

1 1 ; 





v/li;it i- tin- i-'.ii.iition of equilibrium ? In !!.: ilmt you may 

 clearly UM<I . tho kuowlod;- .lowing goo- 



.! pruM-inles U necessary. 



UOPIBTIKS 07 A PARALLELOGRAM AMD TR1ANOLB. 

 1 . The area oj a triangle is A . ralletogram which 



haa -Utilise for one side," . rtea parallel 



to tluit bate for tlui tide opposite. Thin 

 appourx I: 



triiint,'l., unl A B o D any parallelogram 

 on A B formed by drawing from A u 

 any two j.a -< AD, BO to me 



parallel D o to A B through v. For, draw 

 VK parallel to A D, and therefore parull.-l 

 to B c, to meet A B in K. Thentho triangle 



is m;ulf illicit' th<> two tnungleS A V B 

 and B v E. But since A K v j > in a parallelo- 

 gram, the triangle AV K (London 111.) H 

 equal to A D V, and is therefore half the 

 illelogram A E v D. So likewise is B V 

 half UK v c- ; iiiul thcnloiv the triangle A VB half A B CO. 



J. I'lia area of a <*, half the product of its 



,>d tin' p<;-i nJicular from its vertex on that base. This 

 follows from the previous principle. Let tho number of inches 

 or tVrt, say inches, in A B (Fig. 58) be 6, and in the perpen- 

 dicular, v E, be 7, and construct on A B a parallelogram, 

 A B c D, whoso sides are parallel to this perpendicular. Such a 

 parallelogram is termed a " rectangle," on account pf its angles 

 being all right angles. Mark out the inches 

 on A B and v E, and draw the dotted lines in 

 the figure parallel to A B and v E, cutting 

 this rectangle into the smaller ones tho sides 

 of which are all equal to one inch, and which 

 are therefore so many square indies. Now 

 there are seven rows of these squares, one row 

 above the other, and there are six squares in 

 each row ; and therefore there are altogether 

 7 times 6, or 42, square inches in the rectangle. 



But tho triangle being half the rectangle, is 



Fig. 58. 



half of 42 square inches, that is, it is, in num- 

 bers, half tJie product of the base and perpen- 



r. Were the numbers 13 and 9, or any other pair whatever, 

 the reasoning would be tho same. 



3. // two triangles stand on opposite sides of a common base, 

 and the line joining tlieir vertices is bisected by that base, the 

 triangles have equal areas. In Fig. 59, the triangles A B o, A B D 

 stand on the common base, A B, at opposite sides, and D c join- 

 ing their vertices is supposed to be bisected at if ; I have to 

 prove that the areas of the triangles are equal. Draw E F and 



H Q through A and B parallel 

 P to D c, and also through D and 

 c draw H E and Q F parallel 

 to A B. Then we have a 

 large parallelogram E F a H, 

 which is divided into four 

 smaller ones by A B and D c. 

 But since D c is bisected 

 at M, making u c equal to 

 M D, and therefore A E equal 

 to A F, the parallelograms 

 A F a B and A E H B are equal to each other. But, as proved 

 above, tho triangles ABC and A B D are half of these parallelo- 

 grams, and therefore are also equal to each other, as was 

 required to be proved. 



Wo now return to our Mechanics, applying these geometrical 

 principles to determine 



THE MOMENTS IN THE LEVER OF FORCES NOT PARALLEL. 



Two such forces, A p, A q (Fig. 60), being supposed to meet at 

 some point, A, to which they are transferred, and there com- 

 pounded into a resultant A R, represented by the diagonal of the 

 parallelogram, A p R Q, and o being a point taken at random on 

 that diagonal, we can prove the following proposition : 



The moments of hoo intersecting forces in reference to any point 

 mi tht-ir resultant are equal to each other. Now the moment of a 

 force in reference to a point, as has been already explainr.l. i- 

 the product of the force by tho perpendicular dropped on it from 

 that point. In Fig. 60, therefore, the moment of A P in ref e- 

 ronoo to o, a point on tho resultant, is A P multiplied into o x, 



CO. 



I 



r 



Ife B. 



the perpendicular from o on A p. Bo likewise is the 

 of A g in reference to o equal to led into o r. the 



corresponding perpendicular. What I have then to prove i* 

 chat these product* or* equal. But they are equal ; for, from 



y |M BdfMMfai al 

 principle above, the 

 areM of the trianglas 

 A OP, A oq, or* half 

 these products; and, 

 by the third, since 

 these triangles stood 



:. ..... 



A o, and the line pq 

 vortices, being a diagonal, is bisected by A R, that U, 



by that base, their areas are equal. The momenta of A p an<l A g, 



uce to o are equal, as I undertook to prove. 

 Now, to apply thw to the lever, using the same figure, lei us 

 suppose the two forces to be A p, AQ, meeting, as I have stated 

 to be necessary, at some point A. Then it is evident, since then 

 is but one point fixed in the body, that there cannot be equi- 

 librium unless the resultant of A p and A Q passes through that 

 point, and is there resisted by the support* that fix it. The 

 fulcrum, therefore, yon see, must be on the resultant, and there- 

 fore taking o to bo the fulcrum, wo must have A p multiplied into 

 o x equal to A Q multiplied into o T, thu 

 the moments of the forces in reference to 

 the fulcrum must be equal. We arrive 

 thus at the two following modes of stating * 

 the condition of equilibrium in a lever, 

 cither of which may bo selected for use as 

 the occasion requires : 



1 . In a lever, the forces not being parallel, 

 the power multiplied by the perpendicular 

 from the fulcrum on its direction is equal 

 to the resistance multiplied by the perpen- 

 dicular on its direction. 



2. Tho power and resistance are to each other inversely as the 

 perpendiculars dropped from tho fulcrum on their respective 

 directions. 



THE WHEEL AND AXLE. 



This useful mechanism, of which several forms ore given in 

 Figs. 61, 62, and 63, is a kind of lever, or succession of levers, 

 revolving round an axis, from which they project at right mgUtf, 

 Corresponding to this central axle 

 line is a cylindrical axle of some 

 thickness, round which winds the 

 rope which bears the resistance, or 

 weight, to be raised. In Fig', 

 the simplest form of the instrument) 

 consisting of on horizontal axle and 

 four levers, which are worked in 

 succession by the power. In the 

 ship's capstan for raising the anchor 

 (Fig. 62), tho resistance acts horizon- 

 tally, a man pushing also horizontally 

 at the end of each lever, the power being multiplied in tho pro- 

 portion of tho number of levers and men. We have in Fig. 63 

 another form, where tho lovers are the spokes of a wheel, and the 

 power A works in succession on them along 

 the tire as they come round. 



The principle in all is the same, whether 

 tho resistance and power be parallel or not, 

 and may be understood from Fig. 64, which 

 represents a transverse section, the outer 

 circle being the wheel and tho inner the 

 axle. Tho central lino of tho axle, which 

 you must conceive perpendicular to tho 

 paper at tho centre of these circles, is the 

 fulcrum, represented by the point o. Tho 

 line A B thus is seen to be the lever, at the 

 ends of which tho power, p, and resistance, 

 w, act; and, as already proved, these forces 

 must bo inversely as O A to o B, which lines 

 are the radii of the wheel and axle respectively. When the power 

 and resistance act parallel to each other this is evident ; but the 

 same holds good were they not so to act, as in the capstan, where 

 the power is continually changing direction as the sailors go round ; 

 for, referring again to Fig. 64, if tho power were to act not in the 



Fig. 62. 



