Just because it's pole dancing doesn't mean it isn't art. In this case, pole dancing also means "requires strength". When I see something like this, I automatically ask the question: what kind of forces would this guy have on his arms? And my second question: where can you get shorts like that? Ok, forget about the shorts, let's look at the physics.

Rigid Body Equilibrium ———————-

A rigid body in equilibrium means that it is neither accelerating (linearly) nor changing its angular motion. I can write this as the following equations:

Considering the pole guy (Carlos) as the rigid object then the net force in all directions (I have just shown two directions) must be zero newtons. For the torque, the net torque about any point must be zero N*m. Since he isn't changing his motion about any point, you can find the net torque about any point.

Now for a force diagram.

Yes. I got out of control on that diagram. OUT OF CONTROL. Maybe I shouldn't have put all those extra labels on there, but I couldn't help myself. Plus, I figured I might need those measurements and I didn't want to redraw the diagram. Now, how will I get values for these things? Some of them I will just have to make a reasonable guess. I can guess the two masses and the length scale. After that, I can load this image into Tracker video analysis tool. Yes, this works for images as well as videos. Here are the values I get.

m 1 = 70 kg.

= 70 kg. m 2 = 55 kg.

= 55 kg. x g = 0.2 m.

= 0.2 m. x R = 0.75 m.

= 0.75 m. y R = 0.65 m.

= 0.65 m. x L = 0.85 m.

= 0.85 m. y L = 0.49 m.

= 0.49 m. θ R = 48.9°.

= 48.9°. θ L = 44.7°.

With this, I can write the two net force equations (one for x and y directions).

It looks like I can solve for the two forces right away without the torque equation (which is crazy, I know).

This seems like cheating, but let me put in some values for the masses and the angles. This gives the following magnitudes of forces in the two arms.

F L = 807 N.

= 807 N. F R = 872 N.

So the left arm is under a compression of a force greater than the guy's weight and the right arm is under a tension greater than his weight. That can't be fun.

Why might this be wrong? This might be wrong because it seems like it doesn't matter where the girl is standing. Clearly this can't be correct. If she move farther away from the pole it has to be harder to hold himself up, right? Yet this expression doesn't have any dependence on her distance.

Torque ——

If these forces are legitimate, then the net torque must also be zero about any point. Without going into much detail, let me give the following definition of torque about the center of mass for this situation (which is partially wrong).

Here F is the magnitude of some force, r is the distance from this force to the origin (center of mass) and α is the angle between the force and r. I will call torques in the counter clockwise direction positive and clockwise torques will be negative. This means that the weight of the girl will have a negative torque. It looks like F L and F R will both be positive - but I will need to check to be sure.

Let's skip all the details. Let me just list the torque from these three forces using the numbers above. Oh, since I am calculating the torque about the center of mass of the guy, his weight will produce zero torque (r is zero meters).

τ R = 120 N*m.

= 120 N*m. τ L = 201 N*m.

= 201 N*m. τ g = -137 N*m.

Obviously, these do not add up to a net torque of zero N*m. Why not? There are several possible causes. First, I might have the center of mass of the guy in the wrong spot. If I move this location a little more to the left the girl would produce more torque about the center of mass (but so might the other forces). The other possible problem is my assumption that the forces from the pole are along the lines of his arms. I think this could be a good approximation, but it doesn't have to be absolutely true. There is a frictional force along the direction of the pole that I didn't take into account.

Still, I think my values are a good first guess.