George Carlin, on Change

change machine

On numerous occasions, I have repeated this experiment, in keeping with the scientific method. I have obtained the same null result as Carlin obtained, each and every time.

An Asymmetrical Static Equilibrium Physics Problem Involving Pulleys and Hanging Masses

An interesting phenomenon in physics, and physics education, is the simplicity of symmetric situations, compared to the complexity of similar situations which are, instead, asymmetrical. Students generally learn the symmetrical versions first, such as this static equilibrium problem, with the hanging masses on both left and right equal. 

static equilibrium pulley setup

The problem is to find the measures of the three angles shown above, with values given for all three masses. Here is the setup, using physical objects, rather than a diagram.

100_170_100

The masses on the left and right are each 100 g, or 0.100 kg, while the central masses total 170 g, or 0.170 kg. Since all hanging masses are in static equilibrium, the forces pulling at the central point (at the common vertex of angles λ, θ, and ρ) must be balanced. Specifically, downward tension in the strings must be balanced by upward tension, and the same is true of tension forces to the left and to the right. In the diagram below (deliberately asymmetrical, since that’s coming soon), these forces are shown, along with the vertical and horizontal components of the tension forces held in the diagonal strings.

static equilibrium pulley setup force diagram

Because the horizontal forces are in balance, Tlx = Trx, so Mlgcosλ = Mrgcosρ — which is not useful now, but it will become important later. In the symmetrical situation, all that is really needed to solve the problem is the fact that the vertical forces are in balance. For this reason, Tc = Tly + Try, so Mcg = Mlgsinλ + Mrgsinρ. Since, due to symmetry, Ml = Mand λ = ρ,  Mr may be substituted for Ml, and ρ may be substituted for  λ, in the previous equation Mcg = Mlgsinλ + Mrgsinρ, yielding Mcg = Mrgsinρ + Mrgsinρ, which simplifies to Mcg =2Mrgsinρ. Cancelling “g” from each side, and substituting in the actual masses used, this becomes 0.170 kg = 2(0.100 kg)sinρ, which simplifies to 0.170 kg = (0.200 kg)sinρ, then 0.170/0.200 = sinρ. Therefore, angle ρ = sin-1(0.170/0.200) = 58°, which, by symmetry, must also equal λ. Since all three angles add up to 180º, the central angle θ = 180° – 58° – 58° = 64°. These answers can then be checked against the physical apparatus.

53_theta_53

When actually checked with a protractor, the angles on left and right are each about 53° — which is off from the predicted value of 58° by about 9%. The central angle, of course, is larger, at [180 – (2)53]° = 74°, to make up the difference in the two smaller angles. The error here could be caused by several factors, such as the mass of the string itself (neglected in the calculations above), friction in the pulleys, or possibly the fact that the pulleys did not hang straight down from the hooks which held them, but hung instead at a slight diagonal, as can be seen in the second image in this post. This is testable, of course, by using thinner, less massive string, as well as rigidly-fixed, lower-friction pulleys. However, reducing the error in a lab experiment is not my objective here — it is, rather, use of a simple change to turn a relatively easy problem into one which is much more challenging to solve. 

In this case, the simple change I am choosing is to add 50 grams to the 100 g already on the right side, while leaving the central and left sides unchanged. This causes the angles where the strings meet to change, until the situation is once more in static equilibrium, with both horizontal and vertical forces balanced. With the mass on the left remaining at 0.100 kg, the central mass at 0.170 kg, and the mass on the right now 0.150 kg, what was an easy static equilibrium problem (finding the same three angles) becomes a formidable challenge. 

100_170_150

For the same reasons as before (balancing forces), it remains true that Mlgcosλ = Mrgcosρ (force left = force right), and, this time, that equation will be needed. It also remains true that Mcg = Mlgsinλ + Mrgsinρ (downward force = sum of the two upward forces). The increased difficulty is caused by the newly-introduced asymmetry, for now Ml ≠ Mr, and λ ≠ ρ as well. It remains true, of course, that  λ + θ + ρ° = 180.

