IN i HI] 





the co-fflcint of friction be &, and tho strain on a ropo 

 >nt wove* a carriage bo BO pounds, what U the weight of tlio 



liors* baa to exert a strain of 110 pouu<ln to pull a wagou 

 i,' 1J tons. Wh.it is tin? co-efflclcut of friction? 



ILLUSTRATIONS OP PBECKDIN.. |-|:IN< I I'LBB. 



Live nowto trace the practical application of UK; pdbMJplN 



l.ii.l down, and the best way of doing thin in to take 



:MH,..II in i.i!:. 'PS and oarefolly examine thorn, and we 



Khali HCO that the same rules will apply to other and more com- 



A Wavy box ia resting on a four-logged table. What are the 

 forces that act on it, and what on the table ? On tho box there 

 are only two its own weight acting downwards, and tho upward 

 ru of tho table, exactly counterbalancing this weight. We 

 turn, then, to tho table, tho forces acting on which are not quite 

 !y determined. There are its own weight and the weight 

 of the box acting through their respective centres of gravity. 

 These are parallel forces, and, as we have seen, have a resultant 

 equal to their num. and acting at a point in tho line joining 

 them, so taken that their distances from it are in the inverse 

 proportion to their intensities. The other force which acts on 

 the table is the resistance of the floor on which it rests, which 

 resistance is transmitted upward through the four logs. If the 

 weight act at a point equally distant from these, each bears an 

 equal share; but if not, it is divided between them in tho inverse 

 proportion to their distances. To make this more clear, we will 

 suppose these distances to be 2, 6, 6, and 8 feet respectively. Find 

 the least common multiple of those numbers, that is, the least 

 number each will divide without any remainder. In this case 

 I . This wo divide successively by the distances, and obtain 

 tho quotients 12, 4, 4, and 3 ; and these numbers represent the 

 proportion in which tho weight is divided between the legs. 



Now suppose the weight of tho table and box to be 207 pounds. 

 Since 12, 4, 4, and 3, added together make 23, tho log 2 feet off 

 supports 12 parts out of every 23, i.e., i of the weight, or 108 

 pounds. Those 6 feet off support 3, or 36 pounds each ; and the 

 other sustains only 5 ^, or 27 pounds. A calculation of this sort ia 

 very frequently required to deter- 

 mine the relative strength the dif- 

 ferent parts of a building should 

 have. 



We will take another cose. A 

 body, o (Fig. 86), rests on an in- 

 clined plane, the angle of which, A 

 c A B, is 30, and the co-efficient of 

 friction is \. What forces act on 

 o, what are their amount, and what 

 other force must be applied to keep 

 it in its place ? We will try and 

 solve these questions. As already 

 seen, three forces act on G its weight acting along a w, the resist- 

 ance of tho piano acting along o B, and the force of friction, which 

 is i of the weight and acts along F a. We found, when consider- 

 ing the inclined plane, that the power necessary to sustain a 

 must bear tho same proportion to its weight that B c does to A c. 

 This, then, is the first thing wo must find out, and we must 

 have a slight acquaintance with mathematics for this ; there is, 

 however, no difficulty in the matter. Produce c B to D, so as 

 to make B D the same length as B c, and join A D. Tho triangles 

 ABC and A B D are exactly equal. For as B c is equal to B D, 

 and each is at right angles with A B, A B c would, if wo were to 

 turn it. over, exactly lie on A B D. AD is, then, equal to A c. Now 

 in any and every triangle the three interior angles are together 

 equal to 180, or two right angles, and in A B c we know that 

 the angle CA B is 30, and ABC, being a right angle, is 90; 

 therefore A c B must be 60, and A D B is equal to it, and there- 

 fore is also 60. The angle B A D ia likewise equal to B A c, and 

 as each is 30, the angle c A D is 60. We see thus that each of 

 tho angles of tho triangle c A D is 60, and therefore they are 

 equal to one another ; and, since the angles are equal, the sides 

 are also equal, for there is no reason why one should be greater 

 than another. The triangle is thus equilateral and equiangular. 

