been very different if we had built up our pillar of cards or slates lying obliquely to the lines of pressure, for then at once there would have been a tendency for the elements of the pile to slip and slide asunder, and to produce what the geologists call ãa faultã in the structure.
Somewhat more generally, if AB be a bar, or pillar, of cross-section a under a direct load P, giving a stress per unit area =ã₤p, then the whole pressure P =ã₤pa. Let CD be an oblique section, inclined at an angle ö¡ to the cross-section; the pressure on CD will evidently be =ã₤paã₤cosã₤ö¡. But at any point O in CD, the pressure P may be resolved into the force Q acting along CD, and N perpendicular to it: where N =ã₤Pã₤cosã₤ö¡, and Q =ã₤Pã₤sinã₤ö¡ =ã₤paã₤sinã₤ö¡. The whole force Q upon CD =ã₤qã₤ôñã₤area of CD, which is =ã₤qã₤ôñã₤aã₤ãã₤(cosã₤ö¡). {686} Therefore qaã₤ãã₤(cosã₤ö¡) =ã₤paã₤sinã₤ö¡, therefore q =ã₤pã₤sinã₤ö¡ã₤cosã₤ö¡, =ã₤ô§ã₤pã₤sinã₤2ö¡. Therefore when sinã₤2ö¡ =ã₤1, that is, when ö¡ =ã₤45ô¯, q is a maximum, and =ã₤pã₤ãã₤2; and when sinã₤2ö¡ =ã₤0, that is when ö¡ =ã₤0ô¯ or 90ô¯, then q vanishes altogether.