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 [quote="pk"]I agree entirely with your comment about units. However, I don't understand what you mean by WRONG equations that give the RIGHT answer. They do both get the same numerical answer, but that doesn't make it the right one. The answer (a distance) is a physical quantity and MUST therefore have a unit.[/quote]
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What if...
Posted: Wed May 07, 2003 6:50 pm    Post subject: 1

What Crysty B is saying is that, provided you know that s and m/s are the units of time and velocity, you can reason that since the velocity is the time times 0.5 m/s2, the equations are correct if you multiply or divide by the m/s2, or any other unit of distance over time squared, which is the only thing the "mathematicians" left out.

Edited to fix what I missed.

[This message has been edited by What if... (edited 05-07-2003 02:53 PM).]
pk
Posted: Wed May 07, 2003 7:21 am    Post subject: 0

Could we? Where do those units come from?
CrystyB
Posted: Sat May 03, 2003 3:54 pm    Post subject: -1

I agree with the phrase "wrong equations with the right answer(s)". But it's not very wrong, it's just a bit incomplete: if we'd know the units were m/s and s, we could adjust the equations as "d=(p/2)v2 / (1m/s2)" and "d=(p/2)(t/2)2 x (1m/s2)" respectively.

[This message has been edited by CrystyB (edited 05-03-2003 11:54 AM).]
pk
Posted: Thu May 01, 2003 11:19 am    Post subject: -2

In a system with the Earth and the Sun, I would have thought whether the distance is to the surface or to the C of M would be irrelevant because the masses would be virtual points in such a system. I emphasised straight line because I thought someone was thinking about tangents etc. which is not what I meant.

ps the question doesn't say there must only be two bodies.
What if...
Posted: Wed Apr 30, 2003 7:06 pm    Post subject: -3

What do you mean "in a straight line"?
Depending on whether the distance is to the center of mass, or to the closest point of the object, you can have different answers. In the Earth-Sun example, the object is farther from the Earth, which is the closest other body, so if you just use shortest distance to another body it wouldn't work. However, it does get closer to the center of mass. Look at reply 16.
Also, the question asks for two-body answers (although by imagining an Earth-Sun-like system with unusual connecting strings, the mentioned answer would still work).
pk
Posted: Wed Apr 30, 2003 2:15 pm    Post subject: -4

I agree entirely with your comment about units. However, I don't understand what you mean by WRONG equations that give the RIGHT answer. They do both get the same numerical answer, but that doesn't make it the right one. The answer (a distance) is a physical quantity and MUST therefore have a unit.
Rollercoaster
Posted: Wed Apr 30, 2003 1:01 pm    Post subject: -5

pk, in general I agree, but the problem is much worse than just a unit 'conversion' problem. The fact is, 12 cm-inches makes perfect sense. It is a unit of area because it represents a unit of length squared, and it is usable in any other calculation. The torque density of a car engine could be quoted in "FOOT-POUNDS per CUBIC INCH", for example. Sure, it could be simplified by converting units, but it isn't necessary. The issue with the original problem is that the FUNDAMENTAL UNITS OF MEASUREMENT are incorrect (a unit of length being equated to the square of time, for example). And that (to me) is a bigger problem.

As "What If" said (and was the crux of my previous post), the correctness of the answer depends on the interpretation. My take is that both scientists derived WRONG equations which lead to the RIGHT answer.
pk
Posted: Wed Apr 30, 2003 7:33 am    Post subject: -6

But there is "an answer" and there is "the right answer"; an unsound method will not necessarily give the right answer. For example, what is the area of a rectangle with sides 3 by 4? 12? Who knows? If the sides were 3 cm and 4 cm then the answer would be 12 cm^2, but you don't know that - it could be 3 cm by 4 inches, in which case the answer of 12 would be in units of cm.inches (whatever that means). Also, 12 cm^2 is the same area as 0.0012 m^2, yet these are numerically quite different. Which numerical answer would be correct? This question makes no sense, because you MUST give the units for the numerical value to mean anything, and the unit of the answer depends on the units of the bits that were multiplied to get that answer.

