The Physics of Spider-Gwen’s Wall Stand
I don’t know much about the second Spider-Verse movie. But I do know that this scene from the trailer shows Spider-Gwen talking to Miles Morales (another Spider-Man). It’s great shot. Right?
But it’s more than an artistic scene. It’s also a great physics problem. So, here’s the question.
What forces would Spider-Gwen need to exert on the wall in order to stand like this? Oh, it’s game on.
Rigid Body Equilibrium
Let’s pretend like Gwen is a rigid object — sort of like a plank. In order to have a rigid object remain at rest, it must have an acceleration of zero (m/s²) AND an angular acceleration of zero (rad/s²). That means that it will stay at rest as well as remain with an angular velocity of zero (rad/s). We call that equilibrium.
Assuming the x-direction is horizontal and the vertical direction is the y-direction, then the following three equations must be true.
Here’s a quick reminder about torque. I like to think about torque as a “rotational force”. Since a net force changes the linear motion of an object, a net torque changes the rotational motion. The torque depends on both the magnitude of a force (F) AND the location (r) where the force is applied with respect to some reference point. That gives the following definition.
Yes, it also depends on the angle ( θ) as you can see above. Oh, one more thing. Really, torque is a vector — but in this case we can treat it as a scalar. However, we have to give it a sign (positive or negative). Let’s use the normal convention that if the torque would cause a counter clockwise rotation, it’s going to be positive and clockwise would be negative.
Free Body Diagram
The next step is to show the magnitude and location of the forces acting on Spider-Gwen. I suspect that many physics students might start with a diagram that looks like this (hint: this is wrong):
There are two forces on Gwen. First, there is the downward pulling gravitational force (mg) where g is the gravitational field (equal to 9.8 Newtons per kilogram). The gravitational force pulls on all parts of Gwen, but we can model this as a single force acting at her center of mass (which is usually just above the waist). Well, there has to be some force pushing up — and that force obviously comes from the wall (F_w). If these two forces are equal in magnitude, then it’s simple. Right? Not so right.
The problem with the force diagram above is that although the forces add up to zero, the torques DO NOT. It doesn’t matter what point you pick to add up the torques — let’s just take a location somewhere between the two forces (for now). In that case, BOTH of these forces would produce a clockwise (negative) torque. There’s no way they could add up to a zero net torque. If these are the only forces, Gwen would start rotating and not look nearly so impressive.
No. There must be some other forces. There must be some type of force to apply a counter clockwise torque so that Gwen doesn’t tip over. To do this, she has to have contact with the wall in more than one location. OK, here is an updated force diagram. Don’t forget that Gwen isn’t just Gwen. She’s Spider-Gwen. That means that there is some type of sticky force pulling on the wall.
Check that out. It’s not just an artistic shot — it’s a shot WITH physics. I love it. But what about these new forces? I’ll be honest, I don’t really know how the spider-grip works. Does the foot just pull towards the wall and create a frictional force or does the grip also exert a force parallel to the surface? I guess it doesn’t matter. In this diagram, I have split this interaction with the wall into two parts — F_w is the upward pushing force parallel to the wall surface and F_s (spider-force) is pulling her foot towards the wall. That leaves the normal force (N) pushing AWAY from the wall.
Notice that Gwen HAS to have more than one contact point with the wall. If she only had one point touching then yes — it’s possible to make a zero torque situation but what about the forces in the x-direction? Ah ha! You need that normal force pushing away from the wall.
Now for the equilibrium equations. I’m going to need some more values to finish this problem. First, the distance from Gwen’s feet to her center of mass. How about L = 1.0 meter (since she’s on her tip toes). Second, I need the distance from her toes to the ball of her feet (the vertical distance on the wall). I am going to go with s = 0.08 meters (rough guess). Finally, I need her mass of m = 55 kg.
With that, I get the following two force equations.
There are three unknowns there (N, F_s, F_w) but only two equations. I can get the third one from the net torque, but I need to pick a point about which to calculate the torque. You can pick any point you like and it will still work. I’m going to go with the point at where her toes are in contact with the wall. That means that both F_w and F_s exert zero torque since their distances from this point are both zero meters.
Now I have the following torque equation.
That’s three equations with three unknowns. Here’s how we figure everything out.
- From F-net-y, F_w is equal to mg (that’s easy).
- Next, from the torque equation, I can solve for N (I know s, m, g, L).
- Finally, using the value of N and the x-net equation I get that F_s = N.
What about the numbers? For that, I’m going to use my favorite calculator — python. Here’s the code if you want to see it.
Using my estimates, Gwen both pushes and pulls on the wall with a force of 6737.5 Newtons (for you Imperial people, that’s 1514.6 pounds).
But wait! There’s more. Could the wall even handle this kind of force? Actually, it’s not the force that breaks or pulls apart something — it’s the pressure. We call this the compressive strength (for squishing) and the tension strength (for pulling apart).
Bricks have a compressive strength of 80 MPa (that’s 80 million Newtons per square meter) and a tension strength of 2.8 MPa. So you see that a brick doesn’t work as well when you try to pull them apart.
What kind of pressure does Gwen exert on the bricks from the front part of her foot? For that, I’m going to need to estimate her contact area. This is pretty tough to estimate, so I’m just going to go with something larger than it should be. How about a rectangular area of 7 cm by 3.5 cm? That would give an area of 2.45 x 10^-3 m² and a pressure of 2.75 MPa (that’s mega Pascals). Oh, look at that. JUST under the breaking point. Gwen must really know her physics. Impressive.
