Hillary Clinton’s Other E-mail Problem

This has nothing to do with those other e-mails tied to Hillary Clinton — the ones which have recently been under official investigation, and in the news. It’s a completely different thing: e-mails sent out by her campaign for the White House, and unrelated to her time as Secretary of State.

Hillary Needs Me

Other e-mails, entirely too much like this, preceded the “I need you” e-mail I received yesterday. I’ve been making fun of them on Facebook for quite some time, but hadn’t brought them to my blog until now. I’m simply using cropped screenshots from my e-mail account for these pictures, and keeping the e-mail senders, subject lines, and dates together, for each e-mail. If anyone wishes to check the authenticity of these e-mails with the Clinton campaign, that’s fine with me. You’ll find that these e-mails are real (or they’ll lie to you; I can’t rule that out). If lies are told, I’ve got the evidence in my e-mail account, as do many others. This is not a complete set, either; it’s just the most recent of these, um, strange e-mails from her campaign minions. 

Hillary Needs Me 2

I don’t know how I got on Hillary Clinton’s presidential-campaign e-mail list, but I am not complaining about it. If I wasn’t on her campaign’s list, after all, I wouldn’t know that all these e-mails are going out, with her name (and sometimes others, as seen above) as the sender, and such things as “re: last night” in the subject line. That would deprive me of this opportunity to use real campaign materials to ridicule a major-party presidential candidate, or, in other words, prevent this particular bit of fun. There were other such e-mails before June 29 — long before, actually — but this is all of this kind of thing I can stand to put on my blog.

To Hillary Clinton (the person, not her campaign staff): Really, H.C.? Do you not monitor your campaign flunkies at all? These e-mails could bring new meaning to the term “madame president,” and I really don’t think they will help you at the ballot box, either.

To Donald Trump, and his ilk: don’t think this means I support you. I don’t.

For whom will I cast my vote, some may wonder? Well, I have it narrowed down to two candidates, but neither of their names appear in this post. For more information regarding where my vote will go, simply click here.

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.

A Logic Problem Involving Marvel Super-Heroes

Movies_Spider_man_Spider_man_DareDevil_Iron_Man_Captain_America_Wolverine_Black_43340_detail_thumb

Iron Man, Daredevil, Spider-Man, Captain America, and Wolverine each have a favorite food, a favorite beverage, own one pet, and have a single hobby. Based on the clues which follow, find out these things:

  • Which hero’s favorite food is (A) pizza, (B) green eggs and ham, (C) apple pie, (D) Chinese take-out, and (E) caviar?
  • Which hero’s favorite beverage is (A) beer, (B) vodka, (C) Coca-Cola, (D) water, and (E) chocolate milkshakes?
  • Which hero owns (A) a black cat, (B) a porcupine, (C) a robot dog, (D) an iguana, and (E) a real dog?
  • Which hero’s hobby is (A) coin collecting, (B) stamp collecting, (C) collecting comic books, (D) collecting seashells, and (E) collecting rocks?

Here are the clues. Answers will be revealed in the comments, but only after someone solves the puzzle (to avoid spoiling anyone’s fun).

  1. Wolverine drinks beer.
  2. Daredevil is blind. The other four heroes can all see.
  3. Spider-Man eats pizza.
  4. Wolverine has a mutant healing factor that allows him to rapidly heal from injuries.
  5. Iron Man is the only one of these five heroes who wears a suit of armor.
  6. The hero whose favorite food is apple pie always eats it with his favorite drink, Coca-Cola.
  7. Iron Man drinks vodka.
  8. All of the heroes who can see refuse to eat green eggs and ham.
  9. Of these five heroes, no one without either a mutant healing factor or a suit of armor would be dumb enough to keep a porcupine as a pet.
  10. Iron Man, an accomplished inventor, refuses to own a pet which he did not build himself.
  11. The hero who eats apple pie doesn’t like chocolate, nor chocolate-flavored anything.
  12. Iron Man has more money than all the other heroes combined.
  13. The hero whose favorite food is pizza does not own a dog.
  14. The seashell-collector is blind.
  15. The owner of a real dog also collects stamps. 
  16. The porcupine-owner doesn’t like apple pie.
  17. Spider-Man likes the black cat, but has to visit the cat’s owner in order to see her.
  18. The richest hero eats caviar.
  19. The coin collector doesn’t like pizza, nor porcupines.
  20. The comic-book collector hates drinking water. He also doesn’t like milkshakes of any kind.
  21. The owner of the black cat is lactose-intolerant, and, for this reason, doesn’t drink milkshakes.

The first person to leave the solutions in the comments wins bragging rights.

