.1948 feet : 1.4029 feet 
: : 7.21 tons : F 
.'. F = .314 tons. Ans, 

(6.) Let V = velocity of body weighing 1 3 lbs. 

t/ = ditto 1 6 lbs. 
and F = centrifugal force of either body in lbs. 

W = i3lbs. 
W= 16 lbs. 

r = 1 1 feet. 

r^= 15 feet. 

Now, W : F : : ^ :/. 
That is, 13 lbs. : F : : ^ : — ; 

Or, 13 lbs. : F : : ^ : — ; 
Or, 13 lbs. : F : : II ^ ; v*; 



.'. v' = 



II F x^ 



13 
Again, W^ : F : : g \f\ 

That iSi 16 lbs. : F : : a : — ; 
Or, i6\b^. ; 1^ \ \ g \ — \ 



4:4 A KEY TO THE MECHANICS. 

Or, i61bs. : F : : 15 y : t/'; 
15 Fx^ 



.-. t/2 = 



16 



Hence, 



v^ _ II F X g ^ 15 Fx^ 
V^ 13 ' 16" 



II . IC II 16 176 

= r--~= — X — = — ^ = .902564 

13 16 13 15 195 ^ ^ '^ 

And .% by extracting the square root we obtain 
- = .9503 Ans. And ~ = '9S^3 

(7.) 2^= 16 seconds. 

r = 1 1 feet. 

4?r^r _ 4x 3.14159* X II 

_^ 4 X 3.14159' X II 9.8695877x11 
256 " 64 

= J = 1-6963353 ^6et. -47W. 

(8.) r = 1 1 feet. 

1 1 minutes = 660 seconds. 
.•.61° : 360° : : 660 seconds : T. 
.". T= 3895.082 seconds, vert^ nearly. 

HencG ^_4^^_ 4 X 9-8695877 X 11 
Hence, /--^ ^8^5:^8^^ 

_ 9.8695877 X 1 1 _ 108.5654647 
1947-541* 3 7929 1 5.94668 1 

= .0000286 feet. Ans. 



A KEY TO THE MECHANICS. 45 

9.) r = 1 1 yards = 33 feet. 

T = 4 minutes = 240 seconds. 

47r«r 4 X 9.8695877 X 33 
"^ T^ 240* 

^ 9.8695877 X 33 _ 9.8695877 X 33 
120^ 14400 

9.8695877 X 1 1 ^ 108.5654647 
"" 4800 4800 

= .0226178 feet. Ana. 

(10.) r= 100 feet 

/= 146 feet. 

But we have also by formula (13) 

47r*r 



•. T» = 



^Ti^r 4 X 9.8695877 X 100 

~T^ h6 



= "73*9 754 _. 27.04, very nearly. 
73 

.-. T= 5.2 seconds. Am, 

(11.) F= 131 oz. 

r = 100 feet, 
T= I hour = 3600 seconds. 

4?r*r _ 4 X 9-8695877 X 100 

-f'^'ji^ = "; 3600* 

_ 9.8695877 _ 9 8695877 
180* "" 32400 

= .000304617 feet 



46 A KEY TO THE MECHANICS. 

^ow,f:g: : F : W; 
That is, .000304617 feet : 32.1948 feet : : 131 oz. : W; 

.-. W = 13845316.5 oz. 
= 865332.2 lbs. = 386.3 tons. Ans. 



Page 68. 



(3.) <Jos 23° = .9205 

sin 23° = .39073 

.•./= .11126 cos H== .11126 X .9205* 

= .11126 X .8473225 

= -09427310135 feet. Ans. 

(4.) , = .11126 cos ^ sin / 

= .11126 X .9205 X .39073 
= .0400165 feet. Ans, 



Page 69. 

(5.) sin S3° 21' = .8022969 

Qos 53° 21' = .5969252 

.'. cos* 53° 21' = .35631969 

and cos 53° 21' x sin 53° 21' = .47891123 

.-./= .11126 X .35631969 = .03964412 feet, 

and = .11126 X .47891123 = .05328366 feet. 

(6.) r = 441000 miles = 2328480000 feet. 
T = 607 hours 48 minutes = 2188080 seconds. 
.'. by equation (13) we have (recollecting 
that TT = 3.14159 and .*. w^ = 9.8695877), 

jfyr^r_^ 4 x 9.8695877 x i^iS/ySoooQ 
-^'"W^" 2188086* 



A KET TO THE MECHANICS. 47 

986.95877x23848 22981 1375.67696 

1 09404^ 1 1 9692 352 1 6 

= .01920017 feet Ans. 

(7.) r = 1 570 miles « 8289600 feet, 

T = 24 hours 5 minutes = 86700 seconds. 

^""^r 4 X 9.8695877 X 8289600 
•*-/=*25r = 86700^ 

^ .098695877 X 331584 
751689 

32725-973679 



751689 



= .043536 feet. Ans. 



(8.) r = 3900 miles = 20592000 feet. 

T=: 23 hours 21 minutes = 84060 seconds. 

49r2r _ 4 X 9.8695877 X 20592000 
•'••^""r« " 840602 

98.695877 X 20592 

4203' 

2032345499184 ,, „fe^^ ^n.. 

17665209 ^ T/;7 

(9.) r = 2050 miles = 10824000 feet. 

T= 24 hours 37 minutes = 88620 seconds. 

^Ti^r 4 X 9.8695877 X 10824000 
•'•/=^T-= 88620^ 

98.695877 X 10824 _ 1068284. 1 72648 
4431* ~ 1963376' 

= .054415 feet. Ans, 

(10.) r = 43500 miles = 2196^0000 ^^^\.. 
T=g hours 56 mmuUs = 'iSl^^ ^^^wA^- 



48 A KEY TO THE MECHANICS. 

4^r*r 4 X 9.8695877281 X 229680000 

986.95877281 X 22968 
3196944 

^ 986.95877281 X 2871 2833558.63673751 

399618 " 399618 

= 7.090668 feet. Ans, 

(11.) r = 39580 miles = 208982400 feet. 
2^= 10 hours 29 minutes = 37740 seconds. 

4w*r _ 4 X 9'8695877 x 208982400 

9.8695877 X 2089824 
*" 1887' 

^ 20625701.2455648 3^^^^ ^n.. 

3560769 ^ '^ ^ 



APPENDIX. 



N. B As proofs of fhe Bules in the Appendix to the Manual may 

not suggest themselves to every reader, they are here supplied. 

RULE I. 

sin (d + e) = sin d cos £ + cos 9 sin f . 

If in this formula we suppose to denote degrees, and e minutes, cos c 
will be very nearly = i, and hence the foundation of Rule I. 

RULE II. 
The proof is obvious from the same formula. 

RULES IIL AND IV. 
cos(9 + e) = cos9cose-sin0sinc=cos0-8indanc very nearly. 

RULES V AND VI. 

tan + tan € 



tan (0 + f) = 



I —tan 9 tan c 
(by dividing the numerator by the denominator.) 

tan £ (i + tan *0) 



tan + 



= tan0 + 



I — tan tan £ 
2 tan£ 



=^10X19 + 



2 cos> 9 (i — tan tan e) 



(i 4-cosa9^coatC^i-^«0'^^»».^^ 



50 APPENDIX. 



2 am e 



= tan H since cos € = i nearly, 

I + cos 20 ' 

And tan tan e may be neglected, as c is very small. 

Rules i° and 2°, — ^When the angle is very small its circular mea- 
sure, sine, and tangent, are very nearly equal, 

tanO + tane tand-i-sinc 
.•. tan (0 + e) = 



I — tan tan £ i — tan sin c 
approximately, since e is small 



Rule L 
First five figs. = .00^51 J sin 27' = .00785 



(2.) _ sin 34° = .55918 ) , , cos 34° = .82904 



.*. sin 34'' 27' = .56569 Product = .006507964 

sin 54*' = .80901 7 , , cos 54" = .58778 
First five tigs. = .00102 J sin 6' = .00174 



sin 54° 6' = .81003 Product = .0010227372 



(4.) sin 69° = .93358 ) ,^ 

First five figs. = .00344 J 



cos 69° = .35837 
sin 33' = .00960 



.*, sin 69° 33' = .93702 Product = .003440352 

Rule II. 

(2.) Given sine = .7 8 23 1 

sin 51° = .77714, the next lower in Table II. 

.00517 = diflerence, 

cos 51° = .62932 and .00517 -f- .62932 = .00821 

which (Table IV.) most nearly corresponds to 28' 

.'. required Z. = 51° 28 

(3.) Given sine = .31420 

sin 18' = .30901 

c0a iS' = .^j'/oj).oo5i9 = difierence. 

