# Solving the helicopter hanging cable problem through differential analysis  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!

• EDIT: I think I know how to solve it, I will post if I get anything concrete.
– Neox
Oct 28 at 11:31
• The second order makes sense since tension is related to slope, and your governing equations balance the change in tension => change in slope. Oct 28 at 16:04
• I don't understand why anybody would choose (A), (C), or (E). In all three of those cases, the rope is vertical at the point where it attaches to the helo. It could only be vertical at that point if the horizontal component of the force applied by the helo is zero. But that could not be the case if the helo is "flying horizontally at constant speed," and "air friction...is not negligible." Oct 28 at 19:50
• @David Waterworth Initially I guessed B because I have Cargobob in GTA V.
– Neox
Oct 28 at 22:34
• @SolomonSlow as a "C" picker I assumed that helicopter downwash would influence the rope, if it were not-a-helicopter I'd probably have picked B. Oct 29 at 20:59

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.

• Nice! I have also obtained that the tension grows linearly and I have used that to get rid of $T$. However, I kept working with differentials and I got a first-order nonlinear differential equation. I will edit the post to show the work, but for now what you did is simple and elegant, enough to convince me that the true correct answer is (B).
– Neox
Oct 28 at 12:06
• I am actually reading your question more carefully... note that I am making a crucial assumption that the drag $D$ is the same for all chunks. But it seems that you are not willing to do that. Oct 28 at 12:39
• Yup. that's what I was going to comment. You (and Veritasium) have made the assumption that the drag force is equal for equal chunks, but the drag force does no depend on the mass, rather on the shape, here the length is crucial. The drag forces will be equal only if the rope will be linear, but we don't know that yet. And It's actually a second-order differential equation that I have no idea how to solve. Maybe I'll try to put it in Python.
– Neox
Oct 28 at 12:57
• Yeah I also get something quite ugly. You could try starting with a small angle approximation, so that $\cos\theta = 1 - \theta^2/2$, which is pretty decent. Only 1% error until $\theta=0.7$ and 10% for $\theta = 1$ (radians) Oct 28 at 13:00
• Nitpick: the final equation actually does have a closed-form answer for $\theta$. The equation is equivalent to $\sin \theta = \alpha (1 - \sin^2 \theta)$, with $\alpha = \gamma/mg$. This equation can then be solved with the quadratic formula. Oct 29 at 18:41

I think the answer you get depends on an assumption regarding aerodynamic forces. Split the rope into infinitesimal segments of length $${\rm d}s$$ and follow along.

1. Assume the aerodynamic force is proportional to the length of the segment $$\Delta s$$ regardless of the slope of the segment. This treats air flowing across the rope and along the rope the same and the rope & speed combination has a fixed aerodynamic weight value $$w_y$$ such that the force balance is In the figure above $$w_y$$ is the weight per length of the rope, and $$w_x$$ the aerodynamic force per length which is horizontal and constant.

In this scenario the force balance is

\begin{aligned} (T+{\rm d}T) \sin(\theta + {\rm d}\theta) - T \sin\theta & = w_x {\rm d}s \\ (T+{\rm d}T) \cos(\theta + {\rm d}\theta) - T \cos\theta & = w_y {\rm d}s \end{aligned}

which is solved for the change in tension and change in angle using a small angle approximations $${\rm d}\theta \ll 1$$, $$\cos \theta \approx 1$$ and $$\sin \theta \approx \theta$$. The solution is

\begin{aligned} {\rm d}T & = (w_x \sin \theta + w_y \cos \theta) {\rm d}s \\ {\rm d}\theta & = \frac{w_x \cos \theta - w_y \sin \theta}{T} {\rm d}s \end{aligned}

It reasonable to assume that $$\tan \theta = \frac{w_x}{w_y}$$ is a stable solution yielding $${\rm d}\theta =0$$ and $${\rm d}T = \sqrt{w_x^2+w_y^2}\; {\rm d}s$$.

From this point there no point to proceed since $${\rm d}\theta = 0$$ means that the angle is constant, and the shape is straight line.