PS it doesn't matter whether it's the centre or the surface (for the other one), because they're in a straight line

[This message has been edited by pk (edited 04-30-2003 03:42 AM).]
What if...
Posted: Tue Apr 29, 2003 6:49 pm    Post subject: -7

Whether or not the units are important depends on whether you just want an answer, or you want the method to be sound. It is a matter of opinion in my view, although I think the mathematicians are very bad and should consider another career.
Whether or not the Earth-Sun thing works depends on whether "closer" is relative to the center of mass or surface of the body.
pk
Posted: Tue Apr 29, 2003 3:51 pm    Post subject: -8

I agree that units are essential in the speed/distance puzzle. When you multiply numbers you are multiplying the numerical values AND the units together (this is why you can't add or subtract quantities that have different units). Often you leave out the units because it's obvious, but here it is critical. Bearing this in mind, if we look at what these two mathematicians have actually done:

the first one has (pi/2)*v^2 metres^2/seconds^2, which is not a distance!

and the second has (pi/2)*(t/2)^2 seconds^2, which is also not a distance!

So they are both wrong! They do, of course, both get the same numerical answer. If you treat it as a semicircle:

(pi/2)(v)(t/2) (metres/second)(seconds)=(pi/2)(v)(t/2) metres ---which is a distance.

By multiplying only the values, the mathematicians are in fact finding the area ON THE PIECE OF PAPER (or, okay, yes, on the screen!) under the semicircle (we know that it is a semicircle in the diagram beacuse we are told so). However, this is NOT the area in V/T space, which is what the diagram is representing, and it is the area IN V/T SPACE under the curve (whatever shape that curve may be [eg circular, elliptical] by virtue of the scales chosen in order to represent it diagrammaticlly) which is the distance required.

That's what I think anyway!

For the gravity one:

Imagine the Earth, Sun, and a third, much smaller body all in a straight line in space. There is a Lagrangian point between the Earth and the Sun, which is a point where the gravitational forces cancel out. Imagine that the third body is a "test mass" (ie. it can experience gravitational forces but doesn't have enough of its own mass to upset the rest of the system) and is on the surface of the Earth to start. Move it away from the Earth and as it moves towards the Sun, and whilst it is between the Earth and the Lagrangian point, it will experience less and less force, until it experiences no force at the Lagrangian point. Is this any good?

[This message has been edited by pk (edited 04-29-2003 12:21 PM).]
mikegoo
Posted: Sat Apr 26, 2003 6:32 pm    Post subject: -9

Disclaimer: Nothing of improtance will be found in this post.

Quote:
Just as you could find the area of a square by squaring the lengths of it, or taking the length of the diaginal and calculating the area of the two trangles, or some other infernally long winded method, you will arrive at the same answer in the end.
Reminds me of the Foxtrot where Jason finds the area of a 4 X 8 rectangle by integrating (y=4 from 0 to 8).
OcularGold
Posted: Sat Apr 26, 2003 3:40 pm    Post subject: -10

Quote:
Of course, if we want to discuss solutions as inelegant as anti-matter or white holes here, we could also imagine, say, hypothetical objects who's masses decrease when other objects are brought nearer to them

the behavior of white holes has been modeled and predicted, although never observed. while i get your point of defining new hypothetical objects etc., that's just not the same thing. btw, anti-matter is NOT hypothetical, i hope that is not what you were implying.

I doubt my solution is the intended one, but it is one, based on already-existing ideas, that fits.

what-if: yes, i think so. for more information on white holes, you may simply want to google for it - there's plenty of info on the web.
wolfinabag
Posted: Sat Apr 26, 2003 2:27 pm    Post subject: -11

Of course, if we want to discuss solutions as inelegant as anti-matter or white holes here, we could also imagine, say, hypothetical objects who's masses decrease when other objects are brought nearer to them.