,oos4S = Quot = am 19' n^axVf , \57 ^aJoV^ V^ .^ 
. •. required £ = iV 19 



APPENDIX. 51 



(4.) Given sine = .80000 

sin 53° = .79863 

cos 53° = .6oi8i).ooi37 = difference. 



.00227 = Qnot. = sin 8' 

.-. required L = 53° 8' 



Rule III. 



First five figs. = .00095 3 sin 27 = .00785 

.'. cos 7° 27' = -99159 Product = .0009566795 

(3.) cos 53° = .60181 ) .„ut-.„t sin 53» = .79863 

First five figs. = .00464 J '^""""^"' sin 20' = .00582 

.-. cos 530 20' = .59717 Product = .0046480266 

(4.) cos 32° = .84805 \ -„u.__t sin 32° = .52992 

First five figs. = .00601 J s^^^^*^"^ sin 39' = .01 134 

.-. cos 32" 39' = .84204 Product = .0060092928 

Rule IV. 

(2.) cos 36° = .80901, the next higher in the Tables, 

given cosine = .80000 

sin 36" = .58778).oo90i = difference. 

.01532 =Quot =sin 52' (by Table IV.) 
.*. required angle = 36" 52' 



(3.) cos 61** s .48481, the next higher in the Tables, 

given cosine ^ '47320 

sin 61° = .87462).oii6i = ^\^«t«c^t^. 
.•. required Wftg\% = 6\'* t^ 



V 



52 APPENDIX. 

(4-> 

dn 69° = .933j8).ooEEi : 

,00944 ^ Quot. = «in 31' 
.■- required angle = 69° 31' 

SolbT. 
(2.) 1X57" = .14° 

How COS 1 14.° = - cos 6€° (iu nipplemeot) 



ibtract, since cos 1 1 



cos £6° =■ .40673 

1) .59017 

Half-mm >= .19663) .00901 1= iln 31' 

.03037=000! 

.0303; I "■ 

.-. ton j7° 31'= 1.J7013 
I X 49" = 98° 
.-. COS 98° = - cos 81° 
1 .00000 J _ . . 
»8,"= ..39,7 {'"""«» 

1) .86083 

.43041) .00989 = ^ 34' 

.01297 = qoot. 



n 49° 34' = 1-1733+ 
2x19° = 38° 
CM 38° = .78801 J ^ 



.S94O0) .ool 56 = Anifi 
.oo8« = QjkA. 



APPfiKBtX 53 



"""°=:^^!«*» 



.*. tan 19° 26' = .35277 

Rule VI. 

(2 ) Given tan = . 87 245 

tan 41'' = .86928 next lower in the Tables. 



Diff. =.00317 
2 X 41' = 8a* 
cos 82* = 



i.ooooo 3 



add 



2) 1.13917 



.56958 
Diff. =.00317 



Product =.001 80, 55686 = sin 6' 

-.*. tan 41° 6' = .87245 

(3.) Given tan = 1. 14376 

tan 48°= 1. 1 106 1 



Diff. =.033 15 

Now, cos 96" = — cos 84° 
I.ooooo #1 . ^ 

cos 84.°= .lOAC'i S ^^^*'*^*» ^"^^® ^^ 96° *"* negative. 



2) .89547 

.44773 

Diff. = .03315 



Product = .0 1484, 2 2495 = sin 5 1' 
.'. tan 48" 51'= 1. 14376 

(4.) Given tan = 1.63422 

tan 58* = 1.60033 



DVff. = .0^-^%^ 

COB 1 16° = — COa ^Ar* 



54 APPENDIX. 



I.OOOOO 

COS 64"'= .43837 



> subtract 



2) .56163 

.28081 
Diff. = .03389 



Product £=.0095 1, 66509 = 8111 3^' more nearly than sin 32' 
.*. taa 58° 33'= 1.63422 
if we use the Tables and Rule» of the Manual. 



SOLUTIONS 



TO 



THE NEW QUESTIONS IN THE THIRD EDITION 
OF THE MANUAL OF MECHANICa 



Page 12. 
(i.) By equations (i) and (2) we have 

Ph = umt 

_P^_750x 2240X 6x ioo_ 75X 224 25x56 
wt 33000x60x12 33x12 "11x3 



• • 



= 42.42 H. P. An^. 

(2.) Here we have, as in the last example, 

Ph = wnt 

Ph coco X Sox 10 X 60 5x4x10x20 
vm 33000x20 II 

= 363 t7- minutes = 6** 3tV"'. -^w** 

(3). Since Ph = wnty we have 

Ph 1C4X C280X 2 

m; = — = ^^ - ^ ^= 1C4X 88x1 

nt I x6o ^^ 

= 2 7 1 04. ft. W)^. Ans, 
e2 



56 A KEY TO THB MECHANICS 

Page 13. 
(4.) Fh = wnt 

wnt 33000x1310x60 33000x1310x10 



• F = 



73 X 6 73 



= 592 191 7.8 lbs. (see remark followiDgEquation(i) p. 10) 
= 592191.78 gallons, since the imperial gallon of water 
weighs 10 lbs. avoirdupois. 



(5.) Ph^wnt 

wnt 17536x4x60 17536 X4X 2 _ 140288 
" P ~ 32130 1071 1071 

= 1 30.99 ft. Ans. 



(6.) Ph = wnt 

Ph 10000x10x40 IOOOOXIOX2 lOOOOO 



t = 



wn 3897x^x20 1299x2 1299 

= 77" very nearly =^ i^ 17". Ans, 



, ^ ^ wnt 467x500x6000 ^1. 

(7.) p=:—^=.^-J- — ^ =467x1 00x6000 lbs. 

^ 467x100x 60^ ^^^.^ ^ 467x100x300 

2240x2 224 

= -i = 62544.6 cubic yards, -drw. 

224 



A KEY TO THE MECHANICS. 57 



Page 14. 

/o V n ^^ 1126x30x420 , ,, 

(8.) P = -— s= ^ ± — = 1 1 26 X 420 lbs. 

n 30 

^ 17 , . , 1126x420x17x8 

= 1 1 26 x 420 X — '— bricks = ^ 

^ 125 125 X 8 

1126x420x17x8 , 

= =ii.26x42x 17x8 

1000 ^ ' 

= 643 1 7. 1 2 bricks. Ans. 



(9.) wnt -v^ n' i 

,_ «ml 2598x2x10x8x60 2598x2x10x60 
w'n' 1126x4 563 

= 5537.4"" = 92'^ 17.4" = 15^ 2'> 17.4", 
since the men work only 6 hourB daily. 



. . , wni 15588x8x24x60 5.106x8 
(10.) n'=— 7-3=-^^ -"^ — s=i— Z 

W't 33000x24x60 II 

41.568 



II 



= 3.779 H. p. very nearly, Ans. 



Page 82. 
(12.) r= 17250 miles = 91080000 ft. 

T=g^ 30™ 6=34200 seconds, 



58 A KEY TO THE MECHANICS. 

4 9r^r 4x9.8695877x91080000 

9.8695877x9108 9.8695877x1012 

^9988-0227524 8 6 ft. ^n.. 

3249 ^^ 



{13.) Here (the radius of the mocMi's orbit) 

r = 59.96435 X '^ ^' = 59.96435 X 3962.8 miles. 

= 59.96435 X 3962.8 X 5280 ft = 1 2546691 14.2304 fk. 

and 2'=27<^ 7** 43™ ii»= 2360591 seconds; 

4 w' r _^ 4 X 9.8695877 X 1 254669 1 14.2304 
•*• /= -jfT" = 2360591* 

49532267429.51300322432 



5572389869281 



.00888887 ft. Am, 



(14.) By the last exercise the moon's distance from 
the earth = 59.96435 radii of the earth ; .•. if x repre- 
sent gravity at the distance of the moon, we have, 

59.96435* : i' : : 32.2 : x 
32.2 32.2 



. . Ju ~^ 



59-96435* 3595-7232709225 

=^.00895508 ft. A«8. 



A KKY TO THE MECHANICS. 50 

Page S5. 
(i.) Here, Z= 72, a= 50, a?= 30, and g^ 32, 
,-. by equation (iS) we have 

t;« = f(a^-rt?^)=H(S0^-30^-) 
4 / V ^¥^ 

= - (2C0O-9O0) = -i — , 

9 9 

(2.) Suppose AB (fig. p. 84) to be the horizontal dia- 
meter, and X the middle point of the arc AB, then 

a^ = AB^= 2^ = 2 X 100^ = 20000. 

Since BX subtends at the centre G an angle = 45°, 
.*. the angle BDX- 22® 30'; and as the angle BXD in 
a semicircle is right, BX=BD sin BDX^ that is 

X = BX = 2I sin 22^ 30' 
= 200 sin 22° 30' = 200 X .3826834 
= 76.53668 and 0?* = 5857.8633854224 

.-. r' ^^-r (a^ -x")^ ^^2}9^ (20000-5857,86) 
I ^ 100 J Jt f 

= .321908 X 14142.14=4552.46800312 
.'. r= 67.47 ft. Am. 



