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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 :: 2 53'': 29.62182 miles, the distance that the body would fall west of Farnley. will

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, A TI and that A BC is the triangle required;

Z DE 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- AB B ments of the diameter # D. EF = the square of the diff. of the sides, and DE=CB the perpendicular; therefore, by the ques. tion, 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 = 4FH 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. Otherwise by Messrs. Butterworth, Whitley, and Winward. ** ANALYSIS. Suppose the things

to done, and that ACB is the tri-4035 N angle required, and draw the lines as in the figure; then, the · angle DEC being = half the given diff. of the angles at the AK

A LV base, and KL=IC being also

FLIP given, the right-angled triangle

E SO DEC is given, and ID is given t

h e basi Again; (AC-CBX CL (IK) OPEL or (AC-CB) X IK? is to be a max.; but it is well known, that (AC-CB) = 4ID. KE, and 4 ID is given, therefore, I K’X KE must be a max.; which will be the case when 2 KE=IK; 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 CLX (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 EKSEI, draw KB 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 case = x; then, in case of an equilibrium, we shall

TWO have the arm AB= arm BD; also the line of direction CD of the hangingcone, will be parallel to AG, and perp. to AF; hence, AF = (x + 144): EF=12::CF=9:DF=. ♡ (x2 + 144) — 2x, and multiplying means and es. tremes, we have, by reduction, x + 168 x — 432, and x = 1.591612.

2. For an equilibrium with the standing cone, the susa pended globe, and the hanging frustum. Put the length of the part to be cut off = 42, then the radius of the less end of the frustum will be , and by art. 656Marrat's Mechanics, FO = 9+ 32 AF = 12.105091 : EF = 12::FO:Fn = 8.921866 + "2.373955 73 9 + 32 tri; consequently the arm Bn'= 1.591612 2.973955 zł

as which may be supposed to be acted upon 9+3% + z2wm miay de supposed to 43" by the wt. of the frustum, which is found to be = • 27 - 23

(put this =F). Again, EF: AF::48: Fr=

T9 +32 +92: Also

.848

4.03503 %; and rad. : Fr::cos. SFr=.9654245: FS= 3.895517 %; consequently the arm BS= 10.513479 3.895517 %, which may be supposed to be acted upon by the wt. of the globe, or 16.888995, which call G, Finally, put the solidity of the standing cone, which is 31.83356587 = C; then, by the property of the lever,

F XBn+GX BS=C x GP; that is, in symbols, .' —.5352 8 — 1.60555 24 — 50.58 23 – 5.65026 %?

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 heing the centre of gravity of the hanging cone. Draw Gn perp. to CP. B O perp. to A B, and produce É e to meet CP inm; then, when the standing coue is just able to sustain the other, the weight of both will act at B, or in the line O B. But A P B 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 Gn=2 PB. Put a =CP = Ee = 12, x = Ce=PB, then CB=va? + x, and by sim. tri. CP:BP::Ce:em=, and CB:CP :: Gm= + Gn=** which must equal 2x,

va’tra and we get x = (12 v 52 —84) = 1.5916, hence AB=CD=2= 3.1832 = the base of each, Gm= 3.2111, and Gn: Gm :: Go = 4: Gr = 7.60555, then Er = 3a + Gr = 10.60555. Let ab E be the part cut off, c its centre of gravity, then the globė must be suspend at b, draw bs | BO, meeting the axis of the suspended cone, in s. Pui y = EP,p .7854,6 Er.

= 10.60555 and d = 3,1832 the diameter of the globe S; then CP: BP::Ep:bprint and Ep: bp::be: ps = hence rs= Er – Ep+Ps= 6-y+ open and rc = 6- *, also the weight of the globe is 2 dp

. dyip ase

Port to , and that of the cone a b E is as P. Butto produce an equilibrium, when the cone a bE is taken away, we must have 242::636–4 +

en therefore 2 d. 2 x (0-y+) (6-39) from which by red. we get 34 — 14.1407 ** - 1200.812 x = - 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 à sec. A tion of the two equal cones through A their axes; GS their centres of # gravity. By Mechanics, the cone ! A B C will just sustain A EF when CO perp. to BC passes through O 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. . V maj DC = CH, and AC=CK. Put AD=EI=a=12, DC = AI=r; then AC = CK =vu + ; and by sim, tri. A DC, SIK, DC= T: AD =*::IS = *- (by art. 656 Marrat's Mechas nics) : =IK: hence A K = ctr=2vat try; and r= 1.5916. Secondly: through g, the centre of gravity of the frustum AFQ R draw mu parallel to BH

CT

2 22 S

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meeting Q Winm, C O'in n, and AC inú. Put AC =
b, = 12.105, CI = d = 10.5134, n = 7854, 55
47 a ro
* = the solidity of the cone ABC,

- 16 na pot = the solidity of the globe, Qa'= x, and g I=y; then the solidity of @ Ř E = 4973, and that of A FQR= s--4401? By sim. triang. a:r::y;"I = Fuza:

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+-; then b:a::gtigmsm

_air_ary-axtrais

a gens By the property of the centre of gravity, g w = gm x W

(aér-ary-a?.x te pax) x W W + frust. A FQR =

(W+S-400x') xbr

by dr _"}=(947

ya

);

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e

on r3

but gn = gu-un
hence nw=CN- (at r — ary --u?r trí) W

(W 48-49) - 6 + . By Mechanics, DC. X cone ABC = CN * (globe W + frust. A FQR); that is g.x.is

p 3 r.

2

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a? x + goz x) x

W br

(w +s 497*). Multiplying each side of this equation-by all, and restoring -- and .W, we get, by due reduction, y + *- (-) ( -4)=-- But, it is evidena,

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