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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 :: 2153: 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.

16. Qu. (71) Answered by Rylando.
ANALYSIS. Suppose the thing done,
and that A BC is the triangle required;
then because the diff. of the segments


C 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


B ments of the diameter HD. EF = the square of the diff. of the sides, d DE=CB the perpendicular; therefore, by the question, HDX DE x E F is to be a max.; or since DH is constant DEX E F must be a max. which will be the case when DE=2 EF. If, therefore, we take EF and DH=FH and through the points E, D, we draw AB and DC FH, and lastly join A, C, B, C, ABC will be the triangle required. Otherwise by Messrs. Butterworth, Whitley, and Winrard. ** ANALYSIS. Suppose the thing done, and that ACB is the tri angle required, and draw the lines as in the figure; then, the angle DEC being = half the given diff. of the angles at the base, and KL=1C being also given, the right-angled triangle DEC is given, and ID is given. De bas Again; (AC-CB) X CL (IK)

Jo8 or (AC-CB)? ~ IK’ is to be a max.; but it is well known, that (AC-CB) = 4I D. KE, and 4 ID is given; therefore, I Ko X K E must be a max. which will be the case when 2 KECIK; 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 Č L X (AC-CB) is to be a max. ; or. CL’ (IK”) (AC-CB)=IK X 4IDX 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 I CE, 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.

two cones.

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

Academy. 1. For the equilibrium of the

Bn D SF Put the radius of each cone's case = x; then, in

po 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 paraltel to AG, and perp. to AF; hence, AF = (x + 144): EF = 12::CF9: DF= 7. (x + 144) - 2r, and multiplying means and extremes, we have, by reduction, x4 + 168 x = 432, and X = 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. 6562

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373 Marrat's Mechanics, FO = 9+


9 + 3% + %2' AF = 12.105091 : EF = 12::FO:Fn = 8.921866 + 2.973955 23 9+ 32 tri; consequently the arm Bn'= 1.591612 — 2.973955 zi 9 +3% + za

which may be supposed to be acted upon the wt. of the frustum, which is found to be = 27-23

(put this =F). Again, EF: AF::45: Fr: .848161


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4.03503 %; and rad. : Fr::cos. SFr=.9654245 : FS 3.895517 %; consequently the arm BS = 10.5134793.895517 %, 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 BS=C X GP; that is, in symbols, -.5352% -1.60555 24 50.58 23

-5.65026 zo 16.9508 % +. 455.758 = 0; and from this equation 4% is found = 8.1144, as required.





Again, by Mr. Gawthorp. Let the annexed figure represent the cones, G being the centre of gravity of the hanging cone. Draw Gn pèrp. to CP. B O perp. to A B, and produce Ee to meet CP in m; then, when the standing colie is just able to sustain the other, the weight of both will act at B, or in the line O B. But A P the weight of the cone A B C acts in the line CP, or at n, and that of CDE acts at G, therefore (the cones being equal) OG= Onor G n = 2 PB. Put a = ECP = Ee = 12, x = Ce=PB, then CB = Na + x?,

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

:: Gm

: Gn=

wbich must equal 2.x, and we get x = V (12 v 52 - 84) = 1.5916, hence AB=CD = 2 * = 3.1832 = the base of each, Gm= 3.2111, and Gn: Gm :: Go= : Gr= 1.60555,


. 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 =Epp.7854,6 Er


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= 10.60555 and d = 3,1832 the diameter of the globe S, then CP:BP::Ep:bp =

dy and Ep: 6p::b


d? y

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3 a?

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3 y


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hence rs = Er - Ep+ Ps= bayt and rc=b

3.4, also the weight of the globe is ma?

4 as 2 d'P, and that of the cone ab Eis as ?" yop.

Butto 3

3 a? produce an equilibrium, when the cone a bE is taken

2 dp. d’yP 3 у away, we must have

:b-y + dy

d? therefore

y 2d % x

(6 4 a?'

4 a? 3 ao from which by red. we get 24. 14.1 407 maio 1 200.812 y = 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 EF represent å section of the two equal.cones through their axes; GS their centres of gravity. By Mechanics, the cone Å.B C will just sustain A E F when CO perp. to BC passes through 0 their common centre of gravity ; B D CHN hence since the cones are equal OG - OS. Draw SK parallel to AD meeting BC, AC produced in H K. Then since OG : OS, by sim. tri. DC=CH, and ACCK. Put AD= EI=2=12, DCSAI=r; then AC=CK = V2 +r; and by sim. tri. A DC, SIK, DC= r: ADa::IS = hep (by art. 656 Marrat's Mecha nics) :

=IK: hence A K = +r=2 va tr; and r = 1.5916. Secondly: through g, the centre of gravity of the frustum AF Q R draw mu parallel to BH

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meeting QW in m, co inn, and AC in 4. Put AC = b, = 12.105, CI =d-3 10.5134, n = 7854, 84 4ñ á pa

16 n god = the solidity of the cone ABC, W = = the solidity of the globe, Qa' = x, and g I=y; then

4 an x3 the solidity of QREN and that of AFQR=

4 anr SBy sim. triang. a :r::y:"= fu;a:

pay =gu; b:r::Cu=at


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=at; r:a::*:=aE..g a ab IE-gl-aE='Q-Y gt=ga tatay

a'r-ary-a'r tris -4.+ **; then b:a::gtigms


br By the property of the centre of gravity, gw =

(adrary-a?x + pax) x W W + frust. AFQR

4 anx
(w+s :) x br.

dr but gn=gu-un= big (2-2");

To ab

(aér-ary ax + poš) x W hence nw - CN


4 an x3

*** br

3 r. + -. By Mechanics, DC X cone ABC = CN * (globe W + frust. A FQR); that is r.*s= (aq a'ry - a? x + god x) x W


6 (w +s_4ans). Multiplying each side of this equation by and restoring s and W, we get, by due reduction, y + ' , ++.

But it is evident,

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