MacGyver Season 3 Episode 22 Science Notes: Mason + Cable + Choices

May 12, 2019rhettallain 2 Comments

Season 3 finale—but don’t worry, MacGyver has been renewed for a 4th season. Boom. Now for some science.

Descender Device

MacGyver needs to get down an elevator cable—to do this, he builds a descender. The basic idea is to “grab” hold of the cable to produce enough friction that it supports a human. That keeps you from falling. Of course you also want to move down, there needs to be some method to “inch” your way down. The one MacGyver builds looks like this.

Here is an early sketch for a type of descender

Of course the problem is that the elevator cable is under tension and very thick. It’s really more like a pole than a rope. That’s why the design in the episode would work better.

Two Bad MacGyver (Mason) Hacks

These aren’t bad hacks—they are hacks from the Bad MacGyver (Mason). First, there is the cable cutter. This is just a bolt cutter connected to an electric motor. That should work.

The other one is the hydrochloric acid in the basement of a building around support pillars. So, would this work? Well, hydrochloric acid does indeed dissolve concrete and cement—it’s not super fast though. Everyone likes to think of acid as being that kind in the movie Alien. It’s not like that.

Of course a pillar isn’t just cement. It has steel rods in there too. But acid will eat through steel as well—again, it just takes a while. But you don’t have to completely dissolve the pillars to cause destruction. Just making them weak could do the job.

Oh, it’s a good thing the hydrochloric acid is in plastic barrels. It would melt steel barrels.

Atwood Machine

MacGyver’s plan is to connect one elevator to the one next to it. When the cable is cut, the two elevators will create an Atwood machine. This is of course a real physics problem.

The idea is to have two different masses connected by a string. This string then runs over a pulley. If the masses are different, the two masses will accelerate (one up and one down) with a constant acceleration. The key is that this acceleration will be much smaller than the acceleration of a free falling object. That’s a good thing since waaaaay back in the day, it was very difficult to measure the motion of an object with a large acceleration.

I think I will save the physics of an “Atwood Machine Problem” (no one really calls it that) for a later post. Instead, here is my calculation.

But wait! There’s more! This calculation would give you the tension in the cable, but once there is a tension the cable would stretch. How much the cable stretches depends on:

Tension
Cable length
Cable diameter
Type of material

So you see that the stretch really depends on two things—the material and the size of the cable. For the material dependence on stretch, we call this Young’s Modulus.

There is one more thing—maximum tension before a cable breaks. This also depends on the type of material and the shape of the cable. Here is a sample calculation.

Too bad MacGyver never got a chance to put these calculations into practice. Of course it’s Mason’s fault.

Recover Serial Number

It is possible to recover a serial number that’s scratched off a metal. Essentially, when the number is stamped into the metal there is a more than just a surface effect. The deeper metal is also changed in some way. Using acid, it’s possible show these differences and find this number. Yes, this is real.

For a circuit chip, the serial number is not likely to be stamped—it will be printed. Still, it’s entirely plausible that you could still recover some type of artifact.

The Ladder Problem

December 11, 2018rhettallain 1 Comment

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.

A friction problem.
An equilibrium problem.
A ladder problem.
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.

Torque produced by balls in Fantastic Contraption

October 27, 2008rhettallain 1 Comment

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 m_b 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

October 23, 2008rhettallain 1 Comment

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” →

A note about numerical calculations

October 22, 2008rhettallain Leave a comment

[In a previous post, I talked about numerical calculations](http://blog.dotphys.net/2008/10/basics-numerical-calculations/). The basic idea is to use the momentum principle and the following “recipe”:

Update the position of the particle
Update the momentum of the particle
Update the force on the particle

Looks great, right? Well, it mostly is great. I want to give a couple of pointers about the last step, update the force on the particle. How and when can you do this? Really, in numerical calculations, you will see two types of forces:

Forces that you can calculate: That looks strange, but it’s true. Maybe you are thinking, can’t you calculate all the forces? – the answer is no. Yes, you can calculate the gravitational force and the electromagnetic force. Also, really all forces you are likely to see are one of those two. You can also calculate the force due to a spring(depends on position), the air resistance force (depends on velocity). These types of forces work well in the above numerical recipe.
Forces that you CAN NOT calculate: These are all the other forces. Typically, these are forces of constraint. Suppose a block slides down a plane. Yes, you can calculate the force the plane exerts on the block, but it depends on things other than just the position of the block. The force the plane exerts on the block is such as to keep the block on the plane. You can not calculate this in the same way as the previous category of forces. Yes, technically the force the plane exerts on the block IS the electromagnetic force. If you want to calculate this force between all the atoms in the two objects, I encourage that.

So what does this all mean? This means that you can not use the above “recipe” for whatever you want. Sorry.

(I have a trick I will show you later)

Spring Motion and Numerical Calculations

October 17, 2008rhettallain 2 Comments

Maybe you know I like numerical calculations, well I do. I think they are swell. [VPython](http://vpython.org) is my tool of choice. In the post [Basics: Numerical Calculations](http://blog.dotphys.net/2008/10/basics-numerical-calculations/) I used vpython and excel to do something simple. I will do that again today (in that this problem could also be solved analytically). However, there is one big difference. This problem has a non-constant forces. Suppose I have a mass that is connected by a spring to a wall. This mass-spring is sitting on a table with no friction.

![Screenshot 27](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-272.jpg)

There is a very interesting property of springs. The more you stretch them, the greater the force they exert (in the usual model of springs). This model works very well.

