Exercise:
Four bricks are to be stacked at the edge of a table each brick exts as far as possible beyond the edge of the table. abcliste abc To achieve this show that successive bricks mus ext no more than starting at the top frac fracfrac and frac of their length beyond the one below. abc Is the top brick completely beyond the base? abc Determine a general formula for the maximum total distance spanned by n bricks if they are to remain stable. abc A builder wants to construct a corbeled arch based on the principle of stability discussed in a and c above. What minium number of bricks each .m long is needed if the arch is to span .m? abcliste

Solution:
The solution of the four stapled bricks looks as in the following schematic picture indicated: center tikzpicturescale. filldrawcolorred!!black .-------.--.-.--cycle; filldrawcolorblack fillred!!yellow -----.---.--cycle; pgftransformyshift.cm pgftransformxshift.cm filldrawcolorblack fillred!!yellow -----.---.--cycle; pgftransformyshift.cm pgftransformxshiftcm pgftransformxshift-.cm filldrawcolorblack fillred!!yellow -----.---.--cycle; pgftransformyshift.cm pgftransformxshiftcm filldrawcolorblack fillred!!yellow -----.---.--cycle; tikzpicture center abcliste abc The construction is stable if the vertical through the center of mass of all bricks together goes through the edge of the table. In the following drawings the red po indicates the center of mass. If there would be only one stone it could be placed in maximum like this i.e. half of its length beyond the tables edge: center tikzpicturexscale filldrawcolorblack fillblack! ---.----.-----cycle; draw -.---..--..--.--cycle; node at .-. fracL; filldrawcolorred fillred . circle .; tikzpicture center The next brick must support the brick above in its center of mass and the vertical through the center of mass of the two bricks together yet easy to imagine since the two-brick-body is fully symmetric see the red dashed line goes through the tables edge: center tikzpicturexscale drawdashed colorred --; filldrawcolorred fillred . circle .; filldrawcolorblack fillblack! ---.----.-----cycle; node at .-. fracL; pgftransformxshift-.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; node at .-. fracL; tikzpicture center As mentioned since the problem is yet fully symmetric it is obvious that the lower brick is one quarter off the edge of the table. The next brick can be placed under the other two such that it exts frac off the edge of the table: center tikzpicturexscale filldrawcolorred fillred . circle .; filldrawcolorblack fillblack! ---.----.-----cycle; node at .-. fracL; node at .. fracL; pgftransformxshift-.cm pgftransformxshift.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; node at .-. fracL; tikzpicture center And the fourth brick can with similar arguments be placed frac over the edge of the table: center tikzpicturexscale filldrawcolorred fillred circle .; filldrawcolorblack fillblack! ---.----.-----cycle; node at .-. fracL; node at .. fracL; node at .. fracL; pgftransformxshift-.cm pgftransformxshift.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; pgftransformyshift.cm pgftransformxshift.cm draw -.---..--..--.--cycle; node at .-. fracL; tikzpicture center abc Yes the top-most brick is completely off the table since the distance tabledge to right-brick-edge of the top-most stone is as can be seen in the last graphics frac + frac + frac+frac frac. abc The total distance spanned by the arch built by n bricks is d fracL + fracL + fracL + fracL + dots fracL left+frac+frac+frac+dotsright fracL _k^n frack abc If the arch is to span pqm the number of bricks needed is fracpqmpq.m numpr. numpr. _k^n frack &approx ln n e^numpr. n . abcliste