Physics Puzzle: Rope Between Two Poles
Posted on December 10, 2007
Filed Under Physics Puzzles |
A good puzzle, someone once told me, is one that can be solved by a bright high-schooler, and yet challenge even a seasoned practitioner. As a puzzle-aficionado, I second that notion, which is why mechanics puzzles are my favorites. It is the most intuitive discipline and the first to be studied in both high school and university, and yet the complexity of the phenomena it describes can often be quite astounding. Here is a puzzle to test your knowledge of mechanics:
A rope hangs between two poles. Can you calculate the tension in the rope at (a.) the lowest point, and (b.) at the points at which it connects to the poles? For the purposes of the question, assume the rope has some mass, m, which is uniformly distributed. Take L to be its length. You may assume you know the distance between the poles.
Note: although my graphical skills don’t necessarily portray this, the problem is symmetrical (i.e. the right and left poles are the same, and the rope is attached to both at the same height … 
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The equation of the rope, with x = 0 at the center, is y = a cosh[x]. (Is there a simple derivation of this fact, using say the extremization of the total gravitational potential energy? I’d think you still need to solve a ODE?) Let’s say the poles are 2 b apart from each other. Since the poles are holding the rope up, the tension in the rope must be equal and opposite to whatever force(s) the poles are exerting on it, for otherwise the last bit of the ends of the rope will start accelerating. Since we have static equilibrium the vertical components of the tension at the end must sum to be equal to the weight: if theta is the angle such that tan theta = dy/dx = a sinh x, and T is the tension at the ends, then T sin theta[end] = T a sinh b / sqrt{1 + a^2 sinh^2 b} = (1/2) m g => T = (1/2) m g sqrt{1 + a^2 sinh^2 b} / [a sinh b]. The tension at the lowest point must cancel the horizontal component of the tension at the end, so it is just T cos theta = (1/2) m g / [a sinh b].