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String and Wood (Posted on 2005-08-08) Difficulty: 3 of 5
You have a block of wood 1 by 2 by 7 units. One of the 2 by 7 faces has 2 nails inserted, the heads and part of the shaft of each nail protruding. If the coordinates of the corners of this face are: (0,0), (0,2), (7,2), and (7,0), then the nails are located at (1,1) and (6,1).

Assume the coefficient of friction between string and wood is zero, and that the diameter of the nails is negligible.

(1) The ends of a non-elastic string of length 13 units are attached, one end to each of the 2 nails. The string is taut. How is this possible?

(2) What about a second piece of taut string, approximately 8.544 units long, also with ends attached to the 2 nails?

(3) What about a third piece of string approximately 22.0880 units long?

See The Solution Submitted by Larry    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Part 3 | Comment 3 of 6 |

Part 3:  22.0880 is not a clear cut length. But taking integer differences from 22.0880 gives 17.0880 + 5.

17.0880 is close to sqrt(292).  No other integer differences come close to being the square root of an integer.

292 = 256 + 26 = 16^2 + 6^2.

Well 16 is the length to go around the block once in the 7 + 1 direction and 6 is the length to go around the block once in the 2 + 1 direction.

The path is tough to articulate. Essentially wrap the string around the block once in the longest direction while wraping it once around the width.  This places the string back at the first nail.  Then use the 5 units of string remaining to head directly to the second nail.

A 1000 words is worth a picture. Anyone up to it?


  Posted by Leming on 2005-08-08 16:07:53
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