EQUILIBRIUM IN GENERAL. 270. Let n be the number of equal sides of the prism, and the inclination required of the plane to the horizon. Then since (supposing the prism to be prevented from slipping, by friction or any other cause,) it will roll or not, according as the to the horizon, drawn from its centre of gravity, falls upon the base, or the base produced. Consequently, the limit of equilibrium will correspond to that position of the which passes through an at the base, and the centre of gravity. Now the centre of gravity being equally distant from each of the angles, it easily appears that the bisects the at the base. Hence But since then angles of the prism are equal, each of them 271. Let the apparent weights be w, w', and the true one I. Then a, b being the corresponding lengths of the arms of the steelyard, we have if w, w' be unequal, which being the case by hypoth. 272. If be the formed by the string at any tack, and W one of the equal weights appended to the string, it is known the pressure (P) on the tack is expressed by is Now since in the problem the at the vertex of the isosceles A 120°, those at the base must each 30°, and.. the angles at the three tacks, viz., 0, 0, 0", are each 120°, or are equal. Consequently, the equat. (1), the pressures P, P,' P", are equal. Also P = 2 W × cos. 60° = W. 273. Let the weight of the lever, (which may be any figure of which we can find the centre of gravity) = w, a its length, and x the distance of the fulcrum from the end, to which w is attached, its distance from the centre of gravity. Then, supposing the weight of the lever placed in its centre of gravity, we have w x y = w x x :.y = x But x + y a given quantity. distance of centre of gravity from the end = m, .. 2xm, and x = m 2 which determines the position of the fulcrum. 274. Since the sphere is supported, the forces which con spire to effect the equilibrium must tend to the same point. Hence supposing the weight (W) of the sphere placed in its centre of gravity, i. e., in its centre, the force of gravity acting from that point, the other forces, viz., the reactions R, R', I planes at the points of their contact with the sphere, must also tend to the centre of the sphere. Hence, putting the inclinations of the planes to the horizon =0,0, and forming a ▲ whose sides are parallel to the directions of the forces, we easily get R: W: sin. e' sin. (+) R': W:: sin. : sin. (+9) and resolving R, R' into their vertical and horizontal components (V, H), (V', H'), we have V R. cos. 0, HR sin. V'R' cos. ', H′ = R′ sin. 6′′ :. V: W. cos. :: sin. e': sin. (0 + 0') H: W. sin. :: sin. ': sin. (0 +0') } } - H': W. sin. :: sin. : sin. (0 + 0) In the problem, 9 = 60°, and 6′ = 30°. ..V: W:: 1:4 quired. V: W:: 3:4} and V: V' :: 1 : 3, the analogies re 275. Let AW be produced to meet BC in (a), and let a prop be placed at (a) instead of the two props at B, C. Then denoting the pressures on the props at A, a, by A, a, we have A:a:: Wa: WA A: A+a :: Wa: Aa :: A BWC: A BAC But A+ a W. .. A: WA BWC: A BAC. and in the same manner it may be shewn that and .. B: W:: A WAC: ▲ ABC C: W:: A WBA: ▲ ABC A: B: C:: A BWC : ▲ AWC : A AWB EQUILIBRIUM IN GENERAL. For the general problem, of which this is a particular case, the reader is referred to Whewell's Mechanics, p. 52. 276. Since the beam is at rest, the forces which effect its repose, viz., the reactions of the planes at the extremities of the beam, acting in directions planes, and the weight of the beam at its centre of gravity acting horizon, must tend to the same points. Hence 0, ' being the inclinations of the planes to the horizon, (a) the length of the beam, m, m' the distances of its centre of gravity from the extremities leaning on the planes, of which 0, ' are the inclinations, and the inclination of the beam to the horizon, we easily get (p, p' being the corresponding perpendiculars,) (m + m')2 = p2 + p'2 - 2pp'. cos. (0 + 0′). But from similar A it appears that cos.24= (m + m') sin. 0 sin.o 0' m2 sin.o 0′ + m'2 sin.20- 2mm' sin. O sin. ' cos. (+0') which gives the two positions required. If the beam be prismatic and of uniform density, or m=m', and moreover if = 6', then cos.2 = 1, and cos. = ± 1, 9 = 0, which shew that the beam is then horizontal. or T, The same principle will also serve to find the position of equilibrium of a beam supported by two given curve surfaces. 277. The forces which produce the equilibrium, viz., the man's weight (W), and the reactions of the beam and swing R, R', being supposed to act from the man's centre of gravity in the directions vertical, and along the rod and string, if a ▲ be formed by lines drawn parallel to those directions, its sides will represent those forces. Now, making the inclination of the beam or lever to the vertical, that of the string in the given position = ß; and the EQUILIBRIUM IN GENERAL. given lengths of the string and roda, b, we have the inclination (9) of the rod to the vertical, expressed by and . R R':: sin. : sin. 0, which, by means of equation (1), gives the values of R, R', &c. 278. Let a, b, be the lengths of the lever, and a the angle between them; also let & be the inclination of a to the verticle. Then the weights P and Q hanging from the extremities of a, b, being in equilibrium, we easily get P x a sin. @= Qxt sin. (a — 0) 279. Since the tension of the string is every where the same, the weight is supported by two equal forces, and.. the direction in which the weight acts, (viz., the vertical passing through its centre of gravity,) bisects the between the parts of the string. Hence a. being the distance of the tacks, the length of the string, a its inclination to the vertical, and the inclination of either part of the string to the vertical, we readily obtain and drawing lines from the tacks making the 0, thus determined, with the vertical, their intersection will give the required position of the weigh 280. VOL. II. Since the tension of the string is independent of its P |