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lidity of K EF, and by sim. solids, &c. as in the other solutions. * This question was answered also by Messrs. Baines,

Bamford, Brewer, Brooke, Bruster, Cummins, Ely, Eyres, Gawthorp, Harrison, Hine, Jones, Macann, Maffett, Nesbit, Rylando, Webster, and Winward. 4. Qu. (78) Answered by Mr. A. Nesbit, Farnley Academy * In the annexed figure, let . HO denote the horizon, EQ... ma the equinoctial, m m the paral. E lel of the sun's declination, S the sun's place when his rays ! were parallel to the front of the house, T his place when they were perpendicular to it, Z the zenith, and P the north pole. Then, in the spherical triangle SZT, we have the first zenith distance SZ48° 5', the second zenith distance TZ= 31° 52, and the 2 SZT = 90°, to find ST = 55° 26', and the TSZ= 39° 52' 32'.

Again, in the triangle SPT, we have the polar distance SP = TP=66° 37', and ST, to find the 2 TSP = 760 52' 2", from which take the 2 TSZ, and we obtain the PSZ=36° 59' 30%. ...

Lastly, in the triangle PSZ, we have the sides SP and SZ, and their included angle, to find PZ = 35 50' 16", the co-latitude of the place, and the < SZP= 109° 23' 43", the sun's azimuth from the north, at the time of the first observation. Hence, the latitude of the place is 540 9' 44" N. and the declination of the front wall 19° 23' 43" from the east or west.

Again, ( by the terrestrial globe) by Mr. S. Jones, Liverpool.

Bring any place through which'a meridian passes, as London, to the brazen meridian, and under the first given altitude, on that meridian, make a mark upon the globe; turn the globe on its axis eastward 90°, (by the question) and under the second given altitude, on the

brazen meridian, make another mark on the globe; bring the north pole towards the southern edge of the horizon, and at the same time turn the globe on its axis till the two marks appear in the horizon, the hori. zon will then cut the brazen meridian at 59degrees, the sun's meridian altitude on the given day, whence the zenith distance 30 degrees, added to the declination = 231 N. gives the latitude of the place = 540 nearly. The globe still remaining in the same position, thrust a quill between the globe and the brazen meridian, to keep the globe from turning on its axis, and elevate the north pole 90°; the meridian (of London) which passes through the first mark, will cut the horizon in S.S. E. E. nearly, which is the declination of the wall.

In the same manner exactly as the first solution, it was answered by Messrs. Baines, Cattrall, Gawthorp, Hine, Kay, Maffett, Rylando, Smith, and Winward. 5 QU. (79) Answered by Mr. Baines, and Mr. Bruster,

Donington. Let the lines be drawu as in the annexed figure, where AJ represents the surface of the liquor before coming to rest, and < BAJ = 43o per question. Then BH (1): HC (3):: BG (2): GE=6= the altitude of the complete cone. Again, by trigonometry, EG: radius:: BG : tan. 18° 20' 5" 49'" = L BEG=<BJI; moreover LAJI=900_43° = 47°, and L AJB - 470

+ 18° 26' 5" 49'' = 65° 26' 5" 490"; bence, S. 4 AJB: AB::S. Z BAJ:BJ = 2,99947. Then BG?+GE =BE = 6.394555, and BE - BJ = 3,325085 = JE. Now by similar triangles EG : AB

G HIB ::EJ:JK = 2,102968 = diameter of the hoof at the section, also BE: EG:: BJ:JI = 2,845547 = perp. height of the hoof. Then, by Hutton or Bonnycastle, A B-JK (AB.JK).

5 AB-JKXIJ XAB X , 2618 - 10-6,09928,

=

2,102968

(5,21905) x 4 x 2,845547 x ,2618 = 15,552 cubic

VOL. II,

inches = content of the hoof AJB, which was the part A drank. But (4 + 2 + 4x2) x,2618 X 3=21,9912 cubic inches = the capacity of the glass.

Therefore we have 21,9912—15,552 = 6,4392=B's share, and

(15,552 : 4d. 243 + or about 48d.

A must pay. As 21,992 : 0d. :: > 6,4392: 10.756 + or about 1 d.

