statics

[StephenPrivitera]

Five identical wood blocks of sides L and thickness H are shifted in one direction to form a leaning tower of the maximum protrusion. How should you stack the blocks to achieve the maximum protrusion? What is the maximum protrusion? What if you had an infinite number of blocks?

For any block the center of mass of the blocks above it must be to the left of the block's right edge. So suppose we have n blocks. Consider the ith block. There are n-i blocks above the ith block. If we denote the protrusion of each block from the right edge of the ith block as xj, then the center of mass of the blocks above the ith block is the sum of all these xj, call it X, since each block has the same mass. We want X = L/2. Since we have no negative protrusions, the maximum protrusion is then L/2. It doesn't matter how they are stacked. This is true even if we have an infinite number of blocks.
Is this right?

BTW, how can I express the fact the CM must be to the left of the right edge using physics? Intuitively I know this is right.
See if I consider torques I can't come up with this result because I don't know where the normal force acts or how great it is. Any suggestions?

[Hurkyl]

The contact forces are spread throughout the region of contact.

[StephenPrivitera]

So how would I determine the torque extered by the normal force?

[Hurkyl]

You don't; at least not precisely. What you are going to do is to estimate it:


a \int_a^b F(r) \, dr \leq \int_a^b F(r) r \, dr \leq b \int_a^b F(r) \, dr


Or, if you prefer a quick and dirty intuitive argument, the net torque given by the contact forces has to lie somewhere within the (convex hull of the) region of contact. Since the center of mas does not lie in there, there has to be a net torque.

edit: I forgot to add that I'm assuming the CoM is outside the region of contact

[Doc Al]

Originally posted by StephenPrivitera
Five identical wood blocks of sides L and thickness H are shifted in one direction to form a leaning tower of the maximum protrusion. How should you stack the blocks to achieve the maximum protrusion? What is the maximum protrusion? What if you had an infinite number of blocks?

I think this is more of a math problem than a physics problem. The physics seems trivial: for stability with maximum projection, the c.o.m. of blocks 1 through N (top block is #1) must rest at (slightly) less than the edge of block N+1.

Find the pattern, starting at the top. Block 1 projects L/2 past block 2; L/2(1 + 1/2) past block 3; L/2(1 + 1/2 + 1/3) past block 4. Etc. For an infinite number, find the limit.

What am I missing?

[StephenPrivitera]

How did you determine the necessary projections?

[Hurkyl]

Where is the center of mass of the top 2 blocks? Make sure it's over the 3rd block. Where's the center of mass of the top 3 blocks? Make sure it's over the 4th block. et cetera...

[StephenPrivitera]

Originally posted by Doc Al
Find the pattern, starting at the top. Block 1 projects L/2 past block 2; L/2(1 + 1/2) past block 3; L/2(1 + 1/2 + 1/3) past block 4. Etc. For an infinite number, find the limit.
Ok, I am still interested in knowing how you so easily knew what the projections must be without any apparent calculation. I got these same results, but I had to calculate the CM one at a time and then add the results together to find the projections.

[Hurkyl]

This is a classic brain teaser problem. [:)]

[Doc Al]

Originally posted by StephenPrivitera
Ok, I am still interested in knowing how you so easily knew what the projections must be without any apparent calculation. I got these same results, but I had to calculate the CM one at a time and then add the results together to find the projections.
First off, I'm just writing down my answer, not my twisted thinking that got me there. Be thankful for that.[8)]

Secondly, the "trick" is to start from the top. Find the first projection. Now treat the top two blocks as a single block of mass 2M: its CM is midway between that of the two blocks separately. Now that CM is placed on the edge of the third block... etc, etc. You will quickly see the pattern.

As Hurkyl said, this problem has been around for quite some time.