therefore multiplying together, and remembering that S= S', we get P BC (ii). Hence the point C in the line AB, through which the resultant acts parallel to each of the forces, divides the line AB into segments which are inversely proportional to the forces. (i) and (ii) determine the resultant completely. 30. If the two forces act in opposite directions the method is very similar: the point C lies It will be observed that (iii) and (iv) are the same as (i) and (ii), if the sign of Q be changed, so that algebraically (i) and (ii) comprise both cases. COR. 1. The position of the point C does not depend upon the direction of the forces. Hence if the directions of the forces be turned through any the same angles in the same direction about the points A, B, the position of C will not be changed. COR. 2. In the case of Art. 30, we have from (iv), If now PQ, we get AB =0, or AC=∞o, and R=0. A system of two equal forces acting in opposite directions and not at the same point is called a couple ;—and the results R=0, and AC∞ with reference to such a system. indicate that a couple cannot be replaced by any single finite force acting at a finite distance. 31. Moment of a Force. The product of a force into the perpendicular distance of its line of action from a given point is called the moment of the force with respect to the point, or the moment of the force about the point. If an axis be drawn through the point at right angles to the plane which contains the point and the direction of the force, this product is called the moment of the force about the axis. Further, The moment of a force about any line is defined to be the product of the resolved part of the force perpendicular to the line into the perpendicular distance between the line and the line of action of the force. This perpendicular distance is the shortest distance between the two lines. The student will be careful to observe that the force and distance here spoken of are expressed numerically in terms of their respective units; and the moment consequently is the product of two numerical quantities. Thus if AB represent a force P, and O be any given point, Op perpendicular to AB, and if m, n be the number of linear units in AB, Op respectively, then will mn be the moment of P about O, or about an axis through O perpendicular to the plane 0 ABO. Also since the area of the triangle ABO = AB. Op, it is obvious that mn = twice the number of units of area in the triangle ABO. We may then represent moments geometrically by areas, and the moment of P about O would thus be represented by twice the triangle ABO: the unit of moment (i.e. the product of a unit of force into a unit of distance) being represented geometrically by a unit of area. Further, the force P would tend to twist the body on which it acts in one direction or the reverse, according as O is on one side of AB or the other. We shall for convenience consider the moment of a force negative or positive, according as it tends to twist the body in the same direction as the hands of a watch revolve, or the contrary. If P, Q be two equal forces acting in parallel but opposite directions-constituting a couple-if C be any point in the plane of the forces, and CBA be perpendicular to their lines of action (fig. Art. 30), we have the moment of the two forces about C=P.AC-Q.BC=P.AB= constant, i. e. the moment of a couple is the same about any point in the plane of the couple. 32. The following proposition is important. The algebraic sum of the moments of two forces acting in one plane about any point in the plane is equal to the moment of their resultant. When the forces are not parallel it admits of a simple geometrical proof. Let AB, AC represent the two forces P, Q, and complete the parallelogram BC, and through the point O draw rOs parallel to AC; then taking those moments to be positive which tend to twist a body in a direction opposite to that of the hands of a watch, sum of moments of P and Q about O =2AAOC-2AABO = = parallelogram Cr-2AABO = parallelogram BC- parallelogram rD – 2▲ABO = 2▲AOD = moment of R the resultant of P and Q. B The above construction will apply if the point O lie within the angle BAC, or the vertically op posite angle. If O lie within either of the supplemental angles of BAC, as in fig. 2, draw Ors parallel to AC, then sum of moments of P, Q about O c 2ΔΑ00+2ΔΑΟΒ = = = parallelogram Cr + 2AAOB = parallelogram CB-parallelogram Bs +2▲AOB = 2 (AABD+▲AOB-ABOD) = 2AAOD = moment of R the resultant of P and Q. 33. It remains to prove the proposition for two parallel forces. Let A, B be two points through which the forces P, Q act, Ca point in the line AB through which the resultant R passes. Take any point 0 and through P it draw Obca at right angles to the directions of the forces; then since the resultant of P, Q passes through C, when the forces P, Q act in the same directions (fig. 1), we have a sum of the moments of P, Q about the point =Q. Ob+P. Oa (1) - moment of the resultant about O. Q.E.D. If the forces act in opposite directions (fig. 2), the student will have little difficulty in proving that Q. Oc-P. Oa = ( Q − P). Ob, which expresses the same proposition in this case. Obs. The point O has been taken in such a position that the moment of the resultant is in each case positive. The proposition is readily proved for any other position of 0. COR. 1. If the point O be taken anywhere in the line of action of the resultant R, the moment of R vanishes, and |