rection of the forces, and assume the point C, in which it from the principle of moments (Art. 36), or, PX CL Q× CN; = P: Q: CN: CL. But, from the similar triangles CNS and LNM, we have, CN: CL:: SN: SM. Combining the two proportions, we have, P: Q: SN : SM. That is, the line of direction of the resultant divides the line joining the points of application of the components, inversely as the components. From the last proportion, we have, by composition, P: Q: P+Q :: SN : SM : SN + SM; and, by division, P: Q: P-Q : SN: SM: SN-SM. When the forces act in the same direction, P+Q will be their resultant, and SN + SM will equal MN. Since P+Q is greater than either P or Q, MN will be greater than either SN or SM, which shows that the resultant lies between the components. - When the forces act in contrary directions, P Q will be their resultant, and SN-SM will equal MN. Since PQ is less than P (supposed the greater of the components), MN will be less than SN, which shows that the resultant lies without both components, and on the side of the greater. Substituting in the preceding proportions P+Q, P — Q, SN+SM, and SN-SM, their values, we have, P: QR:: SN: SM: MN. That is, of two parallel forces and their resultant, each is proportional to the distance between the other two. Geometrical Composition and Resolution of Parallel Forces. 41. The preceding principles give rise to the following geometrical constructions: 1. To find the resultant of two parallel forces lying in the same direction: Let P and Q be the forces, M and N their points of application. Make MQQ, and NP' P; draw P' Q', = P'Q', cutting MN in S; through S draw SR parallel to MP, and make it equal to P+Q: it will be the resultant. For, from the similar triangles P'SN and Q'SM, we have, Mo R Fig. 23. P'N: Q'M:: SN: SM; or, P: Q:: SN: SM. After the construction is made, the distances MS and NS may be measured by a scale of equal parts. EXAMPLE. Given P 9 lbs, Q = 6 lbs., and MN = 30 in. Re 2. To find the resultant of two parallel forces acting in opposite directions: Let P and Q be the forces, M and = = Q; N their points of application. Prolong For from the similar triangles SNA and SMB, we have, AN: BM : SN: SM; or, P: Q Given P = 20 lbs., Required SN. M Fig. 24. : SN: SM. = We have R 208 12; hence, from Proportion (8), 12: 20 :: 18: SN; .. SN 30 in. Ans. 3. To resolve a given force into two parallel components lying in the same direction, and applied at given points: Let R be the given force, M and N the given points of application. Through M and N draw lines parallel to R. Make MA = R, and draw AN, cutting Rin B; make MP = SB and NQ BR; they will be the required components. S Mo R Fig. 25. For, from the similar triangles AMN and BSN, Given R24 lbs., SM = 7 in., and SN = 5 in. Re 4. To resolve a given force into parallel components lying in opposite directions, and applied at given points. Both points of application must lie on the same side of the given force. Let R be the given force, M and N the given points of application. Through M AN: BN SN: MN; or, AN: R:: SN: MN. But, from Proportion (8), we have, PR: SN: MN; .. AN= P, and AB .. EXAMPLE. Given R = 24 lbs., SN = 18 in., and SM = 9 in. Re p" 5. To find the resultant of any number of parallel forces. Let P, P', P", P'"', be such a system of forces. Find the resultant of P and P', by the rule already given, it will be R' = P + P'; find the resultant of R' and P", it will be R" = P + P' + P"; find the resultant of R" and P'", it will be R = P+P' + P'' + P'". If there is a greater number of forces, the operation of composition may be continued; the final result will be the resultant of the system. If some of the forces act in contrary directions, combine all which act in one direction, as just explained, and call their resultant R'; then combine all those which act in a contrary direction, and call their resultant R''; finally, combine R' and R' by a preceding rule; their resultant R will be the resultant of the system. R" Fig. 27. R If R' = R", the resultant will be 0, and its point of application will be at an infinite distance. In this case, the forces reduce to a couple, the effect of which is simply to produce rotation. Lever Arm of the Resultant. 42. Let P, P', P", &c., denote any number of parallel forces, and p, p', p', &c., their lever arms with respect to an axis of moments, taken perpendicular to the common direction of the forces; denote the lever arm of the resultant of |