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?

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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 C0, must lie on top of the upper right corner of the brick 1, denoted as point A1, so that the torque around point A1 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 A2. Brick 0 can be considered as a point of mass M located at C0 right on top of point A1. Brick 1 can be considered as a point of mass M located at point C1, the center of mass of brick 1. The horizontal distance between C1 and A1 is L. Thus A2 must be at the middle point of the horizontal distance from point C1 to point A1. In other words the horizontal distance between A2 and A1 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 C0/1, must lie above point A2 so that the torque generated by the combined system around point A2 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 C0/1 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 Ak-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 C0/1/../k-2 that is just above point Ak-1. Brick k-1 can be considered as a point of mass M located at its center of mass point Ck-1. The horizontal distance between point Ck-1 and point Ak-1 is L. Thus for those two mass points to balance above point Ak, the horizontal distance between Ak and Ak-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.