80 



THE POPULAR EDUCATOR. 



Fig. 87. 



W 



and towards the base of the wall ; and we easily see that the 

 more nearly vertical the ladder is, the greater is the former as 

 compared with the latter, and therefore the less the amount of 

 friction which is required to keep it in ita 

 place. 



Here is another case, involving the same 

 principle. A bracket, A B (Fig. 88), projects 

 from a wall, to which it is fastened by a 

 screw. A strut, A c, supports the outer end, 

 and a weight, w, rests on it. In what direc- 

 tion is the strain on the screw ? The three 

 forces here are, gravity acting along w G, 

 the thrust of the beam, A c, and the strain 

 on the screw. The two former act throngli 

 the point o, and the direction of the third is 

 found by drawing a line from this point to 

 the screw. We may take a line, o c, of such a 

 length as to represent the weight, and re- 

 solve it into o b and o a, one acting along the strut, the other 

 perpendicular to the wall. These will represent two forces, 

 which are together equivalent to o c, and of these o 6 will be 

 overcome by the pressure of the strut, and the other force, o a, 

 tends to draw the screw from the wall A portion, however, 

 of the pressure of the strut will be borne by the screw, and 

 these two forces combining produce 

 the resultant, which acts on the 

 screw towards the point o. 



It is frequently very important to 

 be able thus to tell in what direction 

 a strain will act, as the strength of 

 our materials must be proportionate 

 to it. In this way the direction of 

 tie-beams and king and queen posts 

 of a roof are determined. We know, 

 too, where to apply struts and braces 

 to the framework of a building, so as 

 to gain the greatest benefit from them. 

 Now there are one or two cases 

 of the composition and resolution of 

 forces that are frequently given as 

 illustrations, and if we clearly under- 

 stand them we shall be able to master 

 most others. The first is that of a kite. 



Let K (Fig. 89) represent a kite. The forces which act on it are, 

 the force of the wind, acting, we will suppose, in the direction of 

 the arrow, the tension of the string, acting along it in the direc- 

 tion K s, and the weight of the kite ; and by the action of these 

 three forces it is kept at rest. We will consider them singly ; 

 the first, we will take the force of the wind. Take K w, of such 

 a length as to represent this force. The kite is always so made 

 as to present a large surface to the wind in proportion to its 

 weight, and the string is fastened to the loop in such a way that 

 it does not hang vertical, but inclined at an angle ; the tail, 

 however, prevents its being so acted on by the wind as to come 

 A 



in the same straight line with the cord. Let us, then, resolve 

 the force of the wind into two, one acting edgewise on the kite, 

 the other perpendicular to its surface. We draw the parallelo- 

 gram K o w A, and thug have the two forces, K o and K A, 

 instead of K w. K o has no effect, as it acts on the edge, and 

 we need, therefore, only consider the part K A. 



Now we will introduce a second force, that of the string. 

 Produce s K backward, draw A B perpendicular to B K, and com- 

 plete the parallelogram D K B A. We can again resolve K A 



Fig. 90. 



into K D and K B. The latter will be expended in stretching the 

 string, and have no tendency to move the kite, and thus we have 

 K D left as the effective resultant of these two forces. We now 

 consider the third, which is the weight of the kite. Draw K G 

 of such a length as to represent this, and complete the parallelo- 

 gram D C G K. We have then K c the resultant of K D and K G, 

 and therefore of all the forces which act on the kite, and this 

 is the direction in which the kite will move, but as it does so, 

 the angle at which it is inclined varies till K D and K G become 

 opposite and equal, and then the kite will remain at rest as long 

 as the force of the wind remains unaltered. 



The other case we will consider is that of a ship, which will 

 sail within a few points of the wind. Let c o (Fig. 90) represent 

 the direction and intensity of the wind, and s V the direction in 

 which it is desired that the vessel should advance. The sail is 

 placed in the dirqction A B, which is midway between that of 

 the wind and that of the vessel. We must, as before, resolve 

 c o into two forces, E o and p o. The part a o, which acts 

 along the direction of the sail, has no effect in moving the 

 vessel ; r o, which acts perpendicularly to the sail, is the effec- 

 tive portion. We must now again 

 resolve this force along two 

 directions, one being that in 

 which the boat moves, the other 

 at right angles to it. We make 

 G o equal to F o, and about 

 it describe the parallelogram 

 H o i o, and thus have two 

 forces, represented by o H and 

 o i, in the place of the original 

 force c o. Now, of these, o I 

 has no tendency to cause the 



vessel to advance ; it acts sideways on the vessel, usually 

 inclining it, and causing a slight motion, but it is resisted by 

 the pressure of the water against the side ; the other portion, 

 o H, represents the portion of the force of the wind which is 

 effective, and produces motion. 



In the same way you can calculate what portion of the force 

 of the wind is effective in turning a windmill. The vanes are 

 always set at an inclination with the plane in which they turn, 

 and you must resolve the force along two directions, one per- 

 pendicular to the surface, the other along it. The former 

 you again resolve, and thus find what part of it produces rota- 

 tion, and what part presses against the face of the mill. 



EXAMPLES. 



1. Forces of 9 and 12 act at right angles ; what is their resultant ? 



2. The resultant of two forces which act at right angles is 10 pounds. 

 One of the forces is 6 ; find the other. 



3. Two men, one on each side of a stream, tow a barge. The angle 

 the two ropes make is 60, and each pulls with a force of 100 pounds. 

 What is the total force exerted on the barge ? 



4. The tension of a wire in a piano is 100 pounds, its length is 5 feet. 

 What force is required to draw its middle point 2 inches out of its 

 position ? 



5. A weight of 90 pounds rests on a plane inclined at an angle of 

 30. The co-efficient of friction is %. What force is required to keep 

 it at rest ? 



ANSWERS TO EXAMPLES IN LESSON XVII. 



1. The forces actiug at the longer end are the power of 10 pounds 

 acting at a distance from the fulcrum of 6| feet, and the weight of the 

 lever, which is also 10 pounds, and acts through its middle point, or 

 2f feet from the fulcrum. The moments on this end are thus 10 x 6j, 

 or 67J, and 10 x 2f or 27J. These make 95 pounds. As w acts at a 



95 

 distance of 1J feet, it must be or 76 pounds. 



2. Since is lost by friction, we may regard the weight as 12 pounds 

 only. Now ^ of the weight of the first pulley is supported by the 

 power, ^ of the next, and ^ and ^ of the other two ; and since each 

 weighs 2 pounds, these amounts are 1 pound, ^ pound, ^ pound, and 

 -J- pound, together Ij. Take this from 12, and we have an effective 

 power of lOjjt remaining ; and as the gain is 16, the weight raised is 

 16 x io|, or 162 pounds. 



3. Friction requires a strain of 9 x 20, or 180 pounds, to overcome 

 it, and T J^ of the weight has to be borne. The strain, therefore, is 

 180 + 448, or 628 pounds. 



4. Friction is here fa of 27 cwt., which equals ^ cwt. The amount 

 of the weight sustained by the horse is 7 1 5 of 27 cwt., or f cwt. The 

 total strain is thus If cwt., or 14t pounds. 



5. The weight of the carriage is 25 x 80, or 2,000 pounds. 



6. The co-efficient of friction is ^y^j, or nearly -fa. 



