Penny Blood: Serial Sensation and Working-Class Entertainment — page 121: what you’re looking at

Machine-transcribed from the original scan — historical spelling and the odd misread are preserved.

1O4 CRUST NOT FLEXIBLE. (oH. 1x. v Rand Q, and, as before, P p the fluid pressure, t the compression, ° 6 the inclination of the tangent to the horizon. Resolving vertically, we have for equilibrium, p cos @ds + tsin 6 — gpkds + Q = 0, and horizontally, p sin Ods —tcos 6+ R=0. Eliminating ¢, pds — gpk cos 6ds + Qcos 6+ Ksin =0......... (1). Now this being true when ds is indefinitely diminished, since Q and R are not thereby altered, we have generally, Qcos@ + # sin d= 0. But at an anticlinal Q and R must be the vertical and horizontal components of the compression at one side of the anticlinal. From the nature of the case the vertical components of the compression on either side of the anticlinal must act in the same direction ; and therefore they cannot be in equilibrium unless they are separately zero. And @ is one of them ; Q=0; . also &sin 0 =0, whence either sin 8 = 0, or R =0. Let us consider these two cases in order, When the element is under the action of the forces Q and R at one extremity of it, and ds is indefinitely diminished, observing that Q cos 6+ Rain 8 =0, we have from (1) in the limit pds — gpk cos @ds = 0, o. pds =] ' grkcos@de commiclsoolkks (CO)