

DYNAMICS. 



tiooi 



' ' " 



1'I.ATE 



I I \1 I. 



Fig. 



Fig. 1. 



Fig. 1. 



Fig. 3. 



ihe direct object of thi* article, \et the following remark * 

 ma\ nut be improper, us tending to illustrate tl: 



li'uo nuitiiin ensue-, still the di.igonal will ex- 



;!ie ten.liney to nit. ' ict-tion of 



that ti-n li-nry. It will express tin- torn- ,:pplied in the. 



.ry dim-lion, which has destroyed thi 1 r.i. 

 whether t!. '.'i- ail active one, or merely tin- rc- 



: (il';i ti\i-d obstacle. In till' case of prc.-u 



will express the pressure that arises again. -t tin- obstacle; 

 a conclusion which we arc justified in draw ing from the 

 analogy established in the (irst Section betwixt pressures 

 iind the motions which they generate, and which, be- 

 is confirmed by experience. The case of impulses 

 is illustrated in MT.U- treatises of natural philosopliy, by 

 a complex piece of machinery. The case of prc.--un - 

 inav be illustrated by the following simple apparatus. 



l.ct the three weights hung around the two fixed 

 pulUvs. as in tile Figure , be in n state of equilibrium ; 

 let Al>. AC. lie taken proportional to the weights N, M, 

 respect i\ ely. which act in the directions of AB, AC; then 

 tbe parallelogram being constructed, the diagonal AD, 

 which is the equivalent according to our proposition, will 

 be found to be vertical, that is directly opposite to P, 

 and also proportional to I 1 . 



Con. I. The equivalent is represented by the third 

 side of a triangle, whose other two sides are drawn in 

 succession parallel and equal to the single forces. This 

 is evident by inspecting the triangle ABD. It is evi- 

 dent also, that the included angle of this triangle is the 

 supplement of that at which the forces act. 



The sides of any triangle perpendicular respectively 

 to those of ABD, will also express the magnitudes of 

 the forces, and of their equivalent ; for such a triangle 

 will be similar to ABD. 



Con. ,:. Ciiven two forces and the angle which their 

 directions make, the equivalent may be found either by 

 constructing the parallelogram, or the triangle, or by 

 trigonometrical calculation. The angle BAD being 

 found, will express the direction of the equivalent. 



It is easy to shew by help of Eucl. 12, 13. 2. that 

 'if a, b denote the two forces, c the angle which they 

 make, and d their equivalent, 



Then d=*/a*+2ab'X. cos. e+6 1 , the radius being 

 unity ; also the sine of the angle which d makes with 



a x sine, c 



n is = ; . 



a 



It is easy to see from 4?. 1 Eucl. that if the angle 

 is right, then d=-/a l + b*. 



COR. 3. \Vhen the two forces act at an angle, the 

 equivalent is less than the sum, and greater than the 

 difference ; and it is more nearly equal to the sum the 

 smaller the angle, so that when the angle vanishes it 

 will become equal to the sum. 



This is evident from the properties of a triangle. 



Con. 4. The simple forces are inversely as the per- 

 pendiculars dropt on their directions from any point in 

 the direction of the equivalent. 



1st, If dropt from D, the AACD= AADB, hence 

 ACxDMrrABxDN. Therefore AC : AB : : DN : 

 DM. 



2d, If dropt from any other point G, it is easy to 

 see by similar triangles, that DN : DM : : GE : GF ; 

 hence AC : AB : : GE : GF. 



CASE 2. Any number of forces applied nt a point. 

 The equivalent will be found by compounding two ac- 

 cording to last case, then compounding the equivalent 

 with a third, and so on. 



If all the forces lie in one plane, it is evident that 



the i:- itiou* way of finding the equivalent will (i 



l>e by drawing ucces.-ively line- equal and parallel to 

 thoM' rcpri'.-cntiug the forces, as in Figure ; and then _'""*' 

 ihe line drawn in complctcing the polygon will be the p, ATE 

 equivalent. i ( \l.l. 



The equivalent of three force.-, is the diagonal of a Fig. i. 

 juirallelopiped, of which the three forces are the adja- 

 cent lineal edp 



AH, AC, AD being the three forces applied at A, Fig. j. 

 the diagonal AE will be the equivalent. 



For joining HE, AG, the equivalent of AC. A I) is 

 Ad, because DC is a parallelogram, anil since the plane 

 in which the parallels AB, EG are, is cut by parallel 

 planes, the lines of common sec-lion BE, AG, will be 

 parallel, (Eucl. 10'. 11.) hence the Figure BG is a pa- 

 rallelogram, therefore AE is the equi\ alent of AH, AG, 

 that is of AH, AC, AD. 



It is ca.sy from the Figure to perceive, by help of 

 (Eucl. 47. 1.) that ii'the purallclopi]>cd is rectangular, 

 and if a, I, c denote the three simple forces, and d 

 the equivalent, then d=y' 1 + 6 1 + c 1 . 



( \ t- K 3. Two parallel forces applied to different 

 points of a body in the same direction. Since the 

 two parallels may be considered as coming from a 

 point infinitely distant, or as making an infinitely small 

 angle with one another, their equivalent will be equal 

 and parallel to their sum by Cor. 3. to ( a-e 1. and it 

 will divide the straight line joining the points of appli- 

 cation in the inverse ratio of the forces. 



For by Cor 4. the forces are inversely as the perpen- 

 diculars on their directions, which, in this case, are as 

 the segments of the joining line. The Figure is suffi- Fig. 6. 

 cient to show this. 



CASE 4. Tno parallel forces applied in opposite di- 

 rections. The equivalent will be equal to their differ- 

 ence, in the direction of the greater, on the other side 

 of it, and divide the straight line produced through the 

 points of application in the inverse ratio of the forces. 



This will follow from last case ; for let NA be a pjg. 7 

 force which, together with CD, balances FE, or, in 

 other words, which prevents the line AD from moving, 

 then, by last case, NA+CD must be =FE, and CD 

 being to NA :: AE: ED, CD will be to FE, as AE: AD; 

 but since NA keeps the line from moving, it must In- 

 equal and opposite to the equivalent which we are seek- 

 ing ; hence, if AG= AN, what we have demonstrated 

 of AN will hold of AG. 



CASE 5. Any number of parallel forces. Compound 

 two, then the equivalent with a third, and so on. If 

 some act in one direction, and others in the opposite, 

 it will be most convenient first to compound all those 

 in one direction, and then all those in the opposite ; 

 and, lastly, compound the two equivalents. It is easy 

 to see from the nature of the process, that the point 

 through which the last equivalent must pass will be 

 the same, so long as the forces and points of applica- 

 tion are the same, although the direction of the for< 

 should be changed. This remarkable point is called 

 the centre of parallel forces. 



CASE 6. Forces applied to diffi-reut points, btu not pa- 

 rallel. If they are in one plane, you may produce 

 them till they intersect, and then compound them two 

 and two. If they are not in one plane, but converge 

 toward one point, you may consider them all as ap- 

 plied at that point, and then find their equivalent, as 

 in Case 2. It they do not converge toward the same 

 point, the case requires the reduction of forces, and 

 will be explained under Art. 4. of next Section. 



