1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Center of gravity of book overhang

  1. May 1, 2005 #1
    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:)
     
    Last edited: May 1, 2005
  2. jcsd
  3. May 1, 2005 #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.
     
  4. May 1, 2005 #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.
     
  5. May 1, 2005 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  6. May 1, 2005 #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!!
     
  7. May 1, 2005 #6

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    .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.
     
  8. May 3, 2005 #7

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Moral : You can build an indefinitely long overhang (as the harmonic series diverges) !!

    Neat !
     
  9. May 4, 2005 #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
     
  10. May 4, 2005 #9

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  11. Nov 30, 2011 #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?
     
    Last edited: Dec 1, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Center of gravity of book overhang
  1. Center of gravity (Replies: 2)

  2. Center of gravity? (Replies: 3)

  3. Centers of gravity (Replies: 1)

  4. Center of gravity (Replies: 3)

Loading...