124 



THE POPULAE EDUCATOR. 



the floor. On constructing, in such a case, the polygon of 

 forces, we should have the figure as represented in perspective 

 below, one of whose Bides, o A, is on the floor, while the others, 

 A K, B BJ, B! B 2 , and B 2 B 3 , are in the air. A figure of this kind 

 is termed a tivisted polygon, as though its sides had been all 

 originally in the same plane, but, by a twist, some of them had 

 .,- been pulled from it. You can 

 see that, since such a polygon 

 cannot be drawn on paper, so 

 as to have the magnitudes of 

 its sides and angles there ac- 

 curately represented, it can be 

 of no practical use in finding 

 resultants. You might make 

 one by fastening five rods of 

 the proper lengths together at 

 the proper angles, but the 

 structure would probably break 

 down before you arrived at 

 your resultant, and at the best 

 the operation would be very 

 troublesome. Calculation alone 

 can help in such cases ; but 



the " twisted polygon " has the educational value of giving the 

 student mechanical ideas. 



EXAMPLES FOB PEACTICE. 



1. Three forces act on a point o A, equal to 3 pounds, o B to 5, and 

 o c to 7. The second lies between tlie other two, making with o A an 

 angle of 30 degrees, and with o c 45 degrees. Find the pounds in the 

 resultant, and the angle it makes with the least force o A. 



2. A roller of a hundred-weight is supported on an incline, the 

 gradient of which is one foot in two, by a force which acts along its 

 slope. Find the magnitude of this force and the pressure of the roller 

 on the plane. 



3. From two points on a ceiling, five feet apart, a sixty-pound 

 weight is suspended by two strong cords, which meet at the point of 

 suspension. The lengths of the cords are three and four feet respec- 

 tively. Find the magnitudes of the forces by which they are strained. 



4. Three weights of three, four, and five pounds are attached to 

 three cords, which are knotted together at their other ends. The 

 two cords bearing the lesser weights are thrown over two pulleys 

 fastened at a distance of 10 feet from each other, and at the same 

 height, into a wall, the greatest weight hanging between them. Find 

 the position in which the cords and weights will settle into equi- 

 librium. 



You will observe that these problems are to be done by rule 

 and compass, etc. We have not yet come to the more effective 

 method of solving them by calculation. The geometric way, 

 however, of drawing and measuring is the best for giving you 

 accurate ideas of the subject, and therefore indispensable 

 in the first stages. The lines you must carefully lay 

 down by a ruler, and the angles by a circular protractor, 

 keeping in mind, as to the latter, that in every right 

 angle there are ninety degrees. The distances representing the 

 forces you must take from an ordinary scale ; and observe, as 

 to this, that you need not make in every case your drawings so 

 large that a ivhole inch be given to every pound of force. You 

 may allow a quarter of an inch to each pound, or hundredweight, 

 or ton, or even a tenth, if the numbers be larg.e. All that is ne- 

 cessary is to keep the proportion of your figures right, whether 

 they be on a large or a small scale, as is done in mapping or 

 drawing plans of buildings. For the above examples a scale 

 of a quarter of an inch for each pound will be quite sufficient. 

 Perhaps for the third example tenths of an inch will best answer. 



In the next lesson the answers to these problems will be given. 

 I now proceed to 



FOBCES APPLIED TO TWO POINTS. 



Three cases present themselves for consideration. 



1. When the lines of direction of the two forces meet within 

 the body. 



2. When they meet without. 



3. When the two forces are parallel to each other. 



First Case. This is easily disposed of. When two forces 

 meet within a body, the point of meeting may be taken as the 

 point of application of both forces, which can there be com- 

 pounded into one ; and the case thus becomes that of a single 

 force applied to a single point. 



Second Case. Here also the two forces may be reduced to 



one ; but, as their directions meet outside the body, it is neces- 

 sary to show that their effect is the same as though the point 

 of meeting was a real point of application. This, in a future 

 lesson, can be demonstrated by a perfect proof ; but, in the mean- 

 time, the following considerations will satisfy you that it is true. 

 Let A P and B Q be the two forces applied to the points A and 

 B (as in Fig. 10), and O the outside point in which their direc- 

 tions meet. Also, let O B be the direction which their resultant 

 would take were the body extended to o and the forces there 

 applied. Suppose now that, in order to 

 extend it, a round bar of iron of uniform 

 thickness is firmly soldered to it, so as 

 to include the line O B within its sub- 

 stance. The body being thus extended, 

 o may be considered a point of applica- 

 tion of both forces, which we may con- 

 ceive to be transferred to it by two 

 thin but strong wires, o A, o B, the mass 



Fig. 10. 



of which is so small that it may be neglected in comparison 

 with that of the body. The forces A P and B Q then evidently 

 become one force, acting along o B on rod and body together, 

 and producing the same effect on both as though they acted 

 at A and B. But the effect taken separately of the resultant on 

 o B, and therefore of A P and B Q, is evidently the same 

 namely, a pressure along its length. Their effects, therefore, 

 on the body itself taken separately must be the same ; and o, 

 although outside, may be considered a point of application. 

 The two forces are reducible to one applied to the body at any 

 point on the line o E within the body. 



TWO PABALLEL FOECES. 



Third Case. The resultant single force can be determined in 

 this case also by the parallelogram of forces, but the proof given 

 by the greatest mechanician of antiquity Archimedes of Syra- 

 cuse is, with a slight alteration, much preferable, on account 

 of its simplicity. I shall first take 

 two equal parallel forces, which act 

 in the same direction. Let A and B 

 (Fig. 11) be the points of applica- 

 tion, and their directions those of 

 the arrow-heads P and Q. Suppose, 

 moreover, that in magnitude they are 

 each one pound, or ounce, or ton say 

 one pound. Now, in the first place, the resultant, whatever 

 it be, must pass through the middle point of A B. The best 

 reason I can give you for this is, that the resultant cannot, 

 since the forces be equal, be nearer to one than to the other. 

 If it were a tenth of an inch nearer to A, it should be also a 

 tenth nearer to B. 



Now, in order to find its magnitude and direction, let us sup- 

 pose that two other forces, A c, B D, each equal to a pound, arc 

 applied to the body along the line A B in opposite directions. 

 These being equal, and therefore of themselves balancing each 

 other, can neither add to nor take from the effect of A P and 

 B Q, which may consequently be considered equivalent to the 

 four forces A P, B Q, A c, B D. Let the two at A be now 

 compounded into one, acting in some direction between them 

 (I care not which), and let the same be done with the two at B. 

 Now produce these resultant directions backwards, until they 

 meet at o, and transfer the resultants themselves to that point. 

 Now resolve them back into their original components, and 

 you have two pounds, O Cj and O D,, acting against each other 

 parallel to A B, and two separate pounds pulling from O down- 

 wards parallel to A P and B Q. The two former cancel each 

 other, and there remain two pounds acting parallel to A p. 

 Hence we can say, that - 



1. If two equal parallel forces act on a body in the same 

 direction, their resultant is parallel to either, and bisects, or 

 divides equally, the line joining their points of application. 



2. The resultant is in magnitude equal to their sum, or tft 

 twice either force. 



As an example to illustrate, take two equally strong horses 

 pulling a carriage ; two equal forces are applied to the splinter- 

 bar, which give one force equal to double the strength of either 

 horse acting at its middle point. When the carriage is backed, 

 these forces are applied in the opposite direction directly to the 

 centre through the pole. 



We are now in a position to find the resultant of any two 