60 A KEY TO THB MECHANICS. 

Page 89. 
( I .) By equation (2 1 ) we have 

^ a 32.182 _ 42.182 X 12 



'^ 3-14159* 9-^695877 

s 3 9. 1 2 8 inches. Ans. 

(2.) By equation (23) we have 

^s=?r* L = 9.8695877 X 39.139 inches. 
= 386.2857929903 inches. 
= 32.1905 feet, very nearljf. Ana. 

(3,) Here, by equation (ai) and the TaWo, p. 64, we 
have 

g 32.2526 32.2526x12. 

L = -^= \^^ \ — ft. = - — ~ — inches. 

9r2 9.8695877 9.8695877 

387.0312 . , . 

= \* \, — = 30.2 144. inches. Ans. 
9.8695877 ^^ ^ 

(4.) Here, by equation (21), aud Tivble, p. 64, we 
have, 

_ a 32.0881 X i^. , 
Z =^ =. ^—^ — 5- — inches. 
^ 9.8695877 

= -^-_21 -iZ_ = 3 g.o 1 44 inches. Ans. 
9.8695877 ^^ ^ 

(5.) By example (2)-jL = 39. 1 39 inches; also T- 2% 
.'. by equation (22) we have, 

= LT^- 39«^39 X 2^*= 156.556 inches 

= 13.046 ft. Ans. 



A KBY TO THE MECHANICS. 61 

(6.) Here Z= 39.139 inches, and T=s^ second; 
.•.Z=iT^= 39.139x^ = 9.784 inches. Ans. 



Page 90. 

(i.) H^re 2X = 2 X 53° 20' « io6° 4c/ 

.% cos 2\ = - cos ( 1 80® - 1 06° 40') = - COS 73® 20' ; 

hence, by equation (24) we have, 

i = 39. 1 1 8 + ife cos 73® 20' 
= 39.1 18 + -Ay X .2868032 = 39.1 1 8 + .02868032 
= 39.14668032 inches. 

By equation (23) we have, 

g=7r^ 2^ = 9.8695877 X 39.146 inches. 
= 386.3548801042 inches, 
= 32.196 feet. 

(2.) 2\ = 2 X 55° 57'= III® 54' 

.'. cos 2\ = -cos (i8o°- III® 54O =-cos 68® 6' 

.-.by equation (24), 

i=39.ii8 + 'Ay cos 68® 6' 
= 39.1 18 + .03729878 =39.15529878 inches; 

and ^ = 9r*X= 9.8695877 X 39.155 
= 386.4437063935 inches 
= 32,2036 ft 

(3.) At the Equator \ = o ,'. cos 2X = i 

and .'. i= 39.118 -1^= 39.018 inches; 
aJso, g^TT^L^ 9-^^9S^n ^ 'i^-^'^'^ 



62 A KEY TO THE MECHANICS. 

Again, at the Pole X = 90°, 

.•. cos 2\ = cos 1 80*^ = - I 

and L = 39. 1 1 8 + Y*^ = 39.2 1 8 inches ; 

^ = 7r*i = 9.8695877 X (39.018+ 2) 

(by the last question) = 385.0915728786+ 1.9739 1754 
= 387.06549041 86 inches = 32.255 ft. 



Page 91. 

(i.) Here n = 86400% the number of seconds in a mean 
solar day; 

<5r = 32.1908 (Table, p. 64) 

5^' = 32.0881 (Table, p. 64, the Isle of Rawak being on 
the Equator). 

Hence, by equation (27) we have, 

# w a -a' 22.1008-32.0881 

n-n'= -x^--^= 43200 x^ — ^ -^^-5 

2 g ^^ 32.1908 

1027 44366400 -. . 

= 43200 X -^- = 11£ — -^= 137.8" Ans, 

^^ 321908 321908 ^' 

(2.) Here 71 = 85945.8 

w' = 85933-83 

and we are to calculate the value of the fraction - — ^ : 

9 

by equation (27) we have, 

n. «/ '^v^"^' . ^"^' 2(n--w') 

'^ g 9 n 



A K£T TO THE MECHAKICS. 63 

2(85945.8 -8 5933.83) 
85945.8 

2X H.9 7^ 23.94 ^ ^ ^^ 

85945.8 85945.8 3590 

(3.) Here n = 86400% and n' - » = 2 J" ; 

g^ — g 
we are required to calculate , .•. by equation (27) 

we have, 

g'-g 2 (n' - w) _ 2 x2 J 
g n 86400 

- - Ans, 



172800 19200 



Page 93. 

(i .) Here l^ = 45 inches, and / = 45 inches. 

32 

I'-l t I I 

= — -^45 = 



/' 32 • ^^ S^xAS H40 
hence by equation (29), we have. 

Acceleration in a day = 43200 x — rj— 

= 43200 X = 30 seconds. Ans, 

1440 

(2.) Here T' - r= 2" = i20», and 1= 39.14 inches (the 
length of the seconds pendulum in London, see p. 88), 
.-. by equation (29) we have, 

120 = 43200 X -y~, .'. I =36ox — r- 



64 A KSY TO THE MECHANICS. 

and i'-i = 4=32dl- finches; 

360 360 18000 

the number of turns (x) of the screw, to which this is 
equivalent is found by multiplying it by 50, 

18000 360 -^ 



Page 94. 

(3.) Here (see last example) X, the length of the se- 
conds pendulum in London, = 39*14 inches, 

n = the seconds in 24** 3" 56.5" = 86636.5' 
and we are required to find I from the equation, 

n*==-j-^ (86400)*; 

/'i728oo\' 

= 39.14 -.213 = 38.927 inches. Ans. 



(4.) By the Table, p. 93, the values of 

/"-/ I 

-J— for 10^ F. in the case of a braaa ^^^uXxmcq. *\% -^ 



A KEY TO THE MEGHAKIC8. 65 



.•. for 1° F. its value is — x 



lo 9600 96000' 
hence by equation (29) we have, 

Retardation = 43200 x -—z — = ^^^^^— 

^•^ 96000 9 

= .45 seconds. Ana, 



5.) By Table, p. 93, 
V-l I 



/ 14400 
.% by equation (29), 



for 10° F 



Acceleration = 43200 x = 3 seconds. Ans. 

14400 •' 

(6.) Here by Table, p. 93, 

, v-i_n\ I _ 13 



I 10° 41000 410000 

.•. by equation (29), 

Retardation » 43 200 x ^ — = i-^ — 

^'^ 410000 41 

= 1.37 seconds. Ana. 

(7.) Here by Table, p. 93, 

V-l 14° I 7 

I 10° 6120 3060D 

.-. by equation (29) we have. 

Acceleration = 43200 x — | — = ~ — ^ 

30000 y^ 

= 9.88 seconds. Auft. 



66 A KBT TO THE MECHANICS. 



(8.) Here increase of temperature 

.*. by Table p. 93, we have, 

/'-Z 17.5° I 7 

I 10** 15600" 62400 

.*. by equation (29) we have, 

1 • J 7 36 X 7 63 

loss m I day = 43200 X r^-^ — = - i.-=«ji 

•^ ^"^ 62400 52 13 

and .*. loss in a week = — x 7 = ^^ 

= 33.92 seconds. Ans, 

(9.) Here we shall use equation (22), viz., Z = X2^, 
where / is the required length, L the length of a seconds 
pendulum in London (39.14 inches, see p. 88), and 

«T ^ X 60 

•'. ^ = 39.14^ 3*75* = 39*^4^ 14.0625 = 550.40625 inches. 

= 45.87 ft. Ans. 

(10.) Here we shall employ equation (20), viz., 

where T is the duration of one vibration in seconds, 
g = 32.1908 ft., and ^= 340 ft. - 6 in. = 339.5 ft. 

.-. r«3.i4j6 ^-1^ = 3.1416 1/(10.5465) 

= 3.1416 X 3.247= 10.2007752 seconds; 
.*. the number of vibrations in half an hour equals 
«ox6o* 1800 , . 

10. 2" 10.2 



SOLUTIONS 



OP 



THE KEW QUESTIONS IN THE FOURTH EDITION 
OF THE MANUAL OF MECHANICS. 



Pages 15, 16. 

(i.) In last figure, page 4, let 

OA = P, 0B= Q, OC = E, and the angle AOB = 0; 

then, by Trigonometry, 

DC* = OA* 4- CA» - 20A . CA cos OAC, but cos OAC 
= - cos AOB, since these L* are supplemental ; also, 

CA = OB = Q, 
.-. R2«P* + Q2 + 2PQcos0. 

