E From B draw BC perp. to the face of the wall BE, which will be the direction of the force at B; also through G, the centre of gravity, draw CGD perp. to the horizontal line AE, then CD is the direction of the weight of the beam; and because these two forces meet in the point C, the third force or push A, must be in CA, directly from c; so that the three forces are in the directions CD, BC, CA, or in the directions CD, DA, CA; and, these last three forming a triangle, the three forces are not only in those directions, but are also proportional to these three lines; viz, the weight in or on the beam, as the line CD; the push against the wall at B, as the horizontal line AD; and the thrust at the bottom, as the line AC. Some of the foregoing problems will be found useful in different cases of carpentry, especially in adapting the framing of the roofs of buildings, so as to be nearest in equilibrio in all their parts. And the last problem, in particular, will be very useful in determining the push or thrust of any arch against its piers or abutments, and thence to assign their thickness necessary to resist that push. The following problem will also be of great use in adjusting the form of a mansard roof, or of an arch, and the thickness of every part, so as to be truly balanced in a state of just equilibrium. PROBLEM 30. Let there be any number of lines, or hars, or beams, Ab, BC, CD, DE, &c, all in the same vertical plane, connected together and freely moveable about the joints or angles A, B, C, D, E, &c, and kept in equilibrio by their own weights, or by weights only laid on the angles: It is required to assign the proportion of those weights; as also the force or push in the direction of the said lines; and the horizontal thrust at every angle. tion of, or paral P lel to, the given lines or beams, viz, ca parallel to AB, and co parallel to BC, and ce to DE, and cf to EF, and cg to FG, &c; also CH parallel to the horizon, or perpendicular to the ver. tical line ang, in which also all these parallels terminate. Then will all those lines be exactly proportional to the forces acting or exerted in the directions to which they are parallel, and of all the three kinds, viz, vertical, horizontal, and oblique. That is, the oblique forces or thrusts in direc- ́ tion of the bars . AB, BC, CD, DE, FF, FG, ca, cb, CD, ce, cf, cg; are proportional to their parallels and the vertical weights on the angles B, C, D, E, F, &c, are as the parts of the vertical.. ab, bò, De, ef, fg, and the weight of the whole frame ABCDEFG, is proportional to the sum of all the verticals, or to ag; also the horizontal thrust at every angle, is every where the same constant quantity, and is expressed by the constant horizontal line CH. Demonstration. All these proportions of the forces derive and follow immediately from the general well-known property, in Statics, that when any forces balance and keep each other in equilibrio, they are respectively in proportion as the lines drawn parallel to their directions, and terminating each other. Thus, the point or angle в is kept in equilibrio by three forces, viz, the weight laid and acting vertically downward on that point, and by the two oblique forces or thrusts of the two beams AB, CB, and in these directions. But ca is parallel to AB, and cổ to BC, and ab to the vertical weight; these three forces are therefore proportional to the three lines ab, ca, cb. In like manner, the angle c is kept in its position by the weight laid and acting vertically on it, and by the two oblique forces or thrusts in the direction of the bars BC, CD: consequently these three forces are proportional to the three lines bD, cb, CD, which are parallel to them. Also, the three forces keeping the point D in its position, are proportional to their three parallel lines De, CD, ce. And the three forces balancing the angle E, are proportional to their three parallel lines ef, ce, cf. And the three forces balancing the angle F, are proportional to their three parallel lines fg, cf, cg. And so on continually, the oblique forces or thrusts in the directions of the bars or beams, being always proportional to the parts of the lines parallel to them, intercepted by the common vertical line; while the vertical forces or weights, acting or laid on the angles, are proportional to the parts of this vertical line intercepted by the two lines parallel to the lines of the corresponding angles. Again, with regard to the horizontal force or thrust: since the the line DC represents, or is proportional to the force in the direction DC, arising from the weight or pressure on the angle D; and since the oblique force DC is equivalent to, and resolves into, the two DH, HC, and in those directions, by the resolution of forces, viz, the vertical force DH, and the horizontal force HC; it follows, that the horizontal force or thrust at the angle D, is proportional to the line CH; and the part of the vertical force or weight on the angle D, which produces the oblique force DC, is proportional to the part of the vertical line DH. In like manner, the oblique force cb, acting at c, in the direction CB, resolves into the two bн, HC; therefore the horizontal force or thrust at the angle c, is expressed by the line CH, the very same as it was before for the angle D; and the vertical pressure at C, arising from the weights on both D and c, is denoted by the vertical line bн. Also, the oblique force ac, acting at the angle B, in the direction BA, resolves into the two aн, нC; therefore again the horizontal thrust at the angle B, is represented by the line CH, the very same as it was at the points c and D; and the vertical pressure at B, arising from the weights on B, C, and D, is expressed by the part of the vertical line aн.. Thus also, the oblique force ce, in direction DE, resolves into the two CH, HC, being the same horizontal force with the vertical He; and the oblique force cf, in direction EF, resolves into the two CH, Hf; and the oblique force cg, in direction FG, resolves into the two CH, Hg; and the oblique force cg, in direction FG, resolves into the two CH, Hg; and