Prob 2
It seems many here quibble about the question of units. There is no fault in the quesiton in this regard, however, since units don't define distance. If 2 objects are 1 meter apart, they are also 0.001 km apart, but the distance is the same regardless. The point being, the two embarassingly dull self-proclaimed "scientsts" (lab technitions, I'd prefer to address them ass) would arrive at the same answer regardless of the method they used to calculate it. Just as you could find the area of a square by squaring the lengths of it, or taking the length of the diaginal and calculating the area of the two trangles, or some other infernally long winded method, you will arrive at the same answer in the end. It's that simple, really.
What if...
Posted: Fri Apr 25, 2003 6:51 pm    Post subject: -12

Would a white hole work? I have almost no knowledge regarding them (despite my interests), so I don't know...
OcularGold
Posted: Fri Apr 25, 2003 5:07 am    Post subject: -13

1. I don't consider going closer to one point of an object yet getting away from the center of mass of the object as actually getting closer to it, since youre getting further away from most of it. but maybe im just being nitpicky about it.

my answer? a white hole with just about anything else would work.

2.
Quote:
And whether or not those equations lead to the right answer when the calculations are carried out is irrelevant - the equations themselves are still wrong because of the mis-matched units.

if you actually solve the integrals, the units work out. calculating the area by using pi r^2 is just simplifying the case, and is essentially the same thing, numerically.

[This message has been edited by OcularGold (edited 04-25-2003 01:15 AM).]
Lucky Wizard
Posted: Thu Apr 24, 2003 5:15 pm    Post subject: -14

Why would it be t/2*pi? I can't see how... ah... I think I see where you're confused.

You are supposed to be finding the area under the curve, not the length/circumference of the curve. The graph is not a picture of the path the car took; it is a graph of the car's velocity (speed) at any point in time. For instance, at the very beginning and the very end, the velocity was zero, since on the graph, a hypothetical marker at which the time was zero or t intersects with a hypothetical marker at which the velocity was zero. And from the graph, you can see that in the middle of the journey, the velocity was v. So if you had a graph of an object's velocity versus time and you wanted the distance the object had traveled in that time, how would you find that? You'd check the area under the curve. (The reason for this is complicated and requires being able to handle stuff like limits, the slope of the tangent line, Riemann sums, and the constant of indefinite integration. For now, simply accept that area has this property.)

So the area is what we're working on, which is why the puzzle specifically asks you to find the area. If the graph were indeed a (semi)circle of radius t/2, and the puzzle asked you to find the circumference, then yes, your formula would be the answer. But the puzzle is to determine the correct area.
ckjgirl
Posted: Thu Apr 24, 2003 2:52 pm    Post subject: -15

for the second one i got a little confused wouldn't it be t/2*pi
What if...
Posted: Mon Apr 21, 2003 6:48 pm    Post subject: -16

Cha's answer is what I meant by B. I like my two mass answer because getting closer to the center of mass can decrease the gravitational force, so it is an answer whether you define closer by distance to the object, or distance to center of mass.
i cant spel!

[This message has been edited by What if... (edited 04-21-2003 02:49 PM).]
Rollercoaster
Posted: Mon Apr 21, 2003 3:22 pm    Post subject: -17

I guess my issue with #2 is that the two answers given were NOT calculated values of distance. They were equations. At least that's how I interpreted their responses. And whether or not those equations lead to the right answer when the calculations are carried out is irrelevant - the equations themselves are still wrong because of the mis-matched units.

What if I said, "What a coincidence - my height is exactly equal to my weight". Does that really mean anything at all? I don't think that statement is usable in any way without more information.
cha
Posted: Mon Apr 21, 2003 6:15 am    Post subject: -18

For 1: There is no need to use two masses connected by a string. One would do. Imagine a large ball of twine with a diameter "d", with a string leading upward for some distance, say "10d". An object that begins at distance "d" to the right of the ball portion, then moves upwards toward the end of the string would experience decreasing gravitational force throughout almost all of the journey.
CrystyB
Posted: Mon Apr 21, 2003 3:17 am    Post subject: -19

i know that the antiparticles behave electrically reversed (so the e and a-p would reject each other, both being negative) but i thought that since an e and an a-e would collide in a little explosion/implosion/whatever, perhaps e and a-p wouldn't want to meet each other... It seems that i was wrong.
Jatan
Posted: Sun Apr 20, 2003 10:26 pm    Post subject: -20

I don't think that the units issue is relevant. In Physics it is a common practice to operate with normalized unit-less quantities, acctually if you want to calculate the example properly and integrate you MUST first introduce normalized t and v (ie. : x = v/vmax, y=t/tmax) Quantities x and y are now without any units. Thus both scientists are right numericaly, yet they are way off if one wants to evaluate the calculation procedure as such.

