Your task in a science contest is to stack four identical uniform bricks, each of length L, so that the top brick is as far to the right as possible without the stack falling over. Is it possible, as FIGURE P12.60 shows, to stack the bricks such that no part of the top brick is over the table? Answer this question by determining the maximum possible value of d.

Verified step by step guidance

1. Understand the problem: The goal is to determine the maximum horizontal displacement (d) of the top brick such that the stack remains stable. Stability is achieved when the center of mass of the entire stack lies directly above the table. Each brick is uniform, meaning its center of mass is at its geometric center.
2. Step 1: For a single brick, the center of mass must remain above the table. If the brick is displaced by a distance of L/2 (half its length), the center of mass will be at the edge of the table. Any further displacement will cause the brick to fall.
3. Step 2: For two bricks, the center of mass of the system is the average of their individual centers of mass. If the top brick is displaced by L/2 relative to the bottom brick, the center of mass of the two-brick system will shift to a point L/4 from the edge of the bottom brick. To ensure stability, the bottom brick must be positioned such that this new center of mass remains above the table.
4. Step 3: For three bricks, the center of mass of the system is calculated by considering the positions of all three bricks. The top brick can be displaced by L/2 relative to the second brick, and the second brick can be displaced by L/4 relative to the third brick. The center of mass of the three-brick system will shift further, and the third brick must be positioned to keep this center of mass above the table.
5. Step 4: For four bricks, the process continues similarly. The top brick can be displaced by L/2 relative to the second brick, the second brick by L/4 relative to the third brick, and the third brick by L/6 relative to the fourth brick. The center of mass of the four-brick system must remain above the table, and the fourth brick must be positioned accordingly. The total maximum displacement (d) is the sum of these individual displacements: d = L/2 + L/4 + L/6 + L/8.

Key Concepts

Center of Mass

The center of mass is the point at which the mass of an object is considered to be concentrated. For uniform bricks, this point is at their geometric center. When stacking bricks, the center of mass of the entire stack must remain above the base of support to prevent tipping over.

Torque

Torque is a measure of the rotational force acting on an object. In the context of the stacked bricks, torque is generated by the weight of the bricks acting at their center of mass. To maintain stability, the sum of torques about the edge of the table must not exceed the opposing torque that keeps the stack balanced.

Static Equilibrium

Static equilibrium occurs when an object is at rest and the net force and net torque acting on it are zero. For the stacked bricks, this means that the downward gravitational force must be balanced by the upward normal force from the table, and the torques must balance to prevent rotation, allowing the top brick to extend as far as possible without falling.