(2.) Using the same figure, let OA = P, OB = Q, 
then OC is equal and opposite to R. 
Now from A AOC we have 

AO : OC : CA = sin OCA : sin CAO : sin AOC, and 
sin OCA = sin COB = sin BOC = sin ok ; 
sin CAO = sin AOB = sin IPQ ; 
sin AOC = sin AOC = sin it ; 
.*. by substitution, 

P ; B : Q = sinQB. : sm^Q,\ %m^^* 



68 A KEY TO THE MECHANICS. 

N. B. — Since the sides of the A OCA are parallel to 
the three forces which are in equilibrium, it is plain 
that when any three forces are in equilibrium the sides 
of any A parcdlel to them, taken in the same order^ are 
proportional to the forces. 

(3.) See(i6Xp. 8, "Key" 

(4.) See (17), p. 9, "Key." 

(SO See(i9), p.9, "Key." 

(6.) See (21), p. 10, "Key-" 

(7.) See (22), p. 10, " Key," 

(8.) Fig., p. 4. Let OC «= 10 lbs., OA » 8 Ibe., then 
L AGO = 36**, apd AC and i. AOB are required. 

OC sin O A C _ sin AOB 
OA " sin AGO " sin AGO 

10 sin AOB 

•'• T " sin 36° 
.-. sin AOB = ^ sin 36° 
=:ix. 5877853 = .-^3473 16 
= sin 47° i/; AOG = 47° 1/ - 36° = 1 1° 17'. 

Again, AC« = A0« + 0C« 

- 2AO . OC cos AOC 
s= 64 + 100 - 2 X 8 X 10 cos 1 1° 17' 
s= 64 + 100 - 2 X 8 X 10 X .9806716 
= 164. - 156.907456 = 7,092544 

.'. AC- 1/(7-09^544) ^ ^•^V 



A KST TO THE MECHANICS. 69 

(9.) Let ABC be a i^ ; P, Q, R, the middle points of 
AB, BC, CA ; and O the centre of the circumscribed 
circle. Since the forces P, Q^ R act perpendicularly to 
the sides through their middle points, their directions 
meet in O. Now by the hypothesis we have 

P : Q : R r: AB : BC : AC 

= sin C : sin B : sin A 

1= sin R^ : sin cit : jsin plt ; 

since C + RQ = 1 8o% &c 
.*. the forces are in equilibrium by Ex. (2). 

(10.) Since the sides of the A CO A are proportional 
to P, Q, B, /. its A* can be found; hence if QP be 
joined, in the A POQ we have Z POQ and L QPO 
= /. QPA' - POC, both of which are known, .% the Z' 
of the A POQ and its side PQ being known, the posi- 
tion of equilibrium is determined. 

(11.) The resolved parts of P and Q along their 
planes are P sin«, Q sin /3, but these are the tensions 
on the string, 

••• Psin« = Q sin/J, 

.*• P : Q :: sin)3 : sin«. 
(12.) We know from Trigonometry that 



sin 105° = cos 15° = i (v^6 + ^z) 
cos 105° = - sin 15°?=- i(V6 - ^2) 

sin 120° = sin 60° = ^ ^ 3 
cos 120** = - cos 60^ = - * 



{ 
{ 

/•sini35°=sin45''=:^ si ^ 

I 008 135® = - ooB4.5°«-\ ^^ 



¥ 2, 



70 A KEY TO THE MECHAHICS. 

Let P = V^3 + I 

Now since the three forces are in eqnilibriumy each 
force is equal and opposite to the resultant of the other 
two; hence, by Ex. (i), we haye 

(i) R' = P« + (y + 2PQcos/Q 

(2) Q« = P2 + R' + 2PRco8i^ 

(3) P« = QP+R»+2QRcos(^R 
(i) becomes, by substitution, 

4 = 4 + 2 ^3 + 6 + 2 ^6(y/3 + 1) cosPQ 

,.eosP^ = .— ^- = -4- 

= -iv^2 = 008135'' 

(2) becomes 

6 = 4 + 2 v^3+4 + 4(^3 + i) cosPR 

A 2 + 24/3 
.-. cos PR = T-, — ^^ = - i = cos 120°; 

hence C^ = 360° - 105*^ - 120° = 135°. 

• 

(13.) Let R be the resultant of all the forces, and m 
its inclination to the first force; also, let Ri be resultant 
of forces resolved along first force, and R2 along third 
force. 

.% J?i = I + 2 cos 60° + 4C0S (90** + 60*^) 

= I + I - 400330**= 2 - 2-^3 
which being negatiye, acta in a direction opposite to z. 



A KEY TO THB MECHANICS. 71 

Again, resolve along tbe third force, 

.*. Rj = 3 + 2 cos (90® - 60®) 
+ 4 cos 60° =3+^/3 + 2 = 5+^/3 

= l/(44+2v^3)= 1/(44 + 3.4^4) = 6-9 lbs.; 
2-2^/3 2-3.464 



also cos « = 



6.9 6.9 

1.464 
^ = - .212174 



6.9 
= cos (180° - 77°450 = cos 102° 15'. 

(14.) OA = 27, OB = 52, OC = 49, taken in the 
order of the letters A, B, C. 

.-. -L AOB = 32^ BOG = 260 and AOG = 58^. 

Let R be required resultant, a the required z. 

Ri = sum of resolved parts in direction of OG, 
R2 = sum of resolved parts X' to 0C« 

.*. Ri = 49 + 52 cos 26^^ + 27 cos 58® 
= 49 + 52 X .8987940 
+ 27 X .5299193 
= 49 + 46.737288 + 14-307821 1 
= 1 10.045 109 1 

R2 = 52 sin 26** + 27 sin 58*^ 

= 52 X 438371 1 + 27 X. 848048 1 
« 22.7952972 + 22.8972987 

= 45-6925959 



72 A KEY TO THE MECHAKIC8. 

.•.R= V{(iio.o5)*+(4j.69)>} 

= \/(i*moo25 + 2087,5761) 

= 1 19.16 required resultant, 

Ri no ^^ 

/. cos « = =— = = .02366 

R 1 19.2 ^ ^ 

= cos 22° 32'. 

( 1 5.) Using the same notation as in the last Example, 
we have 

Z AOC=i7^+S2^==69°. 

Ri = 29 + 16 cos 52** + 3 1 cos 69^^ 
= 29 + 16. X .61566 + 31 X .35837 
= 49.96. 

R2 = 16 sin 52** + 31 sin 69® 
= i6x .788 + 31 X .93358 
= 41.54. 

.•.R=V{(49.96)'+ (41-54)') 

« ^(2j^g6,ooi6'{' 1725.5716) 

» 64.98 lbs. 

••^'* = R""6:f:i8''' * 

= cos 39° 45' 

(16.) The ring will plainly rest at C, the middle point 
of the cord 2/, and if the vertical through C meet the 
line AB in D, then AD = DB ^ a, and 

ZACD = BCD = 
.'. sin ff= y and cos 9 = ^ 



/ ^ I 



A KKY TO THE BCECHAKIC8. 73 

Now the sum of the tensions of CA, CB, resolved ver- 
tically, = W, that is, 

»Toos0«W 

W ^ WZ 

"" 2 cos ^ 2V(Z* - a») 

(17.) Fig., p. 102. 

Suppose G the centre of the wheel, which is also its 
centre of gravity, through which the weight W and the 
force F act ; the former vertically, the latter horizon- 
tally. Through X draw a JL' to the tangent at B, 
meeting it in A, and join XG. Let XA be the obstacle 
= h. The resistance M of the obstacle must pass through 
G, since there is equilibrium, and W, F, G being pa- 
rallel to the sides GH, HX, XG, of the A GUX, taken 
in order, must be proportional to them, 

.-. W:F::GH:HX, 

but HG = r - A, and 

HX= y/{^iai,}5X>)^y/[h{2r-h)} 

.-. W:F::r-A: v'(2rA-A«) 

r- h 

(18.) Since the forces P, Q, B equilibrate at a point, 
they are proportional to the sides of a A, drawn parallel 
to them, and in the same directions ; but if a, &, c be 
the sides, and A, B, C the z.' respectively opposite to 
them, we have, by a common trigonometrical formula, 

• A ^/[{a^\'h + c){a■k■h-c){b■k■C'■a) {c-{-a-h)\ 

Slu XX ^ ' - ■ ■ • 

200 

Now A is Z. included by 6, c, and rQ is L included by 
P and Q, 

.'. sinPQ 



74 A KBT TO THB MECHANICS. 

Pages jo, ji, ji, 

(i.) Let a, &, and c denote the tensions of AC, BD, 
and CD respectively. 