1. But the more interesting scenario is that where the aerodynamic forces acting on the rope are not horizontal because there are different laws governing the air flow along the rope and the air flow across the rope. The flow along the rope I name "skin friction" and I assigned the function $$f_{\rm F}(v_{\rm along})$$ of force per unit length. The flow across the rope I name "pressure drag" and I assigned the function $$f_{\rm D}(v_{\rm across})$$ of force per unit length. The decomposition of speed depends on the slope with $$v_{\rm along} = v \sin \theta$$ and $$v_{\rm across} = v \cos \theta$$.

In this scenario the force balance is

\begin{aligned} (T+{\rm d}T) \sin(\theta + {\rm d}\theta) - T \sin\theta & = (f_{\rm D} \cos \theta + f_{\rm F} \sin \theta) {\rm d}s \\ (T+{\rm d}T) \cos(\theta + {\rm d}\theta) - T \cos\theta & = (w_y - f_{\rm D} \sin \theta + f_{\rm F} \cos \theta) {\rm d}s \end{aligned}

which is solved for the change in tension and change in angle as follows

\begin{aligned} {\rm d}T & = ( w_y \cos \theta + f_F ) {\rm d} s \\ {\rm d}\theta & = \frac{f_D - w_y \sin \theta}{T} {\rm d}s \end{aligned}

Note that $$f_{\rm D}$$ and $$f_{\rm F}$$ depend on the angle via the speed component relationship. The above will give a straight line answer if $$f_{\rm D}-w_y \sin \theta=0$$.

A little more detail here is that most likely $$f_{\rm D} = \tfrac{1}{2} c_{\rm D} \rho D (v \cos \theta)^2 = h_{\rm D} \cos^2 \theta$$. As such the stable angle solution is found by solving

$$h_{\rm D} \cos^2 \theta - w_y \sin\theta = 0$$

the solution I got using the tan-half angle substitution is

$$\tan\left( \frac{\theta }{2} \right) = \frac{w_y}{2 h_{\rm D}} + \sqrt{ 1 + \left( \frac{w_y}{2 h_{\rm D}} \right)^2 } - \sqrt{ \frac{w_y}{h_{\rm D}} \left( \frac{w_y}{2 h_{\rm D}} + \sqrt{ 1 + \left( \frac{w_y}{2 h_{\rm D}} \right)^2 } \right) }$$

So I guess the summary is that in both scenarios it is possible to get a straight line shape, albeit at different angles.

• if you take $~\sin(d\theta)=d\theta~,\cos(d\theta)=1~$ and solve the two equations for $~T~$ and $~\theta~$ with $~ds^2=0~,dt^2=0~$ you obtain that $~\tan(\theta)=\frac{w_y}{w_x}~=\frac{dy}{dx}$ the linearization is now correct I got also this solution
– Eli
Oct 31 at 16:23
• @Eli - yes that is what I did, not what I wrote necessarily. :-) Nov 1 at 13:03

The shape is the same if we imagine that the helicopter is stationary, but there is a wind that produces a force $$F$$ per meter on the cable, to the left.

If the cable is at an angle $$\theta$$ to the horizontal a distance $$x$$ down the cable, of length $$L$$ and of mass per unit length $$m$$ and $$T$$ is the tension at $$x$$,

then for a first approximation

$$T\sin\theta = m(L-x)g\tag1$$ and $$T\cos\theta = F(L-x)\tag2$$

dividing $$\tan\theta = \frac{mg}{F}\tag3$$ and that's constant, leading to a constant angle $$\theta$$, so the straight line, answer B

If we add on a term to take into account that the drag depends on the angle, we can swap equation 2) to 4)

$$T\cos\theta = F(L-x)\times \sin\theta\tag4$$

After dividing and using $$\sin^2\theta + \cos^2\theta = 1$$

we get $$\cos^2\theta+\frac{mg}{F}\cos\theta-1=0 \tag5$$ this can be solved to again give a constant $$\theta$$, independent of $$x$$, answer B.

