Page:Encyclopædia Britannica, Ninth Edition, v. 15.djvu/771

Rh MECHANICS 739 equilibrium form, assuming, by the axiom above, that it will remain in equilibrium, and then treating the question by the methods employed for a rigid body; (3) by employ ing the energy test of equilibrium as in 198. 259. We exemplify each of these methods in the specially important case of the ordinary catenary. Common A uniform chain hangs between two fixed points, find ate- the tension at any point and the curve in ivhich the chain
 * iar ^- hangs.

First Method. Let /j. be the mass of unit length of the chain; T the tension at the point x, y, z ; s the length of the chain to x, y, z from some assigned point ; and let the axis of y be taken vertically. Then we have for the equilibrium of the element 5s, considered as a material particle, ds ds ds T^-fT^ + ^T^VVo. ds ds ds ds j J Omitting the terms which cancel one another, and dividing by these become ds ds d f^dz From the first and third it follows that dz/dx is constant, i.e., the chain hangs in a vertical plane. We may take it as that of xy, and the equations are reduced to the first two. The first gives T dx_ rfs showing that the horizontal component of the tension is constant throughout the whole length of the chain. Substituting for T in the second, it becomes . . dsdxj T The quantity on the right is evidently of [L&quot; 1 ] dimensions, because that on the left is so ; and thus we may write -- or Hence a is the length of a portion of the chain whose weight is equal to the constant horizontal component of the tension. The equation now becomes ^LlL = l l^- = / + ( dx 2 a dx a v dx Integrating, we have +V^i + (JO -cr. dx V dxj If we now assume that the axis of y passes through the point at which the chain is horizontal, we have at that point x = 0,-~- = 0, and therefore C = l. Thus / v Taking the reciprocal of each side, we have Subtracting, _ dx dx no constant being added if we assume the axis of x so that y=a, x = together. This is the equation of the curve required. If we take the sum, instead of the difference, of the above equa tions, we find n ds 2 ^-=&quot; + ; dx so that a s being measured from the axis of y. But the tension at x, y is, as above, Hence the tension at any point of the chain is equal to the weight of a length of the chain equal to the ordinate at that point. Thus if a chain of finite length be laid over two smooth parallel chain rails, its ends, when the whole is equilibrium, will be in the ove r horizontal line corresponding to the axis of x in the above investi- vara n e i gation, the middle part of the chain forming part of the catenary. J. a il s When a given length 21 of chain rests in equilibrium on two smooth parallel rails at the same level, we have therefore s + y = l, while x = b, the half distance between the rails. By the above expressions for y and s in terms of x, this leads to the equation which determines a when b and I are given. From this equation, as the minimum value of the left-hand side occurs when a = b, we see that the least length of chain for which equilibrium is possible under the conditions is equal to g times the distance between the rails. 260. Second Method. The chain being in equilibrium, suppose any finite arc of it, as PQ (fig. 64), to become rigid. The forces acting on PQ are three the ten sions T^ and T 2 at its ends, and its weight V W acting at its centre of inertia. But three forces in equi librium are in one plane ( 221). Hence the curve is in one vertical plane. Also, as the two tensions are not parallel to one another, the lines of action of all three forces meet in one point, are simply Fig. 64. Hence there is no couple, and the conditions T 2 cos 0. 2 -TiCos ^ = 0, T S! sin^ 2 -T 1 sin6 1 -W = 0; where O v 2 are the inclinations to the horizon of the tan gents at the ends of the portion solidified. The first asserts that the horizontal part of the tension is the same atP and Q, i.e., all through the chain. The second asserts that the difference of the vertical parts of the tensions at any two points is equal to the weight of the part of the chain between them. By proper mathematical methods these data lead to the results already obtained. In fact the equations we have just obtained are the first integrals of the equations in 259. 261. Third Method. The potential energy of the chain is with the sole condition / ds s dx= constant , dx the limits of integration being fixed, and the same for each expres sion. Hence, by the rules of the calculus of variations, we have the same equations as before. 262. Still supposing gravity to be the only applied force, there Mass of are many forms of important questions which can be solved by either chain to of these methods. We will take some simple, but varied, examples, hang in Find how the mass of unit length of a chain must vary from point given to point so that the catenary may be an assigned plane curve. curve. We have, as before, d_( T ^ ds but our second equation is now ds ds where fj. is an unknown function of s. The third equation, being of the same form as the first, shows that the curve is in a vertical plane, and is thenceforth unnecessary.