 Wo havo nnw found out what \v wanted ; for, if B c be repre 

 sented in length by 1, c D will be 2 ; and A c is equal to c D 

 therefore it is also 2.. and the proportion BO bears to AC ia 1 



Fig. 86. 



to 2, or J. On an incline of 30, then, the power moat be half 

 the weight; but, iu :..n caws, friction sustains one-fourth, and 

 therefore a power must be applied, acting in tho direction o p, 

 and equal to one-fourth of the weight, in order to maintain 

 eijiiiliLrium. 



Wo have now to solve the remainder of the question. We 

 know how much aw, or, and o P are ; but we want to know 

 what portion of the weight in borne by the plane, that is, what 

 proportion A B, which represent* the resistance of the plane, 

 beam to A c. To find thin, we need another very important 

 geometrical proposition, which you will find fully proved in 

 Lessons in Geometry, Problem XXX., Vol. I., page 337. 



In every right-angled triangle tho square described on the 

 Bide opposite the right angle is equal to the sum of the squares 

 on tho sides containing it. If we represent the sides by numbers 

 expressing their lengths, the rule holds equally true. Suppose, 

 for example, we measure off from one of tho sides containing 

 the right angle a length of 4 inches, and from the other a length 

 of 3 inches, we shall find the line joining these two points ia 

 equal to 5 inches. 



The square of 4 (which means 4 multiplied by itself) is 16, 

 and the square of 3 is 9. These added together moke 25, which 

 is the aquare of 5. The usual way of writing this is 4*4-3* = 5r. 

 In this way, if we know the length of any two aides of a right- 

 angled triangle, we can always calculate the third. Now in the 

 case we are examining, we know that the side A c is equal to 2 

 and the side B c to 1 ; but the square of A C is equal to the sum 

 of the squares A B and B c ; the square of A B must, therefore, be 

 equal to the difference between those of A c and B c. Now these 

 are 4 and 1 ; the square of A B is, then, equal to 3, and the length 

 of A B must be represented by the quantity which, multiplied by 

 itself, will make 3. This is called the square root of 3, and is 

 written \/3. By arithmetic we can easily find exactly what this 

 number is, but you can see that it is very nearly 1J. The pro- 

 portion, then, of A B to A c is If to 2, or 7 to 8, and the plane 

 sustains a pressure equal to about J of the weight. We have 

 thus discovered the magnitude of all the forces as required. 



When, as in our last lesson, we have resolved all the force? 

 acting on a body along two lines at right angles, we can in thi- 

 way find the magnitude of tho resultant without the trouble 

 and possible inaccuracy of actual measurement. Suppose we 

 have a remainder of 12 pounds acting along one of the lines, 

 and one of 5 pounds along the other, the resultant will be equal 

 to V 12 2 +~5 2 ; that is, to the square root of 144 + 25, or 169, 

 which ia 13. In the same way we can solve many questions 

 frequently met with. Here is aa example. Two forces act on 

 a body; the resultant is 34 pounds, and one of the forces ia 16 

 pounds; what ia the other? We first find the square of 34, 

 which is 1,156 ; from this we take the square of 16, or 256, 

 and we have left 900. The square root of this ia 30, and this 

 accordingly is the intensity of the other force. 



ANSWERS TO EXAMPLES IN MECHANICS.-XV. 



1. A power of 201 pounds. 



2. He must pull with a strain of ^ of a ton, or 89f pounds. 

 8. It would support a resistance of 616 pounds. 



4. A force of nearly 10 pounds must be applied, the gain being 

 2x3} feet divided by J inch, which equals 301$. 



5. The pressure will be 3,394$ pounds. 



6. The difference between the threads ia T f a of a foot. The gain is 

 therefore If x 2 x 3} x 110 or, 1,210. 



JO y 



7. 135 pounds. The gain is - or 270. 



I 



N.B. In the foregoing, friction was not taken into consideration. 



CORRESPONDENCE IN FRENCH. I. 



Aa a suitable pendant to our " Lessons in French " and " Bead- 

 ing in French," wo now bring under the notice of our readers a 

 valuable series of model business letters in English and French, 

 relating to the various transactions of commercial life. 



Under each heading the student will first find a model letter 

 couched in language appropriate to the subject under considera- 

 tion in English. Immediately after is given, in every case, a 

 close but idiomatic translation of the English model letter into 

 French. 



