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.

The Ladder Problem

I like to solve physics problems.  Here is one for you (I just made it up).

A 4 meter ladder leans against a frictionless wall at a 30 degree angle.  The mass of the ladder is 10 kg.  A human stands 1 meter up the ladder and has a mass of 70 kg.  What is the minimum coefficient of static friction between the floor and the ladder so that the ladder doesn’t slip?

Here is the solution—in video form.

But wait! There’s more.  Let’s do another problem.  

Suppose you have the same ladder and the same human.  However, in this case the coefficient of static friction between the ladder and the ground is 0.55.  How far up can the human move before the ladder slips?

I like this question because I don’t know how the answer will turn out.  That makes it fun.  So, let’s do it.

But what kind of problem is this?  I’ll make this a multiple choice question.  Here are your options.

  1. A friction problem.
  2. An equilibrium problem.
  3. A ladder problem.
  4. A work-energy problem.

Your answer?  Go ahead and answer.  It’s important to think about things like this if you want to become an expert problem solver.

Did you pick?  OK, I’ll tell you that there will be quite a few students that will say this is a friction problem because it has a coefficient of friction.  That’s not untrue—but it’s not a good way to classify the problem.  You could also say this is a “ladder problem”—again, not untrue but not helpful.

This is an equilibrium problem.  We are trying to find the point at which the ladder slips—the point it leaves equilibrium, but it’s still sort of in equilibrium.

For an object in equilibrium, there are two main ideas—especially for a rigid object.  First, it must have zero net force.  Second, it must have zero net torque about any point.  In two dimensions, I can write these two conditions as the following three equations.

F_\text{net-x}=0

F_\text{net-y} = 0

\tau_\text{net-o} = 0

Let’s talk about the torque stuff.  I don’t want to get into a whole thing about torque, so let me just say that torque is like a “rotational force” and it can be calculated as:

\tau = Fr\sin\theta

In this expression, F is the applied force, r is the distance from the point at which you are calculating torque, and \theta is the angle between r and F.  This equation is equivalent to F-perpendicular times r or F times r-perpendicular.

Oh, torques that would make an object rotate clockwise are negative.

One last thing about torques—especially the sum of the torques.  If an object is in rotational equilibrium about some point (point “o”) then it is also in rotational equilibrium about any other point.  When you set up the torque equation, you can pick whatever point you like to sum the torques—it’s your choice (but choose wisely).

Now we can start setting up some equations for equilibrium.  I’ll start with a force diagram for the ladder.

I know that’s a little busy—but it will have to do.  Here are some comments.

  • There are two normal forces.  One from the wall (labeled 1) and one from the floor.
  • There are two gravitational forces—and this is a cheat.  There is a gravitational force on the ladder and it is as though it is one force acting at the center of mass for the ladder. If the ladder has a uniform density, the center of mass is the center of the ladder.
  • The other gravitational force is fake.  This force m_2\vec{g} is there to represent the weight of the human.  But that gravitational force acts on the human, not the ladder.  The ladder pushes UP on the human the same as the weight.  Since forces come in pairs, an upward pushing force from the ladder means there is a downward pushing force from the human on the ladder.  It’s equal to mg, but not mg.
  • The friction force is parallel to the floor.  The maximum magnitude of this friction force would be: F_f = \mu_s N_2 where \mu_s is the coefficient of static friction.

Now for some equations.  First, this is the sum of the forces in the y-direction.

F_\text{net-y} = 0 = N_2 - m_1 g-m_2g

Just a quick reminder.  These are not vectors.  These are components of vectors in the y-direction.  That’s why the two gravitational forces have a negative sign.  I guess I can simplify this a little bit.

N_2 = (m_1 + m_2)g

Next is the sum of forces in the x-direction.

F_\text{net-x} =0=N_1 - F_f

Since we are at the point of maximum friction, I can include the expression for the frictional force in terms of the coefficient.

N_1 = \mu_s N_2

Notice that I know the value of N_2 since it only depends on the mass of stuff—but I don’t know N_1.  I’m going to need another equation. That’s where the sum of the torques comes in.  

In order to write down the sum of the torques, I need to pick a point about which I calculate the torque.  I’m going to go with the bottom of the ladder.  At this point, there are two forces applied.  Since they are applied at the point about which the torque is calculated, they contribute zero torque and they won’t appear in the equation.  Winning.

Here is the sum of the torques about point O (at the bottom of the ladder).

\tau_\text{net-o} = 0 = m_2gs\cos\theta +m_1g\left(\frac{L}{2}\right)\cos\theta - N_1 L\sin\theta

What do I want to solve for here?  I want the distance the human goes up the ladder.  This is the value “s” in the expression above.  Really, I have all the values I need in that expression to solve for s.  I just need to set N_1 to the maximum frictional force.  But that would be boring.

Instead, let me make a plot of the frictional force as a function of human distance up the ladder.  That will be more fun, right?

Here’s what I get.

From this plot, the human can go up 1.16 meters before the required frictional force exceeds the maximum.  Oh, here is the code for that plot.

Now that you have the code, you can change stuff—like the angle of the ladder or the mass of the human or whatever.  

The end.

MacGyver Season 3 Episode 6 Hacks

Getting through a giant chunk of cement with acid.

Really, this isn’t a “Mac Hack” since MacGyver didn’t do it.  Instead it was someone else.  She used muriatic acid to help get through a tunnel that was plugged up with cement (or concrete—I always get those two confused).

