where [rho], [rho]´ are the radii of curvature of the two curves at J, [phi] is the inclination of the common tangent at J to the horizontal, and h is the height of G above J. The signs of [rho], [rho]´ are to be taken positive when the curvatures are as in the standard case shown in fig. 49. Hence for stability the upper sign must obtain in (8). The same criterion may be arrived at in a more intuitive manner as follows. If the body be supposed to roll (say to the right) until the curves touch at J´, and if JJ´ = [delta]s, the angle through which the upper figure rotates is [delta]s/[rho] + [delta]s/[rho]´, and the horizontal displacement of G is equal to the product of this expression into h. If this displacement be less than the horizontal projection of JJ´, viz. [delta]s cos[phi], the vertical through the new position of G will fall to the left of J´ and gravity will tend to restore the body to its former position. It is here assumed that the remaining forces acting on the body in its displaced position have zero moment about J´; this is evidently the case, for instance, in the problem of "rocking stones." The principle of virtual work is specially convenient in the theory of frames (§ 6), since the reactions at smooth joints and the stresses in inextensible bars may be left out of account. In particular, in the case of a frame which is just rigid, the principle enables us to find the stress in any one bar independently of the rest. If we imagine the bar in question to be removed, equilibrium will still persist if we introduce two equal and opposite forces S, of suitable magnitude, at the joints which it connected. In any infinitely small deformation of the frame as thus modified, the virtual work of the forces S, together with that of the original extraneous forces, must vanish; this determines S. As a simple example, take the case of a light frame, whose bars form the slides of a rhombus ABCD with the diagonal BD, suspended from A and carrying a weight W at C; and let it be required to find the stress in BD. If we remove the bar BD, and apply two equal and opposite forces S at B and D, the equation is W·[delta](2l cos[theta]) + 2S·[delta](l sin [theta]) = 0, where l is the length of a side of the rhombus, and [theta] its inclination to the vertical. Hence S = W tan [theta] = W·BD/AC. (8) [Illustration: FIG. 50.]