[Source of image: http://www.hdwallpaperpc.com/show-wallpaper/Spider_man_DareDevil_Iron_Man_Captain_America_Wolverine_Black_43340.html].

Circumparabolic Regions Inside a Unit Circle

circumparabolic regions

A circumparabolic region is found between a circle and a parabola, with the circle being chosen to include the vertex and x-intercepts of the parabola used, with the circle, to define the two circumparabolic regions for a given parabola-circle pair. There are four such regions shown above, rather than only two, because two parabolas are used above. The formulae for the parabolas, as well as the circle, are shown.

A puzzle which I will not be solving, I suspect, until I learn more integral calculus: what fraction of the circle’s area is shown in yellow?

Geometry Problem Involving Two Circles (See Comments for Solution)

This is a puzzle I made up not long ago. After trying to solve it for a bit (no success yet, but I haven’t given up), I decided to share the fun.

A small circle of radius r is centered on a large circle of radius R. It is a given that 0 < r < R. In terms of r and R, what fraction of the smaller circle’s circumference lies outside the larger circle?

two circles

I am 90% certain there is an extremely simple way to do this, using only things I already know. It’s frustrating that the answer isn’t simply leaping out of the computer screen, at me. For simple math problems, that’s what usually happens . . . so either this is merely deceptively simple, or I am missing something.

The Cone Problem (The Easier Sequel to the Hemisphere Problem)

cone

That hemisphere problem (see previous two posts) was quite difficult. I’m going to unwind a  bit with the much easier cone version of the same problem: at what height x above the ground, expressed as a fraction of h, must a cone of height h and radius r be cut, in order for the two pieces produced by the cut to have equal volume? The fact that a path down the lateral surface of a cone is a straight line, not a curve, should make this much easier than the hemisphere problem.

Since the volume of a cone is (1/3)πr²h, and the smaller cone created above the cut would be half that volume, it follows that

(1/3)πr²h = (2/3)π(r′)²h′                [equation 1]

By cancellation of (1/3)π, this equation becomes

r²h = 2(r’)²h’               [equation 2]

Also, based on divisions of the cone’s altitude, we know that

h = h′ + x                [equation 3]

Furthermore, since the problem asks that the height x be expressed as a fraction of h, we can let that fraction (a decimal between zero and one) be represented by f, so that

x = fh               [equation 4]

Also, by using similar right triangles’ corresponding legs, we know that

r/h = r′/h′                [equation 5]

which rearranges to

rh′ = r′h                  [equation 6]

There is a proportionality constant in play here, p, defined as the fraction of the length of one part of the larger cone which equals the length of the corresponding part of the smaller cone. As equations, then,

r′ = pr         and          h′ = ph              [equations 7a and 7b]

Also, because p is the fraction of h which is h′, and f is the fraction of h which is x, and h = h′ + x, it follows that

p + f = 1                  [equation 8].

Next, by substituting equations 7a and 7b into equation 2 for r′ and h′, we know that

r²h = 2(pr)²ph               [equation 9]

Which reduces to

1 = 2p³               [equation 10]

When equation 10 is solved for p, it becomes

p = (1/2)^(1/3)                [equation 11]

And, since equation 8 states that p + f = 1, it follows that f = 1 – p, and f is the fraction we seek. By substituting equation 11 for p in f = 1 – p, the following value for f can be determined:

f = 1 – (1/2)^(1/3)               [equation 12]

This leads to the following cleaned-up solution to the problem, shown in standard exact form, and with a decimal approximation as well.

f

The cut, therefore, should be made approximately 20.6% of the way from the bottom to the top of the full cone.

To check this answer, I need only find the volume of the smaller cone, times two, and show that it equals the value of the larger cone.

2(volume of smaller cone) = (2/3)π(r′)²h′ = (2/3)π(pr)²ph =

(2/3)πp³r²h = (2/3)π(cube root of ½)³r²h = (2/3)π(1/2)r²h = (1/3)πr²h,

which is the volume of the full cone, as it should be. The problem has now been solved, and the solution f (by way of p, which equals 1 – f, by a rearrangement of equation 8) has been checked.

Working Towards a Solution of the Hemisphere Problem

The hemisphere problem referred to here was described in the previous post. To reword it somewhat, consider this hemisphere, half of a sphere of radius r. The orange cross-section is a circle parallel to the hemisphere’s yellow, circular base.

hemisphere

We are to find the height of the yellow section with the orange circular top (which I shall call x), as a fraction of r, such that the yellow and red sections above have equal volumes.

Since the volume of a hemisphere is (2/3)πr³ (that’s half a sphere’s volume), each of these two sections must have half the hemisphere’s volume, or (1/3)πr³.

