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to be equal, which is manifest inconsistency." Anal. p. 136, &c.

Again, put r = radius of a circle, s the sine of an arc of it, c the cosine, and t the tangent of that arc ; then, by similar triangles, we have c:8::::t (Vide the Elements of Plane Trigonometry) and t ; in this expression let c be diminished beyond all bounds, that is, let it become equal to nothing, for this will by no means destroy the equation, and we shall have t =; but, from the nature of the circle, when the cosine = o the sine is = radius ; therefore, t=", or t xo=p2; whence it follows that one of these conclusions must take place when the cosine is = 0; viz. either that the square of the radius of a circle is then = 0, considering t as some finite quantity, or else that the square of the radius is =0X infinity, when t becomes infinite.

If the former take place, that is, if when t is any finite quantity, we may be allowed to say, which is evidently the case, that txo=0 (as demonstrated above) it is demonstrated beyond all contradiction that go = 0.

And since r may be of any assignable magnitude whatever, it follows that o is = the square of any number whatever, or that nothing is = to the area of any square, let the length of its side be whatever it may. Dec. 28, 1811.

Analyst Minor.

OBSERVATIONS ON A CURIOUS PROBLEM. The following Problem was proposed in the Palladium for

1777, by Mr. Judson, of Beverly. A cone of marble 20 feet high, and diameter of its base 6 feet, is standing upright in a gentleman's garden, upon a horizontal plane : required the position of a rope, with one end fixed at the vertex of this marble cone, so that the least force or weight that can be applied to the other end of the rope shall be sufficient to overturn it. . In the Palladium for 1778, the following Solution was

given by Mr. Jos. James, of Stoke Bishop. Let A BC represent the cone of marble. Put a = CD = 20 feet, and x = DE, then (47 eu. 1.) v (x? + aạ) = CE; and since, by mechanics, the application of a weight, or power applied to the rope, fixed at A the vertex of the cone, is the angle which the line of direction of the power makes with the cone, we have by trig. (r? +.a) = CE:1::DE = x: (x2 + a)

= the angle DCE, which by the question must be a maximum. In fluxions and reduced, x= 13.2278 fere = length of the rope making an angle, with the top of the cone = 33° 28'. This solution was confirmed by Mr. George France and Mr. W. Veck, of Cosham. In the corrections, in the Palladium for 1779, Newtoniensis (probably Heath) observes, that the rope fixed to the top ought to be drawn parallel to the horizon, to pull the cone over with the least force.”

" Mr. Judson also observes, that the solution by Mr. Jumes, cannot be universally true. For, if the alt. of the cone had been 6, and the diam. of the base. 20 feet, then the rope would have fallen within the cone's base, which is an impossibility. And as the slant height of the cone approaches nearer to a horizontal line, while the diam. of the base increases, the solution will still be impossible, &c. A similar question was lately published in the Rockingham, a Hull paper, and two solutions to it were inserted, notwithstanding which it was afterwards published again in No. 18, Whiting's Scientific Receptacle, under the following form,

By Mr. Thomas Wilson, of Hull. A piece of marble. in the form of a right cone, the diameter of whose base is 6 feet, and altitude 20 feet, being placed, on its base, upon an inclined plane, it is proposed to raise the lowest part till the base be parallel to the horizontal, and to retain the cone in that position by a rope fixed to its vertex. Required the direction of the rope, that the weight, or power, necessary to hold it in that position, may be a minimum, and what that weight, or power, will be ?

In the first solution, which appeared in No. 19 of Mr. Whiting's Receptacle, there is a trifling numerical error, which we shall correct, and which will render the solution complete; for, as it is drawn from first principles, and is besides as easy as the subject will admit, we shall insert it here, hoping that the matter may be set at rest, especially as our solution coincides with the very elegant one that follows it.

Let abc represent a section of the cone through its axis, o its centre of gravity, ABC the inclined plane, and let the other lines be drawn as in the figure. Then, since the cone is moveable about the point b, as a centre, the 2 centre of gravity, o, will describe e ann the circular arc nom as the cone A revolves about b; therefore, by art. 156, Marrat's Me. chanics, in every position of the cone, the direction of the power, when least, must be parallel to a tangent at o, or perp. to bo produced. Again, since o is the centre of gravity of the cone, it may be considered as the place of the weight, acting upon the lever bd, or on the bended lever cob; now the weight of the cone is 284 cwt. very nearly, which put= W, and draw bP perpendicular to ab; let b P denote the weight of the cone, then by the resolution of forces, Pd will denote the force in the direction Pd; therefore Pb:Pd:: (by sim. trian.) ob:pb::W: W x pb __ 284 * 3 _

= 146.11684 = the relative weight ob 5.83095

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at o, on the lever db; where dP is perpendicular to db. But, art. 135, ut supra, db = 18.693348 : 146.1 1684 ::bo = 5.83095 : 45 cwt. 2 qrs. 8.71136 lbs. =w, or the power which must be applied at c, in the direction dP, to keep the cone in the required position.

When the centre of gravity O coincides with the point m, the direction of the least power, to keep the cone in that position, is parallel to the horizon, and the power necessary to sustain the cone is indefinitely small, or nothing. When o coincides with n, the direction of the least power is perpendicular to the horizon.

Otherwise by Mr. J. Cottom, of Hull. It appears to me, that the simplest method of solving the question is to continue cd and ba till they meet in c; then by the properties of the lever of the second kind, and oblique forces, we shall have P xeb x nat. sin. Le=pbx W, or P x 36 x .514495 = 3 x 284, from whence P=.

P _ 3 x 284 cwts. = 45 cwts. 2 qrs. 8.71136 Ibs.; and as the cone is to be held in the same position, it is immaterial at what part of the line ce the cord is supposed to be fastened.

What Mr. Bronwin has said, in No. 20, of the Scien. tific Receptacle, is quite unintelligible. For boe is a bended lever, the weight being at o, and it is equivalent to the straight lever bod, as is proved in art. 142, Marrat's Mechanics.

W. M.

45.57778

364 x.514495

Prinied by Whitlingham and Rowland, 103, Goswell Street, London:

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COMMUNICATIONS
ON PRIZE SUBJECTS PROPOSED IN NO. V.

ARTICLES I. and II.
CHE communications which have been received, on

I these two Articles, are not judged sufficiently correct for publication ; at the same time, knowing that our juvenile friends are capable of better things, we beg to recommend to them the revision of their respective productions; and shall leave these articles open until the next number.

ARTICLE III. Translated by Miss M. R. Hough, aged 14 years,

Percy Street, Bedford Square, London, who is re

quested to accept the silvER MEDAL. CONSIDER this death as to the types which marked it; by the signs that portrayed it; by the ceremonies that represented it; by the prophecies that foretold it.

Consider this death by the lightning and thunderbolts which were hurled on the head of Jesus Christ. Behold his soul overwhelmed with sorrow; his blood shed on the ground; the cup of bitterness which your Saviour drank to the dregs : listen to the calumnies, the accusations, the unjust sentences; behold his hands and feet nailed to the tree, his body which is soon a universal wound, at the cross the exasperated and unruly populace augmenting the horror of his punishment; behold

VOL. II.

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