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?
(In reply to
Part 2 by Leming)
Assuming that we're looking directly at the face with the two nails in
it, I'll label the verteces from the upper-left clockwise as A, B, C
and D. [Compare to (0,2), (7,2), (7,0) and (0,0).] Similarly the
respective hidden verteces, hidden behind these corners are W, X, Y and
Z. [WZ goes from (0,-3 to 0,-1) and from (-1,2 to -1,0). YZ goes from
(7,-1) to (0,-1) Bear with me for now if you don't understand this.]
From M, draw a line through (0,0.625) on AD to (-1,0.25) on WZ.
From N, draw a line through (3.333,0) on CD and (0.667,-1) on YZ to
(0,-1.25) on WZ which is the same point as (-1,0.25). I created a
cutoff of the block and string paths for all three steps in a Microsoft
Word document, but I don't know where I could host it. If you would
like it, email me at charley.grevers on gmail.
Edited on August 10, 2005, 5:22 am
|
Posted by Charley
on 2005-08-10 03:16:14 |