A-Class Visual Design

Slide Projection

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Four identical bricks are placed on top of each other at the edge of a table. Is it possible to slide them horizontally across each other in such a way that the projection of the topmost one is completely outside the table? What is the theoretical limit to the displacement of the topmost brick if the number of bricks is arbitrarily increased?

Consider n bricks stacked. In order for the upper n - 1 bricks to be stable, the center of gravity of the n - 1 bricks must be behind the end of the nth brick (count n from the top brick), Let x(n-1) be the distance from the end of the top brick to the CG of the n - 1 bricks. The center of gravity of the n bricks measured from the front end of the top brick is

distance of CG of top (n - 1) bricks times weight of the top n - 1 bricks

x(n - 1)*(n - 1)*W

plus

distance of CG of brick n times weight of one brick

[L/2 + x(n - 1)]*W

all divided by the total weight

n*W

x(n) = [x(n - 1)*(n - 1)*W + L/2*W + x(n - 1)*W]/(n*W)

x(n) = x(n - 1) + L/2

for n = 1 (one brick) x(1) = L/2

for n = 4

x(4) = L/2 + L/(2*2) * L/(3*2) + L/(4*2)

x(4) = L/2*(1 + 1/2 + 1/3 + 1/4) = L/2*(12/12 + 6/12 + 4/12 + 3/12) = L/2*(25/12 = L*25/24

for the whole stack to be stable, the CG of the whole stack must be at the edge of the table, a distance x(4) from the front end of the top brick; the back end of the top brick is x(4) - L behind the front end, so the back end is L*25/24 - L =

1/24*L in front of the table edge.

ARGUS 300 Slide Projector

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