This is very exciting: a former student of mine, Anthony Attia, has submitted a solution to the Tarzan rope problem I posted some weeks ago. Anthony was in my Mechanics class at Vanier College in 2016. He is now pursuing undergraduate studies in mechanical engineering and simultaneously doing a stage at my former employer, MDA Space.
As is the case with some students, Anthony and I have stayed in touch since he graduated from college. This post, however, is the first one in more than ten years of this blog's existence that someone other than me has written; it is about time. Watch as Anthony analyzes a uniform rope, pinned at the top and vertically suspended, subjected to a horizontal uniform wind.
The following text appears here with Anthony Attia's consent:
When faced with a complex physical phenomenon, it is quite common to simplify the problem to a point where an analytical solution can be formulated. The simplification is done by stating assumptions throughout the approach. The more assumptions we take, the more likely our approximated answer will diverge from the true value. As students of science, it is our duty to ensure that we are equipped with enough knowledge to apply the proper assumptions.
Tarzan’s rope problem can be as complex as we want it to be. We can treat the rope as either flexible or rigid, we can treat the wind force as a function of time or a constant, we can consider the effects of cold temperature on the characteristic properties of air or we can neglect them. For the sake of maintaining my sanity and that of the reader’s, we shall treat the rope as a pinned rigid body who is subjected to a constant drag force that is acting in the horizontal direction. An important fact about assumptions is that there cannot be an incorrect one per say, however, every single one of them must be justified.
In my preliminary analysis, I will assume the rope to be rigid, effectively assuming that the profile of the rope will be linear when displaced. Generally, this assumption would not be valid with a rope, but I will make it anyway and check the extent to which it was good later.
With that in mind, we can begin trying to find the velocity of the wind, by relating the drag force FD and the weight W. Consider the model below, which depicts the scenario:Given that the net drag force is acting on the center of gravity in the horizontal direction and the weight is acting in the vertical direction, the ratio of these forces, FD/W, ends up being equal to tan(θ). We can take the sum of all torques about the pin and put them equal to zero. Then, using the following definitions, we may express the wind speed as a function of the other parameters.
The wind speed is then given by:
Say, however, that we now want to treat the rope as a flexible body; how would we proceed? Before answering that question, we must properly understand the behavior of weight and drag. In the previous figure, the drag force was lumped into a single vector whose line of action passes through the center of mass of the rope. Let us do a quick thought experiment: if we were walking headwind, would our entire body feel pushed by the drag force or just a single point? The answer is the former. So, why did we draw a single vector? That vector is actually the resultant or net drag force acting on the rope. If we were to properly illustrate the aerodynamic force that the body is subjected to, we would have to draw many smaller vectors that are acting on the entire exposed surface. These types of forces are called distributed load: though they act on every point of the body, we may sometimes use a single vector to represent the resulting effect (note that gravity is similarly distributed and then a resultant is used). Every segment of the rope has a mass equal to dm and the sum of all segment masses will yield the total mass m. Now, to solve the flexible body problem, we must assess a differential segment dm that is exposed to a differential drag of dFd by drawing its free body diagram.
Newton’s second law in x and in y yields:
These equations simplify to:
Equalizing the two equation we get: