and give an example in numbers when the radius is 10, and the arc contains 147°), 28', 48" I. 16. Qu. (112) By Mr. J. Butterworth, Haggute. Given the base, one of the angles at the base, and the sum of the perpendicular and line bisecting the base, to construct the plane triangle... .. 17. Qu. (113) By Mr. G. Harvey jún. Plymouth... Given, the vertical angle, the line' which bisects it and' meets the circumference of the circumscribing circle, and the sum of the cubes of the sides, to construct the plane triangle. Le 18. Qu.(114) By Mr. R. Vayheeg, Plymouth. ; Given the base, and the difference of the adjacent angles, to construct the plane triangle, when the sum or difference of the squares of the lines drawn from the extremities of the base, to the centre of the inscribed circle, is equal to a given space. ips 19. Qu. (115) By Mr. Putsey. ! . The height of my garden wall, which stands on bo. rizontal ground, is 8 feet, and the length of a ladder, which is placed in a vertical position against it, is 18. feet; now, admitting the foot of the ladder to be drawn along the ground, directly from the wall, it is required to determine the greatest projection of the top of the lad. der over the other side of the wall, and also the area of the curve which is the locus of the top of the ladder in its descent to the top of the wall: 20. Qu. (116) By Mr. A. Hirst; Marsden... * If to the base of a hemisphere of tead, a cone of cork, of a proper length, having the same base, be cemented, it will constitute the toy called a tumbler or mountebank, which will always recover an erect position, however it may be thrown or laid on a table ; required the length of the greatest and least axes of the cone of cork, and also the proper length; so that it may rise in the most favourable manner possible: the "radius of the hemisphere being one inch, and the specific gravity of lead and cork 11325, and 240 respectively. The Prize as before. OBSERVATIONS TO ELUCIDATE THE PRINCIPAL PROPERTIES OF THE LEVER. BY MR. A. HIRST, MARSDEN... As most writers have found some difficulties in demonstrating the fundamental properties of the lever, if you think the following worth noticing, it is at your service. The idea it contains is new to me, and may be so to several other of your readers, to others it may not; it is, however, different from the demonstrations usually given in books on Mechanics; and as every mode of contemplating truth strengthens the intellectual faculties, your inserting it may do some good, but cannot possibly do any harm. Let the circle ABC be .con- j..!! sidered as a heavy ring of uniform matter, equally diffused through its whole circumference; it will A then be at rest, if supported at its centre E; this principle will readily be admitted. If, also, the veh B !': weight of the ring be supposed to be separated into a number of equal weights, placed uniformly at equal distances in the circumference, there will still be an equilibrium ; first considering their number ás infinite, and afterwards as finite. Consider, now, the matter as all collected equally into the three points A, B, C, equally distant, that is, at the distance of 120°, and by what has been said, there will still be an equilibrium. Therefore, a weight at A supports the weights B and C, which are together = 2 A ; but, the weights B and C, acting at the points B and C, may be considered as açting at D; hence, a weight = 2 A acting at D, will balance, or keep in equilibrium, the weight A acting at A. But it may be easily proved, that A E= 2ED, for, ED= cosine 60° = sine 30o = { rad. = {EF, therefore, a weight at D, will be in equilibrium with half its weight at A, when the distance A E is the double of E D. Ănd, with some trifling modifications, the method may be extended to any other case. .: GEOMETRICAL PROBLEMS BY THE COMPASSES ALONE. PROBLEM I. To divide the circumference of a given circle into four equal parts, or to find the chord of 90°. ASSUMING a point in the circumference of the circle, as one extremity of a diameter, apply from thence the radius three times along the circùmference, and thus we get the other extremity of the diameter, and we also mark off the arches of 60° and 120°. With the chord of 120° as a radius, from the two extremities of the diameter just determined, describe two arches, intersecting one another. The distance of this intersection from the centre of the given circle, is the chord of 90°, or of the fourth part of the circumference; and taken in the com.' passes, serves to divide it into four equal parts.. The truth of this is easily made evident. If the radius of the given circle = 1, the square of the dia. meter = 4; and since the lines.drawn from the extremities of the diameter, to any of the points of division in the circumference, contain a right angle, and are one of them the chord of 60°, and the other of 120°, the sum of their squares 4; from which, taking away the square of the chord of 60°, or of the radius = 1; there remains the square of the chord of 120° equal to 3. The square of the line drawn from either extremity of the diameter, to the intersection of the two arches, is therefore = 3; and this line being the hypothenuse of a right-angled triangle, having the right angle at the centre of the circle, if we take away from the square of it the square of the radius, the remainder 2 is the square of the distance of the centre from the intersection of the arches. That distance is, therefore, = w 2. But the chord of 90° is also an 2 when the radius is unity. Therefore, &c, 2. E.D. a PROBLEM II. parts. Tooted From the intersection determined above in problem 1st, with a radius = 1 (that of the given circle) describe an arch intersecting the given circle. This intersection will bisect the arch of 90° above determined. The demonstration is evident. , PROBLEM III. To divide the circumference of a circle into five equal parts, or to construct a regular pentagon. Having assumed one extremity of a diameter, and de. termined the other, as in problem ist; and also the arch of 90°, or the middle of the semicircle ; from this last point set off on each side an arch of 60° (by taking the radius in the compasses ;) these together are = 120°. From the extremities of the chord of 120° thus found, with the chord of 90° for the radius, describe two arches intersecting one another on that side of the chord of 120°, where the diameter already determined lies. The distance of this intersection from either of the extremities of the diameter, is the chord of 72°, or the side of an equilateral pentagon inscribed in the circle. The demonstration cannot easily be given here; but it depends on this, that the side of the equilateral pentagon, the radius being 1, is = 75-75 - This is very nearly the same construction as is given by Ptolemy in his Almagest, for finding the chord of the fifth part of the circumference. ... From Mascheroni's Geometrie du Compas. • • ERRATA IN NO. VI. Page 1.45, line 7, for word, read words. ...... 145, ... 10, before the word Chaldee, insert 22. ...... 145, ... 11, for tela, read Tena. ...... 146, ... 14, for Sabbate, read Sabbat. 6, for Fades, read Faden. ...... 147, ... 13, for Modes, read Moder. 1, for Ædder, read Ædder. 8, for Biter, read Biter. 193, ... 6, (of the Antiquarian Answers) for Lorimer, read Lorum. ...... 212, ... 1, for Rowell, read Nowell. Printed by Whittingham and Rowland, 103, Goswell Street, London, • • THE Juvenile Department. COMMUNICATIONS ARTICLE XIII. TRANSLATED by Master W. Kelsal, Liverpool; who is requested to send for any book he may think proper, the price of which does not exceed one guinea. VIRGIL. ECLOGUE iv.-POLLIO. VOL. II. |