This sounds like a harder problem than the 圆的面积最大. Can not believe this is only +++. Just some ituitive idea: as QFT mentioned in the 圆的面积最大, we may assume there is a pressure applied to every point of the rope (fense). In `good' area, the pressure is 2, in 'bad' area, it is 1. Since the tension along the rope is the same every where, this indicates that: In both the good area and bad area, the rope form an arc, and the good area arc has a radius which is half of that of the bad area arc. Then we can setup an equation to calculate the radius'..... 本贴由[QL]最后编辑于：2005-12-16 14:46:26 |
I assumed 圆的面积最大 is known here. So we know the fence consists of three arcs of different radius, although it may still be a ++++ problem. The problem is to prove the radius of the lower arcs is twice that of the upper arc. I don't know how difficult that will be, so the +++ is an estimate. I am aware your approach, but it is not very convincing to assume the pressure has a 1-1 relationship with price. |
I know, the argument is not convincing to me either, just an intuition. You are right, in the bad area, we should consider two arcs, cause at tree, it may not be smooth. So, for other folks, constant and I have reduced the difficulty of the problem to ++, good luck. |
It makes more sense to think in following way why good area has only the half radius of the bad one. suppose the length is variable. If the system is stable, it must be at its potential minimum. in this case the potential is the negative of the weighted area, which is covered by the rope under the force mentioned by QL. Thus half radius in the good area. |
对面积为A，周长为s的圆，有dA/ds = r。假设三段圆弧组成的栏杆形成了稳定状态，且半径分别为r1, r2。因为周长是固定的，如果坏地上的栏杆长度增加了ds，好地上就要减少ds。因为是稳定状态，如果坏地面积增加了dA，好地就要减少(1/2)dA。所以 r1/r2 = (dA/ds) / ((1/2)dA/ds) = 2。 虽然还不是很严格，至少是不用物理了。 |
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