The above problem is taken from the 2014 $F=ma$ contest.
This question generated a great deal of controversy. At least two test takers challenged the
answer, one who even tried to do the experiment. I’m told that several different “Ph.D”
physicists declared that the correct answer was X, but, interestingly enough, couldn’t agree
on what X should be.
In the YouTube poll the majority of people have chosen (C), however the correct answer is (B).
Veritasium took it to the next level, he rented a helicopter to reveal the real answer to the problem.
I would like to solve this problem using differential analysis, analogous to the catenary equation solved here.
First of all, the helicopter is travelling at constant speed, so we are dealing with an inertial reference frame. If the rope appears stationary to an observer in the helicopter, then the net force is null. The only forces to analyze are the gravity, the air resistance, and last but not least the tension in the rope. I have chosen a coordinate axis with the origin at the lower end of the rope, but it should be unimportant. The rope has unknown shape, so we are restricted on the assumptions we can make about the geometry of the rope — we don't know that the correct answer is (B) yet.
Consider a small mass element $\Delta{m}$ in the rope. Let's write all the forces acting on $\Delta{m}$:
Gravity pointing downwards. The mass is proportional to the linear density and the length.
$$\vec{F_g}=-\Delta{m}g\hat{j}=-\lambda\Delta{s}g\hat{j}$$
Air resistance pointing backwards. The drag force is proportional to the cross sectional area, so proportional to the length of the vertical projection of $\Delta{m}$:
$$\vec{F_d}=-\frac{1}{2}\rho v^2C_dA\hat{i}=-\alpha\Delta{y}\hat{i}$$
Tensions:
$$\vec{T}(x)=-T(x)\cos\theta(x)\hat{i}-T(x)\sin\theta(x)\hat{j}$$
$$\vec{T}(x+\Delta{x})=T(x+\Delta{x})\cos\theta(x+\Delta{x})\hat{i}+T(x+\Delta{x})\sin\theta(x+\Delta{x})\hat{j}$$
Writing $\vec{F_\text{net}}=\vec{0}$ on the horizontal and vertical component:
$$T(x+\Delta{x})\cos\theta(x+\Delta{x})-T(x)\cos\theta(x)=\alpha\Delta{y}~~~(1)$$
$$T(x+\Delta{x})\sin\theta(x+\Delta{x})-T(x)\sin\theta(x)=\lambda\Delta{s}g~~~(2)$$
Divide both equations by $\Delta{x}$ and take $\lim\limits_{\Delta{x}\to0}$ to get to the differential form:
$$\frac{d(T\cos\theta)}{dx}=\alpha\frac{dy}{dx}$$
$$\frac{d(T\sin\theta)}{dx}=\lambda\frac{ds}{dx}g$$
In the catenary example we were lucky because we were able to easily eliminate $T$ from the equations, but here I can't find a way to simplify the system.
Also $\cos\theta=\frac{dx}{ds}$ and $\sin\theta=\frac{dy}{ds}$ are problematic.
How can I continue? Is this even going anywhere?
EDIT:
Equation $(1)$ tells that $T(x)\cos\theta(x)=\alpha y$, and also $T(x+\Delta{x})\cos\theta(x+\Delta{x})=\alpha(y+\Delta{y})$, taking into account that $T(0)=0$.
Substituting in equation $(2)$ leads to
$$\alpha(y+\Delta{y})\tan\theta(x+\Delta{x})-\alpha y\tan\theta(x)=\lambda\Delta{s}g$$
Again, dividing by $\Delta{x}$ and taking $\lim\limits_{\Delta{x}\to0}$ to get the differential form:
$$\alpha\frac{d(y\tan\theta)}{dx}=\lambda g\frac{ds}{dx}$$
Note that $\tan\theta=\frac{dy}{dx}$ and $ds^2=dx^2+dy^2$ according to the Pythagorean theorem.
$$\Rightarrow\alpha y\frac{d^2y}{dx^2}+\alpha\Big(\frac{dy}{dx}\Big)^2=\lambda g\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}$$
$$\Leftrightarrow\alpha yy''+\alpha(y')^2=\lambda g\sqrt{1+(y')^2}$$
The differential equation in the general form is:
$$yy''+(y')^2=C\sqrt{1+(y')^2}$$
Almost diagonal!
Best Answer
I think I got the solution. I chose (C) at the poll, but the correct answer is (B).