What if...
Posted: Sun Apr 20, 2003 7:47 pm    Post subject: -21

quote:

Though one can quibble that this is only an "apparent" effect, well, so is gravity, which is not really a force.

Since when? In relativity it is not treated as a conventional force, but this does not mean it is merely an "apparent" force, or somehow illusory. The same goes for the effect you mention, assuming the particle has zero diameter.
Whether or not the unit thing matters depends on how strict you want to be about your answer. I like the combination v and t because it gives correct answers and
is not dependent on the scale, but if you need a straightforward answer, either given answer will work.
Sorry, CB, but antiparticles work pretty much the same as regular matter as far as gravity is concerned, so an electron and antiproton wouldn't work. Their electrical interaction, though, would work opposite.
CrystyB
Posted: Sun Apr 20, 2003 5:05 am    Post subject: -22

#2
Quote:
Or we can say that they are both wrong because each of them finds his answer from an equation that doesn't consider the physical units, and that each of their expression screw up with the units.
Not necessarily. I used to forget about the units all the time when i was in fifth grade! That didn't make my calculations wrong...

Quote:
If we square time, the units are s^2. If we square velocity, the units are m^2/s^2. Neither result in a distance.
But we square the ammound of s's / the ammount of (m/s)'s. That results in a number. And then we stick a unit in its back, and make it right, if we're lucky w/ the choosing of that unit or if we already knew the units along the axis.

#1 wouldn't an electron and an antiproton do the trick?
TomP
Posted: Sun Apr 20, 2003 3:30 am    Post subject: -23

Question: Can you find an example where the gravitational force decreases when two bodies are brought closer together?

Example: if a point "particle" were brought near an object shaped like a tire/donut along a line extending through a point that is in the middle of the tire/donut (equidistant to the edges), then the closer the point particle got to the center the less "force" would be "felt" because in the exact middle of the tire/donut, the apparent "force" would be null. This would be true of both the point particle and the tire/donut. Though one can quibble that this is only an "apparent" effect, well, so is gravity, which is not really a force.
Lepton
Posted: Sat Apr 19, 2003 3:08 am    Post subject: -24

30 - it'll be gone in no time
Nienscecco
Posted: Sat Apr 19, 2003 1:42 am    Post subject: -25

By the way, How many posts does it take to level up from Icarian member to another title? Just wondering...
Nienscecco
Posted: Sat Apr 19, 2003 1:26 am    Post subject: -26

yes, we have to assume the semicircle pattern. It's part of the given info. Anyway, it's also part of what makes this problem a puzzle. Remove the notion of semicircle geometric representation and this will no longer be a puzzle, it will be a plain physics - cinematic problem including some algebra to express the solution.

The "essence" of solving a puzzle concerns "that tricky thing" that makes everyone wondering...or put in evidence a nonsense fact of reality contronted to a farely known kind of representation.

In this case, it is basically a conflict between the "geometric representation" of the problem and its application to fit the reality.

And reading everyone's posts, I think we've made the complete résumé of the problem and we already have the solution. It's all about understanding what the geometric representation of the problem lacks of...and then explain why both scientists can be right at the same time and also what is not very "scientifically correct" about their answer, considering the units into the equation...

Anyway, I'm hurried to see the official solution though.

[This message has been edited by Nienscecco (edited 04-18-2003 09:39 PM).]
pokerfaced
Posted: Fri Apr 18, 2003 6:57 am    Post subject: -27

I think that what is being said is that the fact that we have been told the graph is a semicircle is enough to assume that 2v and t are numerically equal. Symbolically:

v = x m/s
t = 2x s

Therefore we would have a semicircle, although the measurements aren't equal, for velocity can never be measured as time. However, the graph itself would represent a semicircle.

However, the area under the curve is meant to represent the total distance traveled. If we square time, the units are s^2. If we square velocity, the units are m^2/s^2. Neither result in a distance.

