Equilibrium of a Particle

The forces P and Q evidently have the same moments about any other point H in OC, the line of action of their resultant.

Conversely, if the resultant moment of two parallel forces about any two points is zero, their resultant must act along a line which passes through those two points, and is therefore definite. We might, in fact, have determined the resultant of P and Q in this way.

45. Now consider the moment of the resultant force about any other point K in the plane of the forces. When the forces P and Q have the same sense (diagram A), their resultant is a force of magnitude (P+Q), and its moment about K is given by shown, along a line AB which cuts the line of action of P and Q; and we may combine, according to the vector law, one of these forces with P and the other with Q.

The resultant of P and S is a force and the resultant of Q and S is a force R2. In general, the lines of action of and R2 will intersect, as shown, at some point 0, and according to the principle of transmissibility this point may be regarded as the point of application both of and R2. If we now replace by forces P and S, and R2 by forces Q and S, acting through 0 as shown, the two forces S will again neutralize. So we are left, When P and Q are opposite in sense (diagram B), their result ant is a force of magnitude (P—Q), and its moment about K is given by 46. In both cases we see that the sum of the moments of P and Q about K (taken with due regard to sign) is equal to the moment of their resultant about the same point. The same theorem is true of two forces which intersect (fig. i6). The moment of P about K is PKL, i.e., it is measured by twice the area of the triangle KOP; and similarly, the moments of Q and R are measured by twice the areas of the triangles KOQ and KOR respectively. Since, however, QR=OP, the triangle KOP is equal in area to the sum of the triangle OQR and KQR : therefore the sum of the triangles KOP, KOQ is equal in area to the triangle KOR; i.e., the sum of the moments of P and Q about K is equal to the moment of their resultant R about the same point.

This theorem can be extended to any number of forces, whether parallel or intersecting. In the case of two parallel forces con stituting a couple (§ 43), the total moment about any point in their plane has a constant value, given by the product of either force with the distance between their lines of action.

Centre of Gravity.-47.

The total weight of a body (i.e., the force exerted upon it by the attraction of the earth) is the re sultant of forces exerted upon all the particles which go to form that body. When its dimensions

are small in comparison with those of the earth, the force on each particle is proportional to its mass, and all the forces may be taken as parallel: under these conditions we may show that the weight of a rigid body acts always through a point fixed in relation to the body.

Let Ox, Oy, Oz be three perpendicular axes fixed in relation to the body, and consider the components in the direction Oz of the forces which act upon three particles A, B, C, of mass MA, MB, MC. Whatever be the direction of the resultant forces on the particles, these components also will be proportional to the respective masses; so we may take them to be given by perpendicular to CA and CB respectively. Hence, when the po sition of G in AB is known, we have only to draw lines OA, OG, OB in three known directions, and to place AB so that it is divided by these lines into segments which have a definite ratio : when this condition is satisfied, AB has the required slope.

Catenaries.-49.

Again, we may employ the theorem to cal culate the curve in which a heavy chain AB (fig. 19) will hang in equilibrium under gravity, and the tensions which will be brought into play. Since the chain is flex ible, the tension at every point will act along the tangent to the curve. Let T be the tension at any point P, and H the tension at the lowest point 0. The portion of the chain which extends be tween 0 and P may be regarded as held in equilibrium by three forces; viz., the tensions T and H, acting in the directions CP, CO, as indicated, and the resultant weight of the portion OP. Since these three forces must be concurrent, the centre of gravity of the portion OP must lie in the vertical line through C; the triangle PCN (fig. 19) will be the triangle of forces (§ 39) ; and if W is the weight of the portion OP, we have Let Ox, Oy be drawn horizontally and vertically through 0, and let s denote the length of the curve measured from 0. Then we have, from (59), Generalizing this result for any number of particles consti tuting a total mass M, we obtain the formula (33) of § 23; and applying the same argument to moments about the axes Oz and Ox, we find that the resultant of all the forces on all the particles, i.e., the total weight, will in every case pass through a point which in that article was termed the mass centre of the system. In its present signifi cance it is termed the centre of gravity of the system, i.e., of the body which the particles compose.