We are given unlimited number of identical, very thin domino

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I think this question is most likely solved by a series, possibly a power series.

We are given an unlimited number of identical, very thin dominoes. The length of a domino is 2 centimeters (cm), and its mass is neglegable (10^-4314 gram per domino).

Suppose we stack the dominoes (assume we are infinitely powerful), one on top of the other, so that the stack advances horizontally in one direction because each is offset a little. The top domino is offset by 1 cm, and, as we move down the stack, the dominoes are offset progressively less from each other so as to keep the center of gravity. The stack is kept stable only by gravity. Assume that there is no air and the surface they are on is flat.

1) Show that the stack can advance infinite distance. HINT: Numerate the dominoes in a stack from the top. Note that a stack of two dominoes can advance 1 cm since the center of the first domino should be above the edge of the second domino. Place the third domino underneath the first two dominoes so that the edge of the third domino is below the center of the stack of the two dominoes that has been built already. Place the fourth domino underneath the stack of the three dominoes. By repeating this precess, one gets the distance advanced by a stack as a series. You must find the general term of the series and explain why.

2) How many dominoes are approximately needed for the stack to advance 10 m? Your approximation should be as small as possible. Eexplain how you have arrived at the approximation.

3) What is the approximate mass of the stack in question (2) in kilograms?
 
What have you tried? How far have you gotten? Where are you stuck?

Please be specific. Thank you.

Eliz.
 
If each domino has an overhang of half the preceeding, then

1+1/2+1/4+1/6+1/8..............


This is a version of the classic harmonic series proven by Nicole d'Oresme.

1+1/2+1/3+1/4+................

It's proof is no more than arithmetic.

Do a google. You should find plenty about it.
 
im stuck finding the general term.... and then knowing what method to use to estimate for what n the sum will be 10meters (1000 cm). if each had an overhang of half the preceding i would think : 1/(2^n).. but that cant be right cuz it doesnt diverge and for no n will it add up to 1000. but the terms are 1+ 1/2 +1/4 +1/8 ... that is where each is half the preceding... but doesnt work...
 
It's a little tricky. If you take a domino and place it on the first one, how far can you slide it before it falls off?. Another way, how far can you make it overhang the rest of the stack?. Half a card length.

Place another. With that top domino pushed out half its length over the second one, push the second domino, how much overhang can you get with the two top dominos?.

The trick is to think of the two top ones as one unit. Where's the center of gravity?. It's halfway along the unit, which is altogether one and half dominos long; so it's 3/4 of a domino length from the leading edge of the top domino. The combined overhang is 3/4 of a domino length.

If you push the 3rd domino to see how much you can increase the overhang, you can push it 1/6th of a domino length. The trick is to see the top 3 dominos as a single domino. The center of gravity is 1/6th of a domino length back from the leading edge of the 3rd domino.

You can find out more by doing research on this. Do a search. There's lots to be found. You'll see what I mean with a picture.

Add up the overhangs:

\(\displaystyle \L\\\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+................\)

Factor out 1/2:

\(\displaystyle \L\\\frac{1}{2}\left(1+\frac{1}{2}+\underbrace{\frac{1}{3}+\frac{1}{4}}_{\text{>1/2}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}_{\text{>1/2}}+............\right)\)

Look at the expression in the parentheses:

Oresme showed that it is divergent. By taking 2 terms, 4 terms, 8 terms, etc. you can group them into an infinite number of blocks of numbers, everyone bigger than 1/2. Therefore, infinite.

I hope I helped and didn't confuse you.
 
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