Page:Encyclopædia Britannica, Ninth Edition, v. 15.djvu/735

Rh MECHANICS 703 R and Q attached to the ends of the strings. But now we see that we must have the limiting condition R + Q&amp;gt;W. This is merely the geometrical condition that Fi Here the magnitudes of all three sides of P/?y are given. Hence its angles are given, and the sole position of equili brium is at once found. 1 -tide 1 25. Now take the case of a particle resting on a surface. fixed As we are concerned only with the portion of the surface j face, immediately contiguous to the position of the particle, we may substitute for it its tangent plane at that point (except, of course, at singular points, where there may be an infinite number of tangent planes; but such cases we do not con sider). Hence the problem reduces itself in all cases to that of a particle resting on an inclined plane. If the plane be smooth, the particle cannot remain in equilibrium unless some force is present to prevent its sliding down. Let us suppose it to be supported by a force, F, acting upwards along the plane (fig. 37). Then we have three forces at work : the weight P acting vertically downwards; the supporting pressure of the plane R, which necessarily acts perpendicularly to the surface; and the third force, just mentioned, which we see by previous considera tions must be in the plane of the other two, and must therefore lie along the line of greatest slope of the plane. We might construct a triangle of forces as in the previous examples, but we will now vary the process, and resolve the forces in two directions at right angles to one another in their common (vertical) plane. Let a be the angle of inclination of the plane to the horizon, then the algebraic sum of the components of the forces must vanish both horizontally and vertically. This gives us the two conditions Fcoso Rsina = 0, Fsina + Rcosa-P = 0, From these we obtain at once F = Psina, R = Pcosa Now the choice of mutually perpendicular directions in which to resolve was at our option, and we see that had we chosen to resolve along and perpendicular to the plane we should have obtained the last two equations, which are equivalent to, but simpler than, the former ones, which were obtained by resolving horizontally and vertically. Theoretically speaking, it does not matter which system we choose; in practice, however, it is well to select the directions which will give the required results in the simplest form. The full value of a proper selection will not be felt till we come to the statics of a rigid solid. notion. If we suppose the plane to be rough, friction alone may suffice to develop the requisite force F, But the utmost value of the friction is ( 120) pJR. Hence the particle will be on the point of sliding if Divide the members of this equation by those of E = Pcoso, and we find = tan a. Hence, so long as the coefficient of friction is greater than the tangent of the inclination of the plane to the horizon, 1 We have assumed here, what is properly part of the results of the third law of motion, that the tension of a weightless string, passing over a smooth pulley, is in the direction of its length, and of the same amount at all points. the friction will suffice to prevent sliding. More and more Angle of is called into play as the inclination of the plane increases, repose, and finally when tan a = M the particle is just about to slide down. This simple idea, taken along with Coulomb s results ( 120), points to a very easy method of determining the coefficient of friction between any two substances. The limiting angle defined by a = tan- J /x is called, on account of this property, the &quot; angle of repose.&quot; 126. Let us now suppose the particle to be, in part, Support supported by an elastic string fixed at a point in the plane, by elastic and lying in the line of greatest slope. (This modification s riug is introduced to show the nature of cases in which there are limits between which equilibrium is possible.) We assume &quot; Hooke s Law,&quot; viz., that the tension of an elastic string, drawn out from its natural length I to length I, is expressed by where E is a definite constant, representing theoretically the tension which would just double the length of the string. Our equations are exactly the same as before, only that F consists now of two parts, one due to friction, the other to the elasticity of the string. Thus R=Pcosa. Now, when sliding is about to commence downwards we have If the particle is about to be dragged upwards, Hence for the two extreme positions of equilibrium I I ^Pcosa + E y-=Psina. V Hence the limiting positions of equilibrium of the particle are given by its distance from the fixed end of the string PI

tending to given points. If I be less than the smaller of these, gravity pulls the particle down :, if it be greater than the larger of them, the tension of the string pulls the particle up. In intermediate positions the full available friction is not called into play. 127- Next, let a small ring P (fig. 38) be attached to one Equili- end of a string. Let the string pass round two smooth ^; ium pulleys B, C, at different points, then be passed through ^ t em the ring, then round two more pulleys D, E, and through of equal the ring again, and so on, the other end being either forces fastened to the ring or attached to a fixed point. It is required to find the position of equilibrium of the ring when the string is drawn tight, by operat ing on the lap of it behind two of the pulleys. This is equivalent, from the physical point of view, to finding the position of equilibrium of a particle acted on by a number of equal forces each directed towards a given point. From the geometrical point _ of view its solution obviously answers the question, &quot;Find the point the sum of whose distances from a number of given points is the least possible.&quot; The points need not lie all in one plane. The solution is, from the polygon of forces, that in