which answer it more or less approximately: i.e. we know of many substances, which will submit to the action of considerable forces without undergoing any appreciable change in shape or size. The results which we shall prove absolutely true for perfectly rigid bodies, will be so approximately for bodies that are approximately rigid.

49. We have already seen that any system of forces acting on a particle is equivalent to a single force, i.e. there is a single force, such that its effect on the particle could not be distinguished from that of the combined forces. The question now presents itself, whether this is so or not when the forces do not all act on a single particle, but on different particles of a system. If the particles form a rigid body, we shall see that under certain circumstances there exists a force, which together with the given forces would keep the body in equilibrium, so that the effect of this force reversed on the body as a whole, is the same as that of the original forces. But it must be remembered that it is only on the body as a whole that the effects are the same necessarily: the internal forces called into play by the single force, are not necessarily the same as those called into play by the system of forces, in fact are generally very different. When we cannot find a single force whose effect on the body as a whole is the same as that of the system of forces, we can always find a different system of forces whose effect will be the same, though they will not generally give rise to the same internal forces. It is usual to speak of the single force, when such a one exists, as the resultant of the original forces, and the second set of forces as equivalent to the first, though it must always be understood that they are so, strictly speaking, only in one sense. when the body is not rigid, a single force, or set of forces, which would, if the body were rigid, be equivalent in the above sense is said to be, one the resultant of, the other equivalent to, the original system of forces.

Even

50. Prop. Any rigid body, under the action of any system of forces, can be fixed by applying single forces of the

requisite magnitude at each of any three given points of the body not in the same straight line, the direction of the force at one point being at right angles to the plane containing the three points, and that of the force at a second point at right angles to the line joining it to the third.

Let A, B and C be any three points of the body. The body can be fixed by the following constraints: imagine a very small spherical socket to be made in the body at A, and a ball just smaller than the socket to be placed in it, and the ball to be fixed. Now imagine a very small hoop with its plane perpendicular to AB, to be fixed round B, and also some obstacle to be placed to prevent C from moving at right angles to the plane ABC. The first constraint prevents the body from moving in any way except by turning about A, and exerts a single force through A as the ball and socket touch in only one point: the second prevents B from turning about A, and therefore from moving at all, so that the body can now only turn about AB; the second force acts at right angles to AB. The third constraint prevents C from moving round AB, and therefore from moving at all, and exerts a single force through C at right angles to the plane ABC. As C cannot turn about AB, it is clear that the body is now fixed.

51. This proposition can be extended to the case in which any of the points A, B, and C are not situate in the body. For we may imagine them to be made so in effect, without introducing any forces external to the whole system, by arranging a system of rigid rods, without weight, rigidly connecting them with the body.

Cor. If the lines of action of the external forces all lie in one plane, the body can be fixed by the application of single forces at any two given points A, B in that plane, each force being in the plane, and the direction of one being at right angles to the line AB.

For by the last proposition, if a third point C be taken in the plane but not in AB, the body can be fixed by the application of suitable forces P, Q, R at A, B, C respec

tively, the direction of R being perpendicular to the plane, and that of perpendicular to AB. The body is now in equilibrium under the action of the internal forces, the original external forces, and the forces of constraint P, Q, R. Hence (Art. 46) the algebraical sum of the moments about AB of all the external forces, including P, Q, and R, is zero: but each of these moments except that of R is zero, since each of the corresponding forces either intersects AB or is parallel to it. The moment of R must therefore be zero, i.e. R itself is zero, since R neither meets AB nor is parallel to it. Similarly we may shew that the moment of about every line through A in the plane is zero, i.e.Q is either zero, or lies in the plane in question. Also by taking moments about lines through B in the plane, except AB, we may shew that either P is zero, or it lies in the plane.

52. Prop. A number of forces acting on a rigid body, their lines of action all being in the same plane, will keep it in equilibrium, provided any of the following sets of conditions hold:

(1) If the algebraical sum of their moments about each of three given points in the plane, but not in the same straight line, be zero.

(2) If the algebraical sum of their moments about one given point in the plane, and of their resolved parts in any two given directions in the plane, be zero.

