Show that it is possible to make a stack of lasagnas of the same size, shape, and mass, so that the top lasagna extends any distance at all beyond the edge of the table without falling off if the stack is high enough. Assume a purely mathematical center-of-mass approach - that is, the lasagnas stack perfectly as intended and nothing breaks their structure. One lasagna used per layer.

With that in mind, how many identical stacks will it take to have the top lasagna on the stack be 10 lasagna-lengths away from the edge of the table the lasagnas are stacked upon?

*
$\small{\text{A lasagna:}}$
*

[missing image: http://d3w5aos35qel1w.cloudfront.net/img/Hearty-Meat-Lasagna-large.jpg]

The answer is 272400600.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

How did you prove that this is the most efficient way to stack lasagnas?

Daniel Liu
- 6 years ago

Log in to reply

The most efficient way? I never did. I just proved it is possible to make a stack extending beyond the table for any distance, if there are enough lasagnas. But the efficiency question is certainly a good one. I guess we can take a look at the plots of the harmonic function for various values of $a$ and see which produces the least $N$ . I'll leave you to it :)

John M.
- 6 years ago

Log in to reply

I'm a little confused about your part that says we can use $\dfrac{1}{aN}$ for any $a$ . Doesn't $a\ge 2$ ?

Also the other concern is that a possible stacking that doesn't use the harmonic sequence pattern might be more efficient, giving a smaller $N$ .

We need to prove the most efficient is when the blocks are displaced in a harmonic sequence pattern. It is obvious that the most efficient happens when $a=2$ since for any bigger $a$ it is "compressed", giving a smaller distance.

Daniel Liu
- 6 years ago

Log in to reply

@Daniel Liu – Oh yes of course $a \ge 2$ . As far as different patterns are concerned, many could work. However, the harmonic with $a=2$ proves to be the most efficient for if we note, any $a<2$ will topple the structure. Hence, any other pattern, harmonic or not, would have to, by a direct comparison test for series, be $\le a=2$ . But we know that $a$ cant be $<2$ , so this takes care of everything.

As far as proving the case for $a=2$ goes, that's too much latex ;) I hope this is what you meant.

John M.
- 6 years ago

Log in to reply

@John M. – I obviously don't really understand this yet, because as far as I can see proving $a=2$ is actually pretty obvious, because if $a>2$ then the sum would be smaller and so would take longer to get to $D$ and thus would be less efficient?

Daniel Liu
- 6 years ago

Log in to reply

@Daniel Liu – Yes it is easy to prove that $a=2$ is the most efficient but I was referring to proving that $a<2$ won't work.

John M.
- 6 years ago

For this one, use the approximate formula for the Harmonic function needed here

$\dfrac { 1 }{ 2 } \displaystyle\sum _{ n=1 }^{ x }{ \frac { 1 }{ n } } =\frac { 1 }{ 2 } \left( Log\left( x \right) +\gamma +\frac { 1 }{ 2x } -\frac { 1 }{ 12{ x }^{ 2 } } +... \right)$

otherwise, even computing time can get insane.

$\gamma = 0.5772156649...$ is the Euler-Mascheroni constant.

We also have to assume that the lasagna pieces are completely frozen solid and machined to perfectly identical blocks of aerospace quality.

0 Helpful
0 Interesting
0 Brilliant
0 Confused

In better words, don't expect this thing to work out IRL xD

John M.
- 6 years ago

Your equation don't look right. Do you mean all the $n$ 's in RHS are $x$ ?

"We also have to assume that the lasagna pieces are completely frozen solid and machined to perfectly identical blocks of aerospace quality." You should file report for ambiguities HAHA

Pi Han Goh
- 6 years ago

Log in to reply

"Assume a purely mathematical center-of-mass approach - that is, the lasagnas stack perfectly as intended and nothing breaks their structure."

Nope. No ambiguities here. ^.^

John M.
- 6 years ago

Log in to reply

Ah I see that you have lawyer-ed up. Nice!

Pi Han Goh
- 6 years ago

Log in to reply

@Pi Han Goh – ALWAYS do. Mr. Mendrin always notes this - especially on my Conceptual Physics set. Which, by the way - a new one coming up soon!

John M.
- 6 years ago

..eh...typo fixed it.

Michael Mendrin
- 6 years ago

×

Problem Loading...

Note Loading...

Set Loading...

Define length of lasagnas to be $L$ . Now, the center of mass is unaffected by any other dimension of the lasagna because the axis of rotation in consideration here is one into the page - thus any other dimension of the object does not contribute to the torque (and hence, center of mass) of the lasagna - only mass.

Let's say we make a stack. We want to find the number $N$ so that the total length of lasagnas that is overhang is $D$ . First, consider a specific case:

Say we want to find the number of lasagnas it takes before the top lasagna extends completely beyond the edge of the table. That is, $D>L$ . Let's say that we stack lasagnas in the following fashion: the top lasagna extends half its length beyond the second lasagna, the second lasagna extends a quarter of its length beyond the third, the third extends one sixth of its length beyond the fourth, and so on.

We want to show that the center of mass of the system of $N$ lasagnas lies above the table; that is, $\bar { x } <L$ . Draw an $xy$ plane, with origin at left the tip of the bottom book. Then, the $x-$ coordinates of the centers of mass of the lasagnas are

$x_1=\frac{L}{2}, x_2=\frac{L}{2(N-1)}+\frac{L}{2}, x_3=\frac{L}{2(N-1)}+\frac{L}{2(N-2)}+\frac{L}{2},$ and so on.

Each lasagna has the same mass $m,$ , so if there are $N$ lasagnas, then

$\bar{x}=\frac{mx_1+mx_2+...+mx_N}{mN}=\frac{x_1+x_2+...+x_N}{N}$

$=\frac { 1 }{ N } \left[ \frac { L }{ 2 } +\left( \frac { L }{ 2(N-1) } +\frac { L }{ 2 } \right) +\left( \frac { L }{ 2(N-1) } +\frac { L }{ 2(N-2) } +\frac { L }{ 2 } \right) +...+\left( \frac { L }{ 2(N-1) } +\frac { L }{ 2(N-2) } +...+\frac { L }{ 4 } +\frac { L }{ 2 } +\frac { L }{ 2 } \right) \right]$

$=\frac{L}{N}\left [ \frac{N-1}{2(N-1)}+\frac{N-2}{2(N-2)}+...+\frac{2}{4}+\frac{1}{2}+\frac{N}{2} \right ] = \frac{L}{N} \left [ (N-1)\frac{1}{2} + \frac{N}{2} \right ] = \frac{2N-1}{2N}L <L$

Thus, the center of mass lies above the table no matter how many lasagnas are added.

Also note that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...=\frac{1}{2} \sum{\frac{1}{N}}$ is divergent (since it is a harmonic series). Hence, we can make the top lasagna extend any distance beyond the edge of the table if we add enough lasagnas.

Finally, observe that we do not have to follow the $\frac{1}{2N}$ scheme. In fact, any value of $a \ge 2$ will work for $\frac{1}{aN}$ . The series is still harmonic, and the center of mass $\frac{aN-1}{aN}L<L$ is still above the table.

I'm getting hungry. In fact, having a lasagna right now :)