**Problem 2B-16**: Bridge building**

We want to build a bridge by stacking up identical bricks as shown in Fig. 1.
The bricks are numbered from the top one as 0, 1, 2, ......, n. We want to make
the horizontal distance between the right edge of the top brick and the right edge
of the bottom brick, denoted as *x*, as large as possible. How should the
bricks be arranged?

Scroll down for
solution

See, for example, Scientific American, November 1964, page 128 for this problem.
How large will *x* be if n = 52? Can you use 52 play-cards to build such a
paper bridge? Of course, you should be a little more conservative than the
theoretical case discussed in the solution when you try to build a paper bridge.

**Solution:**

Each brick has a length of 2L and a mass of M. First let us consider only 2 bricks,
no. 0 and no. 1 bricks as shown in Fig. 2(a). The center of mass of the 0-th brick,
denoted as C_{0}, must lie on top of the upper right corner
of the brick 1, denoted as point A_{1}, so that the torque
around point A_{1} generated by brick 0 will vanish. This
means that brick 0 extends beyond by the right edge of brick 1 by a distance of L,
or *x* = L.

The case of 3 bricks is depicted in Fig. 2(b). The arrangement of brick 0 and brick 1
is the same as in Fig. 2(a). We must consider the torques generated by brick 0 and
brick 1 around the upper right corner of brick 2, denoted as point A_{2}.
Brick 0 can be considered as a point of mass M located at C_{0} right on
top of point A_{1}. Brick 1 can be considered as a point of mass M located at
point C_{1}, the center of mass of brick 1. The horizontal distance
between C_{1} and A_{1} is L. Thus A_{2}
must be at the middle point of the horizontal distance from point C_{1} to point
A_{1}. In other words the horizontal distance between A_{2}
and A_{1} is L/2. Thus *x* is extended by L/2, and the total *x* is L + (L/2).
The center of mass of combined brick 0 and brick 1, denoted as
C_{0/1}, must lie above point A_{2}
so that the torque generated by the combined system around point A_{2} becomes zero.

The case of 4 bricks is shown in Fig. 2(c). We already know that the center of mass of brick 0
and brick 1 combined, denoted as C_{0/1 lies on top of point
A2 from the discussion of Fig. 2(b). Brick 0 and brick 1 can be
considered as a point of mass 2M located at C0/1 that is just above
A2. Brick 3 can be considered as a point of mass M located at its
center of mass point C2. The horizontal distance between
C2 and A2 is L. Thus the horizontal distance
between A2 and A3 must be L/3, and the horizontal distance between
C3 and A3 is 2L/3 so that the torque generated by
the combined system of brick 0 and brick 1 around
point A3 cancels the torque generated by brick 3 around
A3. This means that x is extended by L/3. The total
x is now L + (L/2) + (L/3).
}

For the case of 0, 1, ..., k bricks, the center of mass of the combined system
of bricks 0, 1, ..., k-2 is
above the point A_{k-1}, the upper right corner of brick k-1. Therefore,
the combined system of bricks 0, 1, ..., k-2 can be considered as a point of mass (k-1)·M located at its
center of mass point C_{0/1/../k-2} that is just above point
A_{k-1}. Brick k-1 can be considered as a point of mass M located at
its center of mass point C_{k-1}. The horizontal distance between
point C_{k-1} and point A_{k-1} is L. Thus for
those two mass points to balance above point A_{k}, the horizontal distance
between A_{k} and A_{k-1} must be L/k. This means that
*x* is extended by L/k, and the total *x* is now
L + (L/2) + (L/3) + · · · + (L/k). Extended to the case of (n + 1) bricks, we get

*x* = L·{1 + (1/2) + (1/3) + · · · + (1/n)}.

The right hand side of the above equation cannot be expressed as an elementary
function. As n → ∞, it diverges to infinity.