B must pay. And thus very nearly it was answered by Messrs. Armitage, Brewer, Brooke, Cattrall, Cummins, Dunn, Ely, Guwthorp, Hine, Jones, Macann, Maffet, Nesbit, Rylando, Smith, Webster, and Winward.

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6. QU. (80) Answered by Mr. Bruster, Donington.
The cylinder being

timer side supposed of equal density throughout, the centre of gravity will of course be the middle point of its axis; and at any given elévation, the line of suspension continued will always pass through the centre of gravity, and make an angle with the middle section equal to the given angle of elevation.

'In the annexed scheme P is the point of suspension, HO and ho are parallels to the horizon, and the angle. of elevation Aho = E BAC = 25° 15'. As rad. : , AB (1) :: tan. (25° 15') < BAC : ,47163 = BC. Hence the point C = 3,47163 = 5,47163 feet from the bottom, or 4,52837 feet from the top end.

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Again, by Messrs. Baines, Brewer, Cattrall, Ely, and

Smith. Let A be, the centre of gravity, and PC the string; then, when the cylinder is at rest, PC will pass through

A; therefore, in the triangle ABC, we have all the
angles, (for the angle B AČ is = angle A ho) and the
side A B, to find BC =,47163; whence FC = 5,47163,
and CF = 4,52837 feet, as required.

This question was answered also by Messrs. Armitage,
Brooke, Cummins, Eyres, Gawthorp, Hine, Jones, Kay,
Maffett, Nesbit, Rylando, Webster, and Winward.
7. Qu. (81) Answered by Mr. Gawthorp, Mr. Nesbit,

and Rylando. Put %= x -1; then < = , and 2% = 2x 2, therefore, 2x+1=2x-1, and (2x + 1) X %= 2x: - 3x + 1; hence the given fluxion will be transformed

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+g-ii whose fluent is 2,30258

Tx log.

2+-=-1x hyp. log. 2 + 1; and by restoring x we have — 1 x h. l. 2+ = for the fluent required.

Nearly thus it was answered by Messrs. Hine, Jones, Maffett, Whitley, and Winward. 8. Qu. (82) Answered by Mr. Gawthorp and Mr. J.

Whitley. Let 2=: (d + v), then, 3 = ; and 5 V (d' -- 4 de +82-) = å av (1 do + $03) = **** Öv (d: + 0); whose fluent, by prop. 73, p. 178, Vince's Fluxions, is found=%2* { V (d? + *) + x h. 1. (x+v(do

his expression ought to vanish when == 0; hence C= _12 x h.l. (d\/2—d); therefore, the correct fluent, generated while z increases from o to d = 10, is equal to (3 d 5+d , dv2.

x h. 1.

T 10*

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(1+x_20) = {18v 5+1)+*2 x h.l.(3+v 10) xfl + v 2)} x = 1,20215 x d = 12,0215 as

required.

And thus it was answered by Messrs. Hine, Maffett, Nesbit, Rylando, and Winward.

Otherwise by Mr. S. Jones, Richmond Academy, Liverpool. With semiaxes CD, CV, each

Id, describe the equilateral hy. perbola HVB, draw the ordinates DH, AB, and join A V; by a known property of the hyperbola, D C A B = AV = (CD’+C A); put DA = 2, then, CA=%~ 1d. and AB= 'tex v'ldi — 4 dx +8z?;) but the fluxion of the area A DHBA = A B x fluxion AD= = x '& * v (d— 4 d2 + 8 8°), consequently the fluent of Š v (d? — 4 dx + 8x1) = * area ADHV BA; as required. 9. Qu. (83) Answered by Mr. Elias Webster, Armley

Mills. Take y = x-r, and x = x+2r; then, the first becomes a square, and it only remains to make r? - ori -6r, and r? — 2 rx — 5 gra squares; to effect wbich, let x? - 6 rx - 6 m? = (x – nr), and we get to n? +6

Xr. This value of x substituted for it in the latter formula, we obtain, by reduction, nt - 473 + 4n? +96 n-72; which must be made a square. Assume p+ 5 for its root, and we shall find p=8, and n =9; therefore, x = 297. 3? Exam. Takér, = 4, then y= 25, x = 29, and 3 = 37; three numbers that answer the conditions of the ques.

2n

4.

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