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Book-stacking Problem

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Apr 5, 2022
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Book-stacking Problem ๐Ÿ“„๐—™๐—ถ๐—น๐—ฒ ๐—Ÿ๐—ถ๐—ป๐—ธ https://drive.google.com/file/d/1qxlrL5wuJu-Xy0UnBPlLMqNMQ7ZOeBUB/view?usp=sharing ๐Ÿ“ซ๐Ž๐ฎ๐ซ ๐…๐ ๐๐š๐ ๐ž: https://www.facebook.com/ScienceWorld-106933907791981 ๐Ÿ“š๐ƒ๐š๐ฏ๐ข๐'๐ฌ ๐๐จ๐จ๐ค๐ฌ ๐Ÿ“• ๐—ช๐—ฒ๐—ถ๐—ฟ๐—ฑ ๐— ๐—ฎ๐˜๐—ต๐˜€: ๐—”๐˜ ๐˜๐—ต๐—ฒ ๐—˜๐—ฑ๐—ด๐—ฒ ๐—ผ๐—ณ ๐—œ๐—ป๐—ณ๐—ถ๐—ป๐—ถ๐˜๐˜† ๐—ฎ๐—ป๐—ฑ ๐—•๐—ฒ๐˜†๐—ผ๐—ป๐—ฑ (https://www.amazon.com/Weird-Maths-Agnijo-Banerjee-Darling/dp/1786072645) ๐Ÿ“™ ๐—ช๐—ฒ๐—ถ๐—ฟ๐—ฑ๐—ฒ๐—ฟ ๐— ๐—ฎ๐˜๐—ต๐˜€: ๐—”๐˜ ๐˜๐—ต๐—ฒ ๐—˜๐—ฑ๐—ด๐—ฒ ๐—ผ๐—ณ ๐˜๐—ต๐—ฒ ๐—ฃ๐—ผ๐˜€๐˜€๐—ถ๐—ฏ๐—น๐—ฒ (https://www.amazon.com/Weirder-Maths-At-Edge-Possible/dp/1786075083/) ๐Ÿ“— ๐—ช๐—ฒ๐—ถ๐—ฟ๐—ฑ๐—ฒ๐˜€๐˜ ๐— ๐—ฎ๐˜๐—ต๐˜€: ๐—”๐˜ ๐˜๐—ต๐—ฒ ๐—™๐—ฟ๐—ผ๐—ป๐˜๐—ถ๐—ฒ๐—ฟ๐˜€ ๐—ผ๐—ณ ๐—ฅ๐—ฒ๐—ฎ๐˜€๐—ผ๐—ป (https://www.amazon.com/Weirdest-Maths-David-Darling/dp/1786078058/) ** The kindle versions are available *** For more details : http://weirdmaths.com/ ๐Ÿ“„๐—ง๐—ฟ๐—ฎ๐—ป๐˜€๐—ฐ๐—ฟ๐—ถ๐—ฝ๐˜๐—ถ๐—ผ๐—ป: How much of an overhang can you achieve by stacking books on a table? Letโ€™s assume that each book is one unit long. To balance one book on a table, the center of gravity of the book must obviously be somewhere over the table. To achieve the maximum overhang of a stack of books, the center of gravity of the stack should lie directly above the table's edge. The maximum overhang with one book is obviously 1/2 unit. For two books, the center of gravity of the first book should be directly over the edge of the second, and the center of gravity of the stack of the two books should be directly above the edge of the table. The center of gravity of a stack of two books is at the midpoint of the books' overlap, or (1 + 1/2)/2, which is 3/4 unit from the far end of the top book. It turns out that the overhangs are related to whatโ€™s known as the harmonic sequence. If there are n books, the maximum overhang is 1 + 1/2 + 1/3 + ยผ all the way up to 1/n, all divided by 2. With four books, the overhang (1 + 1/2 + 1/3 + 1/4)/2. And because this is greater than 1, no part of the top book is directly over the table. If we want an overhang of 2 book-lengths it turns out that weโ€™d need 31 books. For an overhang of 3 book-lengths the number of required books jumps to 227. And if you were ambitious enough to go for a six book-length overhang, youโ€™d need 91,380 books stacked in the optimal way to achieve your balancing act. #book_stacking #block_stacking #problem

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