The point C is kept in equilibrium by the three 
forces a, P, c acting in the directions of CA, Cc, and 
CD, and the sides of the A CD£? are parallel to these di- 
rections, and .'. proportional to the forces: hence 

a :P:c=:Crf:Drf: CD. (i) 

Similarly D is kept in equilibrium by h^ Q, and c acting 
parallel to sides of A CDc. 

.-. 5 : Q : c = Dc : Cc : CD. (2) 

From (i) we have 

Z- ^ 
c " CD' 

From (2), c Cd 

Q " Cc 

P ^_ jM CD 
•• 7'Q" CD* Cc ' 

that is, P Dd . 

Q Cc 

(2.) From equations (i) and (2) of last question, 
we have 

a Cd , Q Cc 

P = D5*^^y=D^ 

— Q _ Crf Cc 
"'•P'T "Dd'Dc 

that is, a Q Cc Cd 

b'F^Dd'Dc' 



A KEY TO THE MECHANICS. 75 



Now, i^ p = p^7 t^en 






(4.) Plate IIL, Fig. i. 

Let AD, BD be the inclined planes, DF horizon- 
tal, A and B the points of contact of the sphere whose 
centre is C, with the planes, K, H', the normal pressures 
of the planes on the sphere at A and B respectively. 
R and R' must act through C ; hence if the vertical AG 
meet BC in G, and the horizontal line in F, R, R' and 
W yrill be proportional to the sides of the A ACG, since 
the sphere is kept in equilibrium by R, R', W. 

.-. R : R: W = AC : CG : GA 
= sin G : sin GAC : sin ACG 
= sin t': sin t : sin (1 + »^, for 
Z ACG = ADB (since quadrilateral ADBC has Z* at 
A and B right) = 1 80^- (t + t') ; 

also z GAC = ADF, for they have a common comple- 
ment DAF, and z G = 180?- (GAC + ACG), 

= i8o°-{t4-i8o^-(t + 0) 

hence R : R' : W = sin t' : sin i : sin (1 + i) gives 

W sin a 



R = 



sin (» + t") 



W sin t 

R = -T 



sin (1 + • 



76 A KEY TO THE MECHANICS. 

(5.) Plate III., Fig. 2. 

Let BC be the vertical wall, AC horizontal, and A 
the fixed peg against which the ladder AB presses, G 
centre of gravity of ladder, and F the point in which 
vertical through G meets BF J.*" to the walL 

Since the wall is smooth, it can only exert a normal 
pressure P on the ladder, and since two of the three 
forces, viz. P and W, which keep the ladder in equili- 
brium, meet in F, the direction of the third, Q, must 
also pass through F. 

Since the A AFD has its sides parallel to the direc- 
tions of P, Q, W, we have 

P : Q: W = AD : AF:FD. 

Now AC = BC tan ^ = FD tan 0, and AD : DC *= w : n 

» ^ 1^ • r^ wi tan _,_ 
.-. AD = • AC= .FD 

^ ^ «». ^ tan __ 

.-. P :Q: W= .FD 

m-\- n 

^ .,«. w* tan 

that IS, P : W « : i 

m +n 

and Q:W = ii.7^' 



.-. P = W . . tan e 

r. TKT If mHan^e 



A KEY TO THB MECHANICS. 77 

(6.) Plate IIL, Fig. 2. 

Suppose B the hinge, F the pulley, and BL the po« 
sition of the door in equilibrium, the L FBA being 
hert = 9'. The moments of P and W about B must be 
equaL 

Now BM -L' to FL is = ^cos ^, since FB = BL = I, 
and JL' from B on direction of W, which acts vertically 
through middle point of BL, is = -^ cos 0. 

.-. P.^cos^=. W.^cos^, 
and .% P = W . ;; r-;. Ana, 

(7.) Plate IIL, Fig, 3. 

Let the beam AB rest on the two planes AM, BM, 
and CG be the vertical through G,it8 centre of gravity 
(.-. AG = GB). AC is normal to AM, and BC -L' to 
BM, meets the vertical AF in H. 

Since the A ACH has its sides parallel to the direc- 
tions of P, P', W, we have 

P : F : W = AC : CH : HA 

= sin CHA : sin HAC : sin ACH. 

Now L CAH a e, since they have a common comple^ 
ment MAF. 

CHA s BCD = t'', since t' and BCD have the common 
supplement, BMD and ACH = 1 80^ 

- (CHA + HAC) = 1 80°- (t + 

.*. P : P': W = sin e": sin i : sin (i + 1") 

and .-.P^J^^^^*' 



F- 



sin (t + t") 
W sin f 
sin (1 + • 



Again, in A BCA we have BCG = t", ACG = 1, and 
BG = AG, 



78 A KEY TO THE MECHANICS. 

, BG sinr 
•*• CG " sin B 

AG sin t 

CG ""sinA 

sin A sin t 



Sin B sm t 

sin (i + e" + B) sin • 
Sin B sm t 

.% sin (f + i) cot B + cos (t + {) = — ; — - 
^ ^ smt 

• • 

that is, sin (i + tan = — : — - - cos (i + 1*). (i) 
^ sint ^ ^ ^ ^ 

But tan ^ = tan (0 - f) 

tan sin (t + t') - tan i" sin (t + H) 
sin (i + t') + tan sin (t + i^ tan { 

. . substituting from (i), 



-T — r - COS (t + t") - tan t" sin (t + t*) 

^ smt ^ 

tan^ = . . ■ 

sin (t + + -: — 7 - cos (t + t') tan t" 
^ sint ^ 

-: — 7 7 I cos (i + O COS t + sm t sm (t + 1 1 1 

smt cost ^ ^ - ^ 

7 t sm (e + t ) COS t - cos (t + t ) sm t ) + -: — =• 

cost' ^ ^ ^ ^ ^ ' smT 

sin t cos % 



sin t cos % sm (t - t ) 



sin t sin t 2 sin t sin t*' 



COS t'' cos { 



A KEY TO THE MECHANICS. 79 

(8.) Plate III., Fig. 2. 

Let A be the point where half the roof AB rests on 
the side wall, and BC the vertical through its summit, 
then weight of AB = ^W, and acts through G, middle 
point of AB, and •*. m = n. We may suppose BC a 
fixed plane, since there is equilibrium ; and if AD re- 
present the normal reaction of BC on AB, then AF 
will represent the reaction of the point A in magnitude 
and direction, but AF resolved in horizontal direction, 

ACis = AD = P; 
and P in question (5) 

= W. tan^, 

m + n 

but here' weight of ^ roof = ^W, and = 90° - ^, also 
w = n ; .'4 Horizontal Thrust = ^W oot0. 



Pages 88, 89, 90, 91. 

(6.) Here sine of inclination of plane = -r, .*. the part 

of W which acts down the plane is TF'. -r, and .*./, the 
acceleration up the plane, is 

T-W.J 

but equation (4) gives 

S = ^yP, and here S = i, 



80 A KSY TO THS MECHANICS. 



• • • ^ ^ ^ ~~~ • 



„ 2 T + W 






(7.) Let ^ denote the time after the first stone is 
projected when the second must be projected, then if 
the ten seconds be measured from the starting of the 
first stone, equation (4 bis) gives for the two stones the 
two equations — 

S «= 50 X 10+ 16 X 100 

S= ioo(io-«)+ i6{io-ty; 

equating these two values of S, we have, after reduction, 

16^ -420^ = - joo. 

The admissible value of / is 

but if the 10 seconds be measured from starting of the 
second stone, the equations are, 

S = 50 (10 + e) + 16 (10 + ty 
S = 100 X 10 + 16 X 100. 

Equating these values of S, we have 

. -185 ±205.; 

The positive value of t is 

/=1.28'. 



A KST TO THE MECHAVICS. 81 

(8.) Resolving the t.wo weights along the planes, we 
have for the statical force which produces the accelera- 
tion /in the two weights — 






W1.7 -W2 



__ h „, h 



_ gh Wi4-W.?i 



(9.) The space described in the first (( - i) seconds is 

■. n 

and the entire sp^e described in the t seconds is 

hence, substituting for 8 from this equation in the for- 
mer, we have 



. ., ...,.(,_i) = (i-iy, 






82 A KBY TO THB MBCHAHIC8. 

T I I 



t = 



A'-^h <'-^)' 



"\^ n 



I 

n 



= n ± 'V^(n(n - i). Ans. 



(lo.) The eqaation (4 bis), viz., 
becomes 300 = 4V + 16x4*, 

4 

(i I.) Let ^1, ^ denote the times required by the first 
and second stones respectively to reach the middle 
point of the tower, then eqaation (4 his) gives 

ijo= loo^i - 16 fi, (i) 

ijo = 100^2+ i6<*2. (2) 

From (i), fi = 2\ or 3^ 

From (2), ti = i^ or - 7V; 

.'. / « <i - <a = ^i - i| = H*. Ans. 

