Lastly, as rad. (1) is to the nat. co-sine of the lat. 53° 47', so is 25038.559 to the te 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 rotation on its axis; and, as 24h.: 14793.788079 :: 53" 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, Putsey, Rylando, Smith, Whitley, and Winward. 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 DE2EF. 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. A A D E B, 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 base, and KLIC being also given, the right-angled triangle DEC is given, and ID is given. Again; (AC-CB) × CL (IK) 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, 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. Again, by Mr. E. S. Eyres and Mr. Gawthorp. Since CI (see last fig.) and the angle CEI are given, EI and FC become known; but CL x (AC-CB) is to be a max.; or CL2 (IK2) × (AC —С B)2 = I 'K' x 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. 191 Bn D SF 1. For the equilibrium of the two cones. Put the radius of each cone's case r; 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) : EF 12:: CF9:DF = √(x2+144) 2x, and multiplying means and extremes, we have, by reduction, † 168 x2 = 432, and x 1.591612. Marrat's Mechanics, FO = 9+ AF 12.105091: EF 2.973955 z3 9+3x+x2 2.973955 z' G 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. -XZ 323 ; consequently the arm Bn A of Also 8.921866 + 1.591612— which may be supposed to be acted upon 9+3% +22 by the wt. of the frustum, which is found to be = 27-23. .848161 (put this F). Again, EF: AF:: 42: Fr= 4.03503 z ; and rad. : Fr:: cos. S Fr= .9654245: FS= 3.895517%; consequently the arm BS 10.513479– 3.895517 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 B n + G x BSC x GP; that is, in symbols, 26. -.5352 85 - 1.60555 z+4 - 5.65026 z2 50.58 z3 16.9508455.758 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 perp. to CP. BO perp. to AB, 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 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 = 2 PB. Put a = = CP Ee 12, x = Ce PB, then CB = √√a2 + x2, and by sim. tri. CP: BP:: Ce:em= and CB: CP a+which must equal 2x, *it apples > :: Gm= √a2+x2 and we get r√(12√5284) 1.5916, hence ABCD 23.1832 — the base of each, Gm 3.2111, and Gn: Gm:: Go = G'n 2 a x2 4. + : G n = a d PV OG : Gr=1.60555, 3 a Gr 10.60555. Let ab E be the then Er 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 dy S, then CP: BP::Ep: bp= and Ep: bp::b 2 a dzy ps = hence rs Er - Ep+ Ps 4a2 day and rcb- 3y, also the weight of the globe is 2 as d'y3 P. But to 2 2d, and that of the cone a b E is as 3a2 3 3y produce an equilibrium, when the cone a bE is taken 2d3p d'y3p away, we must have :b−y+ 3 3 a2 4. dy d therefore 2px (b−y+dy) 4 a2 4 a" 3 (b_34) ::b a P: b−y + } da P 3 a1 14.1407 x from which by red. we get r1. 1200.812=- 12963.6813 and x the height of the cone ab E, which must be cut off. 8.114 feet, = G Another Answer by Mr. J. Whitley. Let ABC, A E F represent a 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 O G OS. Draw SK parallel to A D meeting BC, AC produced in H K. Then since OGOS, by sim. tri. DC=CH, and AC = CK. Put AD EI a 12, DC AI=r; then AC CK sim. tri. A DC, SIK, DC= r: AD = a :: IS = (by art. 656 Marrat's Mecha a; and by a2 R uns a W nics): =IK: hence AK = ++r=2√@+r2; t r 4 and r = 1.5916. Secondly through g, the centre of gravity of the frustum AFQ R draw mu parallel to BH ** meeting Q W in m, CO in n, and AC in u. 驚 3 "} the solidity of the globe, Qax, and g I=y; the solidity of QRE= 4an x3 S- By sim, triang. a:r: 3 r by b::y: = gu; br:: Cu = a+ a r2 y ab IE * a:r I. gI—a Ea—y TX :x:at; r:a::*: a a W+frust. A FQ R but gnga-un= hence nwCN ay-dr b ax and that of A FQ R = dr T by a (a2r — a ry · ··gt=ga+ata-y ar a2 r—a ry—a2x + r2x br -+; then b: a::gt;gm= r By the property of the centre of gravity, gw= gmxW (a2 r ——— a n y — a2 x + r2 x) × W = 4 a n x3 (w+s 3er Ty a ry y :- = tu; a: a ax ↑ Put AC .7854, s = 16 nr3 23 T then :un= dr aЕ..ga➡ 1. x br (ay) — dr); b a2 x + r2 x) × W + (W + S 4 an3) br 3 r. . By Mechanics, DC x cone ABC CN + * (globe W + frust. A F QR); that is rx s (a3 r — a r y — a2 x + r2 x) x W + (u v — dr) br anx 3r b equation by W a (w+S-4). Multiplying each side of this and restorings and W, we get, by due reduction, y + — 1 — 1 x → (5 — dr) X a (1 + — — 422) = a~~. But it is evident, |