This means that although both are numerically right, neither is right in a practical sense.
What if...
Posted: Thu Apr 17, 2003 9:24 pm    Post subject: -28

Cruciverbalist, meters can't equal seconds squared, because this would mean that when you square a time, you would suddenly get a distance! Fundamental units like meters and seconds do not have such a relationship, and what relationship they do have involves constants like the speed of light.

Nienscecco, I agree w/ everything, except the scale thing is an issue because if the intervals are 10 seconds on the x-axis, and the velocity has 1 m/s intervals, then what looks like a semi-circle would really be a semi-ellipse, and the two mentioned formulas would be wrong numerically, even. The v times t solution would survive these troubles, though.

[This message has been edited by What if... (edited 04-17-2003 05:32 PM).]
The Cruciverbalist
Posted: Thu Apr 17, 2003 8:48 pm    Post subject: -29

~decides, against his better judgement, to make another comment about a math-ish puzzle and more than likely make himself look like a fool~

I can't see why m/s can't equal s. When I simplify, I get m=s2. It seems to me that any number and its square would fill the bill here.

------------------
Um... Seriously. - Homestar Runner

There needs to be a better word for weird. - Strong Sad
Nienscecco
Posted: Thu Apr 17, 2003 7:52 pm    Post subject: -30

ZutAlors! Right, I agree. Thanks for leading me back on the right trail. I was indeed thinking of g = GMm/r˛. And it is ovious that it doesn't apply in the "hole inside the earth" situation. Sorry for my silly comment

Now let's discuss this scale matter for the #2 :

I really think that we have to take the given information from the graph as is. Anyway, the big bug according to everyone (including me at the beguining) seems to be that v cannot = t/2 because of the units : m/s cannot = s. My point is : whatever scale factor that we apply to the 2 axes of the graph, that won't solve the bug with the units. Scale is definitly not an issue here.

The v = t/2 only applies to the numeric values. And it is not true to say that we destroy this equation by trying to work with the units.

Lucky wizard was right with his comments on the Ellipse : Area = pi * a * b where a and b are the 2 axis of the ellipse.

Since time and speed are of different nature, the correct way to calculate the surface and fit the units in is to calculate the surface of the semicircle like if it was an semi-ellipse. So : 1/2 * Pi * 2v * t .

Conclusion :

We can either say that both scientists are right because in this case, v = t/2 numerically and they both calculate the surface of a semicircle in a correct manner : they both find the correct numerical value for the answer.

Or we can say that they are both wrong because each of them finds his answer from an equation that doesn't consider the physical units, and that each of their expression screw up with the units.

Anyhow, to calculate the surface of the semicircle as a semi-ellipse will result in finding the same numerical value that both scientists will also find.

BUT!!! and there is the very question : RollerCoaster is absolutly right about the fact that there is an infinite acceleration problem and that both scientists are discussing an impossible matter in the first place...

Sorry if my novel was too long

[This message has been edited by Nienscecco (edited 04-17-2003 04:02 PM).]
What if...
Posted: Thu Apr 17, 2003 6:58 pm    Post subject: -31

Since people aren't points either, if your center of mass corresponded with the Earth's, parts of your body would be pulled outward (because the mass is around you). Your center of mass, of course, would not move, because (assuming the Earth is almost perfectly spherical) you would not be pulled more in any direction, so net force is zero. Otherwise, this embedded in a sphere solution is the same as having two equal massed objects (very distantly connected) on a line (A), and an object exactly in between them feeling no force:
code:

A. _________
| |
| |
_| |_
|_| 1 |_|

-----------------
B.
2
_ 1
|_|-------

In the second picture is the simplest solution to the problem based on shortest distance rather than distance to center of mass. The line coming out of the mass (box) in B is a stick having very little mass. My second solution was a combination of A and B.

Edited to fix picture.