Moreover, the top (red) portion is a “spherical cap,” described here on Wikipedia, as was pointed out to me, by a friend, on Facebook. On that Wikipedia page, you can find this diagram, as well as the formula shown below for the volume of the purple spherical cap in the diagram.

spherical cap

spherical cap volume

Now, as our goal is to find x, as described at the top of this post, it important to remember that r = x + h, where r is the radius of the original sphere (and height of the hemisphere), h is the height of the spherical cap, and x is the height of the hemisphere, after the spherical cap is removed. We now have two expressions for the volume of the spherical cap: (1/3)πr³ (because it is a fourth of the volume of the original sphere), and (1/6)πh(3a² + h²) from the Wikipedia article on the spherical cap (so all of this assumes, then, accuracy in that Wikipedia article). Setting them equal to each other,

(1/3)πr³ = (1/6)πh(3a² + h²)

Next, I’ll clean this up by multiplying left and right by 6/π, to cancel fractions and π from both sides.

2r³ = h(3a² + h²)

A right triangle exists in the blue-and-purple figure above, and the unlabeled leg is x, the problem’s original goal. I’ll add an “x” to this diagram.

spherical cap with xUsing this right triangle, and the Pythagorean Theorem, it can be seen that a² = r² – x². Also, since r = x + h, it follows that h = r – x. By substitution for a and h, then,

2r³ = h(3a² + h²)

becomes

2r³ = (r – x)[3(r² – x²) + (r – x)²], which then distributes to

2r³ = (r – x)(3r² – 3x² + r² – 2rx + x²), which expands as

2r³ = 3r³ – 3rx² + r³ – 2r²x + rx² -3r²x  + 3x³ – r²x + 2rx² – x³, which simplifies to

0 = 2r³ – 6r²x + 2x³, which becomes, by division:

0 = r³ – 3r²x + x³.

I’m trying to find x, as a fraction of r, meaning that x = kr, and I want k. On that basis, I’ll now substitute kr for each x in the last equation above.

0 = r³ – 3r²(kr) + (kr)³, which then becomes

0 = r³ – 3kr³ + k³r³, or

3kr³ = r³ + k³r³, and then dividing by r³ yields

3k = 1 + k³, which can be rearranged to

3k – k³ = 1, which factors,on the left side, to yield

(k)(3 – k²) = 1.

For k and (3 – k²) to have a product of one, they must be reciprocals. Therefore, 1/k = 3 – k². I can then graph y = 1/k, as well as y = 3 – k², and find the solution by seeing where the graphed functions intersect above the k-axis, with a “k” value between zero and one, since no value of k less than zero or greater than one would make sense, as a solution to the original problem. (I’ll be using k coordinates, and a k-axis, in place of the usual x coordinates and x-axis.) Here’s the initial graph:

graph1

The only intersection in the specified range of zero to one is between point A and point B, so I brought them together, as closely as I could get Geometer’s Sketchpad to let me.

graph2

With points A and B almost on top of each other, k = 0.34958 by one equation, and k = 0.34820 by the other. To two significant figures, then, I can conclude that the horizontal cut in the original problem should be made 35% of the way from the base of the hemisphere to the hemisphere’s top.

Using a graphing calculator, a more precise answer of 0.34729636 was obtained. I’d still like to have an exact answer, but this will do for now.

—-

Later addition: a helpful reader led me to a Wolfram Alpha site where I could get an exact answer, as well as a decimal approximation with a greater degree of precision. In the pic below, I have omitted the two solutions of the third-order polynomial which are not the one solution of interest. Here’s the one which is:

wolfram

Now, however, I have another mystery: how can an exact answer, with all those imaginary units in it, have a real-only approximation? To this question, at least for now, I don’t even have the beginnings of an answer.

—-

Even later post-script: I have been assured by friends on Facebook that the imaginary units in the above exact solution somehow cancel, although I must concede that I still do not see how, myself. I’ve also been shown another way to express the solution, for 2sin(π/18) is also ≈ 0.3472963553338606977034333. This surprised me, due the the lack of any explicit appearance by a π/18 (= 10°) angle in the original problem, and the fact that no trigonometric functions were used to solve it.

The Hemisphere Problem (See Next Post for the Solution)

hemisphere

A hemisphere rests with its circular base on a horizontal, level surface, and is to be cut into two pieces of equal volume. If the hemisphere’s radius is r, at what fraction of r above the floor should the horizontal cut be made?

[Solution in next post: https://robertlovespi.wordpress.com/2015/04/06/working-towards-a-solution-of-the-hemisphere-problem/]