- Dec 30, 2014
Dan, why approximate the arc length as we are giving that length. It is 80 m long.
I used the word "may" intentionally -- something felt wrong, because my conclusion was the opposite of my original thought. It turns out that I made a mistake in the calculation; it should have been 40 - pi/4 T + T, which is more than 40, not less than 40.If we take the cable's thickness (T) into account, then when the poles are two thicknesses apart (assuming that is measured between the near sides), we also have to take into account the semicircle formed by the lowest part of the cable (assuming it is the centerline of the cable that is exactly 80 meters long, and it is sufficiently flexible to bend that sharply). I figure the cable will then take up a length of 40 - pi/2 T + T, so the lowest point on the cable will be a little higher than 10 meters. That is, the bend takes up more cable than it contributes to the vertical fall.
So there may in fact be no solution (unless you borrow a physicist's zero-thickness cable).
On the other hand, do we know how the cable is attached at the top?
\(\;\)The cable might be attached to the "front" of each pole (similarly to having rectangular-solid poles). If such poles were brought together, the cable would hang in front of them, instead of inbetween.… distance between the two poles is 0. I disagree … I think [it's] twice the thickness of the cable …
Clever thought! Good for you. However the picture does not show that.\(\;\)The cable might be attached to the "front" of each pole (similarly to having rectangular-solid poles). If such poles were brought together, the cable would hang in front of them, instead of inbetween.