![Screenshot 28](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-282.jpg)

This is known as Hooke’s law. I have written it as a scalar for simplicity. The “k” is called the spring constant. It is a measure of how “stiff” the spring is. The value “s” is the amount the spring is stretched. Typically, there is a minus sign in front of the ks to indicate that the force is in the opposite direction that the spring is stretched. Really, in a scalar equation this is rather silly to include (but everyone does anyway).

**Question: What will the motion of the mass be like if I pull it back and then let go?**

Although this can be determined analytically, I am going to first calculate this with vpython. I will try to show all the details so that you can reproduce this also. If you have not already installed [vpython](http://vpython.org), do that now (don’t cost nothing).

Continue reading “Spring Motion and Numerical Calculations” →

Basics: Numerical Calculations

October 14, 2008rhettallain 4 Comments

**Pre Reqs:** [Kinematics](http://blog.dotphys.net/2008/09/basics-kinematics/), [Momentum Principle](http://blog.dotphys.net/2008/10/basics-forces-and-the-momentum-principle/)

What are “numerical calculations”? Why are they in the “basics”? I will give you really brief answer and then a more detailed answer. Numerical calculations (also called many other things – like computational physics) takes a problem and breaks into a WHOLE bunch of smaller easier problems. This is great for computers ([or a whole bunch of 8th graders](http://blog.dotphys.net/2008/09/computational-physics-and-a-group-of-1000-8th-graders/)) because computers don’t mind doing lots of little problems. Why are they “basic”? Well, most text would say they are not basic. I disagree. I think this is a legitimate method for solving problems. In particular, this is a great way of solving problems that can not be solved analytically (meaning solving one hard problem).

**Numerical Calculations are Theoretical Calculations**

Let me just get this out of the way. Numerical calculations and analytical calculations are really in the same “class”. Often people will lump numerical in with “computational experiment” but that is a really bad thing to do. Some others will claim that there are three different “paths” to discover stuff in science: theory, experiment, and simulations. Simulations are the same thing as numerical calculations which are the same as theory. ([I wrote a letter about this in the American Journal of Physics](http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000076000009000797000001&idtype=cvips&gifs=yes))

**Example Problem**

Let me start with a problem that can be solved analytically. Suppose I have a ball of mass 0.5 kg and I throw this straight up with a speed of 10 m/s. How high will it go?

Continue reading “Basics: Numerical Calculations” →

Basics: Forces and the momentum principle

October 10, 2008rhettallain Leave a comment

**Pre reqs:** [Free Body Diagrams](http://blog.dotphys.net/2008/09/basics-free-body-diagrams/), [Force](http://blog.dotphys.net/2008/09/basics-what-is-a-force/), [Kinematics](http://blog.dotphys.net/2008/09/basics-kinematics/)

The time has come to look at things that are NOT in equilibrium. The most basic question to ask yourself is: *”What do forces do to an object”*? Aristotle would say that forces make things move. Constant forces make things move constantly. Actually, Aristotle said there were two types of motion:

Natural motions: These motions don’t need anything to happen, they just do. Example: a rock falling. You don’t need to do anything to it. Example: fire rising. It just rises. (there was more to it than that, but you get the idea).
Violent motions: These motions are due to some interaction that forces them from their natural state. The natural state of a cart is to be at rest. If someone pushes on it, it will move. When you stop pushing (stop the violent motion) it returns to its natural state – at rest

I am talking about Aristotle, because these basic ideas are what most people think. If you push something it moves. If you stop pushing, it stops. And these people are correct. The problem is that there is always this extra force that no one thinks about – friction. Without friction, the rules change.

**New Rules (Newtonian ideas)**

If you push something with one force, it changes velocity. If you stop pushing, it stays at a constant velocity.

If you want to test your feelings for force, [try this force game I made on Scratch](http://scratch.mit.edu/projects/rhettallain/285748). The idea is that you need to move the box to the red circle. The arrow keys exert a **force** on the object.

Continue reading “Basics: Forces and the momentum principle” →

Force Rocket Game Level 2

October 7, 2008rhettallain Leave a comment

Popular demand (from a few people – you know who you are) requested another level for the force game. I am open to naming this game. Ideas for levels have been suggested also. There is a new rule change. You must be stopped (or almost stopped) in the red circle.

Learn more about this project

Basics: Friction

October 7, 2008rhettallain 7 Comments

**Pre reqs:** [Free body diagrams](http://blog.dotphys.net/2008/09/basics-free-body-diagrams/)

Friction is an interaction between two objects in contact that opposes relative motion of those two objects. It is not something fundamental (like gravity, or electromagnetic force), but it comes up enough that it will be worthwhile to talk about it. Let me start with a simple example. Suppose I have a book on a table. Here is the free body diagram for the book:

![Screenshot 27](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-27.jpg)

Simple enough – right? There are two forces on the book. A contact force (the table pushing up) and a long range force (the gravitational force of the Earth pulling down on the book). These two forces have the same magnitude, so when added together, they give a total of zero vector. This means the book is in equilibrium.

Now, what if I push on the book from the side? Suppose I push with 1 Newton. If the book is still in equilibrium, what does that mean? It means the free body diagram must look like this:

![Screenshot 28](http://blog.dotphys.net/wp-content/uploads/2008/10/screenshot-28.jpg)

If the book is still in equilibrium, then the force of the table on the book (due to friction) would have to have the same magnitude as me pushing on the book. Note: Even though my push and friction are equal and opposite, these are not Newton’s third law force pairs – I talked about that in the free-body-diagram post.

Continue reading “Basics: Friction” →

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