16 



MECHANICS. 



Let the weight of the entire beam be G. 

 Through m let a line aw a' be conceived 

 to be drawn perpendicular to the direc- 

 tions of the weights, and therefore hori- 

 zontal. Through c draw c d perpendi- 

 cular to a a', and therefore parallel to 



the weight of the beam G acting with 

 the leverage bm. Hence it follows, 

 that, all other things being the same, the 

 sensibility of a balance will be increased 

 by diminishing the weight of the beam, 

 and by diminishing also the distance 



Sa and S a', and parallel to the same mg between the centres of gravity and 



lines let g b be drawn. The machine is 

 thus maintained in equilibrium on its 

 centre m by three weights, viz. W in the 

 direction a S, W in the direction S' a', 

 and G acting at the centre of gravity 

 (Treatise I. chap, iv.) of the beam, and 



motion, for that will evidently diminish 

 the leverage m b. 



Another cause which resists the effect 

 of W is the relatively increased leverage 

 of W. This will be diminished by 

 diminishing m d (d being the middle 



in the direction bg. The weight W tends point of a a') or cm, that is, by dimi- 

 to turn the system round in one direction, nishirig the distance of the line which 

 and the weights G and W tend to turn joins the points of suspension from the 

 it round in the other direction ; and centre of motion. 



(40.) To those students who have 

 attained a slight knowledge of the most 

 elementary parts of mathematical 

 science, the following investigation of 

 the circumstances which regulate the 

 sensibility of the balance will be more 

 satisfactory than the popular explana- 

 tion which we have just attempted to 

 give. Those students who are not 

 parallel, d must be the middle point of familiar with elementary mathematics 

 the line a a', and therefore a m will be will pass over this article, 

 obviously less than a' m, and therefore, in 

 the present position of the beam, Wacts 



since these tendencies mutually coun- 

 teract each other, the product W x a m, 

 of the weight W and its leverage a in 

 must be equal to (W'xa'm + Gxbm) 

 the sum of the products of the weights 

 W and G multiplied into their respective 

 leverages a' m and b m. 



Since c is the middle point of the line 

 S S', and since Sa, cd, and S' a' are 



Let the angle omv, by which the 

 index is deflected from the vertical po- 



with a less leverage than W. A little sition, be D. Let gm = d, cm = b, and 

 attention to the figure will show that, as S c or S' c= a. Since the angles m c d 

 the weight W depresses the beam, it and mgb are each equal to D, by the 



continually loses its leverage, and that, 

 on the other hand, W is continually 

 gaining leverage on W. Besides this, 

 the weight of the beam G, which, in the 

 unloaded position, had no leverage, 

 gains leverage continually as W de- 

 presses S, and conspires with W in 

 resisting the descent of S by the force 

 of the heavier weight W. This being cos. JJ = < 

 the state of the forces, they must come therefore 

 to equilibrium after the beam has turned 

 from its unloaded position. Now, the 

 greater the deviation of the beam from 

 its unloaded position, when it attains 

 the state of equilibrium, the greater will 

 be its sensibility, the difference of the 

 weights being the same. The sensibility 

 then, it is evident, will be increased by 

 diminishing those causes which resist 

 the deflection of the beam from its un- 

 loaded position by the heavier weight W. 

 One of the principal of these causes is 



parallels, and the angles d and b are 

 right, we have dm=b sin D, and b m = 

 dsin D. Also, since the angles formed 

 by the lines S S' and a a' are equal to 

 that formed by m v and m o, which are 

 respectively perpendicular to the former, 

 that angle is equal to D, and since the 

 angles at a a' are right, we have S S', x 

 a', or a a'- 2 a cos. D, and 

 ad=acos. D. Butam=ac? 

 md=acos.D b sin D, and a' m= 

 ad+md=a cos. D + 6 sin d. 



Now, suppose that the weight W con- 

 sists of two parts, one equal to W', and 

 therefore the other E to the difference 

 between W and W : the combined 

 effect of that part of W which is equal 

 to W', and of W itself, will be the same 

 as if 2 W acted at the middle point c 

 of the beam, and with the leverage m d. 

 Hence the condition of equilibrium will 

 be 



, or 

 E (a cos. D b sin D) = 2 W b sin D + Gt/ sin D. 



Dividing this entire equation by cos D, observing that sm S m = tan. D 

 we have E ( a b tan. D ) =2 W b tan. D + G d tan. D. 



= Ea. 



tan. D 



G d + (2W/ + E) b' 