so on continually, the horizontal force at every point being expressed by the same constant line CH; and the vertical pressures on the angles by the parts of the verticals, viz, aн the whole vertical pressure at B, from the weights on the angles B, C, D and bн the whole pressure on c from the weights on c and D; and DH the part of the weight on D causing the oblique force DC; and He the other part of the weight on D causing the oblique pressure DE; and Hf the whole vertical pressure at E from the weights on D and E; and нg the whole vertical pressure on F arising from the weights laid on D, E, and F. And so on. So that, on the whole, aн denotes, the whole weight on the points from D to A; and нg the whole weight on the points from D to G; and ag the whole weight on all the points on both sides; while ab, bɔ, ne, ef, fg express the several particular weights, laid on the angles B, C, D, E, F, Also, the horizontal thrust is every where the same constant quantity, and is denoted by the line CH. Lastly, Lastly, the several oblique forces or thrusts, in the direc tions AB, BC, CD, DE, EF, FG, are expressed by, or are proportional to, their corresponding parallel lines, ca, cb, CD, ce, cf, cg. Corol. 1. It is obvious, and remarkable, that the lengths of the bars AB, BC, &c, do not affect or alter the proportions of any of these loads or thrusts; since all the lines ca, cb, ab, &c, remain the same, whatever be the lengths of AB, BC, &c. The positions of the bars, and the weights on the angles depending mutually on each other, as well as the horizontal and oblique thrusts. Thus, if there be given the position of DC, and the weights or loads laid on the angles D, C, B ; set these on the vertical, Dн, Db, ba, then cb, ca give the directions or positions of CB, BA, as well as the quantity or proportion CH of the constant horizontal thrust. Corol. 2. If CH be made radius; then it is evident that Ha is the tangent, and ca the secant of the elevation of ca or AB above the horizon; also нb is the tangent and co the secant of the elevation of cb or CB; also HD and CD the tangent and secant of the elevation of CD; also He and ce the tangent and secant of the elevation of ce or DE; also Hƒ and of the tangent and secant of the elevation of EF; and so on; also the parts of the vertical ab, bD, ef, fg, denoting the weights laid on the several angles, are the differences of the said tangents of elevations. Hence then in general, 1st. The oblique thrusts, in the directions of the bars, are to one another, directly in proportion as the secants of their angles of elevation above the horizontal directions; or, which is the same thing, reciprocally proportional to the cosines of the same elevations, or reciprocally proportional to the sines of the vertical angles, a, b, D, e, f, g, &c, made by the vertical line with the several directions of the bars; because the secants of any angles are always reciprocally in proportion as their cosines. 2. The weight or load laid on each angle, is directly proportional to the difference between the tangents of the elevations above the horizon, of the two lines which form the angle. 3. The horizontal thrust at every angle, is the same constant quantity, and has the same proportion to the weight on the top of the uppermost bar, as radius has to the tangent of the elevation of that bar. Or, as the whole vertical ag, is to the line CH, so is the weight of the whole assemblage of bars, to the horizontal thrust. Other properties also, concerning the weights and the thrusts, might be pointed out, but they are less simple and elegant than the above, and are therefore omitted; omitted; the following only excepted, which are inserted here on account of their usefulness. sin. bcd sin. D Corol. 3. It may hence be deduced also, that the weight or pressure laid on any angle, is directly proportional to the continual product of the sine of that angle and of the secants of the elevations of the bars or lines which form it. Thus, in the triangle bCD, in which the side bD is proportional to the weight laid on the angle c, because the sides of any triangle are to one another as the sines of their opposite angles, therefore as sin. D: cb :: sin. bcd: bD; that is, bD is as x. cb; but the sine of angle D is the cosine of the elevation DCH, and the cosine of any angle is reciprocally proportional to the secant, therefore bD is as sin. bcD x sec. DCH X cb; and cb being as the secant of the angle bcн of the elevation of be or BC above the horizon, therefore bD is as sin. bcD x sec. bсн x sec. DCH; and the sine of bCD being the same as the sine of its supplement BCD; therefore the weight on the angle c, which is as bD, is as the sin. BCD x sec. DCH X sec. bcн, that is, as the continual product of the sine of that angle, and the secants of the elevations of its two sides above the horizon. = Corol. 4. Further, it easily appears also, that the same weight on any angle c, is directly proportional to the sine of that angle BCD, and inversely proportional to the sines of the two parts BCP, DCP, into which the same angle is divided by the vertical line CP. For the secants of angles are reciprocally proportional to their cosines or sines of their complements: but BCP сbH, is the complement of the elevation bcн, and DCP is the complement of the elevation DCH ; therefore the secant of bсH x secant of DCH is reciprocally as the sin. BCP X sin. DCP; also the sine of bCD is the sine of its supplement BCD; consequently the weight on the angle c, which is proportional to sin. bcD x sec. bcн x sec. DCH, is also proportional to whole frame or series of angles is balanced, or kept in equilibrio, by the weights on the angles; the same as in the preceding proposition. sin. BCD sin. BCP X sin. DCF when the Scholium. The foregoing proposition is very fruitful in its practical consequences, and contains the whole theory of arches, which may be deduced from the premises by supposing the constituting bars to become very short, like arch stones, so as to form the curve of an arch. It appears too, that the horizontal thrust, which is constant or uniformly the same |