Center of gravity of book overhang

In summary, the conversation discussed the maximum overhang possible for one, two, three, and four books placed at the edge of a table. The condition for the books not to fall off is for the center of mass to be placed over the edge of the table. For one book, the maximum overhang is .5L. For two books, the overhang of the top book would be .75L. For three books, the center of mass would be at .9166...L. As the number of books increases, the overhang approaches the maximum limit of L. However, this is not a stable equilibrium and the stack of books would easily topple with the slightest disturbance.
  • #1
adamc637
11
0
If you put a uniform block at the edge of a table, the center of the block must be over the table for the block not to fall off.

If you stack two identical blocks at the table edge, the center of the top block must be over the bottom block, and the center of gravity of the two blocks together must be over the table. In terms of the length L of each block, what is the maximum overhang possible?

Repeat for 3 and 4 books.

This problem is really confusing to me. For one book, the maximum overhang is .5L.

Say that weight of the book = w.

If I take the torques from the right edge of the furthest book,

[tex]\sum \tau = 0 = -w * .5L + F_n * .5L[/tex]

For two books...

[tex] \sum \tau = 0 = 2w * (answer L) = w * (?L) + w*(?L)[/tex]

I don't know how to get the coefficients in the question marks... Actually, is that even the right equation?

Thanks for your help!

o:)
 
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  • #2
For one book, the center of mass must lay on the table.

For the second book not to fall off, its center of mass must lay over the first book's.

For the combined, the net center of mass must lay over the table for the system not to fall. Compare the torques. The torque about the edge, holding the left must keep the torque holding the right. You applied this successfully for the first book, use the same idea behind the first to the second. You'll see a pattern forming.
 
  • #3
Well, that's my problem, I don't know how to apply the equilibrium torque concept to get the total length :cry: . Once I think about the two books on top of each other, I don't even know where to start for values of L.
 
  • #4
adamc637 said:
Well, that's my problem, I don't know how to apply the equilibrium torque concept to get the total length :cry: . Once I think about the two books on top of each other, I don't even know where to start for values of L.

This problem requires almost no computation. The condition for one book to not fall off the table is the same condition for the second book to not fall off the first book. If you arrange two books in that manner, where will their center of mass be located? How close can that center of mass be placed to the edge of the table?
 
  • #5
OH! MY! I get it! I've been trying to do it with a lot of math all this time, but now I realize the answer. Thanks whozum and especially you OlderDan!

So the first book would hang off .5L, the second would hang off .75L, the third .875L... yeah!
 
  • #6
adamc637 said:
OH! MY! I get it! I've been trying to do it with a lot of math all this time, but now I realize the answer. Thanks whozum and especially you OlderDan!

So the first book would hang off .5L, the second would hang off .75L, the third .875L... yeah!

.5L would be correct for a single book, and .75L would be correct for the overhang of the top book for two books. That is all you need for the problem. The extension to three books would not be .875L. It is actually more than that. The CM of two books overlapped by .5L would be .75L from either end. For a third book, that point would have to be directly above the overhanging edge of the book on the table. Look again at where the center of mass for the three books would be.

Measured from the farthest point of overhang, the CM of the top book would be at .5L, that of the second book would be at L, and that of the third book would be at 1.25L The CM is at the average of those three, or .9166 . . .L. As you add books, the progression is 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/6 (not 1/8), 1/2 + 1/4 + 1/6 + 1/8 = 1.04, etc

Yes, you really can have books completely outside the edge of the table if you have enough of them to build it up.
 
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Likes RJLiberator
  • #7
Moral : You can build an indefinitely long overhang (as the harmonic series diverges) !

Neat !
 
  • #8
Would the series not "converge"? There has to be a limit where you are placing the top book so infitesmally close to the one underneight. I did this project a while back in my first physics class, and got the formula.. well kind of. But the way mine worked out, I would have figured as 'n' went to infinity, there woulda been a limit. But as I said I never got the complete equation, and it was a messy one at that hah
 
  • #9
Hybird said:
Would the series not "converge"? There has to be a limit where you are placing the top book so infitesmally close to the one underneight. I did this project a while back in my first physics class, and got the formula.. well kind of. But the way mine worked out, I would have figured as 'n' went to infinity, there woulda been a limit. But as I said I never got the complete equation, and it was a messy one at that hah

As a practical matter, it would converge. If you could place every book exactly right the distance of the overhang would be

[tex]
D = \frac{L}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}}
[/tex]

with p = 1, and you would have unstable equilibrium of the stack. The slightest torque from a fly beating its wings would be enough to topple the thing. The series diverges for p < 1, but for p > 1 it converges. If you made even the slightest reduction from the maximum overhang, that would be equivalent to making p > 1 and the series would converge.
 
  • #10
Intuitively, it seems to me that any stack of books would collectively constitute one object, and that the center of that object's mass would have to lie over the table.

This would require either all books to be stacked with a minimum of 0.5 of their mass over the table, or for the books to be stacked in a staggered fashion such that an average of no less than 0.5 of their mass lie over the table.

Would anyone be kind enough to explain what is wrong with my thinking here?
 
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Related to Center of gravity of book overhang

1. How is the center of gravity of a book overhang determined?

The center of gravity of a book overhang is determined by finding the point at which the weight of the book is evenly distributed. This can be calculated by measuring the length and weight of the overhang and using the formula for center of mass.

2. Why is the center of gravity important for book overhang?

The center of gravity is important for book overhang because it determines the stability of the overhang. If the center of gravity is not positioned correctly, the book may fall or become unbalanced, potentially causing damage to the book or surrounding objects.

3. How does the center of gravity change with different types of books?

The center of gravity can change with different types of books due to variations in size, weight, and shape. For example, a larger and heavier book will have a lower center of gravity compared to a smaller and lighter book.

4. Can the center of gravity of a book overhang be adjusted?

Yes, the center of gravity of a book overhang can be adjusted by changing the position of the book or adding support underneath the overhang. This can help to improve the stability of the overhang and prevent it from tipping over.

5. What are the potential dangers of an unstable center of gravity in book overhang?

An unstable center of gravity in book overhang can lead to the book falling or tipping over, potentially causing damage to the book or surrounding objects. It can also pose a safety hazard if the book falls on someone or causes other objects to fall. Additionally, an unstable center of gravity can cause strain on the book spine, leading to potential damage over time.

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