POSITIVE AND If EG ATI VE MOMEXT8. ] MECHANICAL PHI LOS OPHY. STATICS. 



701 



But C P' B' Q' (Fig. 38) is a parallelogram, and there- 

 fore P' R' = C Q'. 



Hence 



DF OF 

 D E ~ C~Q' 



or, The perpendicular 



from the point D on the direction of the force Q _ P 

 from the point D on tue direction of the force P Q 

 Consequently, the perpendicular from the point D on 



the direction of the force Q multiplied by Q, equals the 



perpendicular from the point D on the direction of the 



force P multiplied by P. 



MOMENT OF FORCES AND FULCRUM. If we 



suppose an axis , H D (Fig. 40), passing through the 



Fig. 40. 



plane in which the 

 forces act, fixed with 

 perfect rigidity, and 

 perpendicular to 

 that plane at the 

 point D, and then 

 placed on a stand, 

 as in the accom- 

 panying figure, so 

 as to allow the plane 

 to move with per- 

 fect freedom about 

 the axis H D, then, 

 since the resultant 

 of the forces P and 

 Q acting at A and 

 B passes through 

 the point D, their 

 joint effect will be counteracted by the reaction of the 

 stand on the axis H D. Neglecting the weight of the 

 plane and the friction of the axis, the plane will be in a 

 state of equilibrium, under the influence of the two 

 forces P and Q, acting at A and B in the directions A P 

 and B Q, together with the reaction of the stand acting 

 on the plane at the point D in a direction opposite to 

 the resultant of the forces P and Q. If the rigid body 

 be reduced to the rigid line A B, A B is called a lever, 

 and the point D its fulcrum. 



Since the conditions of equilibrium, or that the re- 

 sultant of the forces B and Q should pass through D, 

 are as above stated, that, 



The perpendicular from D on the d : rection of the force 

 Q multiplied by Q, equal* the perpendicular from D on the 

 direction of the force P, multiplied by P ; it follows that 

 the force P may be Fig. 41. 



replaced by a force 

 S without altering 

 these conditions, 

 provided that the 

 perpendicular from 

 D, DK on the di- 

 rection K S of the force S multiplied by S, be equal to 

 the perpendicular from D on the direction of the force P 

 multiplied by P (Fig. 40). 



The writers on Mechanics have applied the term 

 moment or momentum to this product of the linear units 

 in the length of the perpendicular from a given point on 

 the direction of a forco, by the units of force in the given 

 force. 



Thus, if the forces P, Q and S be represented in mag- 

 nitude and direction by E P, F Q, and S K, and D E, 

 D F and D K be the perpendiculars on those directions, 

 P E x D E, Q F x b F, and S K x D K will be the 

 respective momenta or moments of the forces P Q and Sj 

 about the point D. 



These products are not only termed the moments of 

 the forces with respect to the point D, but also the 

 moments of the forces with respect to axis H D, perpen- 

 dicular to the plane in which they act. 



The student mast be careful not to confound these 

 statical momenti, or momenta, with the momenta of dy- 

 namics ; the momcnttim of dynamics being the product 

 of the mass of a body by its velocity. 



If the force P alone were to act on the plane at the 



point A, it would evidently twist the plane about the 

 fixed axis H 1), and Q acting alone at B, would twist the 

 plane about H D in a direction opposite to that in which 

 P would do so. The tendency of P therefore to twist 

 the plane about the axis H D, or point D, is counter- 

 acted by the tendency of Q to twist the plane in the 

 opposite direction, in the case where equilibrium 

 exista. 



We have seen, however, that the tendency of Q to 

 twist the body about the axis H D, will be counteracted 

 by any other force whose statical moment, with respect to 

 H D or D, is equal that of the force P. 



Hence we may consider the statical moment of a force 

 about a given point or axis, as a measure of its tendency 

 to twist the plane to ivhich it is applied, about that point 

 or axis. 



POSITIVE AND NEGATIVE MOMENTS. The 

 moments of those forces whoso tendency is to twist a 

 body about the axis or fulcrum, in the direction in which 

 the hands of a watcli move, are termed positive ; those 

 whose tendency is in the opposite direction, are called 

 negative. 



It will readily be seen, that if the directions of the 

 forces P and Q in Proposition X. had been parallel to 

 one another, the demonstration there used cor.'.d not be 

 applied ; as, in that case, the directions of the forces 

 could not be produced till they met. 



PROPOSITION XI. 



To find the magnitude and direction of tlir, rcsnltftnt of 

 two forces acting on different points of a rigid body, the 

 directions of the forces being in ihe same plane, but 

 parallel to each other. 

 Let P and Q (Fig. 42) be the two forces, A and B the 



two points in the plane. 



Fig. 42. 



The force P being represented in magnitude and 

 direction by the line A P, and Q by the line B Q. 



Then A P and B Q are parallel to each other. 



Join A B. 



The conditions of equilibrium will not be altered if 

 any arbitrary force S be applied to A in the direction 

 A S, and to B in the opposite direction B S. 



Let these equal and opposite forces be represented in 

 magnitude and direction by A S and B S. 



Consequently A S = B S. 



Complete the parallelogram S B Q T and the parallel- 

 ogram S A P R. 



Join A R and B T. 



Then A R=R represents the resultant of the forces P 

 and S in magnitude and direction, and may be substi- 

 tuted for these forces. 



Also B T=T represents the resultant of Q and S in 

 magnitude and direction, and may be substituted for 

 them. 



Produce A R and B T to meet in the point C. 



Through C draw C D parallel to A P or B Q, meeting 

 A B in D. 



The force R acting at A in direction A R may be 

 transferred from R to C in the line R A C, and repre- 