(12.) Let ^1, Ai denote the times in which the first 
stone falls through 50 and 400 ft, and t « t, - fi time 
in which second stone falls through 400 ft. 

.'. 50 = 16 fi .•. ti = i^2y 
400 = i6<»2 .'• Ai = 5, 
snd .•. / = /a -*i = S - 4V*> 

= f(4- V2); 



A KEY TO THE MECHANICS. 83 



henoe by (4 bis) we have 


400 = V<+ i6e. 


.17 400 - ifi*" 
..v= ^ y. 


400- 25 (18 -8v^2) 


f(4-V2) 


+ 4-V2 


= ^(iW* + 4), 


= V(i5x 1.4142 + 4), 


= 72.037 ft. Atu. 



(13.) Let t s inclination of plane, I = its length, and 
.'. / sin t s its height, t = time of falling through I sin t, 
and •'. nt = time of running down L 

,', fss g sin t - acceleration down inclined plane. 

28 

,: by Equation (4), viz., ^ = ^^ we have 

2 Z sm c 
<* = -^ for falling through height. 

n^^ = — : — ; for runninff down plane. 
^sm» '^ 

i.lsmi . 2I 

,; X nr = ; ; 

g gsim 

.'. sin' » = -ji or sin i = -. Ans. 



84 



A KEY TO THE HEGHANICS. 



(14.) Let /= acceleration down plane, i « time of d«* 
scribing each of the n parts ; 

a^M s sum of first m parts. 

'. By equation (4) we have 

but I = i/(fK)» .-. ife = 1 
and .'. a?^ - a?^.i = — (2w - i) = the m** part. 

The successive parts are now found by making m = i, 
2, 3, &a up to n, and are 



n" 



(15.) Here we have 

/ = Vt + igaini.f 



y g 



+ A sin f 



^ Z ~ A sin I _ ^ (/ - A) sin t) 

y g 



Ans, 



(16.) Yehcitj acquired in falling tbrougb h is 



A KEY TO THB MECHANICS. 85 

heDoe, substituting for V in the answer to the last, 
j- V, we have 

.'. f A = ^ - A sin t ; 
but A = / sin t^ and 

,*. ^ sin f = I - sin* •*, 

or, sin t = ~ — ^ «= -J or - 2, 
4 

- 2 must, of course, be rejected; 

.'. sin t = ^ = sin 30°. Ans, 

(17.) Plate III, Fig. 4. 

I shall first prove the following geometrical theo- 
rem: — 

If ABB"A" be a quadrilateral, and AC : CB - A"C" 
: CB"; AA': A'A"= BB': B'B", then the right lines 
A'B\ CC- are also divided in C, so that 

A'C':C'B'=AC:CB, 
CC : C'C"= AA': A'A'' ? 

Draw CE parallel to AA", CG parallel to BB", and 
A'D, A"E, B'F, B"G all parallel to AB. Also join 

EG meeting A"B" in a?, 
DF meeting CC" in y. 

Since A^E = AC and B"G = BC, and the A- A'icE, 
B"jpG are similar, A"a? : B"x = A"E : B^'G = AC : BC, 
but A"C" : B"C" is in same ratio, .*. x coincides with 
C", and EC": C"G = AC : CB. 

Again, CD = AA', DE = A'A", CP = BB^ FG « B'B", 
and AA' : A'A''= BB' : B'B", .-. CD : DE = CF : FG, 
and .'. DF is parallel to EG-, heii<i^ l)\j \\i^ r^^^* \^*^<^ 
= A^D.-B'F and the C A.'T)y wi^ ^'^^' ^^^ ^05:^^. 



86 A KEY TO THS MECHANICS. 

.-. L A'yD = B'yF, and as DyF is a right line, .-. A'yB' 
must also be a right line ; hence the point y must co- 
incide with C, and .-. A'C : C'B' = A'D : B'F = AC : CB, 
and CC : C'C = CD : DE = AA' : A' A", which was to 
be proved. 

The physical question now follows at once, for, let 
the vertical plane in which the two bodies move inter- 
sect the inclined planes in AA'^ BB'^ and let A and B 
be the initial positions of the bodies, and A' and B^ A'' 
and B'' any other simultaneous positions; C and C 
being the positions of the centres of gravity correspond- 
ing to AB, A"B". 

Now since the accelerations down the planes are uni- 
form, AA' : A' A" = BB' : B'B", Also, by a property 
of centre of gravity, AC : CB = A"C'' : C"B''. 

.-. by preceding theorem, A'C : C'B' = AC : CB, 
.'. C is another position of centre of gravity, which .*. 
moves with a uniform acceleration in the right line CC'^ 

N. B. — The above is only a proof of a particular case 
of the general question, viz., when the two bodies move 
in the same vertical plane along the two inclined planes; 
but the proof of the general question, viz., when the 
bodies move in any right lines whatever under the ac- 
tion of uniform accelerating forces, starting simvX- 
taneously from a state of rest, may be completed as 
>; follows : — 

If AA'^ BB'' be not in the same plane, then, since 
A''£, B''G are parallel to the same line AB, they are 
parallel to one another (by Euclid, B. xi. p. 9), and 
therefore EG and A''B" lie in the plane of A"E and 
B''G ( by Euclid, B. xi. p. 7), and consequently intersect 
.*. CC lies in the plane ECG, and .'. intersects DF. 

Hence, if any number of bodies start simultaneously 

from rest along right lines under the action of imiform 

accelerations, their common centre of gravity will de- 

scribe a right line, and will move with a uniform acce- 

Jeratj'on, 

This appears by taking two of t\ie "bo^e^ \Q^^'0[«!t 



A KBY TO THB MBCHAVIGS. 87 

then these two and a third body, and so on. The 
reader who is acquainted with analjrtic geometry will 
readily perceive how the question, in its most general 
state, may be solved by referring the motions to three 
rectangular axes, 

(i8.) Let d e diameter, and v = velocity acquired in 
falling down vertical diameter,, then 

^cos^ B chord; 

^ cos ^ =3 acceleration down chord; 

A* ^ ^ ^ 

- ]«= 2^ COS^.OCOS^ 

.'.zgcosO.d QOB0^-~T,2g.d 

.'. cos*^ - -T, and cos^ = -. Ana, 
fr n 



(19.) Let = inclination of required radius (r) to 
vertical diameter, then g cos = acceleration down ra- 
dius ; hence, substituting in the equation S = ^ff^ 
we have 

2r==^g.^ 

r = i^cos^.<* 

.*. 2 cos ^ = I, and cos ^ = -^ s cos 60^ 

.'. a 60°. Ans. 

(20.) Let s inclination of required diameter (zr) 
to vertical diameter, ti «= time of describing the oblique 
diameter, tz =» time of describing its first half, t^^ti 
- ^ s time of describing its last hslC ^ ixsoA ^i^astsxhr 
lag vertical diameter. 



88 A KIT TO THB KSCH&WICS. 



but <l - <2 = ^ 

... / ^ /-^^= i^ 

"ygcosO ^gcosO yg 

( 3V2zA\ 



= 1.5 - I.4I42I35 =.0857865 

= COS 85^ 4' 43". J. Ana, 
(21.) Let A s height, < « time of descent, 

/.^ - I = i V6, and ^ == J--— ^ 

= 7IV6 ° ^ "^ ^^^^ " ^ + 2.44948 

e 5 . 44948 seconds, 

and h = i6t*« 16 (15 + 6y/6) 

a 47 J. 1 5 feet 

(22.) Let (Va the height above A, which the body at 
tains; then ^t « time of describing x from rest, 

.'. putting V e o in equation (j bis). 



A KEY TO THB MECHANICS. 89 



we Imve 



o = V»- tgih + a?) = V»- e^H; 

also ^ *= i^ (i 0* ^y equation (4), 

.'. 0? ■= \gl? and H = A + a;r=A + |^, 

(23.) In time t first body falls tbroagh A, .*. < = / — ; 

S g 

Ith 
kence the first body is in motion for n + / — seconds. 

Equating these two values of S» we have 

/. V«- + ^/{tghy Ana. 
n 

(24.) The resolved part of Q down the plane is Q . -j ; 

.*. the statical force which produces the acceleration 
/in Q up the plane is 

P + Q 

Let a B distance required from bottom of plane, t; » 
velocity at thb point. Now Q would continue to move 



90 A KBY TO THE MECHANICS. 

uniformly with velocity t; when P is detached, were it 
not for the retarding force of gravity acting upon it, 

viz., g.j\ hence by (5 bis) we have for the motion of 

Q, after P is detached from it, 

o = V* - tg -J (I - x), («) 

Also equation (5) gives for the motion of Q through a;, 

p-qi 

v^ s= 2fx = 2gx . (/3) 

P + Q 

From (et) and (/3) we have 

X L^l.(l-x) 

P + Q I 

P + Q hi 

P h + 1 



(25.) Let x= time after projection of the first body 
when second must be projected, then by (4 bis) we have 

S = Vi« + ^g^ for ist body, 

S = V2(< - a?) + iflf (« - xy for 2nd body; 

equating these two values of S, we have 

^(t-xy+Y,{t-x)::^Y,t + ige 

., ^ -V2±i/(V2'+2yVi^ + ^<') 

• • I "~ •*/ — . 



