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Lastly, as rad. (1) is to the nat. co-sine of the lat. 53° 47′, so is 25038.552 (the circumf. of the earth) to 14793.788079 miles, the space through which the town of Farnley, is carried in 24 hours, by the earth's diurnal rotation on its axis; and, as 24h. : 14793.788079 :: 253": 29.62182 miles, the distance that the body would fall west of Farnley."

This question was answered also by Messrs. Baines, Brooke, Cattrall, Gawthorp, Hine, Hirst, Maffett, Putscy, Rylando, Smith, Whitley, and Winward.

A

E

16. Qu. (71) Answered by Rylando. ANALYSIS. Suppose the thing done, and that ABC is the triangle required; then because the diff. of the segments of the base, and the diff. of the angles at the base are given, the diam. HF is. given. But the rectangle of the seg- A ments of the diameter HD. EF the square of the diff. of the sides, and DECB the perpendicular; therefore, by the question, HDX DE x EF is to be a max.; or since DH is constant DE' x EF must be a max. which will be the case when DE=2EF. If, therefore, we take EF and DHFH and through the points E, D, we draw AB and DC perp. to FH, and lastly join A, C, B, C, ABC will be the triangle required.

F

Otherwise by Messrs. Butterworth, Whitley, and Winward. ANALYSIS. Suppose the thing

done, and that ACB is the triangle required, and draw the lines as in the figure; then, the angle DEC being half the given diff. of the angles at the A base, and KLIC being also given, the right-angled triangle DEC is given, and ID is given. Again; (AC-CB) × CL (IK)

Α

or (AC—CB)2 x IK is to be a max.; but it is well known, that (AC-CB) 4 ID. KE, and 4 ID is given; therefore, IK' x K E must be a max.; which will be the case when 2 KEIK; therefore, K E and IK are given, and the construction is obvious.

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Again, by Mr. E. S. Eyres and Mr. Gawthorp. Since CI (see last fig.) and the angle CEI are given, EI and EC become known; but CL × (AC—CB) is to be a max.; or CL (IK) x (AC-CB)'IK' × 4 ID x EK is to be a inax. ; in which case EK= KI; whence this construction. Make IC diff. segments of the base, and having constructed the triangle ICE, set off EK EI, draw K B perp. to DE, meeting the circle passing through DCE in A and B, join AC, BC, and ABC is the triangle required.

17. Qu. (72) Answered by Mr. W. Putsey, Pickering Academy.

1. For the equilibrium of the

two cones.

Put the radius of

each cone's casex; then, in

case of an equilibrium, we shall have the arm A B

arm BD; also the line of direction CD of the hanging cone, will be parallel to AG, and perp. to AF; hence,

AF√(x2+144):

F

Bn D SF

هلان

EF 12:: CF9: DF =
√(x2 + 144) — 2x, and multiplying means and ex-
tremes, we have, by reduction, x+168 x 432, and
1.591612.

2. For an equilibrium with the standing cone, the suspended globe, and the hanging frustum. Put the length of the part to be cut off 4%, then the radius of the less end of the frustum will be, and by art. 656

323

Marrat's Mechanics, FO=9+9+3% + z = "

Also

9+3%+x2' AF 12.105091: EF 12 :: FO: Fn = 8.921866 + 2.973955 23

9+32 +z; consequently the arm Bn 1.591612+22

2.973955 z 3

9+3+ which may be supposed to be acted upon

by the wt. of the frustum, which is found to be = (put this F). Again, EF: AF:: 42: Fr=

27

.848161

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4.03503 z ; and rad.: Fr::cos. SFr= .9654245: FS= 3.895517%; consequently the arm BS 10.5134793.895517 z, which may be supposed to be acted upon by the wt. of the globe, or 16.888895, which call G, Finally, put the solidity of the standing cone, which is 31.83356587 C; then, by the property of the lever, Fx Bn +GX BSCX GP; that is, in symbols, .5352 85 1.60555 24 50.58 23 5.65026%2 455.7580; and from this equation 4% 8.1144, as required.

16.9508 % is found

-

Again, by Mr. Gawthorp.

Let the annexed figure represent the cones, G being the centre of gravity of the hanging cone. Draw Gn perp. to CP. BOperp. to A B, and produce Ee to meet CP in m; then, when the standing cone is just able to sustain the other, the weight of both will act

at B, or in the line OB. But A P B

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d

the weight of the cone ABC acts in the line CP, or at n, and that of CDE acts at G, therefore (the cones being equal) OG On or G n=2PB. Put a = :CP

Ee 12, x Ce PB, then CB

and by sim. tri. CP: BP:: Ce: em

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a2 + x2,

and CB: CP

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and we get = √ (12 √ 52 -84)= 1.5916, hence AB=CD=2x = 3.1832 = the base of each, Gm=

G'n

3.2111, and Gn: Gm :: Go=

2

Gr= 1.60555,

then Er= + Gr 10.60555. Let ab E be the

3 a
4

part cut off, c its centre of gravity, then the globe must be suspend at b, draw bs || BO, meeting the axis of the suspended cone, in s. Put y Ep, p.7854, b

Er

10,60555 and d= 3,1832 the diameter of the globe

S, then CP: BP:: Ep: bp=

dy

and Ep: bp::b,

2 a

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d2y

3

-y+

d and rcb-, also the weight of the globe is

4a2

as

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2 d3p, and that of the cone ab E is as. Butto

3

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produce an equilibrium, when the cone a bE is taken 2d3p d'y3p

:

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therefore 2px (b−y+224)

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diy 4 a2

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3

=

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3a2

X

from which by red. we get x1- 14.1407 x2

1200.812 = 12963.6813 and x

8.114 feet,

the height of the cone ab E, which must be cut off.

Another Answer by Mr. J. Whitley.

Let ABC, A E F represent à section of the two equal cones through their axes; G S their centres of gravity. By Mechanics, the cone ABC will just sustain A EF when CO perp. to BC passes through O their common centre of gravity; BDCHN hence since the cones are equal OG OS. Draw SK parallel to AD meeting BC, AC produced in H K. Then since OGOS, by sim. tri. DC CH, and ACCK. Put

=

2

K

R

W

AD EI=a = 12, DC = AI=r; then AC CK ar; and by sim. tri. A DC, SIK, DC= T: AD = a :: I S = (by art. 656 Marrat's Mechas

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=IK: hence AK==+ r = 2 √ √ @ + r2;

4 r

+r=2√a2+r2;

and r = 1.5916. Secondly through g, the centre of gravity of the frustum AFQ R draw mu parallel to BH

meeting QW in m, CO in n, and AC in u.
b, = 12.105, CI=/d 10.5134, n =
the solidity of the cone ABC, W

4 na r2

3

Put A C 7854, 8 16 mp3

3P

the solidity of the globe, Qax, and gIy; then

the solidity of Q R E =

3 r

4 an x3
3 r

, and that of AFQR =

4 anx3

S

By sim. triang. a:r::y: = Fu;

ry

a

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a

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ax

'b

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b::y: =gu; br :: Cu=a+ Un= +

a

TX
a

ja: r: : x: at; r:a::x:a E.. ga =

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By the property of the centre of gravity, 'g w

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* (globe W + frust. A F QR); that is rx s =

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by due reduction, y + — — 1 x → (y — d2)

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