# The physics of Michael Jackson’s moonwalk

Note: Originally posted on June 2009

Was the moonwalk fake? No, not the Apollo landings. I am talking about Michael Jackson’s moonwalk. You got to admit, he had a big impact on a lot of stuff and this is my way to give him respect – physics.

I am sure you know about the moonwalk. Maybe you can even do the dance move yourself, but how does it work? First, here is a clip of MJ doing his stuff.

As a side note, I can’t remember where I saw it but there was a great discussion of the history of the moonwalk. If I recall correctly, some were saying Michael didn’t create this move. One thing is for sure, he made it popular. Now for the physics.

The key concept here is friction. Friction is actually uber-complicated, but a simple model works for many cases. Static friction is a force exerted on an object when it is in contact with some surface but those two surfaces do not move relative to each other. Kinetic friction is a force exerted on an object when the two surfaces are moving. Suppose I have a block at rest on a table and I pull it with a slowly increasing force. This is what it would look like:

Two key things from this graph. As you pull on the stationary block, the block doesn’t move. If I pull with 1 Newton, and it doesn’t move then the frictional force is 1 Newton. If I then pull with 2 Newtons and it still doesn’t move, the frictional force is 2 Newtons. The static frictional force does what it can to make the thing not move – but not more than it can. This leads to the static friction model of:

$F_\text{static} \le \mu_s N$

In this model, the force is less than or equal to the product of some coefficient (that depends on the two types of surfaces) and the normal force (how hard the two surfaces are pushed together). The direction of this frictional force is parallel to the surface in the direction that prevents the object from sliding.

The other key feature in the graph is the small jump down when the thing starts to slide. This is because the coefficient of kinetic friction is typically smaller than that for static friction. Also, if the object is sliding, the frictional force is constant.

$F_\text{kinetic} = \mu_k N$

Back to Michael and the moonwalk. The key here is: how do you make one foot slide and the other not slide? If both feet are stationary, then this is dealing with static friction. I could make the frictional forces on these two feet different by changing my center of mass. Here is a free body diagram:

Since he is not accelerating up and down, the following must be true:

$N_1 +N_2 = mg$

These are the forces in the y-direction. They must all add up to zero so that:

$F_\text{net-y} = ma_y = 0$

There is another condition that must be satisfied. Since he is not rotating, the total torque about any point must also add up to zero. If you want more info on torque, check out this post (Note: link doesn’t work). But for this post I will just say that torque is like the ‘rotational force’. It depends on the point about which you want to rotate and is essentially the force applied times the perpendicular distance to the point of rotation. For the free body diagram of Michael, I have chosen one of his feet to be the point about which he is not rotating (I could chose any point). This makes 3 of the forces have zero torque (N2, F2 and F1 have zero torque because the perpendicular distance to point O is zero). Here I labeled the other important distances:

The only two forces that exert torque about O are the weight and the N1 force. They have opposite directions of torque because they would cause rotation in different directions. This along with the previous equation gives:

$mgr_1 = N_1r_2$

$N_1+N_2 = mg$

Eliminating mg, and solving for N1, I get: (I know the indices for the forces and distances don’t match)

$N_1 = N_2\left(\frac{r_1}{r_2-r_1}\right)$

If his center of mass is in the middle, then r2 – r1 = r1 and the two normal forces would be equal (as you would expect). If the center of mass is more towards the foot on the right, then r2 – r1 is less than r1 and N1 will be larger than N2. This will make the frictional force on the foot to the right greater and the other foot slide.

Well, what if r1 is greater than r2? One of two things would happen. Either he would fall over, or there would have to be a force pulling the foot on the left down. This is similar to Michael Jackson’s trick in “Smooth Criminal”.

Here he used special shoes that connect to the floor so that he could do this. More details on this page.

Ok. So that is how Michael gets one foot moving. How does he keep one foot sliding and the other not sliding? It is really the same thing as above except that he can increase the force on the moving foot a little bit more since it is sliding. Sounds easy, but Michael could really make it look cool.

Finally, I just want to show another demo that is essentially the same idea.

You can find more details on the meterstick demo in this blog post.

# MacGyver Science Notes Season 3: Episode 10 Matty + Ethan + Fidelity

Using a drone to lift a human.

OK, maybe this isn’t exactly a Mac-hack since he didn’t build it.  But can you use a drone to lift a person?  Oh yeah—this is real.

The basic idea of a drone is that it provides upward lift by “throwing” air down.  In order to conserve momentum, the downward force on the air is equal to an upward lift.  This means a couple of things:

• Faster air gives greater lift (because the air has greater momentum).
• Larger rotor areas give greater lift (because there is more air thrown down).
• The power required to hover is proportional to the air speed to the third power.  That means you don’t want to use fast air.
• Instead, you want big rotors with slower air.

Here is a post with a bit more explanation. But it is indeed possible.  Oh, the drone MacGyver uses could work, but it would be better if the rotors were a little bit bigger.

Cricket Ball Flash Bang

MacGyver tosses a ball into a room.  It then explodes with a flash to stun the people inside.  Is this possible?  Of course—it’s possible to even make your own flash bang.

Since the build for this device is off screen, let’s just leave it like that.  But it’s clearly possible.

Water Cooler Bomb

MacGyver takes a water cooler bottle and puts some stuff in it.  He then rolls it into a room and it explodes.  Again, stunning the people inside.  Yes, very plausible.

Instead of talking about explosives, here is a related demo using a cooler water bottle.  It’s the woosh rocket.  Check it out.

When you ignite the ethanol, it quickly uses up the oxygen in the bottle (because of the neck).  This causes a type of fluttering with the oxygen being used up and then entering the bottle.  It’s cool.

DIY Jaws of Life

OK, these aren’t actually jaws of life.  MacGyver gets some metal pieces to build a device to pry open a door using an electric drill as a motor source.

Here is a very rough sketch of how this would work.

OK, that’s not exactly the same—but it’s the same idea.  The spinning drill turns the threaded screw and pushes the metal bars apart.  The pivot makes the metal bars push apart on the other side.

Since the door side of the pivot is shorter than the drill side, the change in distance on the door side is smaller.  Why does this matter?  This matter because this is the way all simple machines work.  You can get a greater output force if you decrease the motion distance on the output.  That means on the drill side, there is a small force moving over a greater distance.

It’s the same idea as a basic lever. It’s how your basic garden sheers work too.

Here is something very similar (in physics at least)—a DIY floor jack.