A KRT TO THE MECHANICS. 91 



Page 116. 



(I.) Here, 1 = 30% r = 6o°, 

.•. by (31) we have 

tan %-e tan r, 

tan t tan 30° 
tan r tan 60° 



or € = 



= ^72=-. Ans. 
1/3 3 

(2.) Here; i « 20°, e = .24 

•'• ^y (3^) *a^ t = e tan r 

tan 20^ 

.% tan r = 

.24. 

.*. log tan r » log tan 20^ - log .24 

= 9.5610659 - T.38021 12 

= 10.1808547 = log tan 56° 36' 

.-. r = 56° 36'. 

Also, by (32) we bave 

t;» = V^ (sin' t + 6* cos' t) 

= io'{.34202oi'+ (.24)' X .9396926'} 

= io*{.ii67 + .0508} 

= lo* X .1675 nearly, 

.•. t; = 10 X .4093 nearly, 

= 4.093 miles per hour nearly. 

(3.) Here t = o, and .•. by (31) r = o ; hence (^2) 
becomes 

i;» = V*a, oit)=Ne \ ^^ 



92 A KBT TO THS MSCHAHIOS. 

but since the yelocities are uniform, and t}^e spaces the 
same, the times must be as v : Y, that is, as e : i, by 
equation (u). 



Page ii8. 



(I.) Here Jf=io, IT^s, e«i, V = 5, V'=3, 
.*. (35) becomes 

^ 10x54-5x3-1x5(5-3) 

10 + 5 

= 4 j^ miles per hour, 

^, iox5 + 5x3 + ix 10(5-3) 

10 + 5 

= 4^ miles per hour. 

(2.) Here e= i, •*. (35) becomes 

10x5 + 5x3-5(5-3) 

10 + 5 

s 3} miles per hour ; 

10x5 + 5x3 + 10(5-3) 

10 + 5 

B §i miles per hour. 

(3.) Here M = 200, M' = 50, V = 20, V = o, e = |, 

••• (3 Ogives 

' MV + eMV (i + I) 200 X 50 

M + M' ~ 200 + 50 

^x40o _6^^^^_96 

25 " 5 ^ 

= 19.2 miles per Tioui. 



A KEY TO THE MECHANICS. 93 

Page 121. 

(i.) In falling through 100 ft the ball acquires a 
Velocity, 

V = ^(2 X 32 X 100) = 80 ft. by equation (9), 
.*. t; = eV = 8oe by equation (30) ; 

hence putting r = o and V = 80 e in (5 bis) we have 

o = (80 «)' - 2 X 34 X 70 

.-, e* = .7 

and e B \/.7 = .836. Ans. 

(2.) If V s velocity acquired in falling through 100 
ft. (^ 80 ft. by last question), the velocity after fourth 
hop is 

t) = e*V = (^)* X 80 s= J ft. by equation (30); 

hence putting v = o and V = 5 in (5 bis), we have 

o = 25 - 2 X 32 S .•. S = JJ 
«= .390625 ft. Ans. 

(3.) In equation (35) make the following substi- 
tutions : 

M = 200, M' = 50, V = 20, V a - 40, e = J, 

_ 200 X 20 + 50 X 40 - J X 50 (20 + 40) 

200 + 50 
= 5 miles, 

, 2bo X 20 + 50 X 40 + j^ X 200 (20 + 40) 
^ 200 + 50 



94 A KBY TO THE MBOHANICS. 

(4.) By the question we have 

v = V sin 45° .'. by (32) 

V» sin» 45° = V (sin* t + tan» 30** cos" t) 

.*. ^ = sin' t + J cos* t = sin* « + J - i sin* i 

.'. sin' t = J and sin t = ^ 

.-.1 = 30°; 

also, by (31), 

tant tan 30® 
tan r = ^ ^ = i 



e tan 30® 



•. r = 4S°. 



(5.) Let a, a' represent the velocity of A before and 
after impact ; 6, 1/ of B after the first and second impact *, 
c, d of C, and d velocity of D after impact. 

Hence, putting 6 = i in (35), and making the other 
necessary substitutions, remembering that B, C, D are 
at rest, at the instant of their first impact, we have 



(A-B)a 



ft = 



y = 



e = 



A + B 



W 



A + B 

(B-C)6 B-C lAa 



B + C B + C A+B 
tBb 2B 2Aa 



(^) 



&^ 



B+C B+C A+B 

(C-D)c C-D 2B 2A0 
C + lT " C + D * B + C * A + B .V 



"C + D 



A KEY TO THE MECHAKIC6. 95 

Also by the question 

Aa' = m' = Cc' = Drf = iAa. («) 

Mnltipljing the values of a\ b\ and c' in (a), (fi), (7) 
by A, B, G respectively, striking out the common fac- 
tor Aa, and equating the results each to ^ by equation 
(£), we have the three following equations : 



(0 







A-B 

A + B 


=i 










B-C 
B + C 


2B 

A + B 


= i 








C-D 
C + D 


23 

B + C 


2C 
"A+B 


= i 


From 


(I) 


we obtain 


B = 


JA. 





(«) 



(3) 



Substituting this value of B in (2), we obtain, after 
some easy reductions. 

Lastly, substituting for B and C in terms of A in (3), 
we have 

3AJJOD .T)_ I A. 

hence the masses of the four bodies are as 

A. 3 A • ^ A • 1 A 

that is, as, 

10 : 6 : 3 ; 1. Aus. 



96 A KEY TO THE MECHANICS. 

(6.) In equation (39) put V = o, M = M', and e = i, 
then 

tan r = 00 and tan K = o, 

.*. r = 90° and r' = o or M' moves in the line of impact^ 
and M at right z.* to it; and this result being inde- 
pendent of t, shows that the L between the paths of two 
such balls after the shock is always 90°. 

(7.) Here t = 90° - 30° = 60% and i' = 90° + 30** =120, 
.*. by (39) we have 

tan (M -f 2M) 2V sin 60^ 

(M-f x2M)2Vcos6o°+2M (l +|^)VC08I20° 

(M + 2M)Vsini2o° 

fork f* ;— . ^ ' 

(2M - §M) V cos 120^ + M (I + I) 2V cos 60° 

.•• tan r = - 1^ tan 60°, and 

tan r' « I tan 60** = tan 68° 56' J4'^ 

.-. r + r'= 180°; 

and hence the direction of the motion of each sphere 
after impact is equally inclined to the common tangent 
at the point of impact. This inclination 

= 90° - r' = 90° - 68° 56' 54'' = 2 io.3'.6'. 

By the first pair of formulas in (38) we have, since 
sin t = sin t' and sin r = sin r', 

, A,-- -rr sin I _, sin 60° 

vel. of M = 2V . -. — ? = 2 V X 



sinr' sin 68° 56' 54" 



^ ^m -^ sint __ 
vel. of 2M1 » V .-: — , = 2V 

smr 

/, 1; s= 2V X .91^ 

2/= VX.92E 



A KEY TO THE MSOHANIGS. 97 

(8.) In (3 J) putting « = i, and V « - V, we have 

(M-MOV-.M^(V + V) M-^3M^ 
^"^ M + M' "*M + M' 

(M-MOV + M(V + V) 3M-M^ 

^" M+TT " M + M' * 

If M remain at rest, then v must = o, /. M = 3M^ 
but if M' remain at rest, t/ must = o, and .*. M' = 3M. 



Page 127. 



(I.) R » — sm 26 = ' sm 72 

9 32 

= 8450 X .9510565 = 8036.427425 ft. 
2V sin « 520 . o 

= 32.5 X .5877853 = 19.10302225 seconds. 

V* C20^ 

(2.) Maximum range = 2A = — = - — « 8450 it. 

And time of flight for this range 

= i 1/(8450) = i X 91.92 =x 22.98 seconds. 



(3.) Here R = J mile = 2640 ft. 

s 2h sin 26 = 4A sin e cos «, 
and greatest height = 120 ft. =« A sin» 6, 

A sin* 6 \^o 



. • 



4A sin e COS e %^^ 



98 A KEY TO THB MECHANICS. 

.'. tan e» iV ftQ<l •*• log tan e 

«= lo + log 2 - log 1 1 = 9.2596373 

/. 6 = 10° 18'. 

