# Stacking Dominos

1. Jun 11, 2009

### Chewy0087

1. The problem statement, all variables and given/known data
When dominos are stacked one on top of each other on the edge of the table, it turns out that the overhangs are related to the harmonic numbers Hn, which are defined as 1 + 1/2 + 1/3 + ... + 1/n: the maximum overhang (over the edge of the table) possible for n dominos is Hn/2.

Using the concept of centre of gravity prove this, feel free to use diagrams to aid your explanation.

3. The attempt at a solution

I'm struggling with this one, and even having looked it up it's not at all clear to me, from what i thought, the centre of mass of all of the books together must be on the pivot point (at least), over the edge of the table, with 1/2 of the first domino over the side, however I really can't justify to myself that placing another domino ontop will keep it balanced, I read somewhere that " The center of gravity of the stack of two books is at the midpoint of the books' overlap" what exactly does this mean? I've tried drawing diagrams, don't understand it.

Thanks in advance for any help & i'll try and scan in a picture although I don't think my diagram will help

Edit: Hmm looking over it again I get the feeling that for any pile the centre of gravity of the ENTIRE pile must be on the pivot point, however how would you go about working this out mathematically for say 10 books?

Last edited: Jun 11, 2009
2. Jun 11, 2009

### LowlyPion

Consider the Σ of the centers of masses of each of the dominoes.

So long as that sum is not past the pivot point (the edge of the table) then they should be stable, even if diminishingly so with each added domino.

3. Jun 11, 2009

### Chewy0087

That's what I thought would be the intuitive solution, however;

Saw this link, and can't really make heads or tails of it, I understand with two blocks, in that the centre of where they overlap is the centre of mass, and i'm happy with doing that mathematically however when it gets to 3 you have to find the centre of three & four and i get stuck.

4. Jun 11, 2009

### LowlyPion

Consider then the center of mass of the x(n) system as

x(n)*n = (L)*(n-1) + L/2*1

With the overhang equal then to L - x(n)

That should simplify then to Overhang(n) = L - (L - 1/(2N)) = 1/(2N)

Doesn't that look to be the Harmonic number / 2 since any element N has overhang 1/(2n)?

5. Jun 12, 2009

### Chewy0087

Sorry i'm having difficulty understanding how you can find the centre of mass mathematically the way you have...could you go through it a bit slower, sorry but i really need to understand this problem =[

6. Jun 12, 2009

### LowlyPion

You recognize that the center of mass of the bottom one must be right at the corner. That is maximum overhang.

For the second one then you have the center of mass of n = 2 as the sum of the distances of the centers of masses from the far end (away from the edge). The CM of #2 is at L/2 from the end and the CM of #1 is acting at L, which is right at the protruding edge - a distance of L away.

The formula for the CM is:

x = 1/M*Σ CM*d

(For convenience use unit mass for the dominoes.)

For N = 2

x = 1/2*(1/2*L + 1*L) = 1/2*(3/2*L) = 3/4*L

That's how far from the furthest from the table edge end the CM of 1&2 is. And the overhang of 2 from the edge then is L - 3/4*L = 1/4*L.

Now you are going to iterate by placing the 1&2 system on 3. It will then be at the overhanging edge of 3. That means that a mass of 2 will now be at a distance of L and you will have the CM of 3 again still at the L/2 distance from the far end.

This CM calculation then looks like:

x = 1/3(1/2*L + 2*L) = 1/3(5/2*L) = 5/6*L
Overhang of 3 then is 1 - 5/6 = 1/6

Generalizing then for any domino element N, you have

x = 1/N*(L/2 + (N-1)*L)

All of the mass of all the dominoes above the bottom one are acting at the overhanging edge - the distance L away. Then you add the CM of the bottom domino and divide by the mass N.

The rest follows as before.

7. Jun 13, 2009

### Chewy0087

This is the bit i've been having trouble with, i can follow your reasoning from here (ish) however what is this formula and where has it come from? =[ i'm sorry if i'm bieng stupid but i can't find it or really make sense of it.

x = 1 / Mass*Sum of Centre of Mass*distance? Or..?

8. Jun 13, 2009

### LowlyPion

9. Jun 13, 2009

### LowlyPion

So yes, basically.

It's the sum of the each center of mass times distances of the center of mass from the one end. All divided by the total mass.

By employing the previous center of mass, in each successive iteration ...

10. Jun 13, 2009

### Chewy0087

Grrrr really struggling to understand this, so if the formula for the centre of mass is;

x (overhang?) = $$\frac{\sum M*Total Overhang}{\sum M}$$ ? Isn't that redundant?

Also then why is it 1 /Mass * Sum of Centre of Mass * Distance for this example?

I know i'm bieng a pain :<

Edit : I see now that the x is the distance to the overall centre of mass, but i don't see how that helps me..

11. Jun 13, 2009

### LowlyPion

No, it's not the mass times the overhang.

The overhang is found subsequently after you find the center of mass x, because since the center of mass is to be at the corner/edge just in balance, then you subtract that point, the x of the center of mass from L the length, because that's how much is sticking out for that element.

12. Jun 13, 2009

### Chewy0087

Okay, I think i'm starting to get to grips with it abit, thanks again for all of the help! =D

So the distance from the start of the first domino to the centre of mass is given by; (for standard unit mass of 1)

*for x1 meaning distance from the start of the first domino to the centre of mass of that block.

$$\frac{x1 + x2 + x3......xn}{n}$$ for n blocks.

From the formula;

$$\frac{\sum (mass(N)x(N))}{Total Mass}$$

Meaning that to work out x1 would be the distance from the start of the first block to the end of the given block / 2?

Have I interpreted this correctly? And if so, I don't see how given this you can find the overhang to be related to harmonic numbers.

Can I just say the help is great, and you don't know how much i appreciate it :P

13. Jun 13, 2009

### LowlyPion

I think it is better if you think it as one block - the bottom block - with the remainder of the blocks - the (n-1) blocks whose masses you have previously determined to have balanced their combined center of mass at the overhanging edge. This avoids having to recalculate with all the masses individually, because the center of mass of the (n-1) are acting at the already determined overhanging edge.

That makes then the general form of the center of mass of the nth block the natural L/2 of that block alone, acting at the center L/2 as you would ordinarily suppose for just that block, and then the remainder of the mass, the (n-1), is acting at the overhanging edge of the block, which is of course a distance of L away from the edge of the block that lays on the table.

With xn in hand, then you know the overhang of the nth block by subtracting from L - the rest of L up to x of it laying on the table.

As shown before then for the nth block, that L - xn = Hn/2

Last edited: Jun 13, 2009
14. Jun 13, 2009

### Chewy0087

Hmmm, i think i see what you're saying...thanks alot for the help! i'll try to work through it a couple of times tonight and if i'm still having trouble i'll post. thanks again for the help!