• This is indeed an approximation if "the angle doesn't very much". But I wanted to approximate the least possible. Thanks for the answer anyways!
– Neox
Oct 28 at 16:10
• @Neox interesting question. We are getting answer B for small angles, i.e. if the weight is large compared to air resistance, and also if air resistance is large compared to the weight, as then 5) gives a sensible solution, Looking at a plot of the solution of 5) gives a problem if mg is similar to 3F. The question designers might not have considered that possibility, they probably only expected the version from eqns 1) and 2) Oct 28 at 16:20
• In the video description there is a link to the solutions of that test. They say that the drag force is proportional to the length of the cable. I disagree with that. The drag force is proportional to the area that the air "sees" when the object is travelling in air. Imagine a thin rectangular body falling straight down. Compare these two situations: the body has a horizontal surface, and the plane is displaced an angle to the horizontal. The contact area will be maximum when the surface is horizontal.
– Neox
Oct 28 at 16:40
• @ Neox yes I agree with that, that's what equation 4) inlcudes Oct 28 at 16:42
• I noticed. Just wanted to point out that they use a wrong logic to arrive at the conclusion.
– Neox
Oct 28 at 16:44

Here is my two cents $$\theta dy$$ is the air drag on the small section, while $$\lambda ds$$ is the gravity on the small section.

$$y'|_{x} = \frac{T_0}{F_0}\tag1$$ $$y'|_{x+dx} = \frac{T_1}{F_1}\tag2$$ $$\theta dy + F_0 = F_1\tag3$$ $$\lambda ds + T_0 = T_1\tag4$$ Plug(1),(2) into (4) $$\lambda ds + F_0 y'|_{x}= F_1 y'|_{x+dx}\tag5$$ Plug(3) into (5) $$\lambda ds + (F_1-\theta dy)y'|_{x}= F_1 y'|_{x+dx}$$ $$\lambda ds- \theta y'|_{x} dy = F_1 (y'|_{x+dx} - y'|_{x})$$ $$\lambda \frac{ds}{dx} - \theta y'|_{x} \frac{dy}{dx} = F_1 y''|_{x}$$ $$\lambda \sqrt{1+(y')^2} - \theta y'^2= F_1 y'' \tag6$$ One solution to this is $$y'= const = \sqrt{\frac{(\frac{\lambda}{\theta})^2+\sqrt{(\frac{\lambda}{\theta})^2+4\frac{\lambda}{\theta}}}{2}}$$, thus a straight line.

• Nice and short which is a bonus! Does the drag force $\theta ds$, depend on the angle of the section in this model - that is a refinement that was being considered in some of the above? Oct 29 at 9:44
• @user4015990 if you take for the drag force instead of $~\theta\,dy~~,\theta\,ds~$ you obtain the solution $~y'=\frac{\lambda}{\theta}~$ this is also my solution. this is the differential equation $\sqrt {1+ \left( {\frac {d}{dx}}y \left( x \right) \right) ^{2}} \left( \lambda-\theta\,{\frac {d}{dx}}y \left( x \right) \right) =0$
– Eli
Oct 30 at 19:49 for a piece of the rope with the length $$~s(x)~$$

take the sum of the forces

$$\sum F_x=T\,\cos(\theta)-F_d+H=0$$ $$\sum F_y=T\,\sin(\theta)-F_g=0$$

the weight force $$~F_g~$$ is

$$F_g=m\,g=\rho\,A\,g\,s(x)=\rho\,A\,g\,\int_0^x\sqrt{1+y'^2}\,dx$$

the drag force is proportional to the rope length $$~s(x)~$$

$$F_d=\kappa\int_0^x\sqrt{1+y'^2}\,dx$$

the tension force H at the end of the rope is zero so

$$\frac{dy}{dx}=\frac{T\,\sin(\theta)}{T\,\cos(\theta)}=\tan(\theta)=\frac{F_g}{F_d}=\frac{\rho\,A\,g}{\kappa}=k\quad \Rightarrow\\ y(x)=k\,x+c$$

hence: the solution is B

Notice that if $$~F_d=0~$$ and $$~H\ne 0~$$ you obtain

$$y'=\frac{F_g}{H}=\frac{\rho\,A\,g\,\int_0^x\sqrt{1+y'^2}\,dx}{H}\\ y''=\frac{\rho\,A\,g}{H}\,\sqrt{1+y'^2}$$

this is the well known chain differential equation

• Seems good, supposing that the drag force is proportional to the length, which I do not believe to be true. Check my comment on @John Hunter's post.
– Neox
Oct 28 at 22:00
• Oh and there is a sign typo in your second equation.
– Neox
Oct 28 at 22:11