Again, R s — sin 2« 

9 

/. V^ = -T- — = 32 X 2640 cosec 20® 36' 

8in26 •' -r 

.*. 2 log V = log 32 + log 2640 + log cosec 20° 36 
- 10 = 5.3804067 
.-. log V = 2.6902033, 
and .'. V = 490 ft. 



Page 131. 

(i.) Here e = 45°, t = 15°, and V = 500 ft 

, r «v r« *V sin (e - O 
.•.by(48). T = -.-4-^ 

16 cos 15° 32 
.'. log T = log 500 - log 32 + log sec 15° - 10 
= 1.2088762 .'. T = 16.176 seconds. 

COS£ 

Also, R = V . T ; = coo T cos 45** sec 15° 

cost -^ 

.'. log R = log 500 + 1.2088762 

+ Jog COS 45® + log sec 1 5° - 20 « 3.7723874 

.-. E = 5920.9 ft, = 1.121 mWft, 



A KBT TO THE MECHANICS. 



ti 



(2.) € = 42^ t = 12° R = 2^ miles = 13200 ft. 

cos e sin (e + t ) 



.-. by (52), E = 4A 



cos' I 



(3-) 



2 V' COS e sis (0 + 1) 
^ cos' t 

.*. V* *= ^gB, cos't sec e cosec (e + t) 

= 16 X 13200 cos' 12® sec 42° cosec 54^ 

= 21 1200 cos* 12° sec 42® cosec 54° 

.'. 2 log V = log 211 2cx> + 2 log cos 1 2^ 

+ log sec 42° + log cosec 54° - 30 

= S.J264716 .-. log V = 2.76323J8 

.% V = 579.74 ft. Ans. 

1= to**, R = 5280 ft. 
T = 20 seconds. 



By (52) we have 

^ 2V' cos e sin (e + 1) 



andby (sO, V = 



COS't 

grcosi.T 



2sin(6 + t) 



2 ^ cos* t T' cos e sin (e + «) 



^ 4 sin* {e + t) 

yT* cos e 



cos*t 



2 sin {e + 1) 
sin (e + t) ^T* 



cose 



2B 



.'. tan« cost = — ~r - «wi\ 

2 tt. 



u 



o 



100 A KBT TO THB MEORARIOS. 

^16x20° . o 40 
.'. tan e cos lo** = r sin 10° - ^J— - sm 10 

S*8o 33 

= 1.038473 
.-. log tan e =» log 1.038473 + log sec 10** 
= 10.0230437 
.'. e«46o 31'. Ans, 



(4) Here 1 = 15^ Vdoooft, 
and for maximum range, 
26 + » = 90*^ .•. 26 = 90° - 15®= 75*^ and a « 37** 30' 

, , . ~ 2V* cose sin (6 + 
*^ ff cos't 

^ 1000^ COS 37° 30' sin 52° 30' 
~ 16 cos' 15° 

.'. log R =6 - log 16 + log cos 37*^ 30' + log sin 52° 30' 
+ 2 log sec 15° - 40 = 4.6249258 
.'. R = 42162.4 ft. i= 7.98 miles, 
T = J '^(maximum range in feet) 
= J \/ (42 1 62,4) = 51.34 seconds^ 

(5.) Since the plane is descending, we must change 
the sign ofy in (44), 

.-. t?» = 25r(A +y) =3 V* + 2gy, 

but^ = R sin 1^42 162.4 sin 15^ » 10912 ft. 

.-, v^ = 1000' + 64 X 10912 = 1698368 

. •. If = ^'(i 698368) SB 1303 it. Ans. 



A KBT TO THB MBCHANICB. 101 



Page 134. 
(..)V=i6ooJi?=..6ooJ^ 

X 1.7325 



96 

2400 v^ 3 2400 



7 7 

= 594 ft iltM. 

(z.) V = .6ooJf =,6ooJ| 

= 400^/2 = 400 X 1.4142 = 5;64.68 ft ^n«. 

(3.) V = .6ooJ|^ = i6ooJi! 

= 1600 X } = 1200 ft. ilfW. 



Pages 134, 13$. 

(i.) By Example (i) preceding, we have 

V = 594 ft 

.-. B «— * ^^ «= 11026 ft = 3^75 J yds. ; 

also, T = iV(Rft-) = i\/(iio26) = 26J sec. 

(2.) Here P is the quantity sought ; 

W = 196 Ibs^ e =5 35^ t = 10° 40', and R = 4000 ft. 

^ X X T> 2V* cos e sin (e-%) 
^ ^ g cos*t 

2 cose 8m^e- i) 



102 A KXT TO THE SEdHAiriC9« 

substituting this ralue in (55), we have 

yR c os' t 3P 

;— 7 3r = 1000' X ^ 



• P = 



^R W cob' t 



6 cos e sin (« - t) x 1600' 

^ 32 X 4000 X 196 cos' 10^40' 
"" 6 cos 35° sin 24° 20' X 1600' 

.'. P = — cos' 10*^40' sec 35*^ cosec 24° 20' 

.-. log P = log 4«9 - log 30 + 2 log cos 10° 40^ 

+ log sec 35° + log cosec 24® 20' = .6696244 

A P = 4.6733 lbs. Ans. 



(3-) y = ar tan e - 



4A cos' 6 

= a? tan £ — r sec' e «= a? tan t r ( i + tan* c) 

4^ 4* 

.% aj* tan' t - 4^ tan « =» - a;* - 4% 

. . tan t = — 

a; 

where a? = f mile = 3960 ft. 
ya2joft. 

Also, V= 1600^1^ = i6oo^A 

^ , V* V ox 1600^ .^ 

and h^ — = jr-^^ — ^iia^oiXK 

zg tij^ 64 X 32 



A KEY TO THB MBCHANIC8. 103 

' 39S0 

_ 2250 ± V(^5^ ^ iiooo - 396*) 

^ 22501^(4793184) ^ 225012189.3 3 
396 " 396 

^ 4439-33 Qj. 6Q'^? 
396 396 

.-. log tan e = 1 1.0496225 or 9.1852788 

.-. « = 84*^ 54' or 8° 42'. Ans. 

(4.) Here t=3o% V»=45oft., 6=0, 

...by(5,),T=^.^i5ii±i>=4ytan30<' 
•^ ^ a cost 16 ^ 



8 



16.2 seconds, 



and R = V. T . r. (See p. 1 28.) 

cost "^ ^ ^ 

. R - 450 T 4SQX75V3 450 X 75 
COS 30° 4^3 4 

= 8437^ ft « 28 1 2^ yds. 

(5.) The ordinates of the summit of the wall on the 
edge of the cliff are — 

p = 500 ft. ^ = 260 ft. 

and the co-ordinates of the point where the ball falls 
are — 

• p' =» 620 ft. q' » T.^iO i\u 



104 A KEY TO THB MECHANICS. 

.*. log tan 6 = 10.1379404 
and by (54), V^ ^2gh^ gp'eec^e 



2 (p tan e- q) 
16 X 500*860*53^19^ _ 
~ 500 tan 53° 19' -260 "" ^'*^' 

.*. V = 165 ft. 

(6.) By (s;) we have 

V= 1600 /—= 200 i/6; 

henoe, substituting 2Y for t; in (44), we have 

_ 3V* _ 18 X 200» 

^"Tg"' 64 
= 1 1 250 ft. = 2. 1 3 miles. Ans. 

(7.) Here R - 50yds. + 6ft. = 156 ft^ 

.*. RB4A sin6 cose; 

that is, 39 =^ ^ ^^^ ^^ ^* (^) 

Also, to IP 150 ft. 

y = 14 ft. 
must satisfy the equation, 

y = a: tan e - 



•. lJ^.^ I50tan6- 



4Aco8^e 
15P* 



'. /4^ co8*€ = 1 50A sine cose - S^^S* ^^^ 



A KEY TO THE MECHANICS. 105 

By substituting from (a) in (/3), we have 

i^Acos^tf = 150 X 39 - 5625 = 225 

14^008^ e 225 

_ . , lAx 13* ^ ^ 

.•. h sin'fl = -^ ^ = 94.64ft 

which bj equation (47) is the greatest height. 

(8.) Since the range of the projectiles on the hill is 
the same, we have, by (52), 

_ cos «! sin (ci + t) cos e^ sin (e^ -f t) 

K = 4^1 — = 4/ia Tz 

^ cos*t ^ cos'« 

.*. hi cos ei (sin Ci + cos ei tan i) 

= Aa cos ej (sin €2 + cos 62 tan t) 

. , Ai sin 2^1-^2 sin 2e2 . 

.'. tan t = *. 7 ; 1 1 — • -o-ns. 

* A2 cos' e^ - ^1 cos* ei 



rr < 
I 

I ) 



I' 



I 



i 



