No, Gotham, That's Not How Tightropes Work

In this scene from Gotham, Bruce Wayne holds a tight rope for Selina. It turns out this might be much more difficult than it looks.


I'll be honest. I don't really watch Gotham, but it looks interesting. It chronicles the events in Batman's city before he became Batman. That's about all I know.

However, when I saw a recent commercial for an upcoming episode, I had to do something. I'm not sure what's going on here, but from my research this appears to be Selina Kyle and Bruce Wayne doing something with a tightrope. Here's the important shot:

From the rest of the scene, it seems that Selina is trying to avoid laser tripwires. Someone uses an arrow to fire a rope across the room into a wall, then Bruce holds the other end as Selina walks its length. But there's a problem with this: No one could hold that cable. Not Batman, and especially not Bruce Wayne.

A Basic Forces Problem

This is all about forces and equilibrium. If Selina is standing at the middle of the cable, all of the forces on her must add up to the zero vector. Let me start with a force diagram showing the vector forces on a stationary Selina:

Fall Sketches 2016 key

Some important things to consider about this diagram and the forces:

  • These are the forces acting on Selina. These are not the forces that Selina exerts on the rope. Since it is Selina that is in equilibrium, only the forces on her matter.
  • There are two tension forces, one from each side of the rope. Since it's the same rope, the magnitude of these forces must be the same (but of course they are in different directions).
  • I have drawn a bowed rope. You can't have a perfectly horizontal rope---as you will hopefully see below.

If I assume that the x-axis is horizontal and the y-axis is vertical, I can write the vector force equation as two component equations. Let me start with the x-forces. Only two forces act in the x-direction---the components of the two tension forces. Since I know the angle between these forces and the x-axis, I can write the following:

La te xi t 1

Notice that since that T1 pulls to the left, it has a negative x-component and T2 has a positive x-component. However, since the magnitude of T1 is equal to the magnitude of T2 this equation is pretty useless. It basically just says that 1 = 1, which we probably already knew.

What about the y-direction? Here is the sum of forces in just the y-direction:

La te xi t 1

In this case, both of the tension forces are in the positive y-direction. But since they have the same magnitude, T1 = T2 so I can just call these T. Now I get the following:

La te xi t 1

This gives a value for the magnitude of the tension. Remember, this is the force that Bruce must pull on the rope. I can give an approximate value to this tension force by estimating the mass of Selina (50 kg) and the bend angle in the rope (I'll say 10 degrees for now). Putting these values into the above equation, I get a tension force of 1,410 Newtons (317 pounds). Yes, that's a huge force. Oh sure, maybe Batman could pull that hard, but remember, this is Bruce Wayne as a boy. I don't think he could hold her up.

But wait, there's more! Clearly, we have other questions to consider. I'll just leave these homework questions for you.

  • What if the rope angle is greater than 10 degrees? Could Bruce hold the cable then? Try a 25 degrees angle and calculate the tension.
  • Calculate the rope angle so that Bruce only has to pull with a tension of 400 Newtons (this would be 90 pounds).
  • There's another problem---friction. Draw a force diagram for Bruce as he is pulling on this rope. Calculate the minimum coefficient of static friction so he stays stationary. Don't forget that the rope also pulls down on Bruce which would increase the normal force and thus the frictional force.
  • Make a plot of rope tension as a function of rope angle.
  • Make a plot of minimum coefficient of static friction for Bruce as a function of rope angle.