MECHANICS. 



Now, suppose we hare an !>or of such forces, say 



>. counting : inwards tint 



iniiMlc, there will l>.i mi ihc middle of tho 



B will In- in the middle n( i middle 



i-.tir <>t' forces. Wliat I. i u'( Tho in ifostwo 

 pounds at t, < , ..oes tho next 



M ^\ outside, and so tho next; and 



~~7 111 2 7 / tnere aro ov iJ ent ly thus six 



/ l_s pounds of resultant at tho cen- 



** tro of A B. Tako any other even 



F 'K- l3 - number, and tho result is tho 



same; and thus, for both odd 



and even numbers, wo arrive at this conclusion : Tho resultant 

 of any number of equal parallel forces acting on a body at equal 

 distances along a line, is equal to their sum, and bisects tho 

 line joining tho points of application of tho extreme forces. 



An instance of this is tho working of a hand fire-engine. 

 Suppose seven men at tho lever on either side, that is, fourteen 

 hands on each lever ; supposing tho men to bo equally arranged 

 and of equal strength, this makes fourteen equal forces applied 

 at equal distances, the resultant of which is the muscular power 

 of seven acting at tho centre on either side. 



Now we shall, without difficulty, find the resultant of two 

 unequal parallel forces. As before, let A and B be their points 

 of application, and let us first suppose that they act in the tame 

 direction. Measuring the forces by pounds, or ounces, or even 

 grains, there are three cases which may occur. The number, 

 eay of pounds, in the forces may be both even, or both odd, or 

 one odd and the other even. 1. We shall take "both even " first, 

 and, for simplification sake, let them be six at A and four at B. 

 Divide now the line A B into ten equal parts, that is, into as many 

 parts as four and six together make. Extend also A B on either 

 aide, as represented (Fig 14) by the dotted lines, and measure 

 oif on the extensions any number of portions you, please, each 



equal to one of th- 



?.. r .. r -7N subdivisions of A B. 

 / T I I 4 7y Beginning at A, sup- 



+ pose you apply apound 



V force at the end of the 



first subdivision to the 

 right, another pound 

 at the end of the third, 



another at that of the fifth, and so on until you come to B. 



lou wi,l find then that there will be a pound at the end of the 



rt division from B. Put pounds now at the end of the first 



division from A on the dotted line, on the third, and on the fifth 



and do the same on the dotted line from B, on the first and third 



Junt all the pounds you have ; they are ten, five inside and five 



Calling the points occupied by the extreme pounds 



1 Q, the resultant of these ten, so distributed at equal 



mces, must pass through the middle, M, of P Q, and be ten 



)unds, by the principle last established. But if we take 



separately the three outside and the three inside A, they make six 



pounds acting at A. Also the two pair on either side of B make 



four pounds at B. The ton pounds at M must therefore produce 



the same effect on the body as the six at A and the four at B 



and therefore must be the resultant of these forces ; that is to 



say, the resultant is the sum of the components. 



But count now the number of subdivisions on either side, 

 from M to A and B. There are four on tho side of A and six on 

 B'S side that is to say, the resultant cuts tho line A B in the 

 proportion of the numbers 4 and 6, with this peculiarity, how- 

 ever, that the smaller number is on the side of the greater force. 

 This is what we might expect, for the resultant ought naturally 

 to tend towards the greater, on account of its preponderance. 

 When a line is cut in this way, the smaller portion being on the 

 side of the greater number of pounds, it is said to be cut 

 inversely as the two numbers that is, in the contrary order. 



. Now let us take the case of two odd numbers ; let them be 

 9 and 7. It is evident that if we put another 9 pounds at A, 

 and i at B, the resultant of this second 9 and 7 should in every 

 respect agree and coincide with that of tho first, and that the 

 resultant of the four should be the sum of two nines and two 

 sevens. But the double S) at A is 18 pounds, and the double 7 

 at B 14 pounds. Tho case, therefore, becomes one of even num- 

 bers, and the line A B, as proved above, must be cut by the 

 resultant in the inverse proportion of 18 to 14. But to divide 

 a line so that there- way bo 18 parts one side and 14 on the 



Fig. 15. 



other becomes, by throwing every two of the subdivision* into 

 one, tho same thing a dividing it so that 9 may be on one side 

 and 7 on the other. In this case then, also, A B is divided 

 inversely as the forces. 



