Question

Consider two identical uniform bricks each of length "L" which are staggered and stacked in parallel with one on top of the other. Find the maximum possible overhang beyond the table's edge the system can have without toppling. Express your answer in terms of "L." Hint: apply the "free-standing body" theorem THREE TIMES.

Please show work

Answer 1

When stacking any brick or brick system, the furthest along you can stack the brick or brick system is at its center of mass

Let's consider the topmost brick. It's a single brick, so it's simple.

It can be stacked with its center of mass (L/2) hanging over the edge of the brick below it.

Now here's where things get tricky. When you try to figure out where the brick below it can be placed, you can't just consider that brick. You need to balance the top TWO bricks according to their combined center of mass.

So, let's figure out where their combined center of mass is.

Since the top brick's center of mass is L/2, the brick below it has a center L/2 away from that. To understand this, try drawing yourself a picture. If the top brick is hanging over L/2 (at its center of mass), the lower brick's center of mass must be L/2 away from that L/2. Since it's L/2 away from a distance of L/2, the distance from the edge of the top brick to the center of mass of the second brick is L/2+L/2 = L.

We can find the center of mass of the two brick system this way: (#masses*distance to center of mass + #masses*distance to center of mass + ...)/number of masses.

In the case of two bricks, this gives us (L/2+L)/2 = 3L/4.

Now let us consider three bricks. The center of mass of the top two brick system is 3L/4.
By the same logic as above, the center of mass of the third brick must be L/2 away from the center of mass of the top two bricks (3L/4 + L/2 = 5L/4).

So now using our center of mass formula, we see that (2masses*com of the top two masses + 1mass*com of the third book)/3 masses = (2*3L/4+5L/4)/3 = 11L/12.

Now lastly let us consider the case in which we have four bricks.

The center of mass of the top three brick system is 11L/12.
The center of mass of the fourth brick must be L/2 away from this (L/2+11L/12 = 17L/12).