(3) If the algebraical sum of their moments about two given points in the plane, and of their resolved parts in any given direction in the plane, not at right angles to the line joining the two points, be zero.

(1) Let A, B, C be the three given points: if the body is not in equilibrium, it can be fixed by applying forces of constraint P and Q at A and B respectively, both in the plane of the forces and Q perpendicular to AB (Art. 51). The whole system of forces including P and Q must satisfy the necessary conditions of equilibrium:

therefore the algebraical sum of their moments about A is zero: but the algebraical sum of the moments about A of the forces excluding P and Q is zero, and the moment of P about A is zero also; hence the moment of the remaining force is zero, i.e. Q itself is zero, as it does not pass through A. Similarly we can shew that the moments of P about both B and C are zero; hence either P is zero, or it passes through both B and C. As A, B, and C are not in a straight line, P must be zero. Hence the body is in equilibrium without any constraint.

Q

(2) Let A be the given point, and B any other point in the plane of the forces: apply forces of constraint P and Q at A and B respectively as in (1). Then we shew as before that is zero. The forces including P must satisfy the necessary conditions of equilibrium: therefore the algebraical sum of their resolved parts in each of the two given directions is zero; but the algebraical sum of the resolved parts of the forces excluding P, in each of these two directions is zero, i.e. the resolved part of P in each of these directions is zero. But as the resolved part of a force is only zero, in a direction perpendicular to the force, P itself must be zero.

Hence the body is in equilibrium without constraint.
Case (3) can be proved in a similar way.

53. Prop. Two equal forces acting in opposite directions along the same straight line on a rigid body, but not necessarily on the same particle, keep it in equilibrium.

This is obvious as the two forces clearly satisfy the sufficient conditions of equilibrium given in the last article.

The

This proposition is essentially the same as the principle known as the Transmissibility of Force, which is generally assumed as an experimental fact, but which we prefer to deduce as above from the Laws of Motion. formal statement of that principle is as follows: when a force acts on a rigid body, it is indifferent on what particle in the line of action it acts, provided that particle is part

of the body, or rigidly connected with it. This follows directly from the proposition just proved. For let A, B be any two particles, in the line of action of the force P, and rigidly connected with the body. We have just proved that a force Q equal and opposite to P, will counterbalance it, provided acts at a point in AB rigidly connected with the body: hence the force P counteracts Q, whether P acts at A or at B. As regards its effect on the body as a whole, we may say then, that it is indifferent at which point we apply the force P. It is however in this sense only, that it is indifferent; if we take into consideration the internal forces brought into play in the two cases, they will probably be very different.

Imagine, for instance, a sphere resting on a smooth horizontal plane; a force of a certain magnitude, and in a certain direction will give the sphere the same change of velocity, whether the force take the shape of a push behind or a pull in front, yet the internal forces in the sphere will be different in the two cases, as in the first case the tendency of the external force is to compress the sphere, whereas it has the opposite tendency in the second case.

The proof of the converse principle, viz. that if it is indifferent at which of two points a force is applied, the line of action of the force must be the line joining them, is obvious from what has gone before.

Ex. 1. A square lamina ABCD is acted upon by a force of 3 lbs. along AB, 2 lbs. along CB, 1 lb. along CD, 2 lbs. along AD, √2 lbs. along CA, and 2 lbs. along BD: prove that it is in equilibrium.

Ex. 2. A weightless rod AB, 10 feet long, has weights of 7 lbs. hung at each end, and one of 11 lbs. at its middle point: a string is attached to a point 2 feet from A and after passing over a smooth peg vertically above the point of the rod to which it is attached, supports a weight of 10 lbs. another string attached to a point feet from B supports in a similar way a weight of 15 lbs. Prove that the rod is in equilibrium. Ex. 3. A rigid rod AB, 20 inches long, is acted upon by the following forces: 3 lbs. at A along BA, √3 lbs. at right angles to AB, at a point 5 inches from A, 6 lbs. at a point 5 inches from B, and making an angle of 60o with the part of the rod on the same side as A, and 4/3 lbs.