94 



TTTR POPULAR EDUCATOR. 



si 





lines, which, as you will see, when many forces have to be com- 

 pounded, would cause much confusion in your figures. 



Let us apply this principle now to compound any number of 

 forces acting on a point. Let there be five, and that will illus- 

 trate the rule as well as a thousand could. Suppose forces, o A, 



o B, o c, o D, o E, ap- 

 plied to the point, O. By 

 the triangular rule, if I 

 N draw A R equal and 



\ parallel to O B, the line 



x v joining O with R is the 

 _O B O resultant of the first two 

 ""*' forces. I shall not actu- 

 ally draw this line, O R ; 

 let us suppose it drawn. 



Fig. 6. Now, if I compound this 



resultant with o c, I have 

 the resultant of three of 

 the forces. But that, by 



the same rule, is got by drawing from R a line R R! equal 

 and parallel to o c. The line o R is this resultant of three. 

 Again we shall not draw it. The resultant of this and o D, 

 for the same reason, would be o R 2 , got by drawing R, R S 

 parallel and equal to O D, and, lastly, the resultant of this 

 and o E would be o R 3 , the line, R 2 R 3 , being equal and 

 parallel to o E. We have thus exhausted all the forces, 

 and evidently O R 3 is the resultant of the whole five. There 

 was here no confusing ourselves with parallelograms ; all wo 

 had to do was to draw line after lino, one attached to the 

 other, carefully observing to keep their magnitudes and direc- 

 tions aright. A kind of unfinislied polygon was thus formed, and 

 the line o R S , which closes up the polygon, joining the last point 

 R 3 , with the point of application, is the resultant in magnitude 

 a^d in direction. Thus you have made another step in advance, 

 and arrived at the Polygon of Forces. You have learned how, 

 by the mere careful drawing of lines, to determine the resultant 

 of any number of forces. All you require is paper and pencil, 

 a rule, compasses, a scale, and a pair of parallel rulers. 



Now, there is one point about this polygon I wish you 

 carefully to note. You will observe that the arrows on its 

 sides, representing the directions of the forces you have com- 

 pounded, all point from left to right, as you go round the figure, 

 turning it with you so as to bring each side in succession to the 

 top. The resultant, however, points in the opposite direction, 

 from right to left, when that side is uppermost. This is as it 

 should be ; the direction of the resultant, as you go round the 

 'figure, must be opposite to those of the components. The use of 

 this you will see in the next lesson. 



Now, let us suppose that, in determining the resultant after 

 this method, as we come to the end of the operation, the end, 

 R 3 , of the last line, R 2 R 3 , chanced to coincide with, or fall upon 

 the point of application, o. The polygon would close of itself 

 without any joining line ; what is the meaning of this ? It 

 means that there is no resultant ; the line, o R 3 , is nothing, and 

 therefore the resultant is nothing, and the forces produce equili- 

 brium. What a valuable result we have arrived at ! A method 

 by which we can, by rule and compass, tell at once whether any 

 number of forces make equilibrium at a point or not. All we 

 have to do is to describe the polygon of forces, and if it closes 

 up of itself, there is equilibrium ; if it does not, there cannot be 

 equilibrium, and the resultant is in magnitude the side which is 

 necessary to close the figure. 



Deferring the further expansion of this subject for a short 

 time, I shall now turn back and apply these principles to a few 

 elementary examples. 



First Example. Three equal forces act at a point in different 

 directions what condition should they fulfil in order to be in 

 equilibrium ? Get your ruler and compass, and commence 

 constructing the figure by which their resultant may be found. 

 From the end of one of the forces you are to draw a line equal 

 and parallel to the second equal force, and from the end of that 

 another line, equal and parallel to the third. You will thus 

 have three lines strung together, all equal to each other. But 

 if the forces are in equilibrium, the end of the last line must 

 fall on the point of application, that is to say, the polygon of 

 forces must close up, and form a triangle. Your construction 

 will then give you a triangle of three equal sides, commonly 

 called an equilateral, triangle. But such a triangle must have 



all its angles equal ; also the angles between the sides of the 

 triangle, or of the polygon of forces, are the angles between the 

 forces themselves, since they are parallel to these forces. This 

 is evident from the properties, 1 and 2, of the parallelogram 

 referred to above ; therefore, in the case we are considering, the 

 three equal forces must act at equal angles, as I showed other, 

 wise must be the case at the close of the last lesson. 



Second Example. Let a weight hang from the ceiling by 

 means of two cords of unequal length, as in Fig. 7. It is 

 evidently at rest. Whatever be the forces called into action, 

 they produce equilibrium. Is there nothing further to ascertain? 

 There is ; it may be desirable to know by how much each cord 

 is strained. Our assurance that the cords will support the 

 weight depends on this knowledge. Let p and Q be the two 

 points of support of the strings which meet at o. Now, what- 

 ever be the strains on the cords, O P, O Q, they make equilibrium 

 with w, the weight. Therefore, if we suppose a length, o A, of 

 o P to represent the strain R, 



on O P, and from A draw a 

 line, A R, parallel to o Q, 

 equal to the strain, O B, 

 on o Q, then, since the three 

 forces are in equilibrium, 

 the line, R o, closing up the 

 triangle must be equal to, 

 and be in the direction as, 

 the third force, or weight, w.- 

 This, then, tells us what to 

 do. Measure on o R upward 

 as many inches as there are 

 pounds in w ; and from R 

 then draw R A parallel to 

 the cord o Q to meet the cord o A. The number of inches to 

 o A will represent in pounds the strain on o p, and those on 

 R A the strain on o Q. All, therefore, that we desire to know ia 

 determined. 



Third Example. A horse pulls a roller up a smooth inclined 

 plane or slope ; what is the force he must exert when he just 

 keeps the roller at rest ? And by how much does the roller 

 press on the plane ? 



Let the horse pull in any direction, o A. Then there will be 

 three forces acting on the roller ; namely, its own weight right 

 downwards, the horse's pull, and the resistance of the plane or 

 slope, perpendicular to itself. There must be this third force, 

 for the other two, 

 not being opposite 

 to each other, can- 

 not make equili- 

 brium. The roller 

 is somehow sup- 

 ported by the plane; 

 and that it cannot 

 be unless by its re- 

 sistance ; and a 

 plane cannot resist 

 except perpendicu- 

 larly to itself. This 

 third force, you 

 thus see, must be 

 taken perpendicular to the piano. It is represented in the 

 figure by o B. Now apply the polygon of forces. Let o c 

 represent the weight of the roller, and from c suppose a 

 line, C R, drawn equal and parallel to o A, the horse's pull. 

 Then, since there is equilibrium, the polygon of forces should 

 close up and become a triangle that is, the line joining is 

 with O should be the pressure, and therefore should be per- 

 pendicular to the plane. What then are we to do ? Take 



c, equal in inches to the number of pounds in the roller, 

 draw then from c a line C R parallel to the horse's pull, to 

 meet the line drawn from the centre o of the roller perpen- 

 dicularly to the plane ; c li so determined will in inches tc! 1 

 the pounds in the house's pull, and O R the amount by which 

 the roller presses the plane. You can easily see from this thL 

 as the slope increases the pull will increase and the pressure 

 diminish. This is what naturally we should expect. The plane 



1 have supposed to be smooth ; for, where there is friction 

 against the roller caused by roughness in itself or in the plane, or 



i in both, the question is much altered, as ic due time y" -will see. 



