The example below is artificially set up to explain the method without involving or requiring knowledge of other details of Archimedes' mathematics.

 Archimedes uses the principle of the balance in the Quadrature of the Parabola (where it is not used with indivisibles) and in the Method (where indivisibles are used). Given weights A and B and a fulcrum F, with A a distance a from the fulcrum F and B a distance b from F: A : B = b : a Archimedes always treats the weights as uniformly distributed over the magnitude, i.e. he allows that bodies may have different specific gravity, but he avoids having a variation in the specific gravity in a single body.  This important simplifying assumption allows him to treat weights as identical with magnitudes. For the purposes of the mechanical methods, this also allows him to apply the principle of the balance to solids, planes, or lines.

In order to illustrate Archimedes mechanical method with indivisibles, I shall derive a grotesquely simple and fabricated example.

 Suppose that there is a parallelogram ABCD with an inscribed triangle AEB, where DE = CE. We shall derive the remarkably basic theorem that triangle AEB is 1/2 the parallelogram.

 Draw lines intersecting from the midpoints of the four sides of the parallelogram  and parallel to the sides. Let the intersecting point be F.  This is in fact the center of mass of the rectangle.  The proof may be imagined by reader.

 Join EF and extend it to G.  Since E is the minpoint of CD while F was on the midpoint of  a line parallel to AB, G is the midpoint line AB. Extend EG to H, so that HE = EG. Consider HG is a balance beam with E as the fulcrum.

 Take any straight line parallel to AB, MNOPR.

 Clearly NP : MR = DM : AD.  It is up to the reader to show this, perhaps using similar triangles. Hence,  NP : MR = DM : AD = EO : AD = EO : EG = EO : AH (since EH = EG)
 Since NP : MR = EO : AH, if we move NP to H (i.e. as LK), then ...

 LK (i.e. NP moved to H) will balance MR left in its original position.

 If we move all the lines in the triangle to H and leave all the lines in the parallelogram in place, then the two figures balance each other. But the center of mass of the parallelogram is F. The parallelogram ABCE : the triangle at H = EH : EF = 2 : 1.