Homework Help: Center of mass of pizza boxes

1. Aug 27, 2010

Mechdude

1. The problem statement, all variables and given/known data
im trying to find the center of mass of two identical strong square pizza boxes of dimension one inch deep by 18 inches wide one is placed on top of the other with the edge of the top box in line with the center of mass of the bottom box ,

2. Relevant equations

n/a , the boxes have uniform distribution of mass .

3. The attempt at a solution

using symmetry considerations , the c.o.m. of the system is at about line 3/4 of the length from the leading edge of the top box .
the purpose of this is to try and solve the "book stacking problem" i want to solve for the first two first.

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2. Aug 27, 2010

Redsummers

If your aim here is to just get c.o.m. of the first two, 3/4 of the distance away from the edge of the upper box is correct.

Now if you want to get a conjecture so that you can determine the distance just by only knowing that there are n boxes (as you said, the 'book stacking problem'). Try fiddling around with the calculation you just used to get 3/4 (i.e. (1+1/2)/2) ). You will find quite straightforward to get an infinite sum here by using harmonic numbers, you could also proof it by mathematical induction.

I don't know if that's what you were asking for, as I understood, you want to get the conjecture for n books in the book stacking problem.

Cheers

3. Aug 27, 2010

Mechdude

thanks , that is what i was aiming at , i got a hint to work it out first for the two , then for three and so on and try to figure out a pattern, (the one you mentioned). now i got another question, how do i figure out the c.o.m. of three books extended of of the table maximizing the over hang. for two i was almost by observation but for three i cant see an obvious method.

4. Aug 27, 2010

Redsummers

Oh good, then you're not really far from getting the conjecture. You just have to think that to get the center of gravity for the three boxes, the same condition you used before (for two) will hold.

I'll give you a way to look at it:
In the first two boxes, what you actually do for them not to fall, is to put the edge of the second box in the center of gravity of the first box (i.e. the edge of the second box goes at the 1/2 of the first box) and then is like you had only one big box (of length 1/2+1/2+1/2; and you know that the center of gravity for simple –homogeneous– bodies is just half the length.
And that's why you get the iteration of (1+1/2+...)/2.

I don't know if that's clear enough, just don't hesitate to ask.

5. Aug 29, 2010

Mechdude

well thanks, but there's something i still quite do not get, its not the math, its the physics, now the stability condition for two boxes is that the trailing edge of the top box is over the c.o.m. of the bottom box now, for three boxes i can not see an easy stability condition, for the sake of argument lets look at a scenario in which we try to solve it recursively from the top, place the top box over the c.o.m of the middle box and then the c.o.m of the two upper boxes is at a distance of 3/4 the length from the leading edge of the top box, where do we place this new c.o.m with respect to the c.o.m of the lowest box?

6. Aug 29, 2010

Redsummers

Not sure if I understand clearly all this c.o.m. issue we have here. But the way as I understand your question, I would say that we place the new c.o.m at (3/4-1/2 = 1/4) from the lowest box's –and previous– c.o.m. But, what's your point here, on searching the difference between old c.o.m and new ones?

Hmm maybe all this considerations of c.o.m is making it more confusing, once you see the physical pattern of how to make the boxes not to fall, it's not a big deal of a problem. It's just that you cannot treat the c.o.m of individuals boxes, because it goes with the whole system altogether. Perhaps I interpreted wrongly, but I gave the result as if you were talking about the c.o.m. of the lowest box as the 1/2 c.o.m (first one).

Correct me if you meant something different. Also, you may want to check http://mathforum.org/advanced/robertd/harmonic.html".

Last edited by a moderator: Apr 25, 2017
7. Aug 30, 2010

Mechdude

thanks i was able to figure out how tho get the c.o.m of there boxes, which was the issue, i did not use the principle of symmetry properly in obtaining the position of the c.o.m. of three boxes (i made an elementary math error). thanks for the link an the effort
cheers

edit:

i arrived at an erroneous value of 7/8 for the overhang instead of 11/12 , till i corrected the distance of the c.o.m of the assembly of three boxes form the c.o.m of the bottom box

Last edited: Aug 30, 2010