Break the cable in $N$ chunks, each of mass $m$. Call $1$ the chunk at the bottom of the rope, and $N$ the one of the top. Draw a free body diagram for chunk 1. There if a vertical force of $mg$ due to gravity, a horizontal force $D$ due to the drag, and a force $\vec T_1$ exerted by the chunk $2$ of rope. Call $\theta_1$ the angle of $\vec T_1$ away from the vertical. Since the chunk is moving at constant velocity by assumption, we have the equations $$ \left\{ \begin{align} mg &= T_1 \cos\theta_1 \\ D &= T_1 \sin\theta_1 \end{align} \right. $$ Let's move to the next chunk of rope. Again, we have a vertical force $mg$, a horizontal force $D$, the force $\vec T_2$ exerted by the chunk of rope $3$ and, this time, by Newton's third law, we also have $-\vec T_1$ exerted by chunk 1. Then by Newton's second: $$ \left\{ \begin{align} mg + T_1\cos\theta_1 &= T_2 \cos\theta_2 \\ D + T_1 \sin\theta_1 &= T_2 \sin\theta_2 \end{align} \right. $$ It shouldn't be hard to convince yourself that the signs are right. Note also we are not assuming $\theta_2=\theta_1$ for now, as that will be our conclusion. Note that we have already found $\vec T_1$, so let's replace it in here: $$ \left\{ \begin{align} 2mg &= T_2 \cos\theta_2 \\ 2D &= T_2 \sin\theta_2 \end{align} \right. $$ Pretty neat, eh? We have an expression for $\vec T_2$ that does not depend on the details of $\vec T_1$. Actually, you can generalise this formula by induction, or simply inspection, to get $\vec T_k$ for all $k=1,...,N$: $$ \left\{ \begin{align} kmg &= T_k \cos\theta_k \\ kD &= T_k \sin\theta_k \end{align} \right. $$ Ok, now were are almost done. Notice how the magnitude $T_k$ of $\vec T_k$ depends on $k$: $$T_k = \sqrt{(kmg)^2 + (kD)^2} = k \sqrt{(mg)^2+D^2} = kT_1.$$ The tension grows linearly in $k$. Cute. Here is the kicker though: $$\tan\theta_k = \frac{\sin\theta_k}{\cos\theta_k}=\frac{kD}{kmg} = \frac{D}{mg}.$$ So the angles $\theta_k$ do not depend on $k$. Since the angle $\theta_k$ how much the line connecting the chunk $k$ and the chunk $k+1$ deviates from vertical, this tells you that the rope is a straight line. $\square$
You can sanity check: if you ignore air resistance, set $D=0$ then $\tan \theta = 0$ and $\theta = 0$: rope is vertical. As $D$ becomes larger and larger, $\tan\theta$ grows more and more, and $\theta$ approaches $\pi/2$, horizontal rope.
EDIT: what happens if we do not assume that drag is constant?
In the above, we assumed that each chunk of rope experiences the same amount of drag $D$. This simplifies matters a lot. In reality, the drag will be proportional to the areas of the projection of the chunk of rope along the plane perpendicular to the movement. In formulas, $$D_k = \gamma(v) \cos\theta_k,$$ where $\gamma(v)$ has dimensions of acceleration, and is a monotone increasing function of the velocity, with $\gamma(0)=0$. Assuming that $D=D_k$ for all $k$ is compatible with taking a small angle approximation.
At first sight, not taking this approximation might seem to allow for a different shape to the rope. I will now argue that this is not the case, using a purely physical argument.
To simplify matters, let's assume that the rope is hanging from a pole in a wind tunnel. Thus the rope is not moving in our frame, and instead we a have a wind at constant velocity $v$. When $v=0$, the rope is vertical. Now, let a little bit of wind in. The small angle approximation is expected to hold, and we have the whole rope at a small angle $\theta$. Each chunk of rope is experiencing exactly the same amount of drag. Now, what happens as we increase $v$ by an amount $\delta v$? Each single chunk of rope experiences a little more drag. But, crucially, they all experience the same increase in drag! So again the rope is in a straight line, with a slightly larger angle. So, no matter how strong the wind is, there is a steady configuration where the rope is straight.
Let me put it in a different way. The force balance for the $k$th chunk of rope is $$ \left\{ \begin{align} kmg &= T_k \cos\theta_k \\ k\gamma(v)\cos\theta_k &= T_k \sin\theta_k \end{align} \right. $$ Now, let's solve all $2N$ of these equations simultaneously by assuming $\theta_k=\theta$ for all $k$. The equations for a chunk $k$ become: $$ \left\{ \begin{align} kmg &= T_k \cos\theta \\ k\gamma(v)\cos\theta &= T_k \sin\theta \end{align} \right. $$ Then take the ratio of these two formulas to get: $$\tan\theta = \frac{\sin\theta}{\cos\theta}=\frac{k\gamma(v)\cos\theta}{kmg} = \frac{\gamma(v)}{mg}\cos\theta.$$ Note that $k$ dropped out of the formula! We can rearrange this to $$\sin\theta = \frac{\sin\theta}{\cos\theta}=\frac{k\gamma(v)\cos\theta}{kmg} = \frac{\gamma(v)}{mg}(1-sin^2\theta)$$ (thanks @MichaelSeifert!), which has exactly one positive solution: $$\sin\theta = \frac{mg}{2\gamma}\left(\sqrt{1+\left(\frac{2\gamma}{mg}\right)^2}-1\right).$$ If $\theta_k=\theta$, each we solve all the equations. $\square$
"Wait!", you might say, "you assumed that the rope was in a straight line, isn't that what you wanted to prove?". Well, yes, I did assume that the rope was in a straight line. But what I proved, is that there a straight line configuration is a static configuration, an equilibrium. In other words, I proved that if the rope is in a straight line at that angle, it will feel no net forces.
To really know that this is the configuration the rope will take, one would need to show that this is a stable equilibrium, so that small deviations will decay and lead there. I'm happy to take that as an assumption.
Sanity check. Let $\alpha = \frac{2\gamma}{mg}$, this is a dimensionless quantity that quantifies how strong is the wind. We can rewrite the angle as a function of $\alpha$: $$\sin\theta = \frac{\sqrt{1+\alpha^2}-1}{\alpha}.$$ With weak winds, $\alpha\ll 1$ and $$\sin\theta \approx \frac\alpha2 = \frac\gamma{mg},$$ which is compatible with the computation before. This is neat, as there we assumed that the drag was independent of the angle, which is a valid approximation only for small angles, i.e. small winds. (By the way, I realise now that I should have put $\gamma=D$)
If instead we have huge winds, $\alpha\gg1$, we have $$\sin\theta \approx 1 - \frac{mg}{2\gamma},$$ This model predicts a maximum angle of $\theta = \pi/2$ for the rope.