3. When the numbers are one odd and the other even, say 4 

 and 7, tho result is tho same. By doubling each force you get 

 8 and 14 pounds, both even numbers ; the line A B is divided by 

 the resultant inversely as 14 to 8, which is the same as 7 to 4 

 inversely as tho forces. 



We have supposed in all these cases that the forces con- 

 tained an exact round number of pounds ; bat what should we 

 do if there were fractions of a pound in either or in both ? I 

 say, reduce the forces to ounces, and work by round numbers in 

 ounces. If there were fractions of ounces, work in grains. 

 You con thus still secure round numbers, and the above proof* 

 will hold good. But what are 

 you to do if there are fractions 

 of grains ? Work them by 

 tenths, or hundredth, or thou- 

 sandth parts of grains, or by even 

 far smaller fractions, and you 

 will still have round numbers, 

 and you can say that the result- 

 ant cuts A B inversely as these 

 numbers, however great they be, 



and therefore inversely as the forces. To trouble yon about 

 smaller fractions would only get you into a cloud of metaphysics 

 for no practical purpose. 



I have proved this important principle only for particular 

 even numbers, 6 and 4, but you will find that the reasoning will 

 be the same whatever bo the even numbers you choose. The 

 rule simply is to divide the line A B into as many equal parts as 

 there are pounds in both forces, and then to distribute all the 

 pounds at A in two batches on either side of that point, and to 

 do the same at B with the pounds there acting, observing to 

 place tho pounds as you go from A or B in any direction, at the 

 first, third, fifth, and so forth, points of division. 



You are now in a position to find the resultant of three or 

 more parallel forces acting, say, at the 

 points A, B, c, D, as in Fig. 15. First 

 join A with B, and cut it inversely as the 

 forces which are there applied : next join the 

 point x so found with c, and cut the join- 

 ing line at Y inversely as the sum of the two 

 first forces to that at c; join this again 

 with D, and cut it inversely as the three 

 first forces to that of D ; and so proceed 

 until you have exhausted all the forces. 

 The point z last found is that through 

 which the resultant of all passes, and is 

 called the centre of parallel forces. 



Suppose, for example, that the centre was required in the case 

 of parallel forces of 1,2, 3, and 4 pounds applied to the four 

 corners of a square board, A, B, c, D (Fig. 16). First divide AI 

 into three parts, and take two next to A and one to B. The 

 point x so found is the parallel centre for these two forces. 

 Join X now with c, and cut x c into six parts (the sum of 1, 2, 

 and 3), and take three next to c and three to X. The centre Y 

 so found, which evidently will be the middle of c x, is the centre 

 of the three. Now join T with D, and divide Y D into ten parts 

 (the sum of 1, 2, 3, and 4), and take four next Y and six next 

 D. This last point, z, is the centre of all the given forces. Try 

 your own hands now on the following Examples, and in the next 

 lesson we shall have for subject the centre of gravity, which is a 

 centre of parallel forces. 



Examples. 



1. Three equal parallel forces act at tho corners of a triangle ; find 

 the centre through which their resultant passes. 



2. A force of a pound is applied to one end of a beam, of three at 

 the other, and of two at the middle ; find the centre of these forces, 

 they being 1 parallel to each other. 



3. A weight of one pound and three-quarters hangs from one em", of 

 a rod which is two feet in length, and of three and a-half from the 

 other ; find tho magnitude of the resultant, and the centre of parallel 

 forces. 



4. A door is seven feet high and three feet wide, and the centres o* 

 its hinges are distant one foot from its ends. A force of twenty-three 

 pounds ic applied along its upper edge, pulling it off its hinges, and 

 one of thirty-seven along the lower. Find the strains on the hinges. 



Fig. 16. 



