to their sum, applied in a contrary direction, to keep it at rest. It is then said to be in a state of equilibrium. 23. If the forces mA, mB, be applied to a particle m in contrary A fig.2. m B directions, and if mB be greater than mA, the particle m will be put in motion by the difference of these forces, and a force equal to their difference acting in a contrary direction will be required to keep the particle at rest. 24. When the forces mA, mB are equal, and in contrary directions, the particle will remain at rest. 25. It is usual to determine the position of points, lines, surfaces, and the motions of bodies in space, by means of three plane surfaces, oP, oQ, OR, fig. 3, intersecting at given angles. The intersecting or co-ordinate planes are generally assumed to be perpendicular to each other, so that xoy, xoz, yoz, are right R B fig. 3. fig. 4. Hence angles. The position of or, oy, oz, the axes of the co-ordinates, and their origin o, are arbitrary; that is, they may be placed where we please, and are therefore always assumed to be known. the position of a point m in space is determined, if its distance from each co-ordinate plane be given; for by taking oA, oB, oC, fig. 4, respectively equal to the given distances, and drawing three planes through A, B, and C, parallel to the co-ordinate planes, they will intersect in m. 26. If a force applied to a particle of matter at m, (fig. 5,) make it approach to the plane oQ uniformly by the space mA, in a given time t; and if another force applied to m cause it to approach the plane oR uniformly by the space mB, in the same time t, the particle will move in the diagonal R R B B C m m fig. 5. mo, by the simultaneous action of these two forces. For, since the forces are proportional to the spaces, if a be the space described in one second, at will be the space described in t seconds; hence if at be equal to the space mA, and bt equal to mA mB ; whence mA= mB b the space mB, we have t = = a a b which is the equation to a straight line mo, passing through o, the origin of the co-ordinates. If the co-ordinates be rectangular, a is the tangent of the angle moA, for mBoA, and oAm is a b right angle; hence oA: Am::1: tan Aom; whence mA = oA× tan Aom = mB. tan Aom. As this relation is the same for every point of the straight line mo, it is called its equation. Now since forces are proportional to the velocities they generate in equal times, mA, mB are proportional to the forces, and may be taken to represent them. The forces mA, mB are called component or partial forces, and mo is called the resulting force. The resulting force being that which, taken in a contrary direction, will keep the component forces in equilibrio. 27. Thus the resulting force is represented in magnitude and direction by the diagonal of a parallelogram, whose sides are mA, mB the partial ones. m B fig. 6. 28. Since the diagonal cm, fig. 6, is the resultant of the two forces mA, mB, whatever may be the angle they make with each other, so, conversely these two forces may be used in place of the single force mc. But mc may be resolved into any two forces whatever which form the sides of a parallelogram of which it is the diagonal; it may, therefore, be resolved into two forces ma, mb, which are at right angles to each other. Hence it is always possible to resolve a force me into two others which are parallel to two rectangular axes or, oy, situate in the same plane with the force; by drawing through m the lines ma, mb, respectively, parallel to or, oy, and completing the parallelogram macb. 29. If from any point C, fig. 7, of the direction of a resulting force mC, perpendiculars CD, CE, be drawn on the directions of the y B B fig. 8. 30. Let BQ, fig. 8, be a figure formed by parallel planes seen in perspective, of which mo is the diagonal. If mo represent any force both in direction and intensity, acting on a material point m, it is evident from what has P been said, that this force may be resolved into two other forces, mC, mR, because mo is the diagonal of the parallelogram mCoR. Again mC is the diagonal of the parallelogram mQCP, therefore it may be resolved into the two forces mQ, mP; and thus the force mo may be resolved into three forces, mP, mQ, and mR; and as this is independent of the angles of the figure, the force mo may be resolved into three forces at right angles to each other. It appears then, that any force mo may be resolved into three other forces parallel to three rectangular axes given in position and conversely, three forces mP, mQ, mR, acting on a material point m, the resulting force mo may be obtained by constructing the figure BQ with sides proportional to these forces, and drawing the diagonal mo. 31. Therefore, if the directions and intensities with which any number of forces urge a material point be given, they may be reduced to one single force whose direction and intensity is known. For example, if there were four forces, mA, mB, mC, mD, fig. 9, acting on m, if the resulting force of mA and mB be found, and then that of mC and mD; these four forces would 2 B fig. 9. y 776 D be reduced to two, and by finding the resulting force of these two, the four forces would be reduced to one. 32. Again, this single resulting force may be resolved into three R B C fig. 10. m A forces parallel to three rectangular axes or, oy, oz, fig. 10, which would represent the action of the forces mA, mB, &c., estimated in the direc tion of the axes; or, which is the same thing, each of the forces mA, mB, &c. acting on m, may be resolved into three other forces parallel to the axes. 33. It is evident that when the partial forces act in the same direction, their sum is the force in that axis; and when some act in one direction, and others in an opposite direction, it is their difference that is to be estimated. 34. Thus any number of forces of any kind are capable of being resolved into other forces, in the direction of two or of three rectangular axes, according as the forces act in the same or in different planes. 35. If a particle of matter remain in a state of equilibrium, though acted upon by any number of forces, and free to move in every direction, the resulting force must be zero. 36. If the material point be in equilibrio on a curved surface, or on a curved line, the resulting force must be perpendicular to the line or surface, otherwise the particle would slide. The line or surface resists the resulting force with an equal and contrary pressure. 37. Let oA=X, oB=Y, 0C=Z, fig. 10, be three rectangular component forces, of which om=F is their resulting force. Then, if mA, mB, mC be joined, om=F will be the hypothenuse common to three rectangular triangles, oAm, oBm, and oCm. Let the angles moA-a, moB=b, moC=c; then X=F cos a, Y=F cos b, Z=F cos c. Thus the partial forces are proportional to the cosines angles which their directions make with their resultant. being a rectangular parallelopiped (1). of the But BQ FX + Yo + Z3. (2). Hence x2+Y®+Z9 cos a+cos b+cos c = 1. When the component forces are known, equation (2) will give a value of the resulting force, and equations (1) will determine its direction by the angles a, b, and c; but if the resulting force be given, its resolution into the three component forces X, Y, Z, making with it the angles a, b, c, will be given by (1). If one of the component forces as Z be zero, then c = 90°, F = √X2 + Y3, X = F cos a, Y = F cos b. 38. Velocity and force being each represented by the same space, whatever has been explained with regard to the resolution and composition of the one applies equally to the other. The general Principles of Equilibrium. R m B 39. The general principles of equilibrium may be expressed analytically, by supposing o to be the origin of a force F, acting on a particle of matter at m, fig. 11, in the direction om. If o' be the origin of the coordinates; a, b, c, the co-ordinates of o, and x, y, z those of m; the diagonal om, which may be represented by r, will be r = √(x−a)2 + (y—b)2 + (z—c)3 fig.11. But F, the whole force in om, is to its component force in hence the component force parallel to the axis or is In the same manner it may be shown, that are the component forces parallel to oy and oz. Now the equation Again, if F' be another force acting on the particle at m in another direction r', its component forces parallel to the co-ordinates will be, |