[This message has been edited by What if... (edited 04-30-2003 02:52 PM).]
ZutAlors!
Posted: Thu Apr 17, 2003 4:26 pm    Post subject: -32

You're saying a couple contrary things, Nienscecco. At the center of the Earth, the net effect of gravity will be zero, but the gravity won't be infinite. I suspect you're thinking of the gravitational law g = GMm/r2. That treats the mass of the Earth as a point mass, and doesn't apply anymore one you're inside the Earth. I believe you could calculate the actual gravitational forces on the corpse by triple integrating a force calculation over the volume of the Earth, but this will come out to be zero. So "the Earth with a hole through it" should still be a valid answer.
Nienscecco
Posted: Thu Apr 17, 2003 3:54 pm    Post subject: -33

I think your example with the hole into the earth is wrong. When the corpse (Hoping the guy is dead when he reaches the center of the earth) is at the center of the earth (and that we consider this place the center of masses also). There will be many gravitational forces applied the corpse. The results of these forces will be equal to zero : meaning that the corpse will stay at the center only due to the gravity. But At this place the gravity will be maximum, infinite in fact... Considering the corpse at 0 distance from the center of gravity of the earth, the force would be infinite (divisionby zero).

So the best possible example is a massive object that is not symetric, with his center of gravity away from a less massive volume...Like What if suggested in the first place.

Please be indulgent with my english, I'm new here and I speak french by birth so it might not be very clear after all

[This message has been edited by Nienscecco (edited 04-17-2003 12:09 PM).]
ZutAlors!
Posted: Thu Apr 17, 2003 11:31 am    Post subject: -34

I think a good arguement for ignoring the "given" information is that the equation is to be determined symbolically. Although the graph implies that v = t/2 there are two problems here: 1) there's nothing preventing the two axes of the graph from being scaled differently, so that even though the distance corresponding to v equals the distance corresponding to t/2, v doesn't equal t/2. 2) Moreover, the equation(s) determined from the graph have no units attached. Obviously, if you alter the units in any way from whatever the graph represents, then you destroy the numerical equality.
CrystyB
Posted: Thu Apr 17, 2003 5:39 am    Post subject: -35

(2) i don't get it: why do you disregard the info given?? it is said to be a semicircle after all, so v=t/2. Therefore both answers are correct. Scaling was already taken into account.

PS Your answers of pi*v*t/4 is also correct - like in the 0.FFFF...=0.9999...=1 debate
ChienFou
Posted: Wed Apr 16, 2003 11:40 pm    Post subject: -36

No.2 pi x v x t / 4. Boring. compare the area of the rectangle with the area of the semi-circle inevitably the engineering solution.
What if...
Posted: Wed Apr 16, 2003 8:42 pm    Post subject: -37

I found another (simpler) solution to one, in which the point (1) w/lesser force is closer to the object and center of mass than the others (2 and 3).

Like a barbell...
code:

2 3

___________ ___________
| | 1 | |
| More |-----------------------------| More |
| Mass |-----------------------------| Mass |
|___________| |___________|

except big enough that we can measure its gravity.

Edited to fix picture.

[This message has been edited by What if... (edited 04-16-2003 04:47 PM).]
What if...
Posted: Wed Apr 16, 2003 7:18 pm    Post subject: -38

Sorry, I was in a hurry before, and somehow removed the very important word:semicircle. Here's my idea.

What I was picturing was a hollow semicircle/hemisphere (they are the same if you just rotate the semicircle), w/nearly all the mass on the ends. If an object is near (or touching) the midpoint of the semicircle (1), then the gravitational force could be smaller than if the object is slightly farther away overall, but nearer the (more massive) end (2). Of course, the shortest distance from the object to the curve can be smaller in the first case (very close to much less massive part) than the second (slightly larger distance to more massive end).
In retrospect, I could have used a much simpler example for this alone, but this example also exhibits the property mentioned by Griffin, for some reason was the only one I mentioned in my previous post, and which I do not think is the key to the puzzle since it does not mention the center of gravity. Since the center of mass of a semicircle is at the center of the corresponding circle(3), if you move the object from near an end(2) to the center(3), it will feel less of a force, despite being nearer to the center of mass. I think the first part (not using center of masses) is the answer because the second is common, and the puzzle just says closer, which does not necessarily mean the center of masses.
code:

________
/ 1 \
/ \
/ \
| |
most mass is here -> | 2 3 | <- and here

Please just pretend the diagram looks good (and like a semicircle)!

Edited to fix some things that made little sense.

[This message has been edited by What if... (edited 04-17-2003 03:04 PM).]