You remember the scene. Darth Vader picks up a rebel soldier by the neck as a means of interrogation. Yes, I know Star Wars Episode VI is quite old. However, I was reminded of the powers of Darth Vader after someone posted photos from a tour of Industrial Light and Magic. In particular, there is this statue of Darth Vader.

Image: Imgur user Risngproducer

The statue shows the full figures of both Darth Vader and the Rebel. There is something not quite right about this. Darth Vader couldn't pick a rebel up like that. Oh, wait - I'm not saying he isn't strong enough. He has those bionic arms and legs. I am saying it wouldn't work this way. Why? It's all about center of mass.

Center of Mass and Stability —————————-

Let's look at some rigid object. Maybe it is a large letter L that is made out of wood. If I place the L like this, everything should be fine.

In case you can't tell, today's post is brought to you by the Letter L. Now, what if I turn this L upside down? It will just fall over, right? Why? Let me draw the L and the upside down L with the forces acting on both objects. When dealing with rigid objects, sometimes we have to show not just the forces but where the forces act on the object.

Oh, one more thing. What about the gravitational force? Where does it act? Really, the gravitational force pull on all parts of the object. However, we can replace all these tiny gravitational interactions with just one gravitational force that acts at a location called the center of gravity. (In a constant gravitational field this is also the center of mass.) Here I will draw the center of mass as a dot.

It might seem obvious, but I am just trying to be clear. The right-side-up L has a wide support. This means that the the ground can push up on it anywhere along this contact area. (It can actually push up in more than one place.) The point is that the surface can be applied right below the center of mass.

The upside down L is different. In this case even if the surface pushes up as far to the left as possible, it still won't be underneath the center of mass. This means that it will have non-zero torque. What is torque? I really don't want to get into this in too much since it can be a surprisingly complicated concept. At the most basic level, torque could be considered to be a rotational force. A net force changes the linear momentum of an object and a net torque changes the angular momentum of an object. Oh drat. I just ramped up the level of complication by introducing angular momentum.

How about this. I can define the torque (in scalar form - but really torque is a vector) as:

What the heck is the r with the perpendicular sign? This is the line from the point where the force is applied to the point where the thing might rotate about. The line must be perpendicular to the force. In this case, I would just use the distance s in the above diagram of the upside down L.

I guess it would be sufficient to just say that the center of mass of an object must be either over or between two points of support. Yes, that would have been easier.

But how do you find the center of mass? Let me consider the simple case of an object with two point masses.

Don't worry about what connects these two masses - it isn't part of the system. There is some force (F s ) that supports this system. If the support force is at the center of mass, the net torque will be zero. Since it is a support force, the sum of the forces in the y-direction must be zero. This means:

If this support force is at the location of the center of mass, then the net torque will also be zero and the system will not rotate. If the net torque is zero about the center of mass, the net torque is zero about any point. Since I put the left mass (m 1 ) at the origin, let me sum the torques about that point. I never said that scalar torques can be positive or negative - sorry about that. If the torque would cause a counter-clockwise change in angular momentum, that will be a positive torque. The other way will be negative.

I probably should have put the origin somewhere other than mass 1, but oh well. I think you get the idea.

Back to Darth Vader ——————-

Let me go ahead and make an assumption. Well, it isn't really an assumption since I asked George Lucas and he said I was right. Here it is.

Darth Vader is strong. He also has bionic arms and legs and stuff. When he lifts someone, he does not use an external application of The Force to prevent him from tipping over.

Note: I was kidding about talking to George. When he reads this post, though, I'm sure he will agree with my assessment.

In order for Darth Vader to lift up the rebel soldier, the center of mass of the Vader-rebel system has to be over Darth Vader's toes. How can this be? Well, no one knows - you can't see Darth Vader's toes in the clip from the Episode IV. Let just make an estimate anyway. Here is a diagram to start with.

I based the distances on the listed height of Darth Vader at 2.02 meters. Then I just guessed at the location of the front of Vader's foot and the locations of the center of masses for the rebel and Darth Vader. If you take some estimates of the rebel dude, I get about 1.8 meters for his total height. This makes him pretty average for a human so I will put his mass at 73 kg.

Now, I can use this to estimate the mass of Darth Vader. Using my values for the distances to the balance point, I can write the following. (Now I am will switch to m r for the mass of the rebel and m v for Vader.)

Solving for the mass of Vader (or as Obi Wan calls him in Episode IV, Darth).

This gives Darth Vader a mass of 89 kg (196 lbs). Well, that's not too bad. In fact, I would guess he would be heavier than that just based on the height of a 2.02-meter-tall human. But wait! I think I had a fairly generous spot for the location of Vader's front toe. I am sure he would want to look all cool and stuff and not put his feet too far forward. What if I moved the location of the center of mass to right under Darth Vader's chin? I'm sure he wouldn't want his foot in front of his chin while standing there interrogating some poor rebel scum guy. Doing this would increase his mass to 236 kg (520 pounds). Yup. That's some serious mass.

But what about the statue? In this case, I can see Vader's feet. The only problem is that he is stating slightly tilted to the camera. I can only guess the distance from Vader's feet to the center of Vader and Rebel. Using these values, I get Vader's mass of 209 kg. That's still pretty heavy. A typical NBA basketball player is about the same height but with a mass of around 100 kg.

One more point. This is the lower estimate of Darth Vader's mass. He could have a mass of 1,000 kg and the scene would still look the same. However, if he had a mass lower than 230 kg or so, he would tip over when picking up the rebel scum.

An Experiment

There was going to be one other part to this post. I was going to hold up one of my kids to show that doing what Darth Vader did would be pretty tough. Well, it's tough alright. Oh - I wasn't picking kids up by their necks - let me just be clear. They can hold onto my arms and I lift them up. The problem was that they are a little heavier than I thought or maybe I am a little weaker. I'm not even going to show you my failed attempts.

However, I did try something that worked. For some reason, the kids made this impromptu see-saw in the back yard. Here is a picture of me and my youngest child standing on the beam such that it is balanced.

Since the beam isn't rotating, the net torque is zero. The pivot point is at the center of mass. The cool thing about a board like this is that you don't need to worry about its mass. The center of mass for the board is already over the pivot point - it doesn't factor into the torque equation. Anyway, you can see that in order for the board to balance, I have to stand much closer to the point of rotation than a child does. If I had the distances from the center of mass (which I wrote down somewhere), I could get the ratio of masses for two people.