Page:Encyclopædia Britannica, Ninth Edition, v. 15.djvu/772

Rh MECHANICS and d^/dy dn dx Suspen sion bridge. Now, from the assigned equation of the catenary, the left-hand member can be expressed as a function of s, and thus we have the solution of the problem. For example, suppose the chain is to hang in an arc of an assigned circle. Referred to the lowest point of the circle, the equation is a- 2 = iay - if, where a is the radius. From this we have dit x s -,- = = tan , cue a y a, T 9 so that M = - sec 2. ag u The tension at the lowest point is thus equal to the weight of a piece of chain like that at the vertex, and of length equal to the radius of the circle ; the mass per unit length becomes infinite at the end of a horizontal diameter. Next find the form of the chain when the mass of any arc is pro portional to its horizontal projection. This is a rough approximation to the case of a suspension bridge where the roadway is uniform, and much more massive than the chains, to which it is attached throughout by vertical ties. The equations are as before, but the additional condition takes the form ds a constant. This gives ^ = ^o dx 2 T and the chain forms a parabola whose vertex is downwards, and whose axis is vertical. Catenary As a final example, we have what is called the catenary of nni- of uni- form strength; that is, the form in ivhich a chain hangs when the form tension at every point is proportional to the breaking stress at that strength, point. Here we suppose the strength to be proportional to the section, i.e., to the mass per unit length. This gives the condition where a = c/g. This gives dy, .... -^=tan - dx a and the complete integral may be written, by proper selection of origin, in the final form u ,--, x = sec a Limit of This curve has obviously two vertical asymptotes dis- span. tant .1 7ra from the axis of y. The quantity a is directly as the tenacity of the material ; and thus we see that there is a limit (even in this simplest case) to the span of a chain, however strong, formed of any known kind of matter. It is a very curious fact that, if we write the equation of this catenary in terms of the arc and the radius of curvature, it becomes identical with that of the common catenary in terms of Cartesian coordinates, horizontal and vertical. For we see at once that so that &quot; = while by the previous equations d&quot;-&amp;gt;l Thus finallv. - p = g +&quot;. Compare 259. 263. When the chain is not uniform, and when it is subject to Catenary the action of other forces than, or besides, gravity, the equations are general. ds) ds~dsj ds where X, Y, Z are the component forces on unit mass. These three equations involve the unknown quantities T, x, y, x, and s only ; for /JL is supposed to be given in terms of s. Two relations only among x, y, and 2 are wanted (for s is known in terms of x, y, z) ; so that the equations are necessary and sufficient to give these relations and the remaining unknown T. The first members of these equations consist each of two terms : dT ii- T i 1-11 dx d&amp;gt;/ Jz multiplied respectively by -&amp;gt;- . ds J J ds ds ds rJ~yi ^-72 )* f7~ r and multiplied by p - p -i p -.-&quot;. ; p ds- ds~ ds ~ p being the radius of curvature of the chain. Hence ( 22) we conclude that, so far as the tension alone is concerned, the forces on an elementary unit of length of the chain are dT/ds in the direction of the tangent, and T/p in the direction of the radius of absolute curvature. These must balance the corresponding components of the external forces on the element. Hence we see that the resultant of the applied forces lies, at every point, in the osculating plane. Thus we have ^ = - M T__ / d*x_ d-i p ds- ds Here S and N are the tangential and normal components of the applied forces per unit length of the chain. But when a unit particle moves in a curve, we have always dv dv ?- 2 r- = v -r- = S, and = 2s , at ds p where S and N are the normal and tangential components of the requisite force. If we write these in the form dv _ S v _ N ds v p v and suppose that the curve in which the particle moves is the same as the catenary above, while the speed at each point haa the same numerical value as the tension, we see that we must have S S N V = - = - S i = = N v T v T or S =-ST, N =-NT. Thus the catenary will be the free path of the particle provided the force applied at any point is equal to the reverse of the product of that acting on the chain by the numerical value of the tension of the chain at that point. Conversely, if we take any case of free motion of a particle, a uniform chain will hang in the corresponding orbit under the action of the same forces each reversed, and divided by the numerical value of the speed at the corresponding point of the orbit. Thus we can at once pass from particle kinetics to corresponding cases of catenaries. In the case of a projectile, the path is a parabola, the force is constant and parallel to the axis, and the speed is as the square- root of the distance from the directrix. Hence, that the parabola may be a catenary under gravity, it must be turned vertex down wards ; and the mass of the chain per unit length at any point must be inversely as the square root of the distance from the direc trix. It is easily found from this that the mass of any arc of the chain must be proportional to the length of its horizontal pro jection, as in the second problem solved in 262. In the case of a planet we have &quot;-/t(2/r-l/a). Hence a chain will hang in an ellipse if it be repelled from one focus by a force varying inversely as the square of the distance, the mass per unit length of the chain being directly as the square root of the distance from that focus and inversely as the square root of the distance from the other. If the chain be uniform^ the law of the repulsive force from the first focus must be 1/vV 3 &quot; instead of I/; 2, where r, r are the distances from the two foci. 2G4-. When a chain or string is stretched over a curved surface, the surface must exert a reaction on it to keep it in its curved form. The preceding investigation has showu that the force normal to a chain per unit length at any point is balanced by T/p per unit of length, which must therefore be the magnitude of the reaction. We Catenary as path of par- tide. Paraboli catenary Elliptic catenary Pressure f cor( l fl su