But yes, muriatic acid will indeed “eat” through cement.  If the goal is to create a hole that will allow a human to get through, it’s not so bad.  You can use the acid to weaken enough of the structure that you can take it out in pieces.  You don’t have to dissolve all the cement.

Now for a bonus homework question—actually an estimation problem.  Suppose there is 10 feet of cement to get through.  If someone uses muriatic acid, how long would this hole take to make?  Go.

Thermite

Thermite is awesome—oh, and slightly dangerous.  Basically, it’s a chemical reaction between two metals in which one of the metals has the oxygen needed for the reaction (like iron oxide).  The key to getting this reaction to work is to have super tiny pieces of metal (like super super tiny).  Really, that’s the tough part.  But once you get that, the thermite gets really hot, really fast.  Hot enough to melt stuff.

Could it be used to close up gun ports in an armored vehicle?  Probably.

Pressure to open an armored vehicle

An armored car has armor.  That’s why they call it an armored car.  The primary role for the armor is to keep out things like bullets and people so that the stuff inside (probably money) is safe.

But what if you seal off the openings and then pump in some air?  The cool thing about air pressure is that even a small pressure increase can exert HUGE forces on a wall.  Let’s say that MacGyver doubles the atmospheric pressure inside the truck so that it goes from 10^5 \medspace \text{N/m}^2 to 2\times 10^5 \medspace \text{N/m}^2.  If you have a wall that has dimensions of 2 meters by 2 meters, there would be a net outward force of 4 \times 10^5 \medspace \text{N} pushing it outward.  For you imperials, that is like 90,000 pounds.

Oh, this is why submarines are cylindrical shaped.  Flat walls bend under great pressure.

DIY balance

I really don’t know how much detail to go over for this hack.  MacGyver builds a simple balance scale to find the weight of some money.  As long as the arms are equal, this should be easy.  Maybe in a future post (on Wired) I will show you a design for a symmetrical balance scale.

Pry Bar Lever

Again, there’s not too much to say here.  MacGyver uses a piece of metal to pry open a door.  This is why humans use crow bars.  In short, this is a torque problem.  The torque is the product of force and lever arm.  So a small force with a large lever arm (MacGyver pushing on the end of the bar) gives a small lever arm and huge force (from the tip in the door).

Ok, I lied.  Torque is way more complicated than that.  Really, it’s a vector product:

\vec{\tau} = \vec{F} \times \vec{r}

That’s a little better.

Oh, there were two good hacks that didn’t make it into the show.  Too bad.

 

 

Torque produced by balls in Fantastic Contraption

The fun part about exploring the physics of [Fantastic Contraption](http://fantasticcontraption.com/) is coming up with new setups to test ideas. Torque is not too difficult to set up. Here is what I did:

![Screenshot 04](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-041.jpg)

In this setup, I have a “turning ball” with a wood stick attached to the side. I increased the length of the stick until the ball does not turn. At this point, the torque from the gravitational force on the stick is equal to the torque from the ball. I can use [Tracker Video Analysis](http://www.cabrillo.edu/~dbrown/tracker/) to find the lengths of the two wood sticks. The torque from each stick will be its gravitational weight times the perpendicular distance to the center of the turning ball.

![Screenshot 05](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-053.jpg)

In order to calculate the gravitational force, I need the mass of each “stick”. [From my previous post](http://blog.dotphys.net/2008/10/physics-of-fantastic-contraption-i/), I found that the mass density per length for sticks was

![Screenshot 171](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-1711.jpg)

where mb is the mass of a ball and U is the diameter of a ball. I also need to find the horizontal distance from the center of the stick to the center of the ball. I will call the top stick 1 and the bottom 2. This gives:

![Screenshot 06](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-063.jpg)

Notice that stick 2 is connected at the same x-value as the ball, so I did not need to add the radius of the ball to its r value. Now I can calculate the total torque:

![Screenshot 07](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-073.jpg)

Although I do have an ok value for U in meters, I do not have a value for the mass of the ball, so no point in multiplying in the constant g. Anyway, let me test this. If this is true, how many balls could I hang right off the circle and lift? In that case, r would be 0.5 U (U is the diameter). So if the torque is around 3, I should be able to lift 6 balls (depending on the mass of string used). Let me try it.

![Screenshot 08](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-082.jpg)

I love it when a plan comes together. Actually, this was a little more than the weight of 6 balls, it also had the short length of water-sticks. But also, according to my calculation, this should not be able to lift 7 balls. Again, success.

Physics of Fantastic Contraption I

One of my students showed me this game, [Fantastic Contraption](http://fantasticcontraption.com/). The basic idea is to use a couple of different “machine” parts to build something that will move an object into a target area. Not a bad game. But what do I do when I look at a game? I think – hey! I wonder what kind of physics this “world” uses. This is very similar to [my analysis of the game Line Rider](http://blog.dotphys.net/2008/09/the-physics-of-linerider/) except completely different.

Fantastic Contraption gives the unique opportunity to build whatever you want. This is great for creating “experiments” in this world.

The first step is to “measure” some stuff. The game includes three types of “balls” and two types of connectors. The balls are:

  • Clockwise rotating
  • Counterclockwise rotating
  • Non-driven

Connectors:

  • wood lines – these can not pass through each other
  • water lines – these can pass through each other, but not the ground

First question: Do the different balls have the same mass? This can be tested by creating a little “balance”

![Screenshot 05](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-052.jpg)

Continue reading “Physics of Fantastic Contraption I”