T, V or Y
A wire is tied between two trees at distance D, and the ends are at a common height above ground. A weight is suspended midway between the trees at distance 'd' below the height at which the ends are attached to the trees.
Which configuration -- T, V or Y-shaped will minimise the total length of wire needed?
Which configuration -- T, V or Y-shaped will minimise the total length of wire needed?
Labels: SharedPuzzle, thinktank
posted by Rajesh Lal at
3:56 PM
10 Comments:
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- is the best, but the answer to the question is either T, V, or Y
I didn't finish!
T
A "T" is impossible, because if the wire weighs anything at all, there will be a downward deflection even if undetectible.
To make a "Y" would need extra wire for the vertical section. That leaves a "V" type, with the wire stretched to almost the breaking point.
Is there's a trick here to make this something more than I have considered? I hope so because otherwise this is to easy.
The answer must be V. It's always shorter between 2 points with a straight line.
From how I read it, it didn't say that the line was taught, so before the weight is applied, the line may sag. Then once the weight is applied to the middle, then it draws the midpoint down to "d" distance below the endpoints.
With a T config, (D/2) + d > sqrt((D/2)^2 + d^2)
With a Y config, a line from the horizontal endpoint to the weight is shorter than the D/2 + d, though this is shorter than with the T config.
Of course, there will still be some sag, even if you can't tell it.
I will assume the tension in the wire is infinite and the wire is mass-less. This means that there will be no sag in the wire.
Let x be the length of the vertical wire. This will restrict x to be between d (x=d means it's a "T") and 0 (x=0 means it's a "V"). Now the length for one slant section of the V and Y (the horizontal for the T) can be found using Pythagoras Theorem.
sqrt((D/2)²+(d-x)²)
Now we can find the length of the whole shape:
L = x+2sqrt((D/2)²+(d-x)²)
we want to find where L is the smallest value so now we take the derivative and solve for the critical point(s).
L = x+2sqrt((D/2)²+(d-x)²)
dL/dx = 1 + 2{0.5[(D/2)²+(d-x)²]^-0.5[2(d-x)(-1)]}
1 = 2(d-x)/[(D/2)²+(d-x)²]^0.5
[(D/2)²+(d-x)²]^0.5 = 2(d-x)
(D/2)²+(d-x)² = 4(d-x)²
D²/4 = 3(d-x)²
(d-x)² = D²/12
d-x = D/sqrt12
x = d-D/sqrt12
Notice that x does not equal 0 and also x is less than d (height); therefore x has to be between 0 and d exclusive. This means the shape cannot be a V (x=0) nor can it be a T (x=d).
In conclusion the shape that will minimize total length is a Y.
Still don't get it? Consider this:
To simplify this question, lets place this shape on a flat surface to ignore gravity. To further simplify, lest make all three points at the corners of a equilateral triangle, the sides of which is D.
Lets consider the T shape, this means the total length of the wire is the base of the triangle and the height (L = D(2+sqrt3)/2 to be precise).
Now consider the V shape. The total length now is just two of the sides (L = 2D)
Finally the Y shape. if you join any point inside the triangles to the corners you would make a Y shape, although it may be lop-sided. Lets take the point right in the middle for convenience. The total length of the shape would be L = D*sqrt3.
As you can see the Y shape is smaller than the T and the V.
Therefore, the Y shape will minimize total length.
The problem says to minimize the length of wire. If the shape is a "T" and the horizontal segment is D and the vertical segment is d.
If the shape is "Y" the segment from tree to tree is D plus an amount due to the slope, PLUS a vertical plus and segement to drop the weight to the distance of d.
The minimum length of wire would be D and d equals zero. This would be imposible, so the shape would be "V" and d almost zero. To minimze the deflection at the weight means the wire would be stretch tight and d would almost be zero, this gives a minimum wire length of just over D, due the the deflection.
Still not convinced? Consider this:
You have the wire in the V position, the height will be d. Now if you pinch off a part of the V in the middle and tie a knot you would make a Y where the knot will be at the vertex of the Y.
Since you pinched off a part of the V, you lost some of the height. But the lower part of the Y is doubled, so some of the length is not used.
So we cut off one side of the bottom loop near the knot. Now we have two pieces of wire.
If you let the cutted loop hang, the height has now increased. In fact the height from the bottom of the cutted loop to the top is actually longer than the height of the V.
This means that for the same amount of wire used for a V, you could make a Y that has a deeper height.
Therefore a Y with the same height as a V will use less wire.
If none of my solutions satisfy you, wait for the answer to be revealed.
V
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