The relation between the weights and their arms of leverage may be demonstrated as follows: *

377 The Lever Principle Demonstrated 311

Fig. 273.

Let A B G H, Fig. 273, represent a beam of homogeneous material, of equal sectional area throughout, and suspended upon an axle or pin at C, its centre. This beam is evidently in a state of equilibrium. Of the part of the beam A D GK, let E be the centre of gravity; and of the remaining part, D B KH, let F be the centre of gravity.

If the weight of the material in A D GK be concentrated at E, its centre of gravity, and the weight of the material in D BKH be concentrated in F, its centre of gravity, the state of equilibrium will not be interfered with. Therefore let the ball R be equal in weight to the part A D G K, and the ball P equal to the weight of the part DBKH; and let these two balls be connected by the rod E F. Then these two balls and rod, supported at C, will evidently be in a state of equilibrium (the rod EF being supposed to be without weight).

* The principle upon which this demonstration is based may be found in an article written by the author and published in the Mathematical Monthly, Cambridge, U. S., for 1858, p. 77.

Now, it is proposed to show that R is to P as CF is to CE. This can be proved; for, since R equals the area A D G K and P equals the area D B KH, therefore R is in proportion to A D, as P is to DB (Art. 359); or, taking the halves of these lines, R is in proportion to A J as P is to LB.

Also, JL equals half the length of the beam; for J D is the half of A D, and DL is the half of DB; thus these two parts (JD + DL) equal the half of the two parts (AD + DB); or, J L equals the half of A B; or, we have -

JD = AD/2; DL = DB/2.

Adding these two equations together, we have - JD+DL = AD/2+DB/2+AD + DB/2

Now,JD + DL = JL, and AD + DB = AB; therefore,

JL = AB/2.

Thus we have A M= JL. From each of these equals take JM, common to both, then the remainders, A J and ML, will be equal; therefore, A J = CF.

We have also MB = JL. From each of these equals take ML, common to both, and the remainders, JM and L B, will be equal; therefore, L B = E C. As was above shown -

R: Aj:: P: Lb.

Substituting for A J and LB their values, as just found, we have -

R: Cf:: P; Ec;

from which we have (Art. 373) -


Thus it is demonstrated that the product of one weight into its arm of leverage, is equal to the product of the other weight into its arm of leverage: a proposition which is known as the law of the lever.

378. - Any One Of Four Proportionals May Be Found

Any three of four proportionals being given, the fourth may be found; for either one of the four factors may be made to stand alone; thus, taking the equation of the last article, if we divide both members by CF (Art. 371), we have -

Px CF = R x EC CF CF

In the first member CF, in both numerator and denominator, cancel each other (Art. 371), therefore -

P = Rx EC/CF

so likewise we may obtain -



EC = PxCF/R.