In both the vertical and horizontal equations, “g,” the acceleration due to gravity, cancels, so Mlgcosλ = Mrgcosρ becomes Mlcosλ = Mrcosρ, and Mcg = Mlgsinλ + Mrgsinρ becomes Mc = Mlsinλ + Mrsinρ. The simplified horizontal equation, Mlcosλ = Mrcosρ, becomes Ml²cos²λ = Mr²cos²ρ when both sides are squared, in order to set up a substitution based on the trigonometric identity, which works for any angle φ, which states that sin²φ + cos²φ = 1. Rearranged to solve it for cos²φ, this identity states that  cos²φ = 1 – sin²φ. Using this rearranged identity to make substitutions on both sides of the previous equation Ml²cos²λ = Mr²cos²ρ yields the new equation Ml²(1 – sin²λ) = Mr²(1 – sin²ρ). Applying the distributive property yields the equation Ml² – Ml²sin²λ = Mr² – Mr²sin²ρ. By addition, this then becomes -Ml²sin²λ = Mr² – Ml² – Mr²sin²ρ. Solving this for sin²λ turns it into sin²λ = (Mr² – Ml² – Mr²sin²ρ)/(-Ml²).

Next, Mc = Mlsinλ + Mrsinρ (the simplied version of the vertical-force-balance equation, from above), when solved for sinλ, becomes  sinλ = (Mrsinρ – Mc)/(- Ml). Squaring both sides of this equation turns it into sin²λ = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml.

There are now two equations solved for sin²λ, each shown in bold at the end of one of the previous two paragraphs. Setting the two expressions shown equal to sin²λ equal to each other yields the new equation (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml)², which then becomes (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(Ml)², and then, by multiplying both sides by -Ml², this simplifies to Mr² – Ml² – Mr²sin²ρ = – (Mr²sin²ρ – 2MrMcsinρ + Mc²), and then Mr² – Ml² – Mr²sin²ρ = – Mr²sin²ρ + 2MrMcsinρ – Mc². Since this equation has the term – Mr²sin²ρ on both sides, cancelling it simplifies this to  Mr² – Ml² = 2MrMcsinρ – Mc², which then becomes Mr² – Ml² + Mc² = 2MrMcsinρ, and then sinρ = (Mr² – Ml² + Mc²)/2MrM= [(0.150 kg)² – (0.100 kg)² + (0.170 kg)²]/[2(0.150 kg)(0.170 kg)] = (0.0225 – 0.0100 + 0.0289)/0.0510 = 0.0414/0.510 = 0.812. The inverse sine of this value gives us ρ = 54°.

Having one angle’s measure, of course, makes it far easier to find the others. Two paragraphs up, an equation in italics stated that sinλ = (Mrsinρ – Mc)/(- Ml). It follows that λ = sin-1[(Mrsinρ – Mc)/(- Ml)] = sin-1[(0.150kg)sinρ – 0.170kg)/(-0.100kg)] = 29°. These two angles sum to 83°, leaving 180° – 83° = 97° as the value of θ.

31_theta_58.png

As can be seen above, these derived values are close to demonstrated experimental values. The first angle found, ρ, measures ~58°, which differs from the theoretical value of 54° by approximately 7%. The next, λ, measures ~31°, also differing from the theoretical value, 29°, by about 7%.The experimental value for θ is (180 – 58 – 31)° = 91°, which is off from the theoretical value of 97° by ~6%. All of these errors are smaller than the 9% error found for both λ and ρ in the easier, symmetrical version of this problem, and the causes of this error should be the same as before.

Beginning the Fractiles-7 Refrigerator Experiment

To begin this experiment, I first purchased two refrigerator-sized Fractiles-7 sets (available at http://fractiles.com/), and then, early on a Sunday, quietly arranged these rhombus-shaped magnets on the refrigerator in our apartment (population: 4, which includes two math teachers and two teenagers), using a very simple pattern.

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Here’s a close-up of the center. There are 32 each, of three types of rhombus., in this double-set, for a total of 96 rhombic magnets, all with the same edge length.

160207_0001

The number of possible arrangements of these rhombi is far greater than the population of Earth.

The next step of the experiment is simple. I wait, and see what happens.

It should be noted that there is a limit on how long I can wait before my inner mathematical drives compel me to play with these magnets more, myself — but I do not yet know the extent of that limit.

Constructing “Nightday” — An Experiment Involving Sleep

Waning-Gibbous-Moon

The last workweek having left me rather tired, I went to bed early Friday, after work, and then, having slept all I could sleep, I then woke back up quite early Saturday morning, before sunrise, and couldn’t get back to sleep. I was tired all day Saturday, but not too weary to think. What I thought was simple: as tired as I am, it sure would be nice to have a three-day weekend this weekend. Next, I thought, yeah, this would be nice, but that won’t make it happen. Finally, I realized that I actually could, perhaps, come up with some hopefully-clever and effective way to get the three-day weekend I want . . . and, having had an idea to do exactly that, I’m trying it right now.

I’ve tested the 24-hour sleep/wake cycle before, trying to find ways to lengthen that period of time. (Ever wanted more hours in the day? Well, I actually tried to make that happen, once, but the results were less than successful.) This time, however, I’m not trying to get extra hours in a day, but an extra day in the weekend — by simply using shorter “days,” and thus making “room,” temporally, to add an extra sleep-period and wake-period into the weekend. So, Friday, I fell asleep around 5:00 pm, and did so without the prescribed medication (which includes sedatives) which I usually take at bedtime, since it wasn’t that late yet . . . so I simply fell asleep because I was tired.

Without the sedatives I am used to taking, of course, I didn’t stay asleep anything like a full eight hours, and instead “popped” back awake at around 9 pm, which was less than an hour ago, as I write this. Rather than sedating and returning to sleep, however, I took the other medication I am prescribed for this time of the day (such as that needed to regulate blood pressure), and then made my “morning” coffee, which I am enjoying now . . . to begin the extra “day” I’m attempting to add to this weekend, between Saturday and Sunday. My hypothesis is that I can deliberately alter my sleep/wake cycle in such a way that I have three (shorter) sleep/wake cycles in two calendar days, thus giving myself a three-day weekend, of a sort, and enjoy the benefits of a three-day weekend as a result. If, come Monday, I feel like I’ve had a three-day weekend — in that I feel unusually well-rested — I will consider this experiment to “create” a working illusion of a three-day weekend, without any actual extra time, to be a success (subject to the opinion of my doctors, to whom I will describe all of this).

I plan to stay awake until roughly dawn on Sunday, and then go to sleep until, well, whenever I wake up. I’ll then have a shortened post-sleep Sunday wakefulness-period, go to sleep at a reasonable hour Sunday night, and get a good, full night’s sleep then, before going to work on Monday.

Right now, therefore, I’m having the middle “day” of what feels, subjectively, like a three-day weekend, and having it at night, between what seems, now, like it was yesterday (the shortened Saturday), and what I anticipate as my shortened Sunday, after I sleep again, tomorrow. Since it’s easier to talk about this extra “day” I’m having tonight if I give it a name, I’m doing so: I’m calling it “Nightday.”

Some readers may object that I’ve merely come up with an overly-convoluted way to analyze a four-hour Saturday-afternoon nap. I’ll concede that they do have a point . . . but if my calling this “Nightday,” and telling myself that I’m enjoying a three-day weekend, actually turns out to help me feel and act more rested next week, then I’ll take those benefits and run with them, regardless of what any critics tell me (unless, of course, my doctors are among those critics). If this experiment has only beneficial results, and passes medical review, then I’ll likely use more Nightdays to get additional three-day weekends in the future, whenever I need, or simply want, them.

~~~

Important disclaimer: nothing in this blog-post should be taken as any form of medical advice, for I am not medically trained. I have taken the precaution of discussing my practice of occasionally inventing and conducting experiments such as this with my own physicians, and will continue to do so. No one should attempt to replicate this experiment without first consulting their own physician(s).

[Image credit: the photo of the Moon shown above was found here — https://photographylife.com/moon-waning-gibbous. It isn’t identical in appearance to the current waning gibbous Moon, having been photographed quite some time earlier, but it is close.]

An Ethical Dilemma Involving a Polyhedral Crystal

I just ordered a crystal rhombcuboctahedron on eBay because I like its geometrical properties, despite the mystical claims in the item listing. I did so with the full knowledge and expectation that these claims are almost certainly false, because, well, they’re mystical claims.

Rhombicubocta

Here’s my ethical dilemma: would it be ethical to lab-test those claims, then post the results in the feedback I leave?

[Image created using Stella 4d, available at http://www.software3d.com/Stella.php. This isn’t a picture of the crystal on eBay; it is made of quartz, and not rainbow-colored. It is of the same shape, however.]

Free the Frozen People!

150127_0000

After seeing this sign in a local grocery store, I carefully searched the entire frozen food section, but I could find neither the frozen Mexican, nor the frozen Asian. Since they were gone, but the sign indicates they were there at one point, I concluded that the experiment was over, and hoped they had thawed out both experimental test subjects, found them still healthy after a few days in cryogenic suspended animation, and sent them home, each with a fat check to compensate them for the huge risk they just took.

However, even with compensation and signed consent forms, I still have certain ethical reservations about scientists performing this sort of experiment on actual human beings. Why not freeze, thaw, refreeze, and rethaw mice, instead? Is PETA really that scary?

Are they still doing these experiments, in my town or elsewhere? If so . . . free the frozen people!

There is one last thing about this whole thing which I just can’t figure out, though, and that’s this: why were they storing their frozen, experimental, human test subjects in the middle of a central Arkansas grocery store in the first place?