MECHANICS. 



19 



= P x p. Now, since the distance w of 

 the weight from the fulcrum always re- 

 mains the same, and since the counter- 

 poise P is not changed, it follows, that 

 in whatever proportion W is increased 

 or diminished, p must be increased or 

 diminished in exactly the same propor- 

 tion, in order to sustain the equality of 

 the products we have just mentioned ; 

 that is, if W be doubled or trebled, p 

 must be likewise doubled or trebled, 

 and so on. 



If, then, it be required to graduate the 

 steel-yard so as to indicate to the exact- 

 ness of ounces, let one ounce be sus- 

 pended from C, and let the counterpoise 

 P be moved towards G until the beam 

 rests in the horizontal position. Then 

 let two ounces be suspended from C, 

 and move the counterpoise P from G 

 until the beam rests as before. Having 

 marked two divisions at the two posi- 

 tions of the counterpoise thus obtained, 

 let the whole length of the arm G B be 

 divided into equal divisions at the same 

 distance, and the number of any divi- 

 sion, beginning from that which is 

 nearest to G, will give the number of 

 ounces which the counterpoise P placed 

 at that division will sustain. 



If the centre of gravity of the beam 

 be not at or under the fulcrum, the gra- 

 duation must be effected differently. 

 First, let us suppose that the centre of 

 gravity of the beam is at D at the same 

 side of the fulcrum with the weight. In 

 that case the end C of the beam will 

 preponderate when unloaded. Let F be 

 the place at which the counterpoise P 

 must be suspended in order to keep the 

 unloaded beam horizontal, and let G F 

 be called/. As before, let the steel-yard 

 be graduated to ounces. The counter- 

 poise P being placed at F, let one ounce 

 be suspended from C, and a division 

 having been marked at F, let the coun- 

 terpoise be moved from G until equili- 

 brium is established. Let the second 

 division be marked at that position. 

 After this, let the divisions towards B 

 >e continued at equal distances, and the 

 number of any division beginning from 

 F will be the number of ounces which 

 the counterpoise P placed at that divi- 

 sion will sustain : for let the distance D G 

 be d, and let g be the weight of the 

 steel-yard : the whole weight acting at 

 D, its effort to depress the arm C G 

 is measured by the product g X d. 

 Also, the effort of W to depress the 

 arm is measured by the product W x w. 

 These two efforts are counteracted by 



the effort of P to depress the arm G.'B, 

 which effort is measured by the product 

 of P, and the distance of P from G. 

 Let the distance of P from F be p, and 

 its distance from G will be p +/, and 

 the product just mentioned is P x Cp+/,) 

 or the sum of the products P x p and 

 Px/. Thus, we have 



Wxiv + gxd=Pxp+Pxf. 



But it was before stated, that P, at 

 the distance /, balanced the unloaded 

 beam ; and therefore P x /, which mea- 

 sures the effort of P to depress the arm 

 GB, is equal to gx d, which measures 

 the effort of the weight of the beam to 

 depress the arm C G. Hence the equals 

 g x d and P x f being taken from the 

 former equals, we have remaining 

 W x w = P x p. 



Hence, in whatever proportion W is 

 increased or diminished, p must be in- 

 creased or diminished; but p is the 

 distance of the counterpoise P from F, 

 that position in which it balances the 

 unloaded beam. From that point F, 

 therefore, the graduation must com- 

 mence. 



If the centre of gravity be in the 

 longer arm, the graduation will com- 

 mence from that point in the shorter 

 arm at which the counterpoise will 

 balance the unloaded beam. For, sup- 

 pose this position to be D, and the 

 centre of gravity of the beam to be F, 

 the distances G D and G F being de- 

 nominated as before, suppose the coun- 

 terpoise first suspended at D ; a weight 

 W being suspended from C, let the 

 counterpoise be moved towards G until 

 equilibrium is established. First let 

 this take place, when P is between D and 

 G, and let the distance of P from D be p. 

 Then we have 



Wxw + Px (d-p)=gxf, 

 or'Wxw+Pxd Pxp = g x/; 

 adding to both these equals P x p, we 

 obtain 

 AV x w + P x d + P x p P x p = 



gxf+ Pxp, 

 orV?xw+Pxd = gxf+Pxp. 



But, since the product P x d repre- 

 sents the effort of the counterpoise at 

 D to depress the arm C G, and G x .; 

 represents the effort of the weight of the 

 beam to depress the arm G B, and these 

 efforts are equal, it follows that the 

 products P x d and g xf are equal. 

 These then being taken from the last 

 equals, there remains 



AV x ic = P x p. 



Hence, in whatever proportion W is in 

 creased